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694248 | |dc-dc-converter|esp8266|repair| | <p>U1 and D2 are probably the switching regulator, buck diode, and the ferrite cylinder "100" is probably 10µH. Q5 is probably some external load switch.</p>
| <p>I have a water-filter board that receives 36V input, and needs to power an ESP-WROOM-02D in 3.3V. The power circuit that converts from the 36V to 3.3V is no longer working, and I am trying to understand what are the different components involved. From a look at this section of the board, I see the following:</p>
<ul>
<li>9971GH transistor N-channel MOSFET</li>
<li>M6 transistor PNP</li>
<li>M7 diode</li>
<li>4 resistors (2x 27.2kΩ, 2x 10.3kΩ?)</li>
</ul>
<p><a href="https://i.stack.imgur.com/3PoCo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3PoCo.jpg" alt="Board power section with labels" /></a></p>
<p>Now I am trying to understand why when I supply the input 36V, I have no 3.3V in the output (from the test point on the board). The basic test with a multimeter on the M7 diode seems to indicate the diode still let current flow in one direction only. Where to take it from there?</p>
<p><a href="https://i.stack.imgur.com/Fm8M1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Fm8M1.jpg" alt="The ESP to be powered." /></a></p>
<p>Adding overall view of the board front and back:</p>
<ul>
<li><a href="https://i.stack.imgur.com/cllBT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cllBT.jpg" alt="Front" /></a></li>
<li><a href="https://i.stack.imgur.com/WSkB8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WSkB8.jpg" alt="Back" /></a></li>
</ul>
<p><em><strong>[EDIT]</strong> Same thing with labels (fixed vocabulary):</em></p>
<ul>
<li><a href="https://i.stack.imgur.com/dDQkC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dDQkC.jpg" alt="with labels" /></a></li>
</ul>
<p><em>(edit added yet more labels)</em></p>
| Step-down circuitry | 2023-12-15T03:44:32.993 |
694262 | |operational-amplifier|oscillator|impedance|virtual-ground| | <ul>
<li><p>The damping factor of the shown CR-network (node "V3") at the frequency with a phase shift of 180deg (fo=916 Hz) is k=0.10725.</p>
</li>
<li><p>Therefore, you need a gain of at least G=-9.32 in order to allow oscillation at 916 Hz. However, due to the large resistor values (Mega-ohm-range) the simple gain expression (-R2/R1) will be, most probably, not accurate enough (the opamps input resistance comes into the game).</p>
</li>
<li><p>Recommendation: Completely drop the 1M resistor and connect the 100k resistor directly to the opamp input. In this case, the oscillation frequency reduces to app. fo=888.3Hz and the required gain is app. G=-9.26.
(feedback resistor: R2=930k).</p>
</li>
</ul>
| <p>In this oscillator configuration, which I got from the textbook "Operational Amplifiers & Linear Integrated Circuits" by James Fiore (page 354, 3rd edition), since the gain of the ladder network is -1/8, the inverting gain is designed using the formula</p>
<p>Rf/Ri= -8,</p>
<p>so that the loop gain be 1, where Rf=8M and Ri=1M</p>
<p>However, the 100k resistor seems to be in parallel with the 1M resistor. The ladder network seems to be relatively unaffected by this since 1M is a pretty large value and maybe neglected. However, it seems to me that the inverting gain would be something like</p>
<p>Rf/Ri +Rf/100k,</p>
<p>with the latter term dominating.
<a href="https://i.stack.imgur.com/5b91Z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5b91Z.png" alt="enter image description here" /></a></p>
| Why is the inverting gain of this phase shift oscillator 8? | 2023-12-15T06:34:22.540 |
694268 | |microcontroller|current|programming|display|real-time| | <p>The way you have worded your question means you have not yet thought through your requirements.</p>
<p>If you want to display real time data, then you display the latest data as soon as you can.</p>
<p>As you can only update the display every 50 ms, this means you compact the current measurements of the previous 50 ms into a single value (maybe average, maybe peak) and throw it at the display.</p>
<blockquote>
<p>How to properly display data to have smooth move of arrow?</p>
</blockquote>
<p>Ah, smooth, what does that mean? Emulate a moving needle meter? Display some decaying average.</p>
<p>Yes, you can do either of those with a little digital filtering. Just define a filter type, and a time constant or two, and use a digital filter on your stream of current readings, and display its output every 50 ms.</p>
<p>Choose a filter type based on how you expect the user is going to use the indication. For instance, VU meters have very different, and specific, dynamics to other types of meters, based on the needs of the audio mixer fraternity.</p>
<p>You can even create non-linear filters, such as used in weighing machines, if that's the behaviour you want. Generally they use short and long time constant low pass filters, displaying the long time constant output for low noise, switching briefly to the short filter output when they differ by more than a threshold, which suggests a step change to the input has occurred, so leading to a faster real time response.</p>
<p>Once you have a basic system, modifying filter type and time constants based on your users' feedback is called 'tuning'. This is an iterative process, and may never terminate!</p>
| <p>I have nextion display on witch is positioned analog scale. It needs to present Amps from device which mcu monitoring. Problem is that current can grow by several tens of amperes in just several milliseconds but refresh command to display can be performed every 50ms. How to properly display data to have smooth move of arrow? Is there some formula or technique to do that? <a href="https://i.stack.imgur.com/KikXo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KikXo.png" alt="enter image description here" /></a></p>
| How to display real time data on analog scale if maximum send rate is 50ms | 2023-12-15T07:43:10.953 |
694270 | |stm32|openocd|gdb|stmcubeide| | null | <p>I have a setup where I bought a STM32F407ZET6 dev board from AliExpress and have been using for almost 6 years using STM32Cube IDE. It was only until i updated my IDE from 1.6 to 1.14 when it started saying that "Could not verify ST Device! Please verify that the latest version of gdb-server is used for the connection". my openocd is remotely running on Raspberry pi where it uses the bitbanging driver to drive the GPIOs for establishing an SWD connection.</p>
<p>This same question has been addressed in several other threads in the context of bluepill (<a href="https://community.st.com/t5/stm32-mcus-boards-and-hardware/how-to-deal-wirh-could-not-verify-st-device/td-p/234230/page/2" rel="nofollow noreferrer">https://community.st.com/t5/stm32-mcus-boards-and-hardware/how-to-deal-wirh-could-not-verify-st-device/td-p/234230/page/2</a>).</p>
<p>On the OpenOCD side, i am getting this error:</p>
<p><a href="https://i.stack.imgur.com/vDlHa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vDlHa.png" alt="enter image description here" /></a></p>
<p>I tried changing the -expected-id $_CPUTAPID to different values suggested in different posts (including 0), but nothing works. Most of the forums talk about this issue in the context of Bluepill; however this is STM32F407 based board.</p>
<p>Can anyone please help me with this problem?</p>
| STM32F407ZET6 could not verify ST device | 2023-12-15T07:53:25.323 |
694272 | |arduino|voltage-regulator|esp8266|battery-operated|nodemcu| | <p>AMS1117 is basically "the cheapest regulator they could find". It has rather high dropout (>1V) which is not a problem when powering with 5V, but it also has slow transient response: when ESP32 goes from low to high power, for example when going out of sleep, the regulator isn't fast enough to catch up, so this causes output voltage to dip, and this sometimes crashes the micro. Its idle current is also very high (5-10mA) which is a problem on batteries.</p>
<p>In your case the simplest solution is to use a more modern, faster LDO with lower dropout, for example <a href="https://eu.mouser.com/datasheet/2/389/ldl1117-1849797.pdf" rel="nofollow noreferrer">LDL1117</a>. It has the same footprint as AMS1117 so you can directly replace it on the NodeMCU board. The tantalum output cap also needs to be replaced with a 10µF X7R/X5R ceramic cap of the same size. This fixes all power supply related crashes.</p>
<p>No matter what LDO you use, I'd still recommend replacing the tantalum cap with MLCC to get lower transient voltage dip on transition from sleep to active. The tantalum cap's ESR (several ohms) causes problems in this case, whereas ceramic capacitor ESR is much lower (few mOhms for 10µF).</p>
<p>ESP32 will run down to 2.5V so if your battery cutoff is 2.75V, with a real LDO as in "low dropout", you don't need to boost voltage. But you do need to be able to detect when the battery is drained and power down, otherwise it will over-discharge the battery and kill it. If you add a battery management / charging chip in your design it will also handle over-discharge protection, otherwise you'll need a supply supervisor or similar chip.</p>
<p>I was recommending LDL1117 because its pinout is the same as AMS1117 which is convenient for a drop-in replacement on NodeMCU boards and very effective at fixing the power supply related crashes. But if you make your own board, you don't have the footprint constraint, so you can pick a LDO with an ENABLE pin, controlled by the battery supervisor chip. Or maybe you will find a battery supervisor with integrated LDO. LDL1117 idle current is a bit high for battery use (typ 250µA).</p>
| <p>I am trying to develop a custom PCB with the esp12-E module. For development and testing, I am using the nodemcu development board ESP8266 V3. I am powering the nodemcu via a 3.7V battery with a maximum voltage of 4.2V and a discharge cutoff of 2.75V. I am giving power directly via the 3v3 pin on the nodemcu and it works fine. Now my problem is that my nodemcu board has an Ams117-3.3 voltage regulator on it that requires an input voltage of 4v-12v. Now I have the battery plugged in for many hours now and at this point the voltage of the battery should be below the 4V minimum required but the nodemcu is still operating normally. Is there some additional electrical component being used on the nodemcu board to boost the voltage?</p>
| Voltage regulation for NodeMcu | 2023-12-15T08:03:39.547 |
694299 | |transformer|winding|verification| | <p>The calculation appears fine. The values put into it require more thought though.</p>
<p>Saturation flux density is 400mT or more, see datasheet: <a href="https://www.tdk-electronics.tdk.com/download/528856/79b67daa1d54253aa2ae371bb98e9629/pdf-n49.pdf" rel="nofollow noreferrer">SIFERRIT material N49 | TDK</a>. Peak flux density may come down, particularly as frequency goes up, due to core loss. For a small transformer like this, and with efficiency unstated, I might be fine running it at 0.1-0.2T.</p>
<p>Likewise I wouldn't worry about litz, at such low current. I would probably reach for 28AWG myself.</p>
<p>If you need particularly low loss (maybe this is more of a signal than power application, I mean, done with a little power, but point being, the purpose is more finesse than raw output), that would be fine, and probably stranding of 36AWG or finer would suffice. (Since you don't mention application, I'm free to suggest either direction.)</p>
<p>Note that V*I in the relation is apparent power, not real power. Since you say it's resonant, you must define the resonant energy* as well.</p>
<p>And to achieve target flux density, you need so-and-so turns, and to achieve target inductance, you need so-and-so inductivity, which is to say, air gap of the core, or effective permeability in any case. And fringing flux around the airgap will put more pressure on the choice of litz (which, the above values will still be fine for, unless the required Q factor is EXTREMELY high, say upper hundreds or more).</p>
<p>*Reactive power is energy, in the same way that torque is N.m and work is N.m but the directions are different. Essentially the (phase) direction is different, so S has units of VA but represents energy instead of power. To get the (peak or circulating) reactive energy, divide by angular frequency. This is true, at the fundamental at least; of course you'll need to use Parseval's theorem for the general (all frequencies) case.</p>
| <p>I am determining the Area-Product of the core required for a transformer to be used within a series resonant converter. The specifications are as follows:</p>
<ul>
<li>Switching frequency = 400 kHz</li>
<li>Primary current = sine wave with a peak value of 500 mA</li>
<li>Power output = 1 W</li>
</ul>
<p>These are my calculations:</p>
<ul>
<li>I decided to use a core made out of N49 (since many such cores are already available)</li>
<li>From the data given on <a href="https://www.mag-inc.com/Design/Design-Guides/Transformer-Design-with-Magnetics-Ferrite-Cores" rel="nofollow noreferrer">this</a> website, I have determined the saturation flux density of the ferrite core as: 0.04 T (approx.)</li>
<li>The winding factor is considered to be 0.2</li>
<li>Since we are dealing with 400 kHz, I decided to use a Litz wire with a 44 AWG strand size which has a bare Copper diameter of approximately 0.051 mm</li>
<li>Since the RMS value of the primary current is 0.353 A, I decided to use a Litz wire with 18 such strands. This may be too much, but it is fine for my initial design (the overall conductor area now comes out to be 0.0368 <span class="math-container">\$mm^2\$</span>).</li>
<li>I calculated the current density of this conductor as follows:
<span class="math-container">\$J = \frac{0.353}{0.0368} = 9.59 A/mm^2\$</span></li>
<li>The Area-Product of the transformer is given by:
<span class="math-container">\$A_cA_w = \frac{VI}{2\; f_{sw}\; B_m\; k_w\; J} \$</span>
This gives a value of <span class="math-container">\$\frac{1}{2 \times 400000 \times 0.04 \times 10^{-6} \times 0.2 \times 9.59} \$</span> = <span class="math-container">\$ 16.293\; mm^4\$</span></li>
</ul>
<p>Is my area-product calculation correct based on the specifications? Is the saturation flux density value as well as the current density, correct? Also, will there be any issue due to more number of strands (more than the required value) within the wire?</p>
| Transformer Design for a Series Resonant Converter | 2023-12-15T11:00:13.157 |
694312 | |charge-pump| | <p>I'll add to the other answers: your diodes have too much leakage.</p>
<p><a href="https://i.stack.imgur.com/syx6d.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/syx6d.png" alt="enter image description here" /></a></p>
<p>Here's a solution, BOM cost should be reasonable...</p>
<p><a href="https://i.stack.imgur.com/TtiSF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TtiSF.png" alt="enter image description here" /></a></p>
<p>Rather than use two doublers, it adds the input voltage several times. It's simpler, because there's no need for level shifting or any active device. You either need two microcontroller pins (V1 and V3) to output opposing phase 100kHz square waves, or one micro pin and an inverter or any other inverting logic gate.</p>
<p>Output impedance is pretty high due to the 4 diodes in series, so it will output 12V when unloaded and 9V at 3mA. This problem is common with all diode based charge pumps, to have low impedance you'd need FET switches. You could regulate it by using the micro's ADC to check the output voltage.</p>
<p>Simulated efficiency is: 66% with 1N4148 or BAT54. But BAT54 (Schottky) gives higher output voltage, so there's more margin. 1N4148 with its higher Vf is borderline to get 9V output.</p>
<p>A tiny boost chip with an equally tiny inductor would no doubt give better efficiency and more features (like regulation). Whether it's worth it depends on the cost of energy (kWh from coin cells costs a lot!)</p>
| <p>I designed this 2-stage charge pump to get about 9V and 3mA (tolerances accepted) from a lithium-ion battery. The output works well and provides the requested voltage and current. I thought it is a very efficient thing till I measured the input current; it was about 45mA! I double checked everything but it seems correct. So for getting 27mW, I'm using ~170mW.</p>
<ul>
<li>PWM frequency = 10 KHz</li>
<li>PWM duty cycle = 50%</li>
<li>PWM top voltage = same as the battery (3.7 V in tests) and measures exactly half of the 3.7 as expected (50%)</li>
<li>PWMs are from 2 different channels of a MCU but they are not phase-matched</li>
<li>The capacitors in parallel are for better ESR and they are both MLCCs</li>
</ul>
<p>What is the problem? Can I make it better to use it in a battery operated device?</p>
<p><a href="https://i.stack.imgur.com/ZM2i0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZM2i0.png" alt="Double-stage charge pump" /></a></p>
| Why does this 2-stage PWM charge pump have a very low efficiency? | 2023-12-15T13:14:40.600 |
694320 | |filter|buck|switching-regulator|ripple| | <p>The purpose of L1 is not principally to be part of a filter, but be the energy storage element that makes the buck converter work. However, it does resonate with C1.</p>
<p>While L1 + C4 <em>look</em> like a filter, it's not a proper classical filter design. An LC filter needs at least one of its ports, preferably both, to be resistive. You have the output loaded with a high impedance (current sink), and the input is driven from a near-zero impedance. Neither is significantly resistive, and the result is a nasty resonance.</p>
<p>Things will improve slightly if the actual real load to the circuit has a resistive element to it. Loss in the inductor will also damp the resonance. Obviously though, any series resistance will reduce the efficiency of the circuit.</p>
<p>The normal solution to this problem is to place a snubber at the output, a series combination of R and C, that absorbs energy from the resonance without drawing a DC current or dropping some of the output voltage. The energy that this resistor dissipates has to come from somewhere, and it's from the input, and reduces the overall efficiency, but you did want to tame that resonance, didn't you? This is the least worst way of doing it.</p>
<p>It's probably not worth designing the values analytically when your load is so unrealistic. Model the load and L1's loss better, then play with some RC values in the simulator that you already have set up, to see what they do.</p>
| <p>I'm currently designing a synchronous buck converter and trying to simulate it in LTSpice. <a href="https://i.stack.imgur.com/l1jro.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/l1jro.png" alt="enter image description here" /></a></p>
<p>I know the purpose of L1 and C4 is to act as a filter to smooth out the output, since it would be rapidly switched on and off. However, the LC filter seems to be inducing a very significant ripple voltage, precisely at the resonant frequency of L1 and C4. This is what the output voltage looks like, measured before I1 (it's trying to maintain 24V):
<a href="https://i.stack.imgur.com/5Aihr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5Aihr.png" alt="enter image description here" /></a></p>
<p>How would I approach fixing this?</p>
| LC Filter Creates Ripple Voltage in Buck Converter | 2023-12-15T13:54:14.050 |
694322 | |transistors|analog|common-emitter|common-collector|common-base| | <p>The bipolar junction transistor works in the other three configurations too. These are obtained by reversing the roles of the collector and emitter in the three classic configurations.</p>
<p>It works, just not very well.</p>
<p>The geometry and the doping are generally optimized for the emitter to inject carriers into the base, most of which make it through and arrive at the collector. In modern transistors, the emitter is usually small and heavily doped, while the collector is the largest region (usually the substrate), and doped much less. If you use it the other way, the base-collector junction is larger and less sharply defined, with a wide region of spatial charge. If injected from the large, fuzzy collector, much more carriers recombine in the base before they reach the emitter. This simply means much lower amplification (1-2 orders of magnitude). In AC, amification is further reduced by the high cacacity of the base-emitter junction (optimized for a narrow, sharply defined region of spatial charge).</p>
<p>I would like to point out that it's counter-productive to think of the transistor as two diodes back-to-back. It is precisely the failure of the base-emitter junction to work as a diode that gives rise to the transistor effect, because the thin base fails to catch and recombine the vast majority of the charge carriers that enter it. Only one of the two junctions is optimized for this "non-diodey" operation.</p>
<p>MOSFETs are four-terminal devices, and they're bidirectional: the source and drain are usually identical, and interchangeable. The difference comes from the fact that, in three-terminal devices, the source is connected internally to the substrate, so there's no way to bias it usefully if reversed. Power MOSFETs often include a reverse diode between drain and source, so they're even less reversible. But the general MOSFET with a separate substrate terminal is, indeed, reversible.</p>
<p>JFETs are generally bidirectional.</p>
| <p>Almost any analog design book or course I have ever seen starts by introducing the three basic single-transistor stages. For example, in a bipolar technology these are the common-emitter, common-base, and common-collector configurations.</p>
<p>For instance, Gray and Hurst's classic book says</p>
<blockquote>
<p>Bipolar and MOS transistors are capable of providing useful amplification in three different configurations.</p>
</blockquote>
<p>Why is this? Given that there are three readily accessible terminals in each transistor and 3 possible functions to be performed (ac ground, input, output), it follows that there are 6 conceivable configurations. Why are we uninterested in the other 3? Is the answer as simple as "experience has shown that they're not useful" or is this something that should be obvious to me a priori?</p>
| Why are only 3 of 6 single-transistor amplifier configurations useful? | 2023-12-15T14:17:26.287 |
694329 | |transistors|simulation|circuits|falstad| | <p>As already pointed out by others in comments ...<br />
With the resistor at the input and wired ground of the 10 V, you should have a circuit like this.</p>
<p>Made with <a href="https://archive.org/details/mc12cd_202110" rel="nofollow noreferrer">microcap v12</a></p>
<p><a href="https://i.stack.imgur.com/I48iz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/I48iz.png" alt="enter image description here" /></a></p>
<p>NB: when the switch is open, values of currents and voltages are not very "meaningful".<br />
The circuit is not really "complete".</p>
<p>Here is an example of some "resistors" added (switch open).</p>
<p><a href="https://i.stack.imgur.com/ycFB9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ycFB9.png" alt="enter image description here" /></a></p>
| <p>I'm trying to simulate a simple signal amplifying circuit using three transistors as shown below:</p>
<p><a href="https://i.stack.imgur.com/jKx2s.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jKx2s.png" alt="Falstad Circuit, closed switch" /></a></p>
<p>(Note that the LED isn't turned on)</p>
<p>But when I open the switch, the current still flows normally from the 9V source on the left, even though the first transistor's base doesn't receive tension. The current has the same value as before if you look at the bottom right corner. It should be zero.</p>
<p>Here's the image:</p>
<p><a href="https://i.stack.imgur.com/N3lWI.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/N3lWI.png" alt="Falstad circuit, open switch" /></a></p>
<p>I'm very new to electronics and studying it on my own.</p>
| Falstad Circuit - Why is the current still flowing through the transistors? | 2023-12-15T15:34:43.390 |
694330 | |circuit-design|pcb-design|simulation|components| | <p>Most ICs are avialable in several packages. It is important to ensure that you select a part/package that is actually available from your suppliers before laying out the PCB. Some of the package variations shown on the datasheet may not actually be available.</p>
<p>For switches, connectors, and other "mechanical" parts, it is desireable to actually have the part in your hand before designing the footprint for it, so you can compare the datasheet's suggested footprint with the actual part.</p>
| <p>I am an electronics hobbyist. I recently designed and ordered PCBs (using KiCad) but I failed twice. I designed amplifier circuits but something or the other went wrong each time. The problems I faced are:</p>
<ul>
<li>I did not have the parts with me before hand. As a result I ended up selecting the wrong footprints and I couldn't find the parts that would fit.</li>
<li>I made mistakes in the schematic and running the electrical rules checker showed everything was correct. I only found out about the mistakes after soldering the parts to the board.</li>
</ul>
<p>The steps I took to avoid these mistakes:</p>
<ul>
<li>I simulate the circuit to avoid making mistakes in the schematic.</li>
</ul>
<p>My questions are:</p>
<ul>
<li>How can I find a reliable spice model library? I am currently using a free library I found on GitHub.</li>
<li>How can I choose among different variants of a part that affect the circuit? For example, for an amplifier operating between -25 V to 25 V there are a variety of N channel MOSFETs that can operate in that voltage range. How would I pick one?</li>
</ul>
| Mistakes to avoid while designing PCBs | 2023-12-15T15:39:24.483 |
694332 | |switches|cmos|switching| | <p>If the supply to 4066 is turned off, you cannot have any voltages on 4066 IO pins for any reason.</p>
<p>If you do, such as even one Arduino pin high, the 4066 is powered through that IO pin and the inbuilt protection diodes.</p>
| <p>I am using a CD4066 IC in the below configuration. The 5V is coming through a LDO. The switches in the CD4066 are being thrown high and low through Arduino digital pins 2,3,4 and 5. Surprisingly I am able to read voltage on the Arduino pin A0 even when the LDO is not powered. Why is that?</p>
<p><a href="https://i.stack.imgur.com/k79ao.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/k79ao.jpg" alt="enter image description here" /></a></p>
| Unknown source of voltage through a CMOS switching IC | 2023-12-15T15:46:41.690 |
694358 | |led|resistors|dc|ltspice| | <p>A few quick points:</p>
<ul>
<li><p>Do use the LTspice help. If you access it using the menu, there is an option for searching and you can use "diode" or just "d" in the search, then look down for something that says "diode" on it and click on that. This will get you to a page with some useful, technical information about diode models. It may still be a bit unfamiliar, but I want to make sure that you think about accessing the help, anyway. Sometimes, regardless of how prepared you are, you can find something helpful there.</p>
</li>
<li><p>LEDs are diodes. But they operate at distinctly higher voltage differences than do diodes used for full wave rectifiers, for example. So you should be warned that any time someone says "diode" it's almost a knee-reaction to ask, "What kind of diode?" before considering using it in LTspice.</p>
<p>There are many kinds of diodes, but the usual suspects are rectifier (power) diodes, signal (not power) diodes, LEDs (higher voltage differences than usual), Schottky diodes (somewhat lower voltage differences than usual, and a little "leaky"), and Zener diodes (usually used with a reverse-voltage applied to them to gain access to their voltage-knee behavior.)</p>
<p>Just be aware that not all diodes are the same and that you need to have some basic idea about which kind you are working with and how to access the behavior you'd like to see over the range of operation you expect.</p>
</li>
</ul>
<p>This brings us back to the LTspice documentation. If you read it, you will see that they discuss two essentially different types of diode models.</p>
<p>This first one is described by the first table in the help on diodes. There are basically just two things you care about with this one: <span class="math-container">\$V_{_\text{FWD}}\$</span> and <span class="math-container">\$R_{_\text{ON}}\$</span>. <span class="math-container">\$V_{_\text{FWD}}\$</span> sets the voltage for the diode when there is no current at all. And <span class="math-container">\$R_{_\text{ON}}\$</span> is multiplied by the LED current to get an additional voltage drop to add to the total voltage drop, which by definition varies with the current. <span class="math-container">\$R_{_\text{ON}}\$</span> gives a bit of variation of the diode behavior that depends on the current through it.</p>
<p>This first model is only <em>good</em> for a short range of planned currents. But this simpler model is also a pretty good model to pick for LEDs. That's because most LEDs are only supposed to be operated over just such a small range of currents. For example, low power, high efficiency LEDs will be operated within a factor of 3 or so of <span class="math-container">\$2\:\text{mA}\$</span>. Low power, standard LEDs will be operated within a factor of 3 or so of <span class="math-container">\$20\:\text{mA}\$</span>. And high power LEDs will be operated usually within a factor of 3 of whatever current they are spec'd for.</p>
<p>So the range of operating current for LEDs are generally limited and that makes this simpler model a good fit for them.</p>
<p>Plus, it avoids the trickery needed to get a good fit using the more complex model of the second one discussed on the LTspice help page. This other complex model is called the Shockley diode model. Because the Shockley model is designed to work accurately over a very much wider range of currents that span over many orders of magnitude change, it is necessarily more complex because over such a wide dynamic range there are more effects that come into play.</p>
<p>Luckily, you don't need this more complex Shockley model.</p>
<p>So let's return to the simple model:</p>
<p><span class="math-container">$$V_{_\text{D}}=V_{_\text{FWD}}+R_{_\text{ON}}\cdot I_{_\text{D}}$$</span></p>
<p>Just doing a quick check, I see that <span class="math-container">\$I_{_\text{D}}=\frac{9\:\text{V}-2\:\text{V}}{350\:\Omega}=20\:\text{mA}\$</span>. So this tells me this is about a standard low power LED.</p>
<p>We could cheat a bit and just set <span class="math-container">\$R_{_\text{ON}}=0\:\Omega\$</span> (or something really small.)</p>
<p>But because I want to teach you to use it a little better, let's assume something a little more meaningful, like <span class="math-container">\$R_{_\text{ON}}=20\:\Omega\$</span>.</p>
<p>At <span class="math-container">\$I_{_\text{D}}=20\:\text{mA}\$</span>, the voltage drop across <span class="math-container">\$R_{_\text{ON}}\$</span> will be <span class="math-container">\$20\:\Omega\cdot 20\:\text{mA}=400\:\text{mV}\$</span>. So to get a total voltage drop of <span class="math-container">\$2\:\text{V}\$</span>, we have to subtract that and find <span class="math-container">\$V_{_\text{FWD}}=2\:\text{v}-400\:\text{mV}=1.6\:\text{V}\$</span>.</p>
<p>So that's the model.</p>
<p>Just hit 'S' in LTspice to get the dialog box and type in as shown:</p>
<p><a href="https://i.stack.imgur.com/AyPBB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AyPBB.png" alt="enter image description here" /></a></p>
<p>Then place that line on the schematic somewhere.</p>
<p>Here's what I did:</p>
<p><a href="https://i.stack.imgur.com/qfKkJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qfKkJ.png" alt="enter image description here" /></a></p>
<p><em>Note that I edited the "D" and changed it to "MYLED" on the diode, itself!</em></p>
<p>And that's all you need to do.</p>
<p>Let's run it:</p>
<p><a href="https://i.stack.imgur.com/00GWu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/00GWu.png" alt="enter image description here" /></a></p>
<p>Note that the diode current is exactly <span class="math-container">\$20\:\text{mA}\$</span>!! And note that the voltage drop across it is also exactly <span class="math-container">\$2\:\text{V}\$</span>.</p>
<p>That's how you can get LTspice to readily handle simple LED cases.</p>
<p>I went to a little extra trouble to create a non-zero value for <span class="math-container">\$R_{_\text{ON}}\$</span>. You don't have to do that. But LTspice doesn't exactly like <span class="math-container">\$R_{_\text{ON}}=0\$</span> and instead applies some numerical tricks that will actually generate numerical oddities when you try and cheat like that. It's better to just pick a reasonable value and go with it. Besides, it also gives your LED something closer to a more realistic behavior when you do that, too.</p>
| <p>I am trying to get this circuit to work correctly. I am following a book, and according to Ohm's law this resistor should be 350 Ω to get 2 V at the LED. I am only seeing around 800 mV.</p>
<p>To get around 2 V at the LED in LTspice I need a 10 pΩ resistor which doesn't make sense. What am I doing wrong? Or is the example in the book incorrect?</p>
<p><a href="https://i.stack.imgur.com/vtfuv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vtfuv.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/OVZtk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OVZtk.png" alt="enter image description here" /></a></p>
| LTspice showing wrong voltage | 2023-12-15T18:20:13.423 |
694360 | |mosfet|analog|solid-state-devices|device|device-characteristics| | <p>Proper mos device modeling in sibthreshold was not available before EKV model. Level3 and bsim2 models both had tens of percent error in gm at threshold. That gave the field a touch of black magic: you n</p>
| <p>In textbooks like the venerable Gray and Hurst or in contexts like my introductory analog design class, it was always emphasized that with MOSFETs we always want our designs to obey some constraint like <span class="math-container">\$V_{ov} > 150\$</span> mV. A somewhat closer analysis of this constraint shows us that this constraint is equivalent to the requirement that we have all transistors in our designs operating well into strong inversion. I have two questions about this:</p>
<p>(1) What is the reason for so stridently wanting to avoid dipping into moderate or even weak inversion with the effect of our small signal on the gate? Given that it's well known that we get more transconductance in weak inversion than in strong inversion, it can't be that the gain is deleterious. Does it have something to do with linearity and distortion of the signal?</p>
<p>(2) What are the main problems with operating in weak inversion more generally? Given that it has higher transconductance, all else equal it seems we would want to operate there. I know some designs do, so I'm wondering what the difficulties are in doing so with <em>all</em> designs. Perhaps Gray and Hurst will answer this later in the text though...</p>
| Why do we usually operate well into strong inversion? | 2023-12-15T18:27:20.947 |
694377 | |pcb-design| | <p>As others have said in the comments, the series gate resistor is much too large. The waveform presented at the gate (V(b)) is a triangle wave centered at about 6 volts which keeps the transistor on at all times, about 1A which is limited by the series resistance added to the inductor model in the SPICE simulation. <br>
The LTspice simulation shows this. The upper graph is with series gate resistor at 10k ohms, the lower graph is with the series gate resistor at 50 ohms. The transducer model is a very simple model and values are chosen to resonate the circuit at around 113 kHz. Depending on your transducer, you may need to add a capacitance across the transducer to properly resonate the circuit.</p>
<p>You also need to add some capacitance on the VCC power rails on the physical circuit, perhaps 100 nF and 100 uF.</p>
<p>An inductor with a specification of 1 W means nothing. Depending on the inductor construction, you could saturate the inductor which will cause problems.</p>
<p><a href="https://i.stack.imgur.com/2xDsa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2xDsa.png" alt="enter image description here" /></a></p>
| <p>I'm developing a humidifier PCB module to power the fogger transducer for mist generation.</p>
<p>Components:</p>
<ul>
<li>12V 2A VCC knput</li>
<li>Bojack 220uH 1W inductor (inexpensive kit from Amazon)</li>
<li>16mm fogger transducer operating at 113kHz</li>
<li>IRFZ44 TO-220 MOSFET</li>
<li>NE555N oscillator</li>
<li>Supporting resistors as per the schematic diagram below.</li>
</ul>
<p>Upon powering up the PCB, the inductor heats up rapidly leading to the burning out. Unfortunately, I couldn't locate a datasheet for the inductor.</p>
<p>Why does it burn out? Is it because it cannot handle the current? Is there a stronger inductor that can handle higher currents that I could use? Is there an issue in the schematic diagram I'm not seeing?</p>
<p><a href="https://i.stack.imgur.com/4gWiq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4gWiq.png" alt="enter image description here" /></a></p>
| Troubleshooting inductor issues in an ultrasonic humidifier PCB | 2023-12-15T21:00:59.373 |
694387 | |analog| | <p>My problem, as discussed in comments elsewhere, was in not understanding what this diagram was conveying. It is of course true that this circuit does not operate as intended in the <span class="math-container">\$R_L \to \infty \$</span> limit. However, saying that misses what the diagram is really a shorthand for: it is depicting a small-signal scenario with respect to everything <em>other</em> than the MOSFET. This is a fairly standard convention in analog circuits textbooks (and, indeed, in Gray and Hurst, from where this book is taken). In this light, there is nothing wrong with this circuit: this <span class="math-container">\$R_L\$</span> is, in reality, the small signal resistance of the transistor current source which is biasing this CD stage.</p>
| <p>As per the title, how can this circuit operate in the <span class="math-container">\$R_L \to \infty\$</span> limit when it seems to mean that current flowing from the supply has nowhere to go? I ask because my text seems to analyze this circuit in this limit without mention.</p>
<p><a href="https://i.stack.imgur.com/h1BSB.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h1BSB.png" alt="enter image description here" /></a></p>
<p>More generally, in an emitter-follower or source-follower amplifier, is the resistor between the emitter or source necessary? Note that as per the diagram, <span class="math-container">\$R_L\$</span> should not be considered (I think) part of the load seen by this stage, but rather part of the stage itself.</p>
| How can this circuit operate in the \$R_L \to \infty\$ limit? | 2023-12-15T22:25:37.623 |
694391 | |batteries|battery-charging|automotive|lead-acid| | <p><strong>YES</strong> you absolutely <strong>should be worried</strong> about exceeding the voltage rating of the battery!</p>
<p>Overcharging Lead Acid batteries will damage them and can cause Hydrogen and Oxygen gas to form, leading to an explosion risk.</p>
<p>You should never, under any circumstances, provide a voltage higher than the rated peak voltage! A charging curve limits the current into the battery <em>until</em> the voltage rises to the peak battery voltage. Then, the voltage is limited to the peak voltage until the current drops (to 3-5% of the C rate for lead acid batteries).</p>
<p><a href="https://i.stack.imgur.com/H2Ub3.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/H2Ub3.jpg" alt="Charge stages of a lead acid battery" /></a></p>
<p>Standard "12V" Lead-acid batteries are six cells; the peak charge voltage is between 13.8 and 14.7V (at 25C, this value is temperature dependent); however prolonged time at this voltage will cause damage. After the current reaches the cutoff point (3-5% of the C rate of the cell) the voltage should be lowered to 13.5V to 13.8V (the "float voltage").</p>
<p>Diagram from the excellent Battery University. Read there article on Lead Acid charging for excellent detailed information <a href="https://batteryuniversity.com/article/bu-403-charging-lead-acid" rel="nofollow noreferrer">BU-403: Charging Lead Acid</a>. Please don't attempt to build a charger or anything like it until you understand in detail the considerations and the risks; <strong>you can seriously injure yourself or others</strong>.</p>
| <p>I'm looking to make a car battery charger for learning purposes. I have read through this <a href="https://electronics.stackexchange.com/questions/398833/car-battery-lead-acid-discharges-much-much-faster-than-it-charges">this post</a> and <a href="https://electronics.stackexchange.com/questions/176642/how-to-attain-a-topping-charge-for-a-lead-acid-battery/">this post</a>.</p>
<p>While I am still confused about certain things, it seems generally agreed upon that a safe charging current is 15 to 20% of the battery's capacity. For the battery I have in front of me right now (60Ah) that would represent 9 to 12A.</p>
<p>I've connected the battery to a bench power supply, and getting it to 10A seems to take about 17V. This seems quite in excess of even the car's ~14.4V when turned on.</p>
<p>To be clear, I know that constant-current bulk charging is only meant to happen to a certain point, and that the higher voltage will damage the battery if it leads to overcharging.</p>
<p>So, if I'm doing constant-current charging on a lead-acid battery, should I be concerned about reaching a voltage that is too high for the battery?</p>
| How much charging voltage can a lead-acid car battery handle? | 2023-12-15T23:44:32.613 |
694395 | |power-supply|identification|surface-mount| | <p>The regulator marked DKAA6 is probably a <a href="https://semic-boutique.com/wp-content/uploads/2021/01/RY7135C-Datasheet-V1.0.1.pdf" rel="nofollow noreferrer">Rychip RY7135/RY7135C</a> or <a href="https://www.hmsemi.com/index.php/Down/down/id/1065.html" rel="nofollow noreferrer">HMSEMI HM2235/HM2235H</a>.</p>
<p><a href="https://i.stack.imgur.com/eTCdT.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eTCdT.jpg" alt="enter image description here" /></a></p>
<p>I found this information in a thread on <a href="https://www.mikrocontroller.net/topic/543997" rel="nofollow noreferrer">mikrocontroller.net</a> where people describe bad performance of boards with the DK* regulator.</p>
<p><a href="https://i.stack.imgur.com/V7Wtt.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/V7Wtt.jpg" alt="enter image description here" /></a></p>
<p>(Source: <a href="https://www.mikrocontroller.net/topic/543997" rel="nofollow noreferrer">mikrocontroller.net</a>)</p>
| <p>I would like to get the chip part number, to get specifications, for a buck 5V regulator module. I have searched the letters and numbers on the chips (DKAA6 or 1AGCN), but no other reference could be found. These two codes can be found on two different boards, of identical design. Sellers of these are not responding to this question, or say they don't know.</p>
<p><a href="https://i.stack.imgur.com/vEKo0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vEKo0.jpg" alt="photo of PCB" /></a></p>
<p>How could I find out what this is?</p>
<p>image <a href="https://rads.stackoverflow.com/amzn/click/com/B09P7TJJGT" rel="nofollow noreferrer" rel="nofollow noreferrer">source</a></p>
| Identify SMD chip on a buck voltage regulator. Top marking DKAA6 or 1AGCN | 2023-12-16T01:19:49.160 |
694412 | |capacitor|safety| | <p>I think winny is likely right regarding the capacitor "remembering" the voltage applied and I'd like to elaborate a little on the mechanism.</p>
<p>Usually we can assume the <a href="https://en.wikipedia.org/wiki/Polarization_density" rel="nofollow noreferrer">polarization density</a> is proportional to the E-field in most dielectrics, i.e. <span class="math-container">\$\mathbf P = \epsilon_0\chi\mathbf E\$</span>, where <span class="math-container">\$\chi\$</span> is the <a href="https://en.wikipedia.org/wiki/Electric_susceptibility" rel="nofollow noreferrer">electric susceptibility</a>. However, it may happen that after a high voltage is applied for a long time, there is an additional residual polarization <span class="math-container">\$\mathbf P_0\$</span> in the direction of the E-field that remains for a long time even after the capacitor field is removed. This is seen in <a href="https://en.wikipedia.org/wiki/Ferroelectricity" rel="nofollow noreferrer">ferroelectric</a> materials, and has also been observed in polymers (see e.g. <a href="https://doi.org/10.1002/adfm.200902129" rel="nofollow noreferrer">N.-J. Yang et al.</a>) and attributed to the segmental motion of the large polymer molecules that occurs during the application of E-field, which then take a long time to randomize after the field has been removed (<a href="https://doi.org/10.1021/ma100209d" rel="nofollow noreferrer">X. Yuan et al.</a>). It could also be caused by the <a href="https://en.wikipedia.org/wiki/Piezoelectricity" rel="nofollow noreferrer">piezoelectric effect</a> if the capacitor temperature is elevated during operation for example, but that's speculation on my part, and is unlikely if the capacitor voltage develops over the course of days or longer rather than in minutes.</p>
<p>The charge on plates of the shorted capacitor with remnant polarization is proportional to the D-field in the dielectric:
<span class="math-container">$$\mathbf D=\mathbf P_0.$$</span>
This means there is charge on the plates even though the capacitor is shorted. After a long time (days/weeks), the remnant polarization decays to zero (or at least a smaller value). However, the D-field remains constant as long as the capacitor isn't leaky. It must then be provided by an E-field, i.e.
<span class="math-container">$$\mathbf D = \mathbf P_0 = \epsilon\mathbf E$$</span>
<span class="math-container">$$\mathbf E = \mathbf P_0/\epsilon.$$</span>
A non-zero E-field means a voltage has developed across the capacitor terminals. Note that the charge on the plates is still the same as before. Apparently there is a name for this phenomenon: <a href="https://en.wikipedia.org/wiki/Dielectric_absorption" rel="nofollow noreferrer">dielectric absoption</a>.</p>
| <p>High-voltage foil capacitors are shipped with shorting brackets to prevent the buildup of dangerous charge. But once in a circuit that is not energized, such a safety shunt will be an awkward complication, because NC switches for several kV are rather exotic.</p>
<p>In the circuit, I could tolerate a self-charging up until about 3 kV without any adverse effect. The passive shunt resistance is on the order of 100 MΩ, so there would be 30 µA of discharge happening at 3 kV.</p>
<p>I am wondering:</p>
<ul>
<li>What is the mechanism for the self-charging?</li>
<li>How much effective charging current could I expect worst case (say the capacitor terminals just end in the air with no additional precautions)?</li>
<li>Is the passive shunt resistance sufficient to prevent a dangerous self charging.</li>
</ul>
| What causes self-charging of foil capacitors? | 2023-12-16T08:13:00.013 |
694422 | |differential-amplifier| | <p>The following is my analysis.</p>
<p><a href="https://i.stack.imgur.com/Ea6i3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ea6i3.png" alt="enter image description here" /></a></p>
<p>Firstly, we analyse the left part of this circuit with the use of KCL.</p>
<p><span class="math-container">$$- \frac{V_{out2}} {R_D} = (V_{in1} - V_p) g_{m1} + \frac{V_{out2} -V_p} {r_{o1}}$$</span></p>
<p><span class="math-container">$$\Rightarrow - V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D}) =
V_{in1} g_{m1} - V_p (g_{m1} + \frac{1} {r_{o1}})$$</span></p>
<p><span class="math-container">$$\Rightarrow V_p = \frac{V_{in1} g_{m1} + V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D})} {g_{m1} + \frac{1} {r_{o1}}}$$</span></p>
<p>Secondly, we analyse the right part of this circuit with the use of KCL.</p>
<p><span class="math-container">$$ \frac{V_{out1}} {R_D} = -(V_{in2} - V_p) g_{m2} + \frac{V_p -V_{out1}} {r_{o2}}$$</span></p>
<p><span class="math-container">$$\Rightarrow V_{out1} (\frac{1} {r_{o1}} + \frac{1} {R_D}) =
- V_{in2} g_{m2} + V_p (g_{m2} + \frac{1} {r_{o2}})$$</span></p>
<p>We could substitute <span class="math-container">\$\frac{V_{in1} g_{m1} + V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D})} {g_{m2} + \frac{1} {r_{o2}}}\$</span> for <span class="math-container">\$V_p\$</span> from what we previously got.</p>
<p><span class="math-container">$$\Rightarrow V_{out1} (\frac{1} {r_{o1}} + \frac{1} {R_D}) =
- V_{in2} g_{m2} + \frac{V_{in1} g_{m1} + V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D})} {g_{m1} + \frac{1} {r_{o1}}} (g_{m2} + \frac{1} {r_{o2}})$$</span></p>
<p>Since we already assume <span class="math-container">\$g_{m1} r_{o1} \gg 1\$</span> and <span class="math-container">\$g_{m2} r_{o2} \gg 1\$</span></p>
<p><span class="math-container">$$\Rightarrow V_{out1} (\frac{1} {r_{o1}} + \frac{1} {R_D}) =
- V_{in2} g_{m2} + \frac{V_{in1} g_{m1} + V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D})} {g_{m1}} (g_{m2})$$</span></p>
<p><span class="math-container">$$\Rightarrow V_{out1} (\frac{1} {r_{o1}} + \frac{1} {R_D}) =
- V_{in2} g_{m2} + V_{in1} g_{m2} + V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D}) \frac{g_{m2}} {g_{m1}}$$</span></p>
<p><span class="math-container">$$\Rightarrow V_{out1} (\frac{1} {r_{o1}} + \frac{1} {R_D}) - V_{out2} (\frac{1} {r_{o1}} + \frac{1} {R_D}) \frac{g_{m2}} {g_{m1}}= V_{in1} g_{m2} - V_{in2} g_{m2} $$</span></p>
<p>We also assume <span class="math-container">\$r_{o1} \gg R_D\$</span> and <span class="math-container">\$r_{o2} \gg R_D \$</span></p>
<p><span class="math-container">$$\Rightarrow V_{out1} (\frac{1} {R_D}) - V_{out2} (\frac{1} {R_D}) \frac{g_{m2}} {g_{m1}}= V_{in1} g_{m2} - V_{in2} g_{m2} $$</span></p>
<p><span class="math-container">$$\Rightarrow (\frac{1} {R_D}) (V_{out1} - V_{out2} \frac{g_{m2}} {g_{m1}})
= V_{in1} g_{m2} - V_{in2} g_{m2} $$</span></p>
<p>For third equation.</p>
<p><span class="math-container">$$- \frac{V_{out2}} {R_D} = \frac{V_{out1}} {R_D}$$</span></p>
<p><span class="math-container">$$\Rightarrow - V_{out2} = V_{out1} $$</span></p>
<p>Substitute <span class="math-container">\$-V_{out2}\$</span> for <span class="math-container">\$V_{out1}\$</span></p>
<p><span class="math-container">$$\Rightarrow (-\frac{1} {R_D}) V_{out2} (1 + \frac{g_{m2}} {g_{m1}})
= g_{m2} (V_{in1} - V_{in2}) $$</span></p>
<p>The gain will be</p>
<p><span class="math-container">$$\frac{V_{out2} - V_{out1}} {V_{in1} - V_{in2}}
=
\frac{2 V_{out2}} {V_{in1} - V_{in2}}
=
\frac{2 g_{m2}} {-\frac{1} {R_D} (1 + \frac{g_{m2}} {g_{m1}})}
$$</span></p>
<p><span class="math-container">$$\frac{V_{out2} - V_{out1}} {V_{in1} - V_{in2}}
=
- \frac{2 R_D g_{m2}} {(1 + \frac{g_{m2}} {g_{m1}})}
$$</span></p>
<p>As we all know <span class="math-container">\$g_{m2} = 2 g_{m1}\$</span></p>
<p><span class="math-container">$$\frac{V_{out2} - V_{out1}} {V_{in1} - V_{in2}}
=
- \frac{2 R_D g_{m2}} {(1 + 2)}
=- \frac{4 R_D g_{m1}} {3}
$$</span></p>
| <p>Content is from the book <em>Design of Analog CMOS Integrated Circuits</em> (<a href="https://electrovolt.ir/wp-content/uploads/2014/08/Design-of-Analog-CMOS-Integrated-Circuit-2nd-Edition-ElectroVolt.ir_.pdf" rel="nofollow noreferrer">https://electrovolt.ir/wp-content/uploads/2014/08/Design-of-Analog-CMOS-Integrated-Circuit-2nd-Edition-ElectroVolt.ir_.pdf</a>) on Page 112.</p>
<p>Due to a manufacturing error, in the circuit of Fig. 4.19, <span class="math-container">\$M_2\$</span> is twice as wide as <span class="math-container">\$M_1\$</span>. Calculate the small-signal gain if the DC levels of <span class="math-container">\$V_{in1}\$</span> and <span class="math-container">\$V_{in2}\$</span> are equal.</p>
<p><a href="https://i.stack.imgur.com/Te2wg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Te2wg.png" alt="enter image description here" /></a></p>
<p>The following is my analysis. Since it's symmetrical, we can use virtual ground to analyze this circuit:</p>
<p><a href="https://i.stack.imgur.com/C3L2S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/C3L2S.png" alt="enter image description here" /></a></p>
<p>First, we only analyze the left part with KCL:</p>
<p><span class="math-container">$$ -\frac{V_{out2}} {R_D} = V_{in} g_{m1} + \frac{V_{out2}} {r_{o1}}$$</span>
<span class="math-container">$$\Rightarrow \frac{V_{out2}} {V_{in}} = -g_{m1} (R_D \parallel r_{o1})$$</span></p>
<p>Now we only analyze the right part with KCL:</p>
<p><span class="math-container">$$ -\frac{V_{out1}} {R_D} = -V_{in} g_{m2} + \frac{V_{out1}} {r_{o2}}$$</span>
<span class="math-container">$$\Rightarrow \frac{V_{out1}} {V_{in}} = g_{m2} (R_D \parallel r_{o2})$$</span></p>
<p>We combine this two equation into one:</p>
<p><span class="math-container">$$\frac {V_{out2} - V_{out1}} {2 V_{in}} = -\frac{1}{2} (g_{m1} + g_{m2}) R_D$$</span></p>
<p>Since <span class="math-container">\$g_{m2} = 2 g_{m1}\$</span>
<span class="math-container">$$\frac {V_{out2} - V_{out1}} {2 V_{in}} = -\frac{3}{2} g_{m1} R_D$$</span></p>
<p>But the book said:</p>
<p><span class="math-container">$$|A_v| = \frac{2 R_D} {1/g_{m1} + 1/ g_{m2}} $$</span>
<span class="math-container">$$|A_v| = \frac{4} {3} g_{m1} R_D$$</span></p>
| Should gain of this differential circuit be absolute value? | 2023-12-16T09:58:15.723 |
694428 | |current|connector| | <p>The two ratings given for a connector apply to two different properties of the connector.</p>
<p>The voltage refers to the highest safe voltage between the conductors in the connector. As the voltage increases above this rated voltage, the risk increases that the insulation will breakdown and the current arcs from one contact to the other.</p>
<p>The other rating is current, and has nothing to do with voltage rating. The current rating comes from that fact that all real world conductors have some resistance, even if it is small. Current flowing through the resistance of the connector will generate heat. The heat energy generated is I^2 * R. The connector can only dissipate so much heat energy, and will consequently warm up. The current rating is the highest current that the manufacturer has calculated to produce a reasonably small temperature rise in the connector. Running more current that that through the connector risks melting the conductors, insulation, or both.</p>
| <p>I'm currently learning electronics and there's something I cannot seem to figure out regarding connectors.</p>
<p>Let's say I have a male + female connector which are rated up to 250 volts and 5 amperes. This should then be capable of handling 1250 watts as far as I know.</p>
<p>Now, when using lower voltages such as 12V, in order to reach 1250 watts, it should be around 104 amperes.</p>
<p>Let's say you make a circuit which uses 12V@10A, which is about double the current rating of the 220V@5A.</p>
<p>When using lower current like here, can you just use that connector because of the wattage conversion since it uses 120W instead of 1250W, or do I need to find a different connector which is rated for that amperage, even though the current is much lower?</p>
| Can you use the same connector when using lower voltages but with higher current? | 2023-12-16T11:45:24.790 |
694430 | |operational-amplifier|amplifier| | <blockquote>
<p>what is the significance of this resistor [R12] between the two op-amps, what difference does it make in terms of performance.</p>
</blockquote>
<ol>
<li>In circuit 2, Input Common Mode rejection occurs right at the input. The current through R12 is determined entirely by the differential input voltage. All the common mode voltage subtracts to zero here. This means that subsequent amplification will not be influenced by input common mode voltage. This reduces the CMRR demands placed on the differential amplifier in Circuit 1.</li>
<li>Noise is reduced in two ways: 1) The high value resistors in circuit 1 are sources of thermal noise. Circuit 2 uses significantly lower value resistors. 2) The differential amp in circuit 2 has unity gain, so noise generated by the input amplifiers is not amplified.</li>
<li>The lower resistance values in curcuit 2 will allow faster response with circuit capacitances.</li>
<li>The differential amplifier in Circuit 1 will have a reduced bandwidth due to gain-bandwidth product (GBW) than the differential amplifier in Circuit 2. The input amplifiers have a wide unity gain bandwidth in Circuit 1, but a GBW restricted bandwidth in Circuit 2. How this affects time response depends on the chosen amplifiers.</li>
<li>Of course changing gain with one resistor (R12) is easier than changing gain with 2 matched precision resistors.</li>
</ol>
<blockquote>
<p>Why would anyone build an instrumentation amplifier when a differential amplifier with separated unity gain amplifiers as inputs seem to do the job well enough?</p>
</blockquote>
<p>If a differential amplifier with separated unity gain amplifiers as inputs <strong>does the job well enough</strong>, then there is no reason to do anything else.</p>
<p>If a differential amplifier with separated unity gain amplifiers as inputs <strong>does not do the job well enough</strong>, then use the instrument amp.</p>
| <p>I have two circuits. One is a differential amplifier with unity gain amplifiers for voltage control, and another is a instrumental amplifier. "Right" and "Left" are two 1V sine waves 180 degrees out of sync, with an offset of 2.5V.</p>
<p><a href="https://i.stack.imgur.com/m2Kdg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/m2Kdg.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/lAyQ9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lAyQ9.png" alt="enter image description here" /></a></p>
<p>I've noticed these two are very similar with the only difference being R12 connecting the two input op-amps. I am wondering what is the significance of this resistor between the two op-amps, what difference does it make in terms of performance. I know that the gain calculations are different, but I don't understand how it affects other performances such as response time.</p>
<p><strong>To Less Determined Readers:</strong> Why would anyone build an instrumentation amplifier when a differential amplifier with separated unity gain amplifiers as inputs seem to do the job well enough?</p>
| Differential Amplifier with Unity Gain Source vs Instrumentation Amplifier | 2023-12-16T11:59:56.117 |
694433 | |led-driver| | <p>That's a 2A switching regulator, whereas typically LEDs for lighting are on the order of ten to hundreds of milliamps per die. So yes, they're expecting you to parallel diodes. Outside of a relatively few, expensive high current diodes that driver will be overkill for series only LEDs.</p>
<p>This might seem strange, but over the years as LED manufacturing has improved bins have become much tighter even on typical products. Most manufacturers will now provide you with tightly matched forward voltages on very inexpensive parts. For example, I just picked a random 65mA LED on digikey that cost 6 cents in units of one (with digikey markup). The manufacturer provided binning 100mV wide by default. In this configuration if you run at the typical current and paired up the min and max Vf for that bin in parallel with zero resistance, you'll still be well under the maximum current. In reality you'd put a few diodes in series before running them in parallel, and the series combination will introduce resistance while individual Vf would tend to average out across the string, so current would be even better balanced. It is mainly when you go to eBay or Amazon as your LED supplier that putting diodes in parallel becomes dangerous.</p>
| <p>I'm using an LED driver to drive a few LEDs in parallel but I'm confused by the datasheet. The component is a <a href="https://www.diodes.com/assets/Datasheets/PAM2863.pdf" rel="nofollow noreferrer">PAM2863</a>. In the description, you can read "designed for driving single or multiple series connected LEDs", which would mean that the LEDs must be in series, but in the typical applications circuit you can see the circuit driving a few LEDs in parallel.</p>
<p>Is there any problem to use a parallel circuit?</p>
| Can I use LEDs in parallel with an LED driver? | 2023-12-16T12:08:33.250 |
694459 | |circuit-design|rectifier| | <p>Should be fine to put the 10 V DC supply in and bypass the rectifier; rectified 9 V AC produces around 11 V peak on the output (formula: 9 V·√2 - 2·Vf), so the circuitry after the rectifier can definitely handle the 10 V input.</p>
<p>My guess is that the diodes burned out because they were being used at a higher average power than normal--since each diode in a single-phase bridge rectifier is only used half of the time, they only need to handle half of the device's average current consumption. By swapping to a DC supply, two of the diodes now need to handle the full current used by the device, double what they did before, and the other two don't see any current.</p>
<p>Remove the burned out diodes (you can remove the good diodes too, they won't be doing anything) and just connect your DC supply to the rectifier's + and - output terminals.</p>
| <p>A few years ago, a family member asked me to look at a set of bookshelf speakers that stopped working. Inside I noticed obvious signs of a cooked component on the internal power supply, which was a separate board from the amplifier. The power supply outputted 9V AC to the amp circuitry.</p>
<p>At the time, I had no substitute 9V AC supply to swap in, so I used a 10V DC supply and soldered it up in place, figuring the extra volt wouldn't hurt and would simply compensate for voltage drop across the bridge rectifier which was no longer needed due to the DC supply. It worked fine for some time.</p>
<p>Recently I inquired about the speakers, and learned that they stopped working again. Inside, I see the 10V DC supply is working fine but 2 diodes of the (discrete) bridge rectifier are burned up on the amp module.</p>
<p>Assuming I want to stick with a DC-rectified supply, and not revert to AC with replacement bridge diodes on the module, what is the best course for the repair?</p>
| Modify or adapt a device using low-voltage AC to accept DC | 2023-12-16T16:39:42.037 |
694467 | |rf|filter|pcb-antenna|microstrip|balun| | <p>Actually, it doesn't, not as shown here, not by my understanding of it anyway.</p>
<p>(Which is partial, mind; I don't have active design experience with antennas, or field simulators. I've not seen references inconsistent with the below, at least. Comments welcome.)</p>
<p>What they have, is a combination matching/balun stub, approx. λ/4 long, which affords feedline choking (stub balun) and probably some impedance matching.</p>
<p>A λ/4 stub acts as a balun, because the common-mode short-circuit impedance of the ground plane, at the base end of the stub, acts like an open circuit λ/4 away; this is the effect of a λ/4 transmission line impedance transformer. Thus the dipole elements see a high impedance, and feedline currents are low.</p>
<p>Mind, that's not <em>quite</em> the whole story, because a λ/4 segment <em>in free space</em> is also a monopole; the ground plane dimensions matter, too (it's not an ideal short). Thus there will be a monopole characteristic to this as well, and some amount of feedline current (at the elements, and spreading out over the plane). Put another way: yes it's a choke, but part of that choke impedance is radiation resistance; it's not an ideal (infinite impedance) choke. (Or we might load the transmission line with ferrite beads, to the same end, but they also have impedance: core loss. No free lunch and all that.) But anyway, such non-idealities are often an acceptable compromise; maybe reducing feedline current by say 20dB is good enough, or of having a similar reduction in error in the radiation pattern (the monopole mode has a radial radiation pattern which would spoil the directivity of the axially-oriented Yagi pattern).</p>
<p>Or at least, <em>it seems like that was the intent</em> -- but they partly to completely flubbed the dimensions; or just didn't care, or, the oversized ground plane covers it all up anyway.</p>
<p>So... this answer unfortunately becomes an intuitive but inexperienced (on this topic) engineer's analysis, which is nonetheless pulling seemingly-reasonable critiques against this thing, and, what <em>you</em> had hoped was an introduction to a neat little transmission line structure, is actually about almost anything <em>but</em> that.</p>
<p>And <em>I want to get to that</em>, but to get there, first I have to address how it's <em>wrong</em>.</p>
<p>See, the actual stub length is more like λ/8 -- it's 16mm long, while λ/4 is 30.6mm. This is more understandable for the differential mode (which is relevant for impedance matching, but not balun action), because dielectric loading has a larger effect there (putting the DM length closer to, say, λ/6, maybe even λ/5). But the common mode doesn't have as much loading -- I...don't have an estimate, but my <em>guess</em> is somewhere in the 10% range, i.e., due to PCB material being nearby, the CM stub length looks more like, say, 17, 18mm?</p>
<p>So it's a lot harder to say it's nearly enough for a proper λ/4 balun function.</p>
<p>So at last: <strong>the taper section.</strong></p>
<p>What they have here is, more or less just a nod to the proper method, but because it's so small (~12mm!), it just, I can't see how it matters either way -- I think they were neither forced to remove nor improve it. And that's assuming they did much if any optimization on this thing at all -- which I kind of suspect they didn't? The fairly flat 10-15dB RL looks pretty clearly like mismatch to me, that could be easily cleared up with slightly different microstrip/stripline dimensions, say; or at least a passive L-match within the plane region. (From what I've read, they should also be able to squeeze another, 3dB or so? from a Yagi of these rough specs, and maybe without compromising much on bandwidth even.)</p>
<p>What a taper section <em>can</em> do, when it's multiple λ/4s long, is afford a smooth transition between asymmetrical (unbalanced) microstrip (or shielded stripline, or coax), and the -- whatever you want to call it: dielectric-supported differential stripline (broadside coupled) without ground; symmetrical microstrip; parallel-plate transmission line; etc. I haven't found a good (easily googleable) name for it yet, honestly. In any case, between an unbalanced or asymmetrical structure, and a balanced (symmetrical) one.</p>
<p>Hand-wavingly, you can imagine it as, at first the image current from the microstrip is well contained within the margins of the ground plane, i.e. the plane is effective shielding, and the mode is unbalanced; as the plane width shrinks, a bit more and more of that image current reaches its edge, and thus projects magnetic field out into free space -- most of which is canceled out by the trace's field, but some of it wraps around both; and vice versa. Eventually, the pair becomes symmetric, wrapping around each other equally; and by distributing the fields lengthwise (again, several λ/4s), common mode current is reduced while we transition from "un" to "bal". The confined (differential) wave energy doesn't see a thing, conditions changed too gradually to notice; and little is coupled into the common mode for the same reason.</p>
<p>The exact profile of this taper, depends on the desired bandwidth. We could taper it in hard steps, which would give transmission line modes -- peaks and valleys every λ/4 as waves constructively and destructively interfere along its length. Consider the case where there's microstrip on wide ground plane (as shown), then a rectangular section λ/4 long, then balanced pair for unlimited length. For some width of that intermediate section's ground backing, there will be the case that it emits, effectively, a quarter of the incident voltage as common mode voltage. The same will be true of the intermediate-to-balanced transition, and so as a wave propagates from "un" to "bal", the two quarter CM voltages reinforce, and a full balun is had. Conversely, at each transition, a CM wave is launched backwards, and the two transitions are 1/2 wave apart so they cancel out (meaning ~no feedline current is impressed onto the ground plane).</p>
<p>We can subdivide the rectangle into smaller sections, and repeat the process. Say we do three sections, and do 1/4 Vcm per step; to a wave of double wavelength, the structure averages out to the same thing as before, but now we have its 2nd harmonic well-behaved as well!</p>
<p>And so on and so forth; as we continue to subdivide and smooth the outline, we can tweak the response for better balance, less radiation, etc. for wider and wider bandwidths. Here's an article claiming, well, over 10:1 I'd say, while their interpretation of the measurements seems a bit optimistic; still, the low RL seems nice, and the amplitude balance isn't terrible. <br />
<a href="https://www.semanticscholar.org/paper/An-extremely-wideband-tapered-balun-for-application-Bah-Qin/c12c1881140a254a8bc50641cc2a57af9b2ebb54" rel="nofollow noreferrer">Bah, Alpha O. et al. “An extremely wideband tapered balun for application in tightly coupled arrays.” 2016 IEEE-APS Topical Conference on Antennas and Propagation in Wireless Communications (APWC) (2016): 162-165</a> Note that their overall length is about a wavelength at 2GHz, which is also where it starts looking good.</p>
<p>...</p>
<p>A final note -- mind that my use of "wrong" is rather soft here: relative to whatever design principles I've picked up on this topic. Again, I've not done designs myself, so I'm just picking at ideas here. I'm also not claiming this post as highly factual, for this reason. Mind even further, design principles are just that, things you can do, or patterns that repeat often enough to seem useful, but <em>the final arbiter is how it measures.</em> On the upside -- the outward effect of such discrepancies, reduces to just feedline current, radiation pattern, and impedance mismatch. And, I suppose, suboptimal overall dimensions (that ground plane section is huge..!).</p>
| <p><a href="https://www.ti.com/lit/an/swra350/swra350.pdf?ts=1702713055579" rel="nofollow noreferrer">This document</a> shows a planar microstrip/stripline balun to convert the balanced feed line input to feed the (unbalanced) dipole on a microstrip Yagi antenna.</p>
<p><strong>Question:</strong> How does coupling to a tapered ground plane provide the means to convert an unbalanced line to a balanced line?</p>
<p><a href="https://i.stack.imgur.com/Z6AFb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Z6AFb.png" alt="yagi pcb balun" /></a></p>
<p><a href="https://i.stack.imgur.com/z7UX0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/z7UX0.png" alt="PCB balun" /></a></p>
<p>(Linked <a href="http://web.ecs.baylor.edu/faculty/baylis/eel4421/baluns-f07.pdf" rel="nofollow noreferrer">from University of South Florida</a>, but this is <em>not</em> a homework question.)</p>
| How does a planar balun work (ie, microstrip <-> stripline)? | 2023-12-16T19:31:07.093 |
694474 | |motor|induction-motor|induction| | <p>The <a href="https://en.m.wikipedia.org/wiki/Induction_generator" rel="nofollow noreferrer">Wikipedia article</a>, at time of writing, seems to contain an error:</p>
<blockquote>
<p>The generating mode for induction motors is complicated by the need to excite the rotor, which being induced by an alternating current is demagnetized at shutdown <em>with no residual magnetization to bootstrap a cold start.</em> It is necessary to connect an external source of magnetizing current to initialize production. The power frequency and voltage are not self regulating.</p>
</blockquote>
<p>[my emphasis]. Arguably this construction is of the type "here are assumptions X, from which we conclude Y", and the error is merely in not making the assumptions contingent ("<em>if assumptions X</em>, then we conclude Y").</p>
<p>This arguably might even be more of a style choice than an outright miscommunication, as listing ones' assumptions indicates to the reader what conditions can contradict the conclusion. But it's not a generous or clear style, as novice readers, who aren't yet comfortable enough with a given subject, won't know when something is a postulate or an assumption, or when to question then (and which ties in with the general subject of critical thinking).</p>
<p>Anyway, my understanding of it is, either due to <em>residual magnetization in the rotor</em>, random electrical noise, or the presence of a starting circuit, and in combination with capacitors to cancel out the stator's inductance, both startup and continuous operating requirements are met, and generation is had.</p>
<p>Mind, I haven't built or installed any such generators myself, so this is hearsay (notice my lack of citations here). From my knowledge of magnetic materials (silicon steels have pretty generous remanence, though that will come down a fair bit when including the effect of air gap, coercivity, and ringdown, but nonzero still does not equal "zero"), it sounds highly plausible.</p>
<p>It probably also matters if the load is small during startup, so it doesn't need to regenerate field into an electrical load at the same time.</p>
<p>As for a mechanical explanation, note that the torque curve is symmetrical around zero slip, so when generating at a given torque, the output frequency slips higher than the rotational frequency, by the same amount it would drift below due to an equal <em>loading</em> torque instead.</p>
<p>It is also important to note that the reactive current of an induction motor is roughly constant (independent of torque, plus or minus), and mainly the active power (real current) varies with load, obviously being proportional around zero (load current for loading torque, generated current for generating torque).</p>
<p>The reactive current is a consequence of stator inductance (essentially the inductance due to the air gap, assuming the rotor is a solid hunk of electrical steel with no squirrel cage; which is what it looks like when it is spinning synchronous to the stator field i.e. at zero slip), plus the in-phase component of the rotor's field with respect to the stator (in the rotating reference frame, which is normally equal by the same reason).</p>
<p>When excitation is provided, the consequence either way is simply an inductive current draw, from the constant-voltage, fixed-frequency source; when self-generating however, the reactive current impresses a phase and thus angle shift on the induced rotor current, damping it out over time. If enough capacitance is connected, to at least cancel out the inductive current, then the loading phase angle acts to advance the rotor's induced current, building it up over time instead.</p>
<p>Thus, the load must remain capacitive as seen by the generator, otherwise a neutral or lagging current phase causes the rotor current to decay over time. A capacitive load gives rising rotor current, up to the limit determined by magnetic saturation of the rotor and/or stator.</p>
<p>Since (stand-alone) output is defined by slip and magnetic saturation, not only is the frequency fairly poorly defined (load dependent), but voltage is frequency (and thus load) dependent as well. Normally, motors are designed quite close to saturation, so the generator-mode output of a common induction motor isn't much more than the nameplate rating, and they are usable this way, if a bit crude.</p>
| <p>I am thoroughly confused again. I am at the point of relegating electricity as a black box, a mysterious phenomena I'll never comprehend. I just saw a YouTuber build an elaborate water mill for generating electricity. The water wheel spins a standard 3-phase squirrel cage motor wired to 3 capacitors (Linked a few seconds of another video strictly for reference). I didn't see a single comment questioning his generator, leading me to question my own understanding of induction motors and electricity in general.</p>
<p>From my simple understanding, a squirrel cage induction motor doesn't use permanent magnets either on the rotor or the stator for inducing current on the windings. Instead, alternating current is applied to the windings to produce a rotating magnetic field which mechanically spins the rotor. In principle, it converts electric energy into mechanical energy, not the other way around.</p>
<p>If my understanding is correct, then a squirrel cage induction motor cannot convert mechanical energy into electrical energy. Unlike a DC motor, it has no permanent magnets and relies on purely on electromagnetism. Spinning the rotor won't induce current in the stator's winding since the rotor doesn't have a magnetic field.</p>
<p>It could be argued the stator windings could be energized to induce a magnetic field to induce current on the rotor. However, the rotor's bars are shorted, and electrically isolated unlike a DC motor uses brushes with no output path.</p>
<p>There's also possibility that the term <em>induction</em> is applied liberally to motors and generators that are not squirrel cage motors.</p>
<p>Is my understanding incorrect?</p>
<p>Video demonstration of an induction motor running as a generator</p>
<p><a href="https://youtube.com/shorts/Zz0GJa_msjY?si=hqeqisEqOx-bOfNn" rel="nofollow noreferrer">https://youtube.com/shorts/Zz0GJa_msjY?si=hqeqisEqOx-bOfNn</a></p>
| Can a 3 phase Induction (squirrel cage) motor generate electricity? | 2023-12-16T20:07:59.550 |
694475 | |amplifier|speakers|common-emitter|biasing|capacitive-coupling| | <p>Apart from the problem of overdriving the amplifier you also have the problem of overloading the amplifier.</p>
<p>In a comment you state that you are connecting the amplifier to a 1 W speaker. Generally a speaker like this will have an impedance somewhere between 4<span class="math-container">\$\Omega\$</span> and 32<span class="math-container">\$\Omega\$</span>. Your amplifier will not be able to drive this efficiently as it has a much higher output impedance, roughly <span class="math-container">\$R_C\$</span>, 1200<span class="math-container">\$\Omega\$</span> (in simulation I get 1.13k).</p>
<p>So how much voltage does that allow the speaker to get? The load forms a voltage divider with the output impedance, imagine your 9 V supply connected to 1200<span class="math-container">\$\Omega\$</span> in series with say 16<span class="math-container">\$\Omega\$</span>. The voltage across the 16<span class="math-container">\$\Omega\$</span> would be:
<span class="math-container">$$
9V\times \frac{16\Omega}{1200\Omega+16\Omega} = 118mV
$$</span>
which gives you less than a milliwatt of power.</p>
<p>You can see this in the AC load line, in the first image the load is 100k and the AC load line (green) extends above and below the Q point quite a bit.
<a href="https://i.stack.imgur.com/1tQyA.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1tQyA.jpg" alt="Amplifier bias light load" /></a></p>
<p>In the second image with a 16<span class="math-container">\$\Omega\$</span> load the AC load line barely gets above the Q point, there's no room for the voltage to swing very far.
<a href="https://i.stack.imgur.com/qnxL2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qnxL2.jpg" alt="Amplifier bias high load" /></a></p>
<p>So even if you solve the overdrive problem by reducing the gain (removing emitter bypass cap) or attenuating the input signal you're still not going to get a useful output into a typical speaker. A low impedance speaker needs to be driven by an amp with a low impedance output, typically emitter followers (common collector) in a push-pull arrangement. A common emitter amplifier such as this is meant to provide voltage gain into a high impedance load.</p>
| <p>I am new to learning about electricity and I was trying to make a common emitter amplifier.</p>
<p>I did some research and I tried to make the circuit. The circuit amplified the signal but the signal was not clear and was cutting out. The signal I am trying to amplify is audio from an iPad headphone jack. The circuit diagram is below.</p>
<p>Does anyone know why this is happening and how I can fix this? Thanks.</p>
<p><img src="https://i.stack.imgur.com/hySYg.png" alt="schematic" /></p>
<p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fhySYg.png">simulate this circuit</a> – Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a></sup></p>
<p>(Vcc = 9 V)</p>
| CE Amplifier problem | 2023-12-16T20:11:24.130 |
694485 | |vlsi|solid-state-devices|device| | <p>I can take a stab at trying to answer your question.
In both digital and analog design it is helpful/often necessary to have a model of the load that any stage is driving. If the load is a linear network, this load can be replaced by an equivalent driving point impedance that the load presents-- the current and voltage relationship at the terminals of the load (Thevenin equivalent).</p>
<p>In the case of digital gates, we would like to model our loading gates with some equivalent impedance that will give the designer a sense of what kind of load the driving gate is being presented. To be useful in a design context, we would like this impedance to be as simple as possible, and usually abstract it into a lumped capacitance to ground. This lumped capacitor to ground is meant to mimic the <em>I-V</em> characteristic that the actual loading gates would present.</p>
<p>From a network theory perspective, the Thevenin equivalent is only applicable if the network we are deriving an equivalent for is linear and interacts with the rest of network exclusively through two terminals. It's equivalence comes from the fact that as long as we maintain the same <em>I-V</em> relationship at those two terminals, the rest of the network "doesn't know" that we have changed anything. This is what we are attempting to do with our lumped capacitor to ground. We are trying to present a load to our driving gate that will mimic the same <em>I-V</em> relationship as the actual loading gates.</p>
<p>The issue is two-fold. First, our network of loading gates is extremely non-linear. The MOS transistors we are using are intentionally driven into the nonlinear regime to provide the "digital" behavior that we want. Second, even if our load network was linear, if our load network contains more than one reactive element (assuming that the initial conditions can be set independently/ more than one pole) then we can not abstract it away into a single lumped capacitor--- we would require a higher-order impedance. Both of these clash with our desire for a tractable load impedance for first-pass design/analysis.</p>
<p>The solution, as you have seen, is to to try and create a lumped capacitance that mimics the loading gates, in the sense that we try and create a lumped capacitance that will present the same <em>I-V</em> relationship to our driving gates that the loading gates would have. Fig 2.11 is showing how the equivalent lumped capacitance to ground from a MOS gate varies significantly depending on the transitions at the other terminals. Intuitively this is because the voltages across the terminal capacitors are different in the different scenarios (in response to the same gate voltage), resulting in different currents, and hence different equivalent capacitances. The approximation being made is that we view the <em>I-V</em> characteristic that the loading gate presents for a given set of drive and transitions internal to the loading gate, and mimic that <em>I-V</em> characteristic with an equivalent capacitor. If we wanted to design conservatively, we could take the worst case scenario, or we could intelligently choose the lumped capacitance based on what transitions we expect at that gate. You cannot, however, completely/accurately describe the dynamics of the loading network with just one capacitor, since the <em>I-V</em> characteristic that the loading gates present is not a linear function, unlike a capacitor which has a linear <em>Q-V</em> characteristic. In essence the lumped capacitor is a compromise we make as designers to make the design process tractable, but we must always keep in mind that it is merely a model that if used incorrectly can lead to erroneous results.</p>
| <p>A MOSFET is, in reality, a four-terminal device with capacitances between each pair of terminals:</p>
<p><a href="https://i.stack.imgur.com/EeTgh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EeTgh.png" alt="enter image description here" /></a></p>
<p>These capacitances are, of course, the standard MOSFET intrinsic and extrinsic differential capacitances (i.e. they are defined as <span class="math-container">\$C = dQ/dV \$</span> where <span class="math-container">\$Q\$</span> is the charge on one of the terminals and <span class="math-container">\$V\$</span> is the voltage difference between said terminal and the other terminal).</p>
<p>That being said, it is well known that when we do digital design, we sweep a lot of this under the rug. Indeed, in their <em>CMOS VLSI Design</em>, Weste and Harris claim that</p>
<blockquote>
<p>(1) It is convenient to view the gate capacitance as a single-terminal capacitor attached to the gate (with the other side not switching [this is in turn equivalent to saying that the capacitance's other node is ground, as discussed <a href="https://electronics.stackexchange.com/questions/685863/why-doesnt-voltage-on-one-terminal-of-a-capacitor-matter">here]</a>). (2) Because the source and drain actually form second terminals, the effective gate capacitance varies with the switching activity of the source and drain. Figure 2.11 shows the effective gate capacitance in a 0.35 um process for seven different combinations of source and drain behavior [Bailey98].</p>
</blockquote>
<p>Figure 2.11 is from the Bailey paper (D. Bailey and B. Benschneider, “Clocking design and analysis for a 600-MHz Alpha microprocessor,” JSSC, vol. 33, no. 11, Nov. 1998, pp. 1627–1633) and is provided below:</p>
<p><a href="https://i.stack.imgur.com/pZqKf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pZqKf.png" alt="enter image description here" /></a></p>
<p>Even after reading the Bailey paper and the sources cited therein, I am still not sure I understand the nature of the approximations being made. I think it is fair to say that there are two approximations being made though. I am thus led to the following two questions. As an example to illustrate, I will imagine an inverter switching to 0 on its input (1 on its output, though this isn't relevant for the question on the gate here) and focus on the NMOS transistor therein.</p>
<p>(1) The first approximation, I believe, is that with respect to the Gate node (i.e. if there is some circuit driving this node) then we can delete the capacitances <span class="math-container">\$C_{gs}, C_{gd}\$</span> and then there is some equivalent capacitance <span class="math-container">\$C_g\$</span> which we can put between the Gate node and ground such that, for a given stimulus from the driving circuit to our inverter, the delay until the Gate node charges and discharges is the same as it otherwise would be if we hadn't made this approximation. My question is whether this understanding is correct (i.e. is it the correct understanding of Weste and Harris's comment I have labelled (1) above) and, if so, how we justify this? Also, is it fair to say that this "equivalence" is only in the sense of producing the correct delay for the Gate, but not that it will have the exact same temporal dynamics (e.g. the rise time may differ even if the delay is the same).</p>
<p>(2) The second approximation, I believe, is that we can use the MOS differential capacitances to determine this delay. Given that the relevant charge-storage elements are nonlinear, I suppose a strictly rigorous argument for the time it would take to discharge the gate is the time <span class="math-container">\$t_d\$</span> at</p>
<p><span class="math-container">$$I_{out}(V_g(t)) = \frac{dQ_{gs}(V_{gs}(t))}{dt} + \frac{dQ_{gd}(V_{gd}(t))}{dt} = \frac{dQ_{gs}}{dV_{gs}}\frac{dV_{gs}}{dt} + \frac{dQ_{gd}}{dV_{gd}}\frac{dV_{gd}}{dt}$$</span>
<span class="math-container">$$ \equiv C_{gs}\frac{dV_{gs}}{dt} + C_{gd}\frac{dV_{gd}}{dt}$$</span>
where <span class="math-container">\$I_{out}\$</span> would is some function of the gate voltage on our inverter depending on the nature of the PDN in the driving circuit. We see that we have reduced the problem to the form involving these nonlinear differential capacitances. This is a differential equation for <span class="math-container">\$V_g\$</span> (uncoupled if <span class="math-container">\$V_s,V_d\$</span> are constant, though that wouldn't be the case in our inverter. The "data-dependence" noted in Fig 2.11 comes, I assume, from the dependence on <span class="math-container">\$V_s\$</span> and <span class="math-container">\$V_d\$</span>?). Is this basically how one uses the nonlinear differential capacitances of a MOSFET to calculate delay and other times of interest?</p>
<p>If there are any sources that really carefully analyze the approximations which get made here, it would be more than enough to point me to that. Sorry if this question is unfocused. I've tried to spend a lot of time thinking about it but there really does seem to be a lot going on with device modelling here, and in digital design texts it's really swept under the rug. In the end I'm just looking to understand "in principle" what's going on. Thank you!</p>
| Justification for equivalent gate capacitance simplification in digital circuits | 2023-12-16T21:17:15.973 |
694488 | |connector| | <p>My <a href="https://connectorbook.com/identification.html?m=NN&n=pin_socket_comm_matenlok_084&s=Commercial%20MATE-N-LOK%205.08" rel="noreferrer">Identiconn utility</a> identifies it as a TE Commercial MATE-N-LOK.</p>
<p>You can buy them here: <a href="https://www.digikey.com/short/dncjj5vn" rel="noreferrer">https://www.digikey.com/short/dncjj5vn</a></p>
<p><a href="https://i.stack.imgur.com/ojDHz.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/ojDHz.jpg" alt="enter image description here" /></a></p>
<p>{<a href="https://www.te.com/usa-en/plp/rectangular-power-connectors-commercial-mate-n-lok/YG2dEX25kx.html" rel="noreferrer">Source: TE</a>}</p>
| <p>I am trying to build a custom pump controller circuit but can't figure out what this connector is called. Can anyone help identify it?</p>
<p><a href="https://i.stack.imgur.com/SEgdM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SEgdM.png" alt="enter image description here" /></a></p>
| What is this connector type called? | 2023-12-16T21:56:33.397 |
694514 | |kirchhoffs-laws| | <p><span class="math-container">\$I_1\$</span> cannot pass through the voltmeter, so all of it must pass through the 4 ohm resistor. (KCL)</p>
| <p>We know that the voltage of the volt meter gives 2V and the ammeter is 2A. We must find I1 and I2, and I'm at a loss as to what the voltage can be used for. Especially considering the correction of this exercice, uses R1 * I1 -6 = 2. Why would I1 pass through R1, wouldn't it be another current completely?</p>
<p><a href="https://i.stack.imgur.com/2Erm4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2Erm4.png" alt="enter image description here" /></a></p>
<p>My mistake, R1 being 4 ohm, R2 1 ohm and we are looking for R</p>
| A student who should know how to do this, kirchhoff's edition | 2023-12-17T02:53:49.777 |
694515 | |circuit-analysis|thevenin|norton| | <blockquote>
<p>Could someone please explain why the 16ohm resistor is ignored here</p>
</blockquote>
<p>Yes. See the steps shown below:</p>
<p><img src="https://i.stack.imgur.com/NY0fT.png" alt="schematic" /></p>
<p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fNY0fT.png">simulate this circuit</a> – Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a></sup></p>
<ol>
<li><p>You may choose one and only one node to call <span class="math-container">\$0\:\text{V}\$</span>. I've chosen <span class="math-container">\$V_\text{B}\$</span>. This does NOT in any way change the circuit. It's just getting rid of a wire on the schematic and using the ground label, instead. Nothing else implied.</p>
</li>
<li><p>There's no need to show the voltage supply itself on the schematic. You can simply label the node with that voltage and get rid of the voltage supply icon. It's just an extra distraction, so why not get rid of it to clean things up a bit more?</p>
</li>
<li><p>Breaking up the node into two, with the exact same voltage doesn't actually change the circuit <em>if both are connected to the same voltage supply</em>. I think you must agree with that. No change. It's just another way of writing the same thing.</p>
<p>But take this just one step further. Imagine creating two separate voltage supplies, each with the exact same voltage and reference point (ground.) Certainly would not do that due to cost and space and weight, etc., of course. But technically <em><strong>could</strong></em> do that. The circuit would still behave the same. So there's no harm in mentally taking this step to simplify the schematic, further. And it won't change the behavior of node <strong>A</strong>.</p>
</li>
</ol>
<p>So you can see that these are two different circuits, each having no impact on the other. Which means you can throw away the one that includes <span class="math-container">\$R_1\$</span> and focus on the other one:</p>
<p><img src="https://i.stack.imgur.com/6toLE.png" alt="schematic" /></p>
<p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2f6toLE.png">simulate this circuit</a></sup></p>
<p>It's just a resistor voltage divider, now.</p>
| <p>I am really confused by the concept of ignoring resistors while doing Thévenin theorem, especially for this problem.</p>
<p>Some online videos, like <a href="https://savvy.engineer/question/Consider-the-following-circuit-The-Thevenin-equivalent-resistance-Omega-at-Points-A-B-is-most-nearly-A-8B-12C-20D-26" rel="nofollow noreferrer">this one</a>, and other examples that I watched seem to suggest otherwise.</p>
<p>Could someone please explain why the 16 Ω resistor is ignored here:</p>
<p><a href="https://i.stack.imgur.com/yDrDG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yDrDG.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/e8Bpr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/e8Bpr.png" alt="enter image description here" /></a></p>
<p>The example link above seems to contradict the solution given by the book.</p>
| Thévenin equivalent voltage: why ignore the 16 Ω resistor? | 2023-12-17T03:20:28.673 |
694519 | |mosfet|leakage-current| | <p>As discussed in the comments, what I was missing is that <span class="math-container">\$V_{gs} - V_t < 0 \$</span> by definition in the subthreshold regime, so it is not correct to say there is a factor of <span class="math-container">\$e^{1/T}\$</span>. Rather, there is a factor of <span class="math-container">\$e^{-1/T}\$</span>. It then follows that the <span class="math-container">\$T^2\$</span> factor out front dominates in any and all but the most pathloogical cases (where maybe, for example, <span class="math-container">\$\mu\$</span> decays very strongly with <span class="math-container">\$T\$</span>).</p>
| <p>As far as I know, the source-drain subthreshold leakage current in a MOSFET is given by</p>
<p><span class="math-container">$$I_{ds} = e^{1.8} \frac{\mu C_{ox} W}{L} \left( \frac{kT}{q} \right)^2 e^{\frac{V_{gs} - V_t}{n \frac{kT}{q}}} \left( 1 - e^{-\frac{V_{ds}}{\frac{kT}{q}}} \right)$$</span></p>
<p>where everything should be clear except perhaps that <span class="math-container">\$n\$</span> is some ideality coefficient. Now it's known that the threshold voltage decreases with increasing temperature, but what I can't see is how this (or any other) effect dominate the two exponentials which involve temperature, which go as <span class="math-container">\$e^{1/T}\$</span> and <span class="math-container">\$(1-e^{-1/T})\$</span>, respectively. Both of these are monotonically decreasing with temperature as far as I can tell. Perhaps someone is able to parse the <span class="math-container">\$T\$</span> dependencies better than I am, however.</p>
<p>I am looking for a mathematical argument based on the formula above.</p>
| Why does subthreshold leakage increase with increasing temperature? | 2023-12-17T04:30:54.990 |
694538 | |mosfet|simulation|buck|spice|proteus| | <p>One problem that is immediately obvious from your schematic is that the arduino does not share a ground with your external circuit. Without such a common ground, (or some special circuitry) you cannot effectively transmit a signal from the arduino to the external circuit.</p>
| <p>This is a circuit that will take any Vin > 5 (through a diode bridge against reverse connection) and buck convert it to 5 volts. But, there is some error related to PMOSFET when I try to simulate it on Proteus.</p>
<p>It gives the error:</p>
<p><a href="https://i.stack.imgur.com/ZmBzq.png" rel="noreferrer"><img src="https://i.stack.imgur.com/ZmBzq.png" alt="Error" /></a></p>
<p><a href="https://i.stack.imgur.com/Si44V.png" rel="noreferrer"><img src="https://i.stack.imgur.com/Si44V.png" alt="Circuit Scheme" /></a>
<a href="https://i.stack.imgur.com/UX1ca.png" rel="noreferrer"><img src="https://i.stack.imgur.com/UX1ca.png" alt="My Arduino code" /></a></p>
<p>What is wrong with my circuit?</p>
| Why doesn't my MOSFET work? | 2023-12-17T11:06:45.450 |
694542 | |circuit-analysis|circuit-design|circuit-protection| | <p>It is a bit strange, but <strong>EPCOS</strong> arrestors marked as red have positive polarity - they are unipolar instead of bipolar. In this case plus from arrestor goes to a plus of capacitor which is fine.</p>
<p>By specification one of two arrestor's electrodes should be 'grounded', connected to the ground. It gives me an idea they are always bipolar. However, technically surge breakdown should be channelled to anything capable for it, so a capacitor will fit this use case as well.</p>
| <p>I am repairing 'Siemens SITOP power 20' power supply. I have to draw its circuit to better understand it and a notice a bizarre design decision. Can you help me to understand it?</p>
<p>It has an arrester which one connection stands on the 'plus' of DC and another connection should be a 'minus'. That 'minus connection is goes through induction coil and connected to a 'plus' (!) of the capacitor. How does it possible?</p>
<p>On photos please note plus and minus from AC-DC bridge and how it is connected to arrester and capacitor.</p>
<p>(220V/380V AC as an input + one neutral, 3 thermistors and 3 varistors as protection layer, next is three-phase rectifier (diode bridge for 3 phase current) which converts AC to DC, next is a filter which 'plus' goes to arrestor)</p>
<p><a href="https://i.stack.imgur.com/FoWII.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FoWII.jpg" alt="1st side" /></a>
<a href="https://i.stack.imgur.com/FP9If.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FP9If.jpg" alt="2nd side" /></a>
<a href="https://i.stack.imgur.com/0hmU5.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0hmU5.jpg" alt="reversed engineered so far circuit, have to do it on plain paper because of lack all required components in top list circuit software" /></a></p>
| Arrester 'minus' connected to the 'plus' of a capacitor | 2023-12-17T13:21:55.670 |
694570 | |power-supply|diodes|reverse-polarity| | <p>This diode configuration relies on having a fuse, PTC or other protective device upstream (i.e. on the supply side). If the supply is reversed then the fuse will blow and the circuit downstream of the diode will only see a small reverse voltage, which is typically survivable.</p>
| <p>Regarding making a dual supply from two single supplies, I see two Schottky diodes are placed between the supply terminals in reverse biased way as in the following example I drew. The diodes are randomly chosen by me just for illustration, so they might not be sized correctly:</p>
<p><a href="https://i.stack.imgur.com/gJlH3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gJlH3.png" alt="enter image description here" /></a></p>
<p>The reasoning behind using such diodes is given as: "The diodes also prevent possible reverse voltage damage if one power supply is off while the others are on."</p>
<p>When I remove PSU1 in the above schematic or make its voltage to zero and simulate it, I don't see any difference such as excessive current etc. I have spent several hours searching for a detailed explanation but couldn't find any.</p>
<p>Can someone show how in a scenario the diodes protect the supplies by maybe simulating or if not by showing the routes of the currents?</p>
| How do the protection diodes actually protect the supplies in this case? | 2023-12-17T18:19:52.200 |
694595 | |identification| | <p>It is a <a href="https://en.wikipedia.org/wiki/Reed_switch" rel="noreferrer">reed switch</a>. It's a switch that closes in the presence of a magnetic field. If you look, you'll find a magnet nearby.</p>
<p>You can buy a replacement here: <a href="https://www.digikey.com/en/products/filter/magnetic-reed-switches/193" rel="noreferrer">https://www.digikey.com/en/products/filter/magnetic-reed-switches/193</a></p>
<p><a href="https://i.stack.imgur.com/7kxr0.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/7kxr0.jpg" alt="Reed switch" /></a></p>
<p>{Source: <a href="https://en.wikipedia.org/wiki/Reed_switch#/media/File:Reed_switch_(aka).jpg" rel="noreferrer">wikipedia</a>}</p>
| <p>Can someone identify this electrical component we found in a Little Doctor LD-250U Ultrasonic Nebuliser.</p>
<p>Seems to be a flat metal wire going through a glass tube.</p>
<p>It is broken and the metals got separated.</p>
<p>The whole thing is just over a centimeter.</p>
<p><a href="https://i.stack.imgur.com/WLKY9.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/WLKY9.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/qm77u.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/qm77u.jpg" alt="enter image description here" /></a></p>
| Help identify small flat wire in glass tube | 2023-12-17T22:53:22.053 |
694604 | |amplifier|biasing| | <p>you can undo a DC bias with a compensating DC bias of the opposite sign, or one of the same sign applied to an inverting input.</p>
<p><img src="https://i.stack.imgur.com/hDdkq.png" alt="schematic" /></p>
<p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fhDdkq.png">simulate this circuit</a> – Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a></sup></p>
| <p>As you know, if you have an amplifier or comparator that can only output 0v to +v, you need to bias the input from -v to +v to +v to +v so that the output will produce the full signal range. But what if you wanted to connect the output of the amplifier or comparator to a speaker? The output range should be from-v to +v. How could you reduce the reference voltage back to a range that the speaker would take? I unfortunately cannot draw a schematic because I’m using a phone, but imagine an amplifier that needs to connect to a speaker, and needs to change the volt range from 0v to 20v to -10 to +10v</p>
| “Undoing” the bias of a resistor | 2023-12-18T01:38:24.093 |
694608 | |rf|remote-control| | <p>As the other <a href="https://electronics.stackexchange.com/a/694630/101063">answer</a> already mentions, adding a CRC checksum would be useful to protect against random noise and/or inteference being detected as a valid message.</p>
<p>From a quick look at the HT12E / HT12D devices used for the encoding / decoding I can't see any mention of a CRC being used on the protocol.</p>
<p>Are you able to change away from the HT12E / HT12D devices and existing 433 MHz RF modules for a microcontroller based approach?</p>
<p>There are are some microcontrollers which have a built-in radio module which can be used on the 533 MHz RF band. E.g. the TI <a href="https://www.ti.com/lit/ug/slau259e/slau259e.pdf" rel="nofollow noreferrer">CC430F613x, CC430F612x, CC430F513x MSP430™ SoC With RF Core</a> MCUs which have a built in sub-1-GHz radio core based upon of the <a href="https://www.ti.com/product/CC1101" rel="nofollow noreferrer">CC1101 Low-power Sub-1 GHz wireless transceiver</a>.</p>
<p>If you use a microcontroller based approach the benefits could be:</p>
<ol>
<li>Can create a custom protocol which includes a CRC to filter out corrupt messages, or even a ECC to be able to correct some errors.</li>
<li>Allows the encoder / decoder and RF modules to be combined.</li>
<li>The same hardware module to be used for both transmit or receive, with software to configure as a transmitter or receiver.</li>
</ol>
<p>The downside is having to program software.</p>
| <p>I'm encountering problems with my RF module circuit for a cold pyro igniter using HT12E:</p>
<p>Unintended Flash on Power On:When powering the receiver, the output flashes without button press.</p>
<p>Continuous Output When Remote is Turned Off:After turning off the remote, the output remains high without button press.</p>
<p>Circuit Components:RF Module,HT12E, HT12D, Relay.</p>
<p>Am facing the problem since circuit construction.</p>
<p>Seeking guidance on preventing unintended output activation. Any insights or solutions appreciated.</p>
<p>Circuit Schematic :-
[<img src="https://i.stack.imgur.com/uSmCU.png" alt="Original Source: Electrogadget][1]" /></p>
<p>Image Source: <a href="https://circuitdiagrams.in/fm-remote-encoder-decoder-circuit/" rel="nofollow noreferrer">Electrogadget</a></p>
<p><a href="https://i.stack.imgur.com/qNaWE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qNaWE.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/YTwMy.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YTwMy.jpg" alt="enter image description here" /></a></p>
<p>Power source:</p>
<p>For transmitter 9v battery connected to 5v voltage Regulator(7805)
Receiver side is connected to 5v mobile charger.</p>
| Issues with unintended output activation in RF module circuit | 2023-12-18T03:19:45.103 |
694620 | |voltage|transistors|voltage-measurement|zener|schottky| | <p>1.2V will not be detected as logic level 1 by the GPIO, so you can use:</p>
<ul>
<li>A transistor to level shift to 3V3</li>
<li>An external comparator</li>
<li>One of the analog comparators inside your STM32</li>
</ul>
<p>Regarding input protection:</p>
<p><a href="https://i.stack.imgur.com/PXZKD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PXZKD.png" alt="enter image description here" /></a></p>
<p>The transistor case is very simple: with an input protection resistor, input voltage can go quite high, limited by resistor value and maximum base current spec. A diode ensures Vbeo does not go below its max rating.</p>
<p>For the others, I've used a 2V7 Zener diode to limit positive voltage. It is also a diode, so it limits negative voltage, but it will be around -0.6V which is out of spec. So Schottky diode D1 could be added. However, the GPIO input voltage spec really is about ensuring the MCU's internal input protection diodes do not burn out from too much current. So if the Zener diode limits voltage to -0.6V, and a few kOhm protection resistor (R2,R4) is added before the MCU's input, then the input current will be limited to a few tens of µA which is safe. In this case the extra Schottky diode is not really needed.</p>
| <p><a href="https://i.stack.imgur.com/GCe8L.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GCe8L.jpg" alt="enter image description here" /></a></p>
<p>I have obtained a voltage signal from a Weigand sensor (a sensor that creates a low voltage signal when placed in a rotating alternate magnetic field), which generates a positive pulse of 1.2 V and negative pulses (≤840 mV) in a rotating alternate magnetic field.</p>
<p>I intend to feed this signal into the GPIO pin of an STM32L4xx microcontroller. While the negative voltage is relatively low and unlikely to impact the GPIO pin, I am considering eliminating these negative pulses for adherence to safety SOPs.</p>
<p>I attempted to reduce the negative signal using a Zener diode circuit but faced challenges due to the ON state voltage threshold of Zener diodes falling within the range of 2-200 V, while the pulse magnitude is only 1.2 V. Subsequently, I explored the use of Schottky diodes, known for their low forward bias voltage (0.2-0.3 V). Unfortunately, I did not achieve positive results. If you have experience with low-voltage drop diodes and can provide guidance on resolving this issue, I would greatly appreciate your insights.</p>
| Eliminating Negative Voltage Pulses (≤ 840 mV) | 2023-12-18T07:55:31.487 |
694627 | |soldering|hot-air| | <p>I'm not saying it can't be done without, but I would <em>really</em> recommend a board pre-heater.<p>
These devices apply lower, non-soldering heat to the whole board to make it easier to reflow the solder. If you consider that the room temperature might be 20 deg C, and you need to get the solder in excess of 210 deg C that's a big jump. All that heat having to come from one hot air source makes it very difficult to get the solder you want - particularly solder that's underneath a chip that will insulate it - hot enough without damaging it. <p>
Whereas if you preheat the whole board, even just to 100 deg C, it's much easier to apply the hot air gun, because you need to use a lot less heat. You can go higher on the preheat if you like, just be mindful of what else is on the board and whether this will be a problem. <br></p>
| <p>I set my hot air device to 320 °C and at 20% flow. Using low melt solder paste it takes quite some time to desolder BGA chips which I think get singed! Increasing to 400 °C is faster but the tiny components nearby get singed.</p>
<p>Can someone experienced with hot air stations suggest settings for me?</p>
| Hot air. Too hot or too slow? | 2023-12-18T09:26:17.050 |
694644 | |pcb|pcb-design|high-speed|routing|stack-up| | <p>The 18μm outer thickness is almost certainly the thickness of the un-plated copper foil (not the "<em>Finished thickness</em>" you've shown in the screenshot of JLCPCB's capabilities)... it'll likely be plated up to approx 1oz / 0.035μm (but this isn't actually specified anywhere in the project), and then plated again with a very thin layer of gold (ENIG, as mentioned in the board layout, just below the table you quoted).</p>
<p>Just below the table is also mention of the suggested material - "ISOLA PR2116", which is likely what the project has used to verify the design - so, I'd suggest you either accept their "Advanced PCB" service (6+ layers, but with incorrect layer heights), or look for an alternate supplier.</p>
<p>As always, impedance controlled traces will need to be reviewed if you want confidence that things will work on a new material and stackup.</p>
| <p>I am looking at an open-source design for an NVIDIA Jetson Nano carrier board made by a company called AntMicro. <a href="https://github.com/antmicro/jetson-nano-baseboard" rel="nofollow noreferrer">See here</a>
The design is an 8-layer design with the following Stack-Up and Build-Up recommendation:</p>
<p><a href="https://i.stack.imgur.com/8AnZU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/8AnZU.png" alt="enter image description here" /></a></p>
<p>And the Stack-Up:</p>
<ul>
<li>L1: Signal (Top Layer)</li>
<li>L2: 5V Power Plane</li>
<li>L3: Signal</li>
<li>L4: Signal</li>
<li>L5: GND Plane</li>
<li>L6: 3V3 + Signal</li>
<li>L7: GND Plane</li>
<li>L8: Signal (Bottom Layer)</li>
</ul>
<p>The only problem with using this design for my custom carrier board are the capabilities of my manufacturer, which in my case, is JLCPCB.</p>
<p><a href="https://i.stack.imgur.com/TmR5P.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TmR5P.png" alt="enter image description here" /></a></p>
<p>In the case of the open-source carrier board, inner copper layers are 0.035mm (~1oz) and the outer copper layers are 0.018mm (~0.5oz).</p>
<p>What alternative PCB Build-Up can I use to be able to manufacture the board. Ideally, I do not want to change too much of the routing and layout to be on the safest side.</p>
| PCB Build-Up and Stack-Up alternative for 8-layer PCB | 2023-12-18T13:24:49.377 |
694661 | |filter|active-filter|band-pass| | <blockquote>
<p><em>I'm trying to design a filter with a bandwidth of 150Hz, center
frequency of approx. 60Hz</em></p>
</blockquote>
<p>If the bandwidth is 150 Hz and the centre frequency is 60 Hz then we can only be talking about a high-pass filter followed by a low-pass filter that have combined 3 dB points at 21 and 171 Hz.</p>
<p>Put another way (and thinking logarithmically), 60 Hz is 2.85 times lower than 171 Hz and, 21 Hz is 2.85 times lower than 60 Hz. And, 171 Hz minus 21 Hz is a bandwidth of 150 Hz.</p>
<p>That's the only realistic way you can define this filter.</p>
<p>And, because of this (the massive asymmetry), you cannot use the Q-factor approximation of centre frequency divided by bandwidth. Reason: the pass-spectrum is so close to DC that the asymmetry of the higher and lower cut-offs are just too different.</p>
<p>Normally we might talk about a selective filter having a centre frequency of (say) 1 MHz with a BW of (say) 10 kHz and then, we can use the Q-factor formula approximation you have used but, it falls-over when we are this close to DC.</p>
<p>So, you have to regard the two cut-off frequencies (low and high) as being formed with a high-pass filter followed by a low-pass filter.</p>
<blockquote>
<p><em>with a roll-off of 40dB/decade</em></p>
</blockquote>
<p>If you want 40 dB/decade on both slopes then, you'll need two 2nd order filters cascaded to achieve this. Design each independently and cascade them. For non-peaking in the passband, use a Butterworth Q-factor of 0.7071 for both high-pass and low-pass filters.</p>
| <p>I'm trying to design a filter with a bandwidth of 150Hz, center frequency of approx. 60Hz, with a roll-off of 40dB/decade. I am following the design of a 2nd order Multiple Feedback Band-Pass filter as discussed in page 315 and 316 of <a href="https://web.mit.edu/6.101/www/reference/op_amps_everyone.pdf" rel="nofollow noreferrer">this reference manual</a>.</p>
<p><a href="https://i.stack.imgur.com/LWCl2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LWCl2.png" alt="enter image description here" /></a></p>
<p>For my application, the Q factor should be 60/150, or 0.4. Looking at one of the resistor calculations,</p>
<p><a href="https://i.stack.imgur.com/6i6AS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6i6AS.png" alt="enter image description here" /></a></p>
<p>R3 requires 2(Q^2)+Am > 0 to function properly, otherwise the value will be negative. This suggests the Q-factor should be adequately large for this design of MFB Band-pass to function. In my case, Am needs to be between 0 and -0.32 (gain is inverted in this circuit), which seems like a very low number and would defeat the purpose of an op-amp based filter.</p>
<p>I am confused, however, since I recall reading(cannot remember where) that MFB Band-pass filters are used for their good response to low Q-factors. In this case, was what I read incorrect? Or is a Q-factor of 0.4 simply too low for the MFB band-pass to handle?</p>
<p>Please let me know if more information is required.</p>
| Multiple Feedback Band-pass behavior at low (<1) quality factor | 2023-12-18T16:20:00.043 |
694663 | |active-filter|multisim| | <ul>
<li>A 2nd order <strong>low</strong>-pass filter has a roll-off above the cut-off frequency of 40 dB/decade</li>
<li>A 2nd order <strong>high</strong>-pass filter has a roll-off below the cut-off frequency of 40 dB/decade</li>
<li>A 2nd order <strong>band</strong>-pass filter has a roll-off above and below the cut-off frequencies of <strong>20 dB/decade</strong></li>
</ul>
<p>Also, you can't use a simple 2nd order filter to achieve grossly asymmetrical cut-off frequencies as implied by this statement: -</p>
<blockquote>
<p><em>I'm trying to design a filter with a bandwidth of 150Hz, center
frequency of approx. 60Hz</em></p>
</blockquote>
<p>The above means a lower cut-off of 21 Hz and, an upper cut-off of 171 Hz = not possible with a single 2nd order filter.</p>
| <p>I'm trying to design a filter with a bandwidth of 150Hz, center frequency of approx. 60Hz, with a roll-off of 40dB/decade. I am following the design of a 2nd order Multiple Feedback Band-Pass filter as discussed in page 315 and 316 of <a href="https://web.mit.edu/6.101/www/reference/op_amps_everyone.pdf" rel="nofollow noreferrer">this reference manual</a>.</p>
<p><a href="https://i.stack.imgur.com/LWCl2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LWCl2.png" alt="enter image description here" /></a></p>
<p>I am simulating the example 2nd order MFB Band-pass, using the values given.</p>
<p><a href="https://i.stack.imgur.com/6i6AS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6i6AS.png" alt="enter image description here" /></a></p>
<p>Despite being a 2nd order filter (so the decade roll-off should be 40dB/decade), the simulated roll-off in Multisim, according to my understanding, is still 20dB/decade.</p>
<p><a href="https://i.stack.imgur.com/ensax.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ensax.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/0KmhQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0KmhQ.png" alt="enter image description here" /></a></p>
<p>I'm not sure what I've misunderstood in terms of core filter concepts, or simulation software. I'm assuming the documentation I've read is correct, so please let me know if more information is required.</p>
| Simulating a 2nd order filter provides a 1st order response | 2023-12-18T16:31:01.870 |
694681 | |rf| | <p>I found a service manual, which is quite thorough, available online at:
<a href="https://www.manualslib.com/manual/917665/Shure-Slx2.html" rel="nofollow noreferrer">https://www.manualslib.com/manual/917665/Shure-Slx2.html</a></p>
<p>Although it doesn't have a full schematic, there are block diagrams and detailed descriptions. As you suspected, the battery is used as the antenna. It is fed by a long chain of passive components, likely the ones you've highlighted in yellow.</p>
<p>Here's a quote from the manual that provides the clue:</p>
<blockquote>
<p>The low pass filter output couples to the <strong>battery antenna</strong> via C641 and L607. Connector CON640 and C613 are only usedfor Japanese (JB) units. Coupling capacitor C610 is used to ensure that both batteries are driven equally</p>
</blockquote>
<p>One would think a battery, with a negative end at PCB ground and a positive end feeding all the circuitry, is a poor choice for an antenna. I'd guess they use RF chokes to isolate the battery assembly from the audio and digital circuitry at UHF frequencies, then feed it with a capacitor.</p>
<p>Your "movement-dependent noise" could originate in components in the audio chain (e.g. a "microphonic" component) or in the RF chain. Tough to troubleshoot without a schematic. Good luck!</p>
| <p>I'm trying to locate the transmitting antenna on a wireless mic, in particular a shure SLX2 J3. I'm unable to find any antenna circuit, although I suspect an RF switch on side 1, a circular part with 414 marking.</p>
<p>All I can figure out so far is that it is using one of the battery ports, or perhaps mic ground as some type of antenna, however these mic modules are somewhat universal across non wireless mics.</p>
<p>Symptoms are noisy movement dependent noise coming from the mic with no mic attached at all. I've tried this on a couple frequencies, using the demodulation feature of my rigol DSA815.</p>
<p>Where might the designers have they hidden the antenna on this? There is no 50R trace anywhere I can see.</p>
<p><a href="https://youtube.com/shorts/L0L9oVp8A80?feature=share" rel="nofollow noreferrer">https://youtube.com/shorts/L0L9oVp8A80?feature=share</a></p>
<p><a href="https://i.stack.imgur.com/6BRWr.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6BRWr.jpg" alt="Side 1" /></a></p>
<p><a href="https://i.stack.imgur.com/6bpeU.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6bpeU.jpg" alt="Side 2" /></a></p>
<p><a href="https://i.stack.imgur.com/qbQgo.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qbQgo.jpg" alt="IR Port" /></a></p>
<p><a href="https://i.stack.imgur.com/kRQQD.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/kRQQD.jpg" alt="Mic port" /></a></p>
<p><a href="https://i.stack.imgur.com/hof5c.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hof5c.jpg" alt="MIC" /></a></p>
<p><a href="https://i.stack.imgur.com/sebYg.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sebYg.jpg" alt="End" /></a></p>
<p><a href="https://i.stack.imgur.com/9HADs.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9HADs.jpg" alt="Assembly side 1" /></a></p>
<p><a href="https://i.stack.imgur.com/hGMqc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hGMqc.jpg" alt="Assembly side 2" /></a></p>
<p>See yellow trace here
<a href="https://i.stack.imgur.com/cQykI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cQykI.jpg" alt="Proposed path" /></a></p>
| Wireless microphone antenna location | 2023-12-18T18:51:21.150 |
694682 | |mosfet|identification| | <p>It's probably a <a href="https://datasheetspdf.com/pdf-file/1094144/UNIKC/PZ2003EEA/1" rel="nofollow noreferrer">Unikc PZ2003EEA</a> P-channel MOSFET. The datasheet doesn't specify the markings, but I found a listing on shopee with a (low quality) picture that shows a "D1" product code.</p>
<p><a href="https://i.stack.imgur.com/NDg3y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NDg3y.png" alt="enter image description here" /></a></p>
<p>(Source: <a href="https://webcache.googleusercontent.com/search?q=cache:qZuKQaSnUf4J:https://shopee.co.id/search%3Fkeyword%3Dd1hw&hl=de&gl=de" rel="nofollow noreferrer">shopee.co.id</a>)</p>
<p>Here is a better picture of another Unikc MOSFET (PE534BA) in the same package that shows the marking style of this manufacturer as reference.</p>
<p><a href="https://i.stack.imgur.com/5Obn1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5Obn1.jpg" alt="enter image description here" /></a></p>
<p>(Source: <a href="https://de.aliexpress.com/item/4000225180822.html?src=google&aff_fcid=91ff05534ce9416aa8d4806902a82bcd-1703155619285-07263-UneMJZVf&aff_fsk=UneMJZVf&aff_platform=aaf&sk=UneMJZVf&aff_trace_key=91ff05534ce9416aa8d4806902a82bcd-1703155619285-07263-UneMJZVf&terminal_id=bf341379b04f4d21b63bbccb56305de5&afSmartRedirect=y" rel="nofollow noreferrer">aliexpress.com</a>)</p>
<p>Note: The MOSFET is also available from <a href="https://www.niko-sem.com/" rel="nofollow noreferrer">Niko-sem</a>.</p>
| <p>The MOSFET is on a laptop motherboard close to the power jack. The part reads:
<code>D1 GNA</code> and <code>2R1H01</code>
It's faulty so I'm looking to replace it.</p>
<p>I added some pictures of the area + the MOSFET. The laptop is an HP Elitebook 850 G1 (I haven't had any luck finding schematics but maybe someone else has more luck). And I found this code on the motherboard: 6050A2560201-MB-A03.</p>
<p><a href="https://i.stack.imgur.com/DTQmH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DTQmH.jpg" alt="photo of device" /></a></p>
<p><a href="https://i.stack.imgur.com/YUHRE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/YUHRE.jpg" alt="photo of PCB" /></a></p>
| Identifying a MOSFET marked D1GNA and 2R1H01 | 2023-12-18T18:52:01.037 |
694683 | |power|laser|optics|ionization| | <blockquote>
<p>Could this illumination be maintained without any threat to the lines/people (i.e.: complete dielectric breakdown)?</p>
</blockquote>
<p>As the other answers (at time of writing) have covered the other aspects ably, I'll concentrate on this point.</p>
<p>No -- as can be seen by any demonstration of a Tesla Coil, high frequencies inevitably cause air to break down into dazzling, chaotic streamers. We have several consequences:</p>
<ol>
<li>The light output is unstable, flickering as some streamers cool off then new paths are created, etc.</li>
<li>The light is harsh, having mostly a bluish color, and considerable UV that would be unsafe for exposure.</li>
<li>Ionization of air leads to the ions snapping back together in whatever way is fastest -- we get chemical reactions, most notably the oxidation of oxygen (ozone, O<sub>3</sub>) and nitrogen (N<sub>2</sub>) into NO and NO<sub>2</sub>. These are quite toxic by themselves, and also further react with water and trace organics to make nitric acid and other smog molecules.</li>
</ol>
<p>Overall, it would be a very unpleasant, to destructive, way to replace lighting. :)</p>
<p>The other sort of thing you're thinking, though, has been done -- electrodeless glow discharge lamps have been a thing for a long time, though subject to various difficulties that have kept them a fairly minor technology. See: <a href="https://en.wikipedia.org/wiki/Induction_lamp" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Induction_lamp</a></p>
<p>(This is mostly a physics/chemistry answer, but that's not a bad thing; chemistry is naught but the motion of electrons on small scales, and chemistry and electronics are sister disciplines under the family of physics. Both go well together.)</p>
| <p>I had a thought experiment I wanted to run past some experts.</p>
<p>It starts with CFL light bulbs. CFL light bulbs are light bulbs that use ~500 VAC @ ~20 kHz or more to excite a gas in a light vacuum.</p>
<p>High Voltage AC (HVAC) lines will often have 100 kVrms or more, at 60 Hz.</p>
<p>This seems like more than enough to ionize CFL lights, though, the atmosphere is not a CFL.</p>
<p>Working in the principal that above a certain total voltage change, the DE/dt of any electric field determines whether a gas will be ionized, it would seem like increasing the frequency of our HVAC lines could cause a natural lighting of sorts through local ionization.</p>
<p>So with the thoughts as above, my question is as follows; could increasing the frequency of our HVAC lines cause said power lines to illimunate? Could this illumination be maintained without any threat to the lines/people (i.e.: complete dielectric breakdown)?</p>
| If we increased the frequency of HV-AC lines, could we get atmospheric air to excite and illuminate? | 2023-12-18T19:02:40.827 |
694687 | |diodes| | <p>The data sheet for a diode will have curves showing the Current .vs. Voltage.</p>
<p>E.g. for from the Vishay <a href="https://www.vishay.com/docs/81857/1n4148.pdf" rel="nofollow noreferrer">1N4148</a> Small Signal Fast Switching Diode datasheet:</p>
<p><a href="https://i.stack.imgur.com/Rhbbt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Rhbbt.png" alt="enter image description here" /></a></p>
<p>Note the use of a logarithmic scales on the Current .vs. Voltage curves since the current–voltage curve is exponential in accordance with the <a href="https://en.wikipedia.org/wiki/Shockley_diode_equation" rel="nofollow noreferrer">Shockley diode equation</a>.</p>
<p><a href="https://en.wikipedia.org/wiki/Diode#Semiconductor_diodes" rel="nofollow noreferrer">Semiconductor diodes</a> has information on the characteristics of semiconductor diodes, which these days as mainly silicon.</p>
| <p><a href="https://i.stack.imgur.com/KXIGw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KXIGw.jpg" alt="enter image description here" /></a>Does a silicon diode drop voltage by the same amount no matter how large the supply voltage can get? I mean let's assume I have conducted the experiment in which I have connected only a diode and
a resistor across a power supply. I will increase the power supply gently at the beginning, and I will notice the voltage across the diode. At the beginning when the voltage across the diode is below .70 V, I notice a one to one correspondence, that is when the supply voltage is 0.2V, the voltage drop across the diode is 0.2V given that zero current passes through the circuit. After we reach 0.7 voltage drop across the diode, the supply voltage is 10 V then, when I increase further supply voltage, like as high as 30 V, the voltage drop across the diode is just .74V increasing only .04 V when the supply voltage increased 3 times?!</p>
<p>The data isn't mine, I have just googled it.</p>
| Does a silicon diode drop voltage by the same amount no matter how large the supply voltage? | 2023-12-18T19:42:34.793 |
694700 | |pcb|usb|rp2040| | <p><img src="https://i.stack.imgur.com/UdZTh.png" alt="screenshot showing the USB C connector" /></p>
<p>The center four pins (highlighted), from top to bottom, are:</p>
<ul>
<li>B7 - Dn2</li>
<li>A6 - Dp1</li>
<li>A7 - Dn1</li>
<li>B6 - Dp2</li>
</ul>
<p>i.e: You've connected the first pair D+ and D- together, and the second pair's D+ and D- together.</p>
<blockquote>
<p>I noticed I made a mistake where I accidentally connected my USB C data P1 to N1 and P2 to N2 but I managed to remove the connection by scratching off the copper trace between them then checking for continuity to verify the disconnection and seem to still have an issue with my USB communication.</p>
</blockquote>
<p>I'm not fully clear on what you've done here, but it won't be possible to make this functional by only cutting traces - you'll at best end up with one leg or the other... don't forget that USB C will only connect one pair at a time, depending on the orientation, so you can't use A7 (Dn1) and B6 (Dp2) together as a pair. Please feel free to clarify if I've made a bad assumption!</p>
<blockquote>
<p>I believe there might be an impedance mismatch</p>
</blockquote>
<p>You should still see the host detect the device, even with a mismatch - <a href="https://www.youtube.com/watch?v=VG5bWzEPfsg" rel="nofollow noreferrer">you can do that with just a resistor</a>. It's possible you'll see this with (what I presume to be) your one-leg-only connection if you rotate the USB C connector.</p>
<hr />
<p>Separately, as feedback, note that your RP2040 has a poor ground connection... the thermals here are doing you a disservice - if you can't remove them, then definitely try to make them larger.</p>
<p><img src="https://i.stack.imgur.com/xt1j6.png" alt="screenshot showing very narrow connections to the wider ground plane" /></p>
| <p>I designed a RP2040 board where the github can be found here <a href="https://github.com/peterk268/Pico-Pal/tree/main" rel="nofollow noreferrer">https://github.com/peterk268/Pico-Pal/tree/main</a></p>
<p>I noticed I made a mistake where I accidentally connected my USB C data P1 to N1 and P2 to N2 but I managed to remove the connection by scratching off the copper trace between them then checking for continuity to verify the disconnection and seem to still have an issue with my USB communication.</p>
<p>I bought an oscilloscope recently to confirm that my crystal is oscillating well at ~12MHz.
But I see that my USB data - is stuck at ~0.3V and the + ~3.3V and I am not sure why.</p>
<p>I understand that a 2 layer board of this size was not the best idea and that I should not have routed any signals on the second layer and need a clean ground reference under the data lines.</p>
<p>There is a ground pour on both sides of my PCB and when I calculated the needed trace width and spacing in Altium to get the 90ohm differential Z0, I used differential coplanar since the ground pour is there and the ~58mil dielectric thickness if I remember correctly as it was a 1.6mm board, this gave me the trace width of ~9.5 mils I believe with 5 mil spacing.
This is something I am concerned about as I believe there might be an impedance mismatch because I see the example board in the hardware design guide use ~30 mil traces but I had not seen that before fabricating and trusted the Altium calculator.</p>
<p>But yea this being my first board with a microcontroller on it, I made mistakes and have learnt but I just want to know why the USB is not even budging.
The internal 1v1 regulator is reading at 1.8V for some reason. I have changed the microcontroller and checked for shorts and unconnected pads and fixed any that were there (thankfully none with ground and power).
I had made a mistake to put a couple of vias close under the microcontroller but those are not shorted which I have checked as well.</p>
<p>Also went the way to solder another board together with fresh components.
I am using the same flash used in the example board which has its own CS pull up. I have a 1k ohm resistor connected to a button to pull down the line when flashing but still it should enter boot loader whenever the flash is clear and I have verified that it is being pulled down.</p>
<p>I have plugged it in on windows and mac and heard the windows usb sound but nothing in device manager and nothing on mac.</p>
<p>Another disclaimer is that the other board I have made where I did not disconnect the USB data shorts read around the same voltage levels so i am not sure if this is something that can damage the rp2040 or perhaps USB communication was never initiated.
I have done another continuity test and confirm that one has the short and the other doesn't (the one I scratched off).</p>
<p>Power supply is great with my 3v3 regulator working as expected.</p>
<p>Any help would be much appreciated as I've been trying to find any more issues in this board that I do not reproduce in another that I am doing a 4 layer SIG GND GND SIG with a larger size and less routing under ICs.</p>
<p>Edit:
Attached picture of working board with USB cable connected
<a href="https://i.stack.imgur.com/3C8H6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3C8H6.jpg" alt="enter image description here" /></a></p>
| RP2040 PCB not communicating over USB | 2023-12-18T22:16:17.977 |
694712 | |pcb-design|solder-mask| | <p>Some engineers like to add extra solder to high-current traces, to increase the trace's cross-sectional area; people argue about whether this has any significant effect, given solder's substantially worse conductivity than copper. (I'm on the side of "just go for 2-ounce copper instead", personally.)</p>
<p>That's what's been done here. By leaving areas of soldermask open, large amounts of solder from the HASL finish and/or wave-soldering processes are left on these high-current traces. The reason for the stepped design going to the relays is presumably because some of the current will divert through each relay, meaning the traces after that point will never need to carry as much current as those before it.</p>
<p>It appears that additional solder was placed on the right side of the relay trace, beyond what HASL or wave soldering would leave. Presumably, this was done by printing extra solder paste on that part of the board when the surface-mount components were reflowed.</p>
| <p>See the annotated image below.</p>
<p>The board takes AC power in, and feeds it to two relays (plus an empty spot where a 3rd relay could go). On the other side, it has two connections labeled <code>AC1</code> and <code>AC2</code>. <code>AC1</code> feeds to the relays, and <code>AC2</code> is connected directly to several other headers all called <code>COM</code>.</p>
<p>I am curious about the unusual solder patterns. <code>AC1</code> starts with heavy solder and gets progressively thinner with each relay. As you can see, the connection to <code>Relay A</code> is stripy. What does the stripy pattern do? Does it have a name?</p>
<p>The <code>AC2</code> <=> <code>COM</code> connections have their own stripy pattern. It's a little harder to see in the picture, but it has a bunch of long, 45º stripes running between all the headers. They're more continuous than the ones to <code>Relay A</code>, looking more like a "grill".</p>
<p>What is up with these traces?</p>
<p>A bit more detail: This board is one of 3 found in a 240V space heater, so it's 240V between <code>AC1</code> and <code>AC2</code>. Most of this board is simply converting down to 5V for a microcontroller (found on another PCB). The relays actually control <em>other relays</em> found on the 3rd PCB, where the heating coils are connected.</p>
<p><a href="https://i.stack.imgur.com/J8A2d.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/J8A2d.jpg" alt="Annotated image of PCB with funky traces" /></a></p>
| What's up with these funky solder/trace patterns? | 2023-12-19T02:11:42.507 |
694714 | |resistors|adc| | <p>The resistors are source terminators. For a source terminator to be effective, the total source impedance must equal the characteristic impedance of the transmission line (the PCB trace in this case).</p>
<p>Most PCB traces on high-speed boards are either stripline or microstrip. Search for these terms to learn more. A typical PCB characteristic impedance is about 50 ohms (single-ended signal).</p>
<p>The total source impedance is equal to the equivalent impedance of the driving IC plus the series resistor.</p>
<p>The impedance of the driving IC is not something that is normally specified in the datasheet. You usually have to derive it from I-V curves or measure it.</p>
<p>So, for your example, these might be the values. Transmission line: 50 ohms. Driving IC: 28 ohms.</p>
<p>50 - 28 = 22 ohm resistor.</p>
<p>Buy the book recommended in my comment. Or, if you don't want to spent $100US, dig around the internet and try to learn it yourself.</p>
| <p><a href="https://i.stack.imgur.com/yOtzd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yOtzd.png" alt="enter image description here" /></a></p>
<p>The evaluation boards I found put a resistor in series on each of the ADC's digital outputs. Can someone please explain to me what is the purpose of these resistors? Is it necessary or can I just not use them in my design?</p>
<p>Also, the two evaluation boards I found use different values. One uses 10 Ω while the other uses 22 Ω. So, If I add this to my PCB, how can I calculate the required resistance value?</p>
<p>For further details:</p>
<ul>
<li>The ADC is <a href="https://www.analog.com/media/en/technical-documentation/data-sheets/ad9226.pdf" rel="nofollow noreferrer">AD9226</a>. It is a 12-bit ADC with parallel CMOS output. Figure 22 of its datasheet (page 23) shows one example of the use of the resistors (RP1 and RP2).</li>
<li>I planned to use <a href="https://digilent.com/shop/arty-z7-zynq-7000-soc-development-board/" rel="nofollow noreferrer">Arty Z7 20</a> to read the ADC's digital outputs. The ADC will be on my custom PCB and connected to the Arty through its headers (I put male headers that are compatible with the Arty's headers). The ADC will be clocked using an external 25 MHz clock generator. The signals' logic level will be 3.3 V.</li>
</ul>
| What is the role of the resistors in this circuit? | 2023-12-19T02:27:43.147 |
694721 | |power| | <p>Keep in mind that on a normal breadboard all five holes in a row (in this case row 7) are shorted together.</p>
<p>In your second and third pictures, no significant current will flow in the LED because both leads are shorted together. It's much easier for the current to flow through the metal strip in your breadboard than through the LED, so nearly all of them do.</p>
| <p>My question is, why does this work:</p>
<p><a href="https://i.stack.imgur.com/e61Gs.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/e61Gs.png" alt="enter image description here" /></a></p>
<p>But not this:</p>
<p><a href="https://i.stack.imgur.com/K7rH2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/K7rH2.png" alt="enter image description here" /></a></p>
<p>My best guess is that since the motor draws current, it forces electrons through the LED as well. But then, this doesn't work:</p>
<p><a href="https://i.stack.imgur.com/Zjznj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zjznj.png" alt="enter image description here" /></a></p>
<p>I'm not an electrical wiz clearly, if there's a simple explanation, I'd appreciate it.</p>
<p>I usually think of electricity as water moving through a pipe, with the pressure being voltage and amount of water being current. I understand this is not an actually accurate way of thinking about it.</p>
| Connecting LED to just positive | 2023-12-19T03:38:17.947 |
694724 | |raspberry-pi|voltage-measurement| | <blockquote>
<p>However, when I shut down the RPi, I notice 60mA - 70mA of current flowing from the battery into the MCP3002.</p>
</blockquote>
<p>The <a href="https://ww1.microchip.com/downloads/en/DeviceDoc/21294E.pdf" rel="nofollow noreferrer">MCP3002</a> Datasheet has the following which shows the input stage for one ADC channel:</p>
<p><a href="https://i.stack.imgur.com/cjyOh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cjyOh.png" alt="enter image description here" /></a></p>
<p>Note the two input protection diodes between the CH<sub>x</sub> channel input and the V<sub>DD</sub> and V<sub>SS</sub> supply rails. Such diodes are normally intended to only conduct (briefly) to protect against ESD.</p>
<p>The resulting combination for the power rails is the following, which shows that when the RPi is shutdown <code>BAT</code> will power the RPi and MCP3002 via the input protection diode <code>D1</code>. The RPi <code>5V PWR</code> and MCP3002 <code>VDD</code> will be the battery voltage minus the 0.6 V forward biased voltage drop of the input protection diode:</p>
<p><img src="https://i.stack.imgur.com/fPRSr.png" alt="schematic" /></p>
<p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2ffPRSr.png">simulate this circuit</a> – Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a></sup></p>
<blockquote>
<p>This clearly looks to be a problem that will drain the battery needlessly at best, and I suspect damage the RPi at worst.</p>
</blockquote>
<p>It will also be causing the MCP3002 input protection diode to conduct 60 mA - 70 mA continuously, and so may also damage the MCP3002. The MCP3002 datasheet doesn't provide an absolute maximum rating for the input protection diodes, but for other devices where the datasheet does provide an absolute maximum rating it is usually only a few milliamps.</p>
<blockquote>
<p>One idea (untested) would be to put a 47k resistor between the CH0 pin and the battery, reducing the current to 90uA.</p>
</blockquote>
<p>Yes, a series resistor would reduce how much current could be sunk through the MCP3002 <code>CH0</code> input when the RPi is off.</p>
<blockquote>
<p>However, I am still unsure as to whether this is the 'right way' to setup this circuit.</p>
</blockquote>
<p>An alternative could be use an analogue switch between the PowerBoost <code>BAT</code> and MCP3002 <code>CH0</code> input. Where the analogue switch would only be on when the RPi is powered and the software wants to take a battery voltage measurement. A GPIO output from the RPi could be used to control the analogue switch.</p>
<p>Not sure about using discrete FETs to create an analogue switch, but there are some ICs such as the <a href="https://www.analog.com/en/products/adg7421f.html" rel="nofollow noreferrer">ADG7421F Low Voltage Fault Protection and Detection, 12 Ω RON, Dual SPST Switch</a> which will be off (leakage current in nano-amps) if no power or the enable input is a logic zero:</p>
<blockquote>
<p>When no power supplies are present, the switch remains in the
off condition, and the switch inputs are high impedance.</p>
</blockquote>
<blockquote>
<p>The switches turn on with a Logic 1 input and conducts equally well in
both directions with an analog signal range of VSS + 0.1 V to VDD −
0.55 V for a 5 V single supply. The digital input is compatible with
1.8 V logic inputs over the full operating supply range.</p>
</blockquote>
| <p>I have a project involving the following components:</p>
<ul>
<li>Raspberry Pi Zero W</li>
<li><a href="https://www.microchip.com/en-us/product/MCP3002" rel="nofollow noreferrer">MCP3002</a> (<a href="https://ww1.microchip.com/downloads/en/DeviceDoc/21294E.pdf" rel="nofollow noreferrer">Datasheet</a>)</li>
<li><a href="https://www.adafruit.com/product/1944" rel="nofollow noreferrer">Adafruit PowerBoost 500C</a> (using a <a href="https://www.ti.com/product/TPS61090" rel="nofollow noreferrer">TPS61090</a>)</li>
</ul>
<p><strong>The Setup:</strong></p>
<p>The MCP3002 is connected to the Pi via SPI0 (pins 19,21,23,24), V<sub>REF</sub> is connected to 5V and V<sub>SS</sub> is connected to GND (pin 39) giving a 0V-5V range of measurement. CH0 is connected to the battery via the Bat pin (<a href="https://cdn-learn.adafruit.com/assets/assets/000/017/555/original/adafruit_products_powerboost500C.png?1403289316" rel="nofollow noreferrer">schematic here</a>) of the 500C.</p>
<p>The Pi is powered by the PowerBoost 500C, which is a USB/Li-Ion battery power supply. The 500C is an all-on-one board using the TSP61090 with a female USB-A jack to power the RPi via the USB micro power jack.</p>
<p><strong>The Problem:</strong></p>
<p>When the RPi is powered on and running, everything works as expected. Reading the value of CH0 returns the correct voltage. However, when I shut down the RPi, I notice 60mA - 70mA of current flowing from the battery into the MCP3002. This clearly looks to be a problem that will drain the battery needlessly at best, and I suspect damage the RPi at worst. What is the right way to prevent this?</p>
<p>One idea (untested) would be to put a 47k resistor between the CH0 pin and the battery, reducing the current to 90uA. However, I am still unsure as to whether this is the 'right way' to setup this circuit. What is the right way to setup this circuit?</p>
<p>[EDIT]: Added the schematic below.</p>
<p>Note that the issue I am seeing is only when the RPi is off. The 500C is still providing power via the 5V pin.</p>
<p><img src="https://i.stack.imgur.com/4Tagc.png" alt="schematic" /></p>
<p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2f4Tagc.png">simulate this circuit</a> – Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a></sup></p>
| Current leakage when measuring battery voltage via Raspberry Pi and MCP3002 | 2023-12-19T04:27:21.713 |
694728 | |voltage|circuit-analysis|sensor| | <p>You are right. You should not use DC to sense the presence of water. In this circuit, the current going through the WA_R2 is in one direction and goes from the Vcc to the base of the transistor. This will lead to ionic migration. The sensor will be covered and eventually stop working.</p>
<p>When the current that goes through the probe is low, the ionic will happen slowly. In your case, if you are not draining any current from AOUT, and for example, using 5V for Vcc, then when the transistor is fully on, the current going through the WA_R3 is about 5mA. If the gain of the transistor, for example, is 100, then the current through the probe is 0.05mA or 50uA. This is not very low!</p>
<p>However, even if you manage it with lower currents, if you are using this for an application where the probe is always in the water and you can not remove the probe, or clean or change it, it's not a good idea to use this circuit.</p>
| <p>I have been interested in this <a href="https://www.waveshare.com/liquid-level-sensor.htm" rel="nofollow noreferrer">liquid level sensor (Waveshare 9525)</a> which has this <a href="https://files.waveshare.com/upload/8/8f/Liquid-Level-Sensor-Schematic.pdf" rel="nofollow noreferrer">schematic</a>.</p>
<p><a href="https://i.stack.imgur.com/QoElN.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QoElN.png" alt="enter image description here" /></a></p>
<p>I need to understand the principal behind its working because I'd like to replicate it. I have two platinum wires which I would like to put to use for it.</p>
<p>The reason I am asking this question is because I thought you could not use DC for liquid level sensor as that would lead to ionic migration and eventual drop in voltage measured, but I guess this sensor is not measuring voltage between any electrodes?</p>
| How does this liquid level sensor work? | 2023-12-19T05:38:05.373 |
694736 | |simulation|altium|spice|heatsink|test-point| | <p>If you need a project with no warnings or errors, -- well, you can still just disable the relevant errors entirely -- but one way would be to make a one-pin model to global ground (<code>0</code>).</p>
<pre><code>.SUBCKT DUMMY 1
RD 1 0 1e12
.ENDS
</code></pre>
<p>something like that would do I guess. Maybe an R isn't even necessary? I'm not actually sure what a valid minimal <code>.SUBCKT</code> is.</p>
<p>Personally, I just ignore the warning. If there is no model attached, there is no entry in the netlist.</p>
<p>Yet another method could be: use one schematic sheet common to two projects, which contains just the active circuitry in question. The upper-level sheet contains non-simulation and mechanical components (and the rest of the project) for the physical PrjPcb, and maybe nothing, or maybe simulation harness stuff (virtual components like simulation sources, function blocks, etc.).</p>
<p>Or for that matter, in this way you can divide anything that you just don't need to simulate, or simulate all at once -- you could divide one project into multiple subcircuits this way, simulated in multiple different projects.</p>
<p>Put another way: this is the schematic equivalent of using one e.g. <code>*.c</code> module in an overall project, while also running the same file in a separate test suite.</p>
| <p>I am setting up a simulation model in Altium with SPICE models and have no idea how to provide a model for a single connection device (connector pin, test-point) or a no-connection device like a heatsink or mounting hole. Altium is saying these parts have no model and the auto-search has not linked anything to them.</p>
<p>I've searched and come up with things like a model with large resistances to ground or components to represent parasitic values, etc., but for a single point connection, how does that work? Also, what if the device has no pins at all, like a heatsink?</p>
<p>An alternative is to tell Altium to ignore them. It may be possible to draw compile masks over them, but there must be a better way, and my circuit has around 150 of them.</p>
<p>So please can I ask what experienced users of SPICE do in this case?</p>
<p>Many thanks in advance.</p>
| Spice model for a heatsink/mounting hole/test-point/connector pin - devices with one or no connection points - using Altium simulator | 2023-12-19T06:59:01.993 |
694739 | |image-sensor| | <p>There is a problem related to hardware but I solve this. When I decrease voltage supply to 3.1V, view is cleared like below.</p>
<p><a href="https://i.stack.imgur.com/myTSz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/myTSz.jpg" alt="enter image description here" /></a></p>
<p>Thank you for your attentions and helping.</p>
| <p>I have a problem about color CMOS analog NTSC image sensor. I use OV7955 image sensor and I connected to the monitor. Monitor displays like this:</p>
<p><a href="https://i.stack.imgur.com/Ip3TM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ip3TM.jpg" alt="enter image description here" /></a></p>
<p>I tried anathor lens and view doesn't change. I thought the problems is related with connectors, I soldered the connections of image sensor again and the output view is like the first picture. How can I solve this problem?</p>
<p>In once, image was cleared like that:</p>
<p><a href="https://i.stack.imgur.com/XHaPE.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XHaPE.jpg" alt="enter image description here" /></a></p>
<p>I don't have any idea, this is how happened. Is there a connection problem are hardware problem? After that I suspended connections but connections are true. I soldered again, renew the connections but view was never changed.</p>
<p>In below, there are some stuff on the camera</p>
<p><a href="https://i.stack.imgur.com/D8s3G.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/D8s3G.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/zBQJI.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zBQJI.jpg" alt="enter image description here" /></a></p>
<p>Green marker's colour on the display is purple. Multimeter's colour is totally green. This is my what I use lens:
<a href="https://i.stack.imgur.com/bl8Sw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bl8Sw.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/uNhMi.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uNhMi.jpg" alt="enter image description here" /></a></p>
| Image Sensors Tint Issue | 2023-12-19T08:21:07.430 |
694745 | |pcb-fabrication| | <p>Ask a fab -- that's really your go-to location for any questions about fab. If they can do it, I mean, that's all you need to know. Capabilities vary from shop to shop, of course, so you may have to ask around a few places to see what, or why.</p>
<p>There is one general reason, however: heavy gold is electroplated. The traces must be shorted together at some point, so that an electrical connection can be made.</p>
<p>Edge fingers are made by extending the board, where the fingers are shorted together, and the whole end is dipped and plated. After, the shorting tab is milled off, and the edges chamfered for easy insertion.</p>
<p>In the middle of the board, you'll probably need to add a shorting feature somewhere, somehow, to allow electrical conduction. Maybe this is milled out by the fab, maybe you can arrange some way to connect them externally (assemble with one connector, plug it in to the bath?), but the more specialized it gets, the more they're going to charge for it.</p>
<p>There is another route:</p>
<p>Have you considered pogo targets, such as these?
<a href="https://www.digikey.com/short/j55dt343" rel="noreferrer">Contacts, Spring Loaded (Pogo Pins), and Pressure | Digikey</a></p>
| <p>I have charging pads at the center of the device which makes repeated contact with a pogo pin. I'm concerned an ENIG pad may wear out over time as the user clicks it on and off the charger.</p>
<p>Can hard gold can be used to selectively plate this pad at the center of the board?
I've only seen hard gold be partially used on a PCB when it's at the board edge, for example in PCIE fingers.</p>
<p>Would hard gold have a higher chance of corrosion if a user sweats onto the pads?</p>
| Can hard gold be selectively used on a PCB in the center of the board? | 2023-12-19T09:50:38.403 |
694750 | |mosfet|nmos|charge|pmos|temperature-coefficient| | <p>According to <a href="https://www.microsemi.com/document-portal/doc_view/14692-mosfet-tutorial" rel="nofollow noreferrer">this</a> application note:</p>
<blockquote>
<p>Gate charge values vary slightly with drain current and drain-source
voltage but not with temperature.</p>
</blockquote>
<p>Their definition of 'slightly' is of the order of 10%, so we might presume any variation with temperature (there surely is <em>some</em>) is of the order of 1%.</p>
| <p>For P- and N-MOSFETs, Vgsth will decrease (in absolute value) when the temperature rises.</p>
<ul>
<li>What is the impact of the temperature on the total gate charge for P- and N-MOSFETs?</li>
<li>What is the order of magnitude of the variation?</li>
</ul>
| How is the total gate charge of a MOSFET affected by the temperature? | 2023-12-19T10:16:05.483 |
694771 | |digital-logic|state-machines|sequential-logic|sequence-detector| | <blockquote>
<p>After detecting "1011", why does the detector go back to B</p>
</blockquote>
<p>The diagram is correct for the non-overlapping sequence. To get into state <strong>D</strong> requires the sequence <strong>101</strong>. If the next input is <strong>1</strong> (resulting in the <strong>1011</strong> pattern), then the sequence is starting from the condition where you detected the 1st bit as <strong>1</strong>, as if you are in the <strong>B</strong> state.</p>
<p>To demonstrate that the diagram is correct, you can run this <a href="https://www.edaplayground.com/x/Zj2E" rel="nofollow noreferrer">Verilog simulation on EDA Playground</a>. It shows several example input data streams, but you can change the input sequence to whatever you want.</p>
<blockquote>
<p>Wouldn't it become overlapping</p>
</blockquote>
<p>It would not be overlapping. An example of overlapping is:</p>
<pre><code>Input : 101010
Output: 000101
</code></pre>
| <p>Assuming that <strong>A</strong>, <strong>B</strong>, <strong>C</strong> and <strong>D</strong> are the four states. When detecting the sequence "<strong>1010</strong>", in the <strong>D</strong> state, if overlapping is not allowed, we have the following state diagram:
<a href="https://i.stack.imgur.com/1DK9w.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1DK9w.jpg" alt="State diagram for "D" only" /></a>
<a href="https://i.stack.imgur.com/Ap5h1.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ap5h1.jpg" alt="Complete state diagram" /></a></p>
<p>After detecting "<strong>1011</strong>", why does the detector go back to <strong>B</strong> instead of having a self-loop to <strong>D</strong>? Wouldn't it become overlapping if we consider the "<strong>1</strong>" as the first bit of sequence "<strong>1010</strong>", because then the input sequence would be "101<strong>1</strong>010"?</p>
| Design of non-overlapping "1010" sequence detector | 2023-12-19T13:56:48.550 |
694774 | |circuit-analysis|current|test|quartz| | <p>The capacitors in the Colpitts oscillator circuit shown are relatively high value and maybe not ideal for modern many-MHz crystals. If you replace them with the crystal tester oscillator circuit typically used in the inexpensive frequency counter modules it may work better with high-frequency crystals:</p>
<p><img src="https://i.stack.imgur.com/Uemwb.png" alt="schematic" /></p>
<p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fUemwb.png">simulate this circuit</a> – Schematic created using <a href="https://www.circuitlab.com/" rel="noreferrer">CircuitLab</a></sup></p>
<p>The transistor shown (C9018) has a higher ft than the one in your schematic (1.1GHz vs. 600MHz) but either should work about as well for most crystals you are likely to find.</p>
<p>If you have a counter or oscilloscope or even an inexpensive DMM that has a frequency counter input, it is better to measure the actual frequency and compare with the markings. The frequency may be off slightly because of the load capacitance mismatch, but it should still be within a very small percentage.</p>
| <p>After working with quartz components in circuits, I planned to make a test on them in order to be sure if they are really still working as they are supposed to be.</p>
<p>I found this circuit on the internet:</p>
<p><a href="https://i.stack.imgur.com/SaodU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SaodU.png" alt="net circuit" /></a></p>
<p>I understand that the trick here is to block the DC current by the capacitors around and if there is an AC current flowing, which is coming from working quartz through the capacitors, then the LED will glow. That means the quartz component is working. If the LED does not glow, the quartz is not working because it is not able to produce AC current.</p>
<p>I thought in practical sense to make my own design for quartz testing, which I have tested in LTspice as working circuit. It is just below.</p>
<p><a href="https://i.stack.imgur.com/DhhJD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DhhJD.png" alt="own circuit" /></a></p>
<p>It works like the first circuit above which I have found on the internet. I soldered the components together to see what the result is going to be, but the result was negative. The quartz, especially the new ones, do not work. I was so frustrated.</p>
<p>Can you tell me what is wrong with the circuit I built?</p>
<p>Edit:</p>
<p>Here is the LTspice file:</p>
<pre><code>Version 4
SHEET 1 880 1000
WIRE -1088 560 -1248 560
WIRE -992 560 -1088 560
WIRE -848 560 -928 560
WIRE -752 560 -848 560
WIRE -1248 576 -1248 560
WIRE -752 576 -752 560
WIRE -1088 608 -1088 560
WIRE -752 640 -848 640
WIRE -752 736 -752 640
WIRE -1088 752 -1088 688
WIRE -752 832 -752 816
WIRE -1088 976 -1088 816
WIRE -752 976 -752 912
WIRE -752 976 -1088 976
FLAG -1248 576 0
SYMBOL SBORKA\\QUARTZ_CRYSTAL\\XTAL_param -768 576 R0
WINDOW 39 -115 77 Left 0
SYMATTR InstName U1
SYMBOL voltage -752 816 R0
WINDOW 123 0 0 Left 0
WINDOW 39 0 0 Left 0
SYMATTR InstName V1
SYMATTR Value 5
SYMBOL res -864 544 R0
SYMATTR InstName R1
SYMATTR Value 47k
SYMBOL res -768 720 R0
SYMATTR InstName R2
SYMATTR Value 1k
SYMBOL cap -1104 752 R0
SYMATTR InstName C1
SYMATTR Value 220p
SYMBOL cap -928 544 R90
WINDOW 0 0 32 VBottom 2
WINDOW 3 32 32 VTop 2
SYMATTR InstName C2
SYMATTR Value 220p
SYMBOL res -1072 704 R180
WINDOW 0 36 76 Left 2
WINDOW 3 36 40 Left 2
SYMATTR InstName R4
SYMATTR Value 100
TEXT -984 984 Left 2 !.tran 0.5ms
TEXT -1144 696 Left 2 ;Led
</code></pre>
| Get practical crystal tester working | 2023-12-19T14:06:07.987 |
694779 | |nodal-analysis|mesh-analysis| | <p>If you change the sense of <span class="math-container">\$i_1\$</span>, you're also changing the voltage produced by the dependent source, so the two cases are not equivalent. They would be if in addition to flipping the sense of the currents, you also change the sign on the dependent source.</p>
| <p>I have applied the mesh analysis to this circuit using both clockwise and counterclockwise directions but I ended with different results.</p>
<p><a href="https://i.stack.imgur.com/slD1B.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/slD1B.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/qgsE6.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qgsE6.jpg" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/v1Cq2.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/v1Cq2.jpg" alt="enter image description here" /></a></p>
<p>Please let me know if there's anything I'm doing wrong.
Also, how can I correctly apply <strong>node analysis</strong> to this circuit?</p>
| How do I apply mesh analysis to this circuit? | 2023-12-19T14:28:44.243 |
694793 | |power-supply|circuit-analysis| | <p>The simplest is to remove AMS1117 and apply your external 3V3 supply to the corresponding 3V3 pin on the connector.</p>
<p>These ESP32 modules sometimes crash due to transient current demand dropping the "3V3" voltage, for example coming out of wifi-sleep causes quick steps in supply current. The fix is to use a better LDO with fast transient response and replace the tantalum cap with 10µF ceramic. For example LDL1117 works flawlessly.</p>
<p>If you remove the LDO this does not apply, but then the only remaining decoupling on the board for 3V3 will be a high-ESR tantalum cap of unknown origin and a 100nF MLCC. This could be a problem if the supply wires are long. So I'd recommend replacing the tantalum cap with a 10µF ceramic, or simply add an electrolytic cap of few hundred µF on the power pins.</p>
<p>The <a href="https://www.silabs.com/documents/public/data-sheets/CP2102-9.pdf" rel="nofollow noreferrer">CP2102</a> USB chip is powered from USB VBUS and creates its own 3V3 supply via an internal regulator, so it will still work fine and you will be able to program the ESP32 and use the serial port normally even with AMS1117 removed.</p>
<p>According to the datasheet, it is allowed to apply logic level voltage to CP2102 input pins even when it is powered down. So when USB is disconnected, even if ESP32 outputs some serial signals, CP2102 will not be damaged.</p>
| <p>I have an ESP32 (ESP-WROOM-32) on a dev board, the same as pictured here:</p>
<p><a href="https://i.stack.imgur.com/BXvVD.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BXvVD.jpg" alt="enter image description here" /></a></p>
<p>It will be used in a project and powered by an external 3.3 V power supply through the 3.3 V and GND pins.</p>
<p>I'll need to communicate with the help of the USB to serial converter of the dev board, through a USB cable. So when the USB cable is plugged, the AMS1117 regulator will be powered by the 5 V line of the USB cable and produce its own 3.3 V supply, on the same voltage rail as the external power supply.</p>
<p>It's not an option to disconnect the external power supply while using the USB port, so I was wondering if it could cause any issue.</p>
<p>From what I understand:</p>
<ul>
<li>If the AMS1117 3.3 V voltage is higher than the external power supply's, some current will go to the power supply and start charging its capacitors at a rather slow rate, which should not be a problem (the regulation will do its job).</li>
<li>If it's the other way around, the small excess voltage will cause the AMS1117's output to be at a slightly higher voltage than it should, and what happens depends on its internals.</li>
</ul>
<p>For the second option, I assume that the AMS1117 can probably take it, but I have nothing to base this intuition on. I searched for more info, but everything I found is about not using Vin together with USB, which is not what I'll be doing.</p>
<p>There is also a diode between the Vout pin and the Vin pin of the AMS1117, so anything over ~4.0 V would be fed back to the Vin pin of the AMS1117.</p>
<p>What's your opinion on this? How can I know if it's going to be a problem or not, short of trying it and see if something burns?</p>
| Dual power supply with AMS1117 and regulated 3.3 V | 2023-12-19T16:01:10.693 |
694800 | |analog|ground|interference|ground-plane|leakage-current| | <blockquote>
<p><em>What could this shift be due to? Is my grounding incorrect? Could my
signals be interfering with something? Could there be leakage current
from somewhere?</em></p>
</blockquote>
<p>The pressure sensor draws about 8 mA from the 5 volt supply. That's a power of 40 mW that will heat the surrounding circuits. That is what I believe is heating-up the circuit board and affecting the temperature reading from your temperature sensor.</p>
<p>You could try and prove this by drawing heat away from the pressure sensor if it's possible to attach a heatsink or, just use an external temperature probe to measure it. Or, remove the pressure sensor chip and test the remaining items on the PCB with and the 5 volts connected.</p>
<p>You could even test the pressure sensor on its own (in air) and see how warm it gets without having the PCB thermally remove heat.</p>
<p>You only have a tiny PCB so it can easily rise in temperature even for a few tens of milliwatts.</p>
| <p>I have a <a href="https://www.infineon.com/dgdl/Infineon-KP236_N6165-DS-v01_00-en.pdf?fileId=db3a30432ad629a6012af68133600b1a" rel="nofollow noreferrer">KP236N6165XTMA1</a> 5V analog pressure sensor and a <a href="https://www.digikey.fr/en/products/detail/sensirion-ag/SHT31-ARP-B/5872293?utm_adgroup=Capacitors&utm_source=google&utm_medium=cpc&utm_campaign=Dynamic%20Search_FR_Product&utm_term=&productid=&utm_content=Capacitors&utm_id=go_cmp-153613254_adg-22469737014_ad-668122506945_dsa-1218196147359_dev-c_ext-_prd-_sig-Cj0KCQiAm4WsBhCiARIsAEJIEzX1U9s7rceFLlIygl2_FSZw_LFY2CspPxxnn-FWvpY1Vs1BavwAgKcaAktJEALw_wcB&gad_source=1&gclid=Cj0KCQiAm4WsBhCiARIsAEJIEzX1U9s7rceFLlIygl2_FSZw_LFY2CspPxxnn-FWvpY1Vs1BavwAgKcaAktJEALw_wcB" rel="nofollow noreferrer">SHT31-ARP-B</a> 3.3V analog temperature and humidity sensor on the same circuit, connected to the same ground plane (see v0.2 schematic bellow).</p>
<p><a href="https://i.stack.imgur.com/ZFenG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ZFenG.png" alt="enter image description here" /></a></p>
<p>The layout on my PCB looks like this (this is v0.1 which I have been testing on, it had its 3.3V power coming from the connector with an additional pin, but in v0.2 seen in the schematic above, the 5V will be converted to 3.3V directly on the circuit):</p>
<p><a href="https://i.stack.imgur.com/NhD9i.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/NhD9i.png" alt="enter image description here" /></a></p>
<p>If I power both sensors, the temperature and humidity analog outputs are incorrect. For example, the temperature readings are 5-6 degrees celsius (+0.3V) above what they should be.</p>
<p>When I remove the ESD diodes and only power the 3.3V, the 3.3V temperature and humidity sensor gives correct analog outputs, measuring around 20 degrees celsius which is indeed the room's temperature.</p>
<p>When I power back the 5V pin, and thereby the 5V pressure sensor, my temperature readings incrementally increase in the next couple of seconds, reaching 25 or 26 degrees celsius. They then go back to normal readings when I disconnect the 5V.</p>
<p><strong>What could this shift be due to? Is my grounding incorrect? Could my signals be interfering with something? Could there be leakage current from somewhere?</strong></p>
| Precautions when using multiple analog devices on same circuit | 2023-12-19T17:13:29.070 |
694821 | |lcd|display| | <p>In video processing, even before you think about a picture shown on any specific type of screen, you have the concepts of brightness and contrast.</p>
<p>You change brightness of a video signal by adding (or subtracting) an offset.
You change the contrast by multiplying the signal with a coefficient.</p>
<p>That is what is done inside a TV video processing path, whether the signal is analog like it was on CRT TVs or digital like with modern TVs.</p>
<p>But the brightness "knob" on a modern TV does not directly control the backlight brightness. That's because of many things like being able to have a wide contrast range by dynamically adjusting both the backlight and the pixel data on screen, or have multiple areas with different range, sometimes called local dimming.</p>
<p>Now, computer screens are not TVs. Some still have both brightness and contrast settings. For a laptop, you generally just want to be able to change how bright the screen is. Changing how bright is does not mean you are actually controlling only the backlight or only altering the video pixel data. The one brightness knob may control both the backlight and pixel data in any way necessary to be able to provide the best picture at different settings.</p>
<p>Also computer screen use standards such as sRGB how the colors and contrast should look like, so there is little motivation to alter the contrast. In video player programs you can change the contrast and brightness for the decoded data. For TVs, they also use standards, but with TVs, consumers are expected to have access to legacy features to adjust their settings to their likings, including contrast, brightness, saturation, hue, and even sharpness and whatnot.</p>
<p>If you simply talk about LCDs such as the ones with HD44780 or similar controller, the contrast is just the LCD glass drive voltage basically.</p>
| <p>I was not able to find an answer via web search. AFAIK LCD displays can control brightness via changing brightness of backlight, in turn via
electric current value (up/down). I don't recall reading about changing properties of screen layers or else by a user. What is the point of having two sliders: brightness / contrast?</p>
<p>On laptops there is usually only brightness but on most TVs and standalone displays I've used both brightness and contrast. What each usually does/change? Why laptops have one slider and standalones two? TIA</p>
| Brightness / contrast controls: what do they do on LCD displays? | 2023-12-19T19:55:15.113 |
694825 | |audio|digital-communications| | <p>This chip requires a system clock and it only supports a system clock that is a multiple of the sample clock.</p>
<p>Supported multiples are 384x, 512x, and 256x.</p>
| <p>I'm trying to use an Iridium Core 9523N module to send the speaker output of my computer over satellite to any telephone. The <a href="https://telemetry.groupcls.com/wp-content/uploads/2018/12/Iridium-9523-9523N_Development-Guide.pdf" rel="nofollow noreferrer">integrator's guide</a> says it accepts PCM-formatted digital audio data, and mentions the TLV320AIC1110 as a possible source, but I find this chip more complicated than it should in this application so I've been looking for a simpler alternative.</p>
<p>I have come across the <a href="https://www.ti.com/lit/ds/symlink/pcm1808.pdf?HQS=dis-dk-null-digikeymode-dsf-pf-null-wwe&ts=1703002921319&ref_url=https%253A%252F%252Fwww.ti.com%252Fgeneral%252Fdocs%252Fsuppproductinfo.tsp%253FdistId%253D10%2526gotoUrl%253Dhttps%253A%252F%252Fwww.ti.com%252Flit%252Fgpn%252Fpcm1808" rel="nofollow noreferrer">PCM1808</a> and it looks like it does the job (to me anyway...), but the datasheet shows a system clock that is also connected to the PCM master, whereas I cannot do that. I am tempted to use the BCK both for the bit clock and the system clock since apparently the only requirement is that BCK and SCKI are synchronized, but I am not sure so seeking confirmation here.</p>
| PCM audio ADC: can the system clock be the fed with the PCM bit clock? | 2023-12-19T20:23:18.173 |
694830 | |motor|electric-machine|synchronous|rotor| | <p>Reluctance motors <em>necessarily</em> have salient-pole rotors, as there would be no angle-varying reluctance (which is the key parameter that makes reluctance motors work in the first place) if the rotor were perfectly cylindrical.</p>
<p>They almost always also have salient stator poles, though this is not an absolute necessity for them to work as it is for the rotor.</p>
<p>Synchronous motors also frequently (maybe even usually) have salient poles, which can improve performance (effectively using it as both a synchronous and a reluctance motor at the same time), but this doesn't mean that reluctance motors don't also have salient poles.</p>
| <p>I am doing research on Syncrhonous Reluctance Motor (SynRM) and the rotor used in it. I am a bit confused in their rotor types. Now according to Google, the rotor types do consists Salient Poles Rotor but according to Chat GPT, salient pole is not used in Synchronous Reluctance Motor but instead is used in Syncrhonous Motor. I know that SynRM does have axial and radial rotors. Does SynRM also has a salient pole rotor and if yes how good and efficient it is?</p>
<p>Edit: I am trying to design a SynRM but I can't seem to get hold of specs for the axial or radial rotor. If you could guide me to some specs then I would like to draw it on AutoCAD and analyze it on ANSYS Maxwell.</p>
<p>One more thing if I design a rotor in ANSYS Electronics Desktop, will I get to know the dimensions? Like the barrier widht and depth and angle of each barrier?</p>
| Rotor types in Synchronous Reluctance Motor | 2023-12-19T20:41:04.137 |
694832 | |integrated-circuit|identification|surface-mount| | <p>It's a <a href="https://www.silergy.com/download/downloadFile?id=3742&type=product&ftype=datasheet" rel="noreferrer">Silergy SY8088I</a> step-down regulator.</p>
<p><a href="https://i.stack.imgur.com/4iK7y.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/4iK7y.jpg" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/t9QUV.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/t9QUV.jpg" alt="enter image description here" /></a></p>
<p>(Source: <a href="https://world.taobao.com/item/673067648914.htm" rel="noreferrer">taobao.com</a>)</p>
| <p>Please help me to identify the IC marked in the picture, the marking states bPDFZ but I couldn't find any relevant information. The board is a TP-Link KC401 home camera.</p>
<p><a href="https://i.stack.imgur.com/SUbTW.jpg" rel="noreferrer"><img src="https://i.stack.imgur.com/SUbTW.jpg" alt="photo of PCB showing unidentified IC" /></a></p>
| Can't identify this IC, package SOT23-5 top mark bPDFZ | 2023-12-19T20:50:27.247 |
694835 | |circuit-analysis| | <p>Well, the circuit we want to analyze is given by:</p>
<p><img src="https://i.stack.imgur.com/aLRy2.png" alt="schematic" /></p>
<p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2faLRy2.png">simulate this circuit</a> – Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a></sup></p>
<p>When we use and apply <a href="https://en.wikipedia.org/wiki/Kirchhoff%27s_circuit_laws" rel="nofollow noreferrer">KCL</a>, we can write the following set of equations:</p>
<p><span class="math-container">$$\text{I}_3=\text{I}_1+\text{I}_2\tag1$$</span></p>
<p>When we use and apply <a href="https://en.wikipedia.org/wiki/Ohm%27s_law" rel="nofollow noreferrer">Ohm's law</a>, we can write the following set of equations:</p>
<p><span class="math-container">$$
\begin{cases}
\begin{alignat*}{1}
\text{I}_1&=\frac{\displaystyle\text{V}_1-\text{V}_3}{\displaystyle\text{R}_1}\\
\\
\text{I}_2&=\frac{\displaystyle\text{V}_2-\text{V}_3}{\displaystyle\text{R}_2}\\
\\
\text{I}_3&=\frac{\displaystyle\text{V}_3-\text{V}_4}{\displaystyle\text{R}_3}\\
\\
\text{I}_3&=\frac{\displaystyle\text{V}_4-0}{\displaystyle\text{R}_4}
\end{alignat*}
\end{cases}\tag2
$$</span></p>
<p>Using <span class="math-container">\$(2)\$</span> we can rewrite <span class="math-container">\$(1)\$</span> as follows:</p>
<p><span class="math-container">$$
\begin{cases}
\begin{alignat*}{1}
\frac{\displaystyle\text{V}_3-\text{V}_4}{\displaystyle\text{R}_3}&=\frac{\displaystyle\text{V}_1-\text{V}_3}{\displaystyle\text{R}_1}+\frac{\displaystyle\text{V}_2-\text{V}_3}{\displaystyle\text{R}_2}\\
\\
\frac{\displaystyle\text{V}_4-0}{\displaystyle\text{R}_4}&=\frac{\displaystyle\text{V}_1-\text{V}_3}{\displaystyle\text{R}_1}+\frac{\displaystyle\text{V}_2-\text{V}_3}{\displaystyle\text{R}_2}
\end{alignat*}
\end{cases}\tag3
$$</span></p>
<p>So, we can solve for <span class="math-container">\$\displaystyle\text{V}_4\$</span>:</p>
<p><span class="math-container">$$\text{V}_4=\frac{\displaystyle\text{R}_4\left(\text{R}_1\text{V}_2+\text{R}_2\text{V}_1\right)}{\displaystyle\text{R}_1\left(\text{R}_2+\text{R}_3+\text{R}_4\right)+\text{R}_2\left(\text{R}_3+\text{R}_4\right)}\tag4$$</span></p>
<p>Now, when using your values we get:</p>
<p><span class="math-container">$$\text{V}_4\left(t\right)=\frac{\displaystyle1000\left(15\cdot6\sin\left(4000\pi t\right)+50\cdot5\right)}{\displaystyle15\left(50+270+1000\right)+50\left(270+1000\right)}\tag5$$</span></p>
<p>Which, when simplified gives:</p>
<p><span class="math-container">$$\text{V}_4\left(t\right)=\frac{2500}{833}+\frac{900}{833}\cdot\sin\left(4000\pi t\right)\approx3.0012+1.08043\sin\left(12566.4 t\right)\tag6$$</span></p>
| <p>I need help to find the method to find the solution of this problem.
In the following circuit, we cancel U1 and we need to find the tension u_4_2 going through R4 contributed by of the source u_2(t).</p>
<p><a href="https://i.stack.imgur.com/tJCWZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tJCWZ.png" alt="enter image description here" /></a></p>
<p>The answer should be: u_4_2(t) = 1.08sin(2 pi f t) V</p>
| Circuit analysis problem contribution from an AC source | 2023-12-19T21:10:39.813 |
694841 | |mosfet| | <p>One of the key factors in using MOSFETs as switches is to find their <span class="math-container">\$\mathrm{R_{DS(on)}}\$</span> rating in the data sheet. There should be a note or a column saying at what <span class="math-container">\$V_{GS}\$</span> that <span class="math-container">\$\mathrm{R_{DS(on)}}\$</span> rating applies. In the case of your NTE2374 that's 10V.</p>
<p>You should then make sure that your circuit supplies an on-state gate-source voltage between that rated <span class="math-container">\$V_{GS}\$</span> and the maximum rating (20V in this case). Note that <span class="math-container">\$\pm 20 \mathrm V\$</span> is practically ubiquitous for modern silicon power FETs. So you'd want to drive this one at between 12V and 18V to give yourself some margin between 10V and 20 volts.</p>
<p>In your case, if you wanted to use a gate drive of 3.3V, you'd want to first shop for "logic level" FETs, then read each data sheet and make sure they have the <span class="math-container">\$\mathrm{R_{DS(on)}}\$</span> you need at a gate voltage of 3.3V. There's lots of "logic level" FETs that only reach their rated <span class="math-container">\$\mathrm{R_{DS(on)}}\$</span> with a gate voltage of 5V, which wouldn't work for you in this case.</p>
| <p>I have been playing around with an NTE2374 to try and make a simple power switch (forgive me if my terminology is off, the MOSFET is supposed to toggle on/off the power to a circuit - that's what I mean by power switch.)</p>
<p>I have configured the MOSFET to be in common source mode, with the gate connected to the control signal, the drain connected to the load in question, and the source connected to ground.</p>
<p>When I use a microcontroller to toggle the signal, the LED's brightness is faint. When using ground/Vdd as the control signal, the LED is much brighter.</p>
<p>What do I need to do to make the microcontroller able to toggle the LED such that the LED is bright (not faint) when the appropriate signal is given?</p>
<p>Here is a simple schematic:</p>
<p><a href="https://i.stack.imgur.com/smgh7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/smgh7.png" alt="enter image description here" /></a></p>
| Can I use an NTE2374 as a power switch? | 2023-12-19T22:07:13.527 |
694848 | |filter|ltspice|sallen-key| | <p>From the standard tables, the circuit configuration you require with the resistor and capacitor values normalised to 1 rad/sec is as below....</p>
<p><a href="https://i.stack.imgur.com/6vHhH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6vHhH.png" alt="Filter design" /></a></p>
<p>First we must frequency scale the capacitors.</p>
<p>C1 = 3.546 F/(2 * pi * 116) = 4.87 mF</p>
<p>C2 = 1.392 F/(2 * pi * 116) = 1.91 mF</p>
<p>C3 = 0.202 F/(2 * pi * 116) = 0.277 mF</p>
<p>Next we must impedence scale all the component values to get realistic values.</p>
<p>Choose a standard value for C1 and calculate the impedence scaling factor</p>
<p>let C1 = 100 nF</p>
<p>Impedence scaling factor = 4.87 mF/100 nF = 48.7k</p>
<p>C2 = 1.91 mF/48.7k = 39.2 nF</p>
<p>C3 = 0.277 mF/48.7k = 5.69 nF</p>
<p>R1 = R2 = R3 = 1R * 48.7k = 48.7k</p>
<p>Final circuit...</p>
<p><a href="https://i.stack.imgur.com/FoXo7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FoXo7.png" alt="Final Filter" /></a></p>
<p>As you can see the output is 0.707 down (-3 dB) at 116 Hz.</p>
<p>Although the component values are not standard values, their values are in the right ball park to be practical.</p>
| <p>I am trying to design a third order Butterworth filter with Sallen–Key topology. I am quite new to this and there is not much only about third order variations.I have the following requirements:</p>
<ul>
<li>Gain: 1 (unity)</li>
<li>Passband (cut off): 116 Hz</li>
<li>Stopband: 500 Hz</li>
<li>Stop band attenuation: ≥ 35 dB</li>
</ul>
<p>I thought I could make the original sallen key configuration into a third order by simply adding an RC circuit at the output, then calculating both of the fc to be 116 Hz. The 2nd order circuit was calculated by normalizing the resistor and capacitor values by setting fc to 1 rad/s, then scaling up (method shown in image). These values where multiplied by m = 116 (resistor) and 1/116 (capacitor), yielding:</p>
<ul>
<li>R1 & R2: 116 ohm</li>
<li>C1:16.7 μF</li>
<li>C2:8.36 μF</li>
</ul>
<p>With this the passband is correct (116 Hz at -3 dB), however obviously the attenuation is not steep enough to yield the passband attenuation. I am not sure on how to configure the third order section and was wondering if I could get some assistance on this. This is a revision piece, where we were told to make a third order butterworth filter with sallen key topology.</p>
<p><a href="https://i.stack.imgur.com/MoeWj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MoeWj.png" alt="filter schematic" /></a><a href="https://i.stack.imgur.com/cntEc.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cntEc.png" alt="sallen key 2nd order calculation" /></a></p>
| Creating third order butterworth filter with Sallen–Key topology | 2023-12-19T23:14:17.473 |
694849 | |mosfet|pwm|inverter|three-phase| | <p>A very common is to use an optocouplers.
You can connect its leds to one pin but it doesn’t allow you make a deadtimes.
If you connect the leds on separate pins the deadtimes can be done.</p>
<p><a href="https://i.stack.imgur.com/uLDJl.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uLDJl.jpg" alt="enter image description here" /></a></p>
<p>Note: Be careful, during Mcu startup without led signals the both mosfets are on. Best is to start Mcu as first and connect the power supply as second.
Fuse the power supply also for mosfet safety.</p>
| <p>I want to create a very simple 3-phase inverter that is easy to understand, e.g. without using MOSFET drivers. I decided to use logic level P-channel MOSFETs on the high side and logic level N-channel MOSFETs on the low side.</p>
<p>I tried to draw a diagram, it went so-so. But you get the overall picture:
<a href="https://i.stack.imgur.com/eWzPO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eWzPO.png" alt="enter image description here" /></a></p>
<p>The gates on the high side MOSFET gates are "pulled-up" to 3v. And the low side MOSFET gates are "pulled-down" to 0v. I have understood that this is to put all the gates to a known closed state.</p>
<p>I think I understand it so far. Now, I want to control the gates from a microcontroller, using PWM signals to the gates.</p>
<p>Since the P-channel MOSFETs opens on <strong>low signal</strong> and the N-channel MOSFETs opens on <strong>high signal</strong> and in each half-bridge either high side or low side should be open, but not both (it would short circuit) - then <strong>can I control both gates by a common PWM signal?</strong> like the diagram below:</p>
<p><a href="https://i.stack.imgur.com/rb6eC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rb6eC.png" alt="enter image description here" /></a></p>
<p>But what is confusing now for me is that I connect the "pulled-up" and the "pulled-down" gates together. Isn't this a problem?</p>
<p>If that does not work, can I control the high side and low side of the half bridge from two identical PWM signals, originating from two different GPIO pins of the microcontroller instead - that way the "pulled-up" and "pulled-down" gate wouldn't be connected, but controlled by an identical PWM signal.</p>
<p>A diagram of this last alternative:
<a href="https://i.stack.imgur.com/yGxZk.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/yGxZk.png" alt="enter image description here" /></a></p>
| Control P-channel and N-channel MOSFETs by common PWM signal | 2023-12-19T23:44:43.793 |
694855 | |power|power-electronics|switch-mode-power-supply|dc-dc-converter|buck| | <p>I found the issue. Turns out the SMT to TH adapter I have been using didn't have long pins to make contact with the breadboard (Silly mistake, I know). Holding down the IC to ensure sufficient contact fixed the issue immediately, confirming that you can experiment with this type of circuit on a breadboard despite it being a SMD.</p>
| <p>Recently I have been experimenting with the <a href="https://www.diodes.com/assets/Datasheets/AP63200-AP63201-AP63203-AP63205.pdf" rel="nofollow noreferrer">AP63203</a>, and I just have not been able to get it to output 3.3V successfully. Currently, the way I have been testing it is using an SMT to TH adapter, which I have soldered the buck to and stuck it on a breadboard along with the <span class="math-container">\$4.7 \mu\text{H}\$</span> inductor and the other capacitors based on the diagram below:
<a href="https://i.stack.imgur.com/wpQw4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wpQw4.png" alt="enter image description here" /></a></p>
<p>I have stuck with the recommended values as outlined in the datasheet, except for the inductor, as I don't have a <span class="math-container">\$3.9 \mu\text{H}\$</span> one.</p>
<p>With the first test, I supplied 24V to the input and noted that the power supply was shoring out. I tested the pins on the buck and realised that pins 4, 3 and 5 (<code>GND</code>, <code>VIN</code> and <code>SW</code>, respectively) are shorted together permanently, whether the buck had power or not.
(See the image below for the breadboard layout, the top left of the IC is pin 1).</p>
<p><a href="https://i.stack.imgur.com/4zn7I.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4zn7I.jpg" alt="enter image description here" /></a></p>
<p>I repeated this with a brand new buck converter and, this time kept the input and output grounds separated, and I found that all 6 pins were now at about 24V (with respect to input ground), and the power supply was not shorting. I also found that <code>GND</code> (output) was not permanently shorted to SW this time around, so I am not quite sure what is going on.
(See the image below for a slightly different layout where the grounds are separated).</p>
<p><a href="https://i.stack.imgur.com/AgbMb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AgbMb.jpg" alt="enter image description here" /></a></p>
<p>I feel like I am making a very obvious mistake somewhere, but I can't see it, have I wired something incorrectly on my breadboard? If anyone could point me in the right direction, that would be greatly appreciated. TIA</p>
| AP63203 Buck not working | 2023-12-20T02:15:37.440 |
694860 | |transistors| | <p>The short answer is no.</p>
<p>Knee voltages are those across a collector-emitter junction when the forward current spikes (think of the graph of an approximate diode model). This is commonly considered 0.7 V, like the voltage drop across the base-emitter junction, for practicality's sake. Actual values vary a bit.</p>
<p>Saturation occurs across the collector-emitter junction at the point that coincides with the top of the load line, which is a graphical means of easily finding a transistor's Q or quiescent voltage/current point. It has collector current on the y-axis and collector-emitter voltage on the x-axis.</p>
<p>The true saturation point is a little below the upper limit of the load line, as the collector-emitter voltage is close to but not exactly zero. The collector-emitter voltage at saturation should inform you of the max collector current. Upon saturation, a transistor experiencing increases to its base current will experience no more increase to its collector current.</p>
<p>Load lines are implemented in this way because they make handy visual references of the main points of a given transistor's behaviour. This allows you to decide how you want to use any given model, whether that's as a switch or an amplifier.</p>
| <p>Is the knee voltage in a BJT transistor is the same as the saturation voltage?<br />
That is, in the cutoff region, is the collector emitter voltage drop the same as the knee voltage, the voltage above which any further increase induces no increase in the collector current?</p>
| Is the knee voltage \$V_{CE}\$ (knee) in a BJT transistor the same as the saturation voltage \$V_{CE}(sat)\$? | 2023-12-20T04:01:01.440 |
694865 | |resistors| | <p>Nothing says that the final band must be gold or silver. Gold just means 5% tolerance.</p>
<p>So orange-orange-brown-violet is a perfectly valid code for a 330 ohm 0.1% tolerance resistor.</p>
<p>The other way does not result into a valid code, because orange is not a tolerance code.</p>
<p>One of your multimeters that resulted in 330 ohms proves it.</p>
<p>The other multimeter measures wrong result for some reason, like low battery. You can double-check by measuring a known resistance with both meters.</p>
| <p>What am I overlooking? My Viz-WD0747 multitester measures a four band resistor--Violet, Brown, Orange, Orange (the 4th band does not look gold at all)--as 330 Ohms. My other meter reads it as 523 Ohms. An online calculator give me yet a very different value.</p>
<p>If I read the colors from the other end, the color code indicates the resistance is 71kOhms that is unlikely, given the reading of the meters. So I have two question:</p>
<ol>
<li><p>Is there a way to determine which end is to be read firstly when the color bands are evenly spaced (not grouped at one end) and without gold or silver to show that the colors must be read from the opposite end?</p>
</li>
<li><p>If more than one ohm meter is used to measure the resistance, and each give a very different reading, do you go with the lowest?</p>
</li>
</ol>
<hr />
<p><a href="https://i.stack.imgur.com/cdGTd.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cdGTd.jpg" alt="photo of resistor" /></a></p>
| How do you identify the value of a resistor? | 2023-12-20T05:13:17.080 |
694868 | |current|boost|load| | <p>The first thing I'll point out is that finding a 1 mH inductor that has decent performance at a switching rate of 500 kHz could be troublesome but, assuming you have one, your duty cycle will be theoretically about 84% with the circuit you show and, you could raise the load resistor value (currently 750 Ω) considerably higher without upsetting the stable output voltage.</p>
<p>In other words, your converter is operating in CCM with 750 Ω load and, will only drop into DCM when that load resistor rises to around 46.6 kΩ. In DCM the output voltage will rise for a fixed duty cycle so, this is to be avoided.</p>
<blockquote>
<p><em>How would I go about ensuring that the load resistance is sufficiently
high enough so that I don't draw too much current from the boost
converter?</em></p>
</blockquote>
<p>I assume you mean that you want a resistor in parallel with the real load so that when the real load (750 Ω) is disconnected, the output voltage is still stable at 75 volts.</p>
<p>20 kΩ would be my suggestion (if I read your question correctly).</p>
<p>If you then realized that finding a 1 mH inductor that had good efficiency at a 500 kHz switching rate might be problematic, you might decide something much lower is acceptable. With a 100 μH inductor, the transition to DCM happens when the load resistor is 4.66 kΩ (theoretical). But, you get a 1.2 watt inherent power dissipation in that resistor.</p>
<p>Of course, you could bite the bullet and use a control loop to set the duty cycle and this means you could enter DCM (on light loads) and still maintain the 75 volt output. Maybe that's a step too-far for you?</p>
| <p><a href="https://i.stack.imgur.com/G4znp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/G4znp.png" alt="enter image description here" /></a></p>
<p>I'm using the following boost converter topology to supply power to a vacuum fluorescent display. The PWM signal is 500kHz at a 80% duty cycle. I'm getting the 75V I need with some headroom to spare. My question is:
<strong>How would I go about ensuring that the load resistance is sufficiently high enough so that I don't draw too much current from the boost converter?</strong></p>
<p>Is it as simple as just slapping on an extra current limiting resistor?</p>
<p>Thank you</p>
| How can I make sure load draws the correct current? | 2023-12-20T05:48:29.650 |
694873 | |operational-amplifier|analog|jfet|integrator|sample-and-hold| | <p>After a bit of debugging it turns out it was the offset pin. Apply voltage to the offset pin offered no change to the circuits function, I applied voltage up to the positive rail with no difference. Leaving the offset pin floating solved the problem and now the integrator resets to 0V every time with no hassle. Thank you all for the help!</p>
| <p>I've recently built a resettable integrator board which uses a low-drift JFET <a href="https://www.ti.com/lit/ds/symlink/opa140.pdf?HQS=dis-dk-null-digikeymode-dsf-pf-null-wwe&ts=1703055768263&ref_url=https%253A%252F%252Fwww.ti.com%252Fgeneral%252Fdocs%252Fsuppproductinfo.tsp%253FdistId%253D10%2526gotoUrl%253Dhttps%253A%252F%252Fwww.ti.com%252Flit%252Fgpn%252Fopa140" rel="nofollow noreferrer">op-amp</a> and a <a href="https://www.ti.com/lit/ds/symlink/lf398-n.pdf" rel="nofollow noreferrer">sample and hold</a> chip. The sample and hold amplifier datasheet showed a resettable integrator configuration which is what I used to design my PCB, although with a different op-amp.<br />
I simulated the circuit in Spice and it worked as advertised with my chosen op-amp. Once assembled the circuit integrated as expected but when reset, it would reset to the -12 V rail and not ground.<br />
I suspected grounding the negative rail of the sample and hold amplifier might fix the problem and to some extent it did. Once grounded the integrator would integrate up and down and would reset towards ground but had a ~2 V offset when reset and instead of integrating up/down to +-12 V the integrator would stop at +8 V and -2 V.<br />
Any ideas on what might be wrong or what could be done? Thanks in advance.</p>
<p><a href="https://i.stack.imgur.com/S86v4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S86v4.png" alt="enter image description here" /></a></p>
| Resettable integrator resetting to the negative rail not ground | 2023-12-20T07:23:34.020 |
694883 | |ltspice|inrush-current| | <p>There are quite a few resonant sub-circuits made of L and C and the PWL input
excites them all.</p>
<p>Bouncing comes from one of the inductors that couples with one capacitor.
The oscillations you see comes from an LC resonant circuit that is weakly dumped by a resistor in the current path.</p>
<p>If L4.1 and L4.2 are a common mode filter you should add the coupling factor to refine your simulation.</p>
<p>A Common Mode Filter is composed by 2 inductors both wound on a Ferrite core. The two inductors are "coupled".</p>
<hr />
<p>Filtering action of noisy currents by the Common Mode Filter:</p>
<p>If inductor 1 creates a magnetic flux, due to some noisy current, inductor 2 receives that flux thanks to the magnetic core.</p>
<p>Inductor 2 receives the flux and creates an opposite flux that is received by inductor 1, which in turns rejects the noisy current.</p>
| <p>I'm trying to simulate the inrush current of a system that I measured with an oscilloscope before.
In my simulation I have a voltage bouncing effect on the wire between the supply and the Rshunt that I haven't with real life measurement.
What causes that bouncing effect?</p>
<p><a href="https://i.stack.imgur.com/HbRx1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HbRx1.png" alt="enter image description here" /></a></p>
<p><a href="https://i.stack.imgur.com/nxNCl.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nxNCl.png" alt="enter image description here" /></a></p>
| LTspice inrush current simulation | 2023-12-20T10:00:04.063 |
694907 | |avr|atmega328p| | <p>I probably wrote cached hex on the avr, this answer is incorrect. I left this answer because there are useful comments under it.</p>
| <p>I have a program written in C. I have 390 byte array defined and a long string value consisting of around 475 bytes. Now I hit a problem if I add more characters to the string value, around 30 bytes, the chip starts to act weird and glitch out. Even if I define a new separate variable, the same thing happens, so it's not code related.</p>
<p>I checked the avr-size and it's returning that I still have a lot of memory left:</p>
<pre><code>avr-size -C --mcu=atmega328p test.o
AVR Memory Usage
----------------
Device: atmega328p
Program: 2079 bytes (6.3% Full)
(.text + .data + .bootloader)
Data: 907 bytes (44.3% Full)
(.data + .bss + .noinit)
</code></pre>
<p>I suspected maybe I had knock-off chips, so I purchased few original ones from local electronics store, but the same exact issue occurs.</p>
<p>The tech specs state that atmega has 2048 bytes of RAM, my string and array together take up about half of that as reported by avr-size. What could be the reason why this is happening? Thanks!</p>
| ATMEGA328 out of memory, but avr-size reports I still have a lot of space | 2023-12-20T12:17:11.833 |
694922 | |pcb|pcb-design|pogo-pins| | <p>It can be. But a bare PCB pad would have to be gold-plated to be suitable; cheap HASL finish will oxidize over time and the contact will degrade. Your cheaper gold finishes (ENIG, ENEPIG) will also degrade over connect/disconnect cycles, as the gold is extremely thin and soft; it'll wear off.</p>
<p>While you could spend a lot of money to have your board plated with hard gold, a better option would probably be to use a small gold-plated brass pogo pin target, like <a href="https://www.digikey.com/en/products/filter/contacts-spring-loaded-pogo-pins-and-pressure/311?s=N4IgjCBcoKxgTFUBjKAzAhgGwM4FMAaEAeygG0QBma%2BANgAYQBdIgBwBcoQBldgJwCWAOwDmIAL6SgA" rel="noreferrer">these</a>. These are soldered onto the board and provide a more durable contact surface than ENIG as the gold is several times thicker (10 μ" compared to ENIG's typical 1 to 3 μ").</p>
| <p>I made a custom PCB which I would like to charge through a (2pin) pogo pin connector. Do I only need to place two SMD pads and glue two magnets on the PCB for it to work?
I only need the USB's 5V and GND. There are some premade charging cables (2pin to USB) I would like to use something like this: <a href="https://i.stack.imgur.com/wUZ0z.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/wUZ0z.png" alt="pogo pin charger" /></a></p>
| Is the PCB side of a pogo pin charging just an SMD pad? | 2023-12-20T14:51:28.990 |
694938 | |voltage|circuit-analysis|impedance-matching|power-factor-correction| | <p>When you have a load that is described as "lagging" (or it has a lagging power factor), it (the load) has an equivalent series impedance that is a combination of inductive reactance and resistance. Under these circumstances the current's phase angle is negative with respect the the applied voltage phase angle (normally considered to be at 0°).</p>
<p>The presence of a transformer might slightly increase the angle but, it will still remain negative with respect the the voltage (the reference).</p>
| <p><a href="https://i.stack.imgur.com/FyaG0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FyaG0.png" alt="enter image description here" /></a></p>
<p>I don't understand why the current angle is negative.</p>
| Why is the current angle negative? | 2023-12-20T17:39:32.370 |
694948 | |low-pass|active-filter| | <p>Here's a clue that might point at a bad model for the InAmp (for example): -</p>
<p><a href="https://i.stack.imgur.com/ayTcY.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ayTcY.png" alt="enter image description here" /></a></p>
<p>It tells me that you have no signal of any reasonable value entering your filter from the InAmp.</p>
<p>Or, it might point to a bad connection of the source to the InAmp inputs. I can't be sure because your source connection isn't shown.</p>
<p>Or maybe the 100 Ω resistor at the output of the InAmp got changed to something too high in value.</p>
<p>Maybe something else but, for sure, the dB scale is firmly in the noise.</p>
| <p>I have a 2nd order Sallen-Key Low-pass filter, cut-off frequency of 170Hz.</p>
<p><a href="https://i.stack.imgur.com/M0cV3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/M0cV3.png" alt="enter image description here" /></a></p>
<p>When attached to a perfect source, this is the bode plot. As expected, 40dB/decade response.
<a href="https://i.stack.imgur.com/lEpRV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lEpRV.png" alt="enter image description here" /></a></p>
<p>The low-pass filter is used as part of a 20Hz to 170Hz band-pass filter.
<a href="https://i.stack.imgur.com/hA9KP.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hA9KP.png" alt="enter image description here" /></a></p>
<p>When attached to a perfect source, this is the bode plot. Again as expected, 40dB/decade response.
<a href="https://i.stack.imgur.com/d7G8X.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/d7G8X.png" alt="enter image description here" /></a></p>
<p>However, I attached the band-pass filter to an instrumentation amplifier that acts as a source.
<a href="https://i.stack.imgur.com/bP9te.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bP9te.png" alt="enter image description here" /></a></p>
<p>This is the bode plot now. 20dB/decade response.
<a href="https://i.stack.imgur.com/so9kr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/so9kr.png" alt="enter image description here" /></a></p>
<p>I attempted to override the high-pass filter, and attached the instrumentation amplifier to the low-pass filter directly. This is the bode plot.</p>
<p><a href="https://i.stack.imgur.com/of494.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/of494.png" alt="enter image description here" /></a></p>
<p><a href="https://electronics.stackexchange.com/questions/530811/sallen-key-low-pass-filter">This post</a> suggests the capacitor values being low as a factor, but the capacitor values here are normal. The cut-off frequency of 170Hz is well within the GBP of the op-amp I'm using, the <a href="https://www.analog.com/media/en/technical-documentation/data-sheets/ADA4091-2_4091-4.pdf" rel="nofollow noreferrer">ADA4091</a>, which is 1.22MHz.</p>
<p>Please let me know if more information is required.</p>
<p>Additional information: The remainder of the circuit, simulating a strain gauge at 100Hz.
<a href="https://i.stack.imgur.com/jSsGq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jSsGq.png" alt="enter image description here" /></a></p>
<p>AC analysis for remainder of circuit.
<a href="https://i.stack.imgur.com/Ztqnq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Ztqnq.png" alt="enter image description here" /></a></p>
<p>PR10 is the low-pass, PR3 and 4 (only 4 visible) is outputs from Left and Right, which is output from a simulated strain gauge. PR8 is a point within the InAmp.</p>
| Loss of 2nd order behavior on 2nd order low-pass filter | 2023-12-20T18:18:29.837 |
694953 | |microcontroller|led|raspberry-pi| | <p>The LED module datasheet says, that the minimum input high level is 0.7 * Vcc. This would be 3.5 V in your setup. So you need some logic level translation from the 3.3 V logic level of the Raspi.</p>
<p>If you disconnect the GND line between Raspi and LED module, the module receives more or less random noise as data. This may destroy the LED inputs or Raspi outputs at any time.</p>
| <p>I am working on a small project where I am using a SparkFun Lumini LED Ring. The power supply for the ring is external, connected via its own cable and power adapter. I have connected the ring to the Raspberry Pi using three wires: DI, CI, and GND. The issue is that the ring does not work when GND is connected. When GND is not connected, the ring does illuminate but not as it should.</p>
<p>The link of the datasheet of the Leds:
<a href="https://cdn.sparkfun.com/assets/d/c/a/4/2/APA_102-2020-256-6.pdf" rel="nofollow noreferrer">https://cdn.sparkfun.com/assets/d/c/a/4/2/APA_102-2020-256-6.pdf</a></p>
<p>the schematic of Leds:
<a href="https://cdn.sparkfun.com/assets/0/4/2/e/a/LuMini_3Inch.pdf" rel="nofollow noreferrer">https://cdn.sparkfun.com/assets/0/4/2/e/a/LuMini_3Inch.pdf</a></p>
<p><a href="https://i.stack.imgur.com/zbbE3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zbbE3.png" alt="it is the same connection but with a LED-Ring" /></a></p>
<p>it is the same connection but with a LED-Ring</p>
| My LED ring is not working when I connect the GND | 2023-12-20T20:41:19.107 |
694965 | |high-voltage|insulation|voltage-multiplier|insulation-voltage| | <p>With any HV transformer/inductor the layering is key, as it impacts the turn to turn stress on the wire, in particular the lead outs. Get this wrong, and no magnet wire will survive.</p>
<p>If the layering is right, the other major problem, especially if you're going up to 50 kV will be corona. If the windings are in free air, they will generate significant ozone which with time will badly attack the insulation.</p>
<p>Short term, you might be fine, but for reliable long term use I would suggest you either need to vacuum impregnate / encapsulate with epoxy, or for a more forgiving setup, oil immerse.</p>
<p>I have made a 100 kV centre tapped transformer wound with standard magnet wire, immersed in oil, and absolutely no issue.</p>
<p>I used a split bobbin approach, so with a centre tapped setup, the centre tap was near the core for both half windings, rather than one having a high voltage to core. The center wire comes out a hole at the side/bottom of the bobbin, not up the inside of the bobbin (as that would place the centre wire and the last layer close to each other).</p>
<p>Then even wound layers for the HV winding, with kraft paper between layers to ensure no mixing of layers. I just trap the paper under a turn and a half of wire to fix it in place as I wind.</p>
<p>This layering is critical as it controls/shapes the interlayer voltage gradient, with each layer effectively isolating the layers on either side. The alternative of random winding gives a significant chance of wires touching which are many turns apart, with resulting high insulation stress.</p>
<p>(There is choice over whether you are better winding alternate layers left-right and right-left, or have a return wire at the end of each layer to keep lay direction the same. Personally I prefer the evenness of alternating which gives a peak potential difference of twice the per layer potential, but others use an end-to-start "return" wire with a goal of reducing interlayer winding stress to the potential of one layer, at the expense on disturbing the uniformity of each layer. )</p>
<p>And finally the oil. I evacuated the tank for a day, then slowly let in the dry transformer oil from the bottom to ensure no voids.</p>
<p>HV lead outs were through homemade tufnol bushes, with the inside ends directly oil immersed, and with corona rings on the outside to limit stress / tracking.</p>
| <p>I have built ~2mH inductors wired on air core plastic bobbins from AWG26 ML240 polyimide magnet wire and tested them at ~25kV. They seem to work fine without shorting through the polyimide insulation.</p>
<p>I am designing a Marx generator and would like to use voltages ~50kV per stage instead of 25kV but I do not have a power supply to test the inductors at this voltage level.</p>
<p>Before I fork out the money to buy/build a 50kV DC power source to test the inductors I was hoping someone would know if the magnet wire in those inductors should work ok at 50kV or if it's a lost cause.</p>
<p>AFAIK polyimide magnet wire insulation has the highest dielectric strength of any magnet wire insulation.</p>
<p>Edit:</p>
<p>I don't know the wire length of number of turns. The wire I used is: <a href="https://www.remingtonindustries.com/magnet-wire/magnet-wire-240-c-26-awg-polyimide-6-spool-sizes-available/" rel="nofollow noreferrer">https://www.remingtonindustries.com/magnet-wire/magnet-wire-240-c-26-awg-polyimide-6-spool-sizes-available/</a></p>
<p>The bobbin I used 18mm ID, 30mm OD, and 18mm height: <a href="https://www.digikey.com/en/products/detail/ferroxcube/CP-P36-22-1S/7034192" rel="nofollow noreferrer">https://www.digikey.com/en/products/detail/ferroxcube/CP-P36-22-1S/7034192</a></p>
| Can ML240 polyimide magnet wire handle 50kV when used in inductors? | 2023-12-20T23:46:05.923 |
694973 | |timer|programming|plc|ladder|hmi| | <p>One way to do this is with a FIFO (First-In-First-Out) structure. On the Do-More PLC, the relevant instructions are FIFOLOAD and FIFOUNLOAD. Because you want the display to update each minute, you need to keep track of runtime during the current minute, perhaps with an accumulating timer (TMRA). At the end of the minute, you unload the data for the minute that happened 24 hours ago (the oldest minute in the structure), load the data for the minute that just passed, and then add up the data for all of the minutes in the structure so you can update the display. There are 1440 minutes in a 24-hour period, so that is how big your data block will need to be.</p>
<p>As an alternative to adding up 1440 values every minute, you can keep the running total in a separate variable, and then each minute you can subtract the value you unload and add the value that you load.</p>
| <p>I'm working on a BRX PLC connected to a C-More HMI. I'm using Do-More Designer to write ladder code. I'm looking for ladder logic that will give me the runtime of a circulating pump in the last 24 hours. I want something that updates every minute and displays as HH:MM on the HMI. I've included a visual representation of what I'm looking for.
<a href="https://i.stack.imgur.com/LRL5n.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LRL5n.jpg" alt="Runtime Visual Example" /></a></p>
| Runtime display for last 24 hours | 2023-12-21T01:42:50.780 |
694980 | |switches|relay|wiring|bad-wiring|fail-safe| | <p>In your original circuit if two or more by-pass switches were left in the enable position (with the key switch off), there is the possibility that a working relay on one of those lines could inadvertently turn on two or more lamps.</p>
<p>If you were to change the arrangement a bit and also use SPDT by-pass switches the relays can be completely isolated and above unintentional operation could be avoided.</p>
<p>In the circuit below each SPDT by-pass switch (S1,S2,S3,S4), will select either the NO pin of a relay or the key switch line.</p>
<p>If the key switch is enabled (closed) each by-pass switch can manually light a single lamp circuit.</p>
<p>If the key switch is disabled (open) any combination of by-pass switch positions will not cause an extra lamp to light up even if a relay were to switch.</p>
<p>The only potential issue is that if a by-pass switch is left in the by-pass position and the key switch is disabled the lamp in that line will always be disabled even if the corresponding relay were to operate.</p>
<p>If you had to have absolute full manual control for each lamp (for example to over ride a relay stuck in the on state or off state), you could go a step further and select SP3T by-pass switches. That would allow you to have three operating modes: A manual on mode, a relay control mode, and a manual off mode.</p>
<p><a href="https://i.stack.imgur.com/MeUum.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MeUum.jpg" alt="enter image description here" /></a></p>
<p>.</p>
<p>If you ultimately need to have full control in addition to preventing any issues with unintentional positioning of the by-pass switches then you should also use four separate key switches. If you don't use 4 separate key switches there is the chance that a working system relay could be inadvertently by-passed. The key switches (shown as "KEY Sn") would need to be SPDT types.</p>
<p>In the second circuit below when the corresponding key switch is disabled (set to the right) the main relay controls that line and any unintended positioning of the by-pass switch will have absolutely no effect on that line or the other lines.</p>
<p>When an only when the corresponding key switch is enabled (set to the left) will the single by-pass switch gain full control of the lamp (load) in that line.</p>
<p>As shown with SP3T by-pass switches (shown as "BP Sn") the by-pass options would include On-Relay-Off. The by-pass Relay position could be useful when someone gets around to changing a defective relay and would like to test it prior to a supervisor resetting the key switch. If one of these options is not needed you could use a SPDT type switch here.<br />
.</p>
<p><a href="https://i.stack.imgur.com/9Q2IH.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9Q2IH.jpg" alt="enter image description here" /></a></p>
| <p>The lamps are normally operated by relays. When the relay or relay driving circuit fails totally, the bypass switches are to be used as fail-safe to operate the lamps. The key operated switch which is normally open (NO) at all the times, to be used when there is a relay circuit failure. The problem here is when say, relay 4 is on, and by chance switch 4 is also on, now it becomes easy to operate the lamps 1,2,3,4 normally with the switches making relay operation useless. The key operated switch has no use as well. Please suggest an alternative wiring using switches only to operate the load in case of relay circuit failure.</p>
<p><a href="https://i.stack.imgur.com/djfOa.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/djfOa.jpg" alt="enter image description here" /></a></p>
| Relay with bypass switches haywire | 2023-12-21T05:14:23.853 |
694983 | |capacitor|identification| | <p>That capacitor matches the definition, at least according to some manufacturer's capacitor markings.</p>
<p>For markings according to Nichicon the top row is lot number, middle row is capacitance and voltage code, and bottom row is capacitor series.</p>
<p>I did not find Nichicon series RA but otherwise the markings are consistent. 22 meaning capacitance in microfarads (uF) and C is voltage code for 16V.</p>
<p>And 20% is quite typical for electrolytic capacitors, it can be assumed by default, but as there is little room for printing anything extra, it's just the capacitor series that has some tolerance so there is no need to separately print it.</p>
| <p>Can anyone tell me the specs of this capacitor?</p>
<p>Specsheet says <code>22MF 20% 16V</code>, but I'm not sure what capacity exactly the <code>M</code> stands for. And it has <code>C</code> or Ohm on the capacitor itself. It is confusing.</p>
<p><a href="https://i.stack.imgur.com/aaCFD.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aaCFD.png" alt="enter image description here" /></a></p>
| Identify this capacitor marked 'ū1 22c RA' | 2023-12-21T07:08:44.397 |
695000 | |system-verilog|hdl|synthesis|optimization|netlists| | <p>Combinational logic optimization in hardware and software is basically the same. It applies algebraic and Boolean expression evaluation to propagate constants and simplify logic equations.</p>
<p>Sequential logic optimization can be very specific to hardware design as well as the target technology the hardware is being mapped to. Even though the input to a DFF might be a constant, the output will not become a constant until the first clock edge. That may or may not be significant depending how power up reset is handled.</p>
| <p>I'm implementing a pipelined CPU in SystemVerilog. I need to propagate datapath signals from one pipeline stage to the next but, of course, not all stages produce the same number of signals.</p>
<p>To keep things tidied up, I group the signals in structs. Since each stage takes the signals from the previous one, adds a few of them, and propagates to the next stage, I have 2 options:</p>
<ul>
<li>Create several different structs with a different number of signals, one per pipeline stage.</li>
<li>Create one big struct that contains all the possible signals, then set the unused ones to 0 in stages that don't use them.</li>
</ul>
<p>I'm not sure which one to pick. The first one has the benefit of using less signals, potentially generating a smaller netlist after synthesis. The second one, however, is simpler/more regular and easier to understand and follow.</p>
<p>If this was a programming language, I would choose the simpler option and let the compiler optimize it. However, can I count on the HDL synthesizer to optimize out the signals that are always set to 0 and never read/used, in an more-or-less equivalent way as compilers do with source code?</p>
| Do HDL synthesizers "optimize code", more or less as compilers do? | 2023-12-21T11:07:26.237 |
695004 | |oscilloscope|safety| | <p>For safety reasons I would recommend using a ground plug adapter that has a banana plug socket.</p>
<p><a href="https://i.stack.imgur.com/vMGDw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vMGDw.png" alt="enter image description here" /></a><br />
<sub>Source: <a href="https://www.hisco.com/Product/09838-15062" rel="nofollow noreferrer">https://www.hisco.com/Product/09838-15062</a></sub></p>
<p>As Lundin said, you have to make sure that it doesn't integrate a MΩ resistor.</p>
| <p>I just bought my first oscilloscope. It's a Rigol DHO804. Quite happy with it, but I have 1 question. It is powered by a USB-C adapter and comes with a seperate PE lead. I can plug this in the back of the scope but then there is the other side of the lead. To what should I connect this to? Attached a photo of the scope and the lead. I have very little knowledge about electronics so the answer might be painfully obvious. Please enlighten me.</p>
<p>EDIT: Manual says "Connect the instrument chassis to the ground using the lead.". That's all.</p>
<p><a href="https://i.stack.imgur.com/sOYfM.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sOYfM.jpg" alt="enter image description here" /></a></p>
| How to connect PE lead of oscilloscope? | 2023-12-21T11:30:33.513 |
695016 | |batteries|battery-charging|charger|lipo| | <p>For Li-Pol and Li-ion batteries, yes. Just make sure they are charged at the same level and are on the same voltage the first time you are connecting them together.</p>
<p>Making them parallel makes them the same as a battery with double capacity. It will just be a little bigger than one battery with the same capacity.</p>
| <p>I am making a wearable device so for better space distribution I would like to use two smaller batteries instead of a big one. I've been using TinyCircuits's ASR00007 (290mA, lipo) battery which has a charger ic recommended. Can I simply connect two smaller batteries to the end in paralell?
(Datasheet to the battery: <a href="https://hu.mouser.com/datasheet/2/855/ASR00007_290mAh-3078659.pdf" rel="nofollow noreferrer">https://hu.mouser.com/datasheet/2/855/ASR00007_290mAh-3078659.pdf</a>)</p>
<p><a href="https://i.stack.imgur.com/QS28h.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QS28h.png" alt="TinyCircuits's ASR00007 battery charger" /></a></p>
| Can I use a single battery charging ic to charge two batteries in parallel? | 2023-12-21T13:34:45.893 |
695018 | |pwm|voltage-divider|plc| | <p>While the comments suggested a resistor divider, from the pump information I'm not sure what the input impedance is.</p>
<p>Given the C2-14D2 supports currents of up to 0.1 A was thinking using a zener diode to supply a 4.7 V high input to the pump PWM input and be able to deliver several mA. This is a variation of the other <a href="https://electronics.stackexchange.com/a/695019/101063">answer</a> which mentions using a logic gate as a buffer:</p>
<p><img src="https://i.stack.imgur.com/EmDxL.png" alt="schematic" /></p>
<p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fEmDxL.png">simulate this circuit</a> – Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a></sup></p>
<p>Where:</p>
<ul>
<li>R1 limits the current used in the divider to under 10 mA. Needs a resistor rated for at least 0.25 W</li>
<li>D1 is a 4.7 V zener diode to give a logic high voltage</li>
<li>R3 serves as a pull-down to allow a logic low level when the PLC output is off</li>
<li>The other components simulate the <em>Equivalent Output Circuit</em> of the C2-14D2 output module in the PLC</li>
</ul>
| <p>I am working on a small dose spraying system for a farming application. I have a Click Plus PLC, part number <a href="https://www.automationdirect.com/adc/shopping/catalog/programmable_controllers/click_plus_plcs_(stackable_micro_modular)/cpus/c2-01cpu-2" rel="nofollow noreferrer">C2-01CPU-2</a>, and it has one IO card so far, a <a href="https://www.automationdirect.com/adc/shopping/catalog/programmable_controllers/click_plus_plcs_(stackable_micro_modular)/cpu_option_slot_modules/c2-14d2" rel="nofollow noreferrer">C2-14D2</a>. I want the PLC to control a Kamoer diaphragm pump <a href="https://www.kamoer.com/us/product/detail.html?id=10056" rel="nofollow noreferrer">KLP180-E-B12</a> using the PWM signal. I was attempting to run the 24 volt DC PWM signal through a Wago DC to DC Converter <a href="https://www.automationdirect.com/adc/shopping/catalog/power_products_(electrical)/dc-to-dc_converters/din_rail_mount/859-801" rel="nofollow noreferrer">859-801</a> to get a 5 volt DC PWM signal, but it doesn't seem to be functioning correctly. I have a feeling the cycle speed is what is causing me issues. I would prefer to not add another card to my PLC for space saving reasons, but if I have to, I can. I feel like a voltage divider should do the trick, but I am a mechanical engineer... not an electrical engineer! Would someone be able to advise me on this?</p>
<p>Pump Info Below:
<a href="https://i.stack.imgur.com/ojFSm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ojFSm.png" alt="enter image description here" /></a>
<a href="https://i.stack.imgur.com/9gEho.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9gEho.png" alt="enter image description here" /></a></p>
| How do I take a 24 volt DC 10-30 kHz PWM signal from a Click Plus PLC and convert it to a 5 volt DC PWM signal to control a pump? | 2023-12-21T13:44:18.487 |
695030 | |vhdl|formal-verification| | <p>You have to use the PSL <code>restrict</code> directive to model your Reset behavior. It is used to say the tool that every "run" has to match the given SERE. Or, with other words, every exploration has to fulfill the behavior given using the <code>restrict</code> directive.</p>
<p>From the book <strong>A Practical Introduction to PSL</strong>:</p>
<pre><code>restrict: This directive restricts a verification tool to
constrain the verification so that it matches a SERE. For
instance, restrict {a ; b ;
c[*]} instructs the tool to consider only traces every prefix of which
matches {a ; b ; c[*]}.
</code></pre>
<p>This should do what you want:</p>
<pre class="lang-vhdl prettyprint-override"><code>INITIAL_RESET : restrict {not rst_n_i[*1]; rst_n_i[+]}[*1];
</code></pre>
<p><code>not rst_n_i[*1]</code> rst_n_i = '0' in the first cycle</p>
<p><code>;</code> Skip to the next cycle</p>
<p><code>rst_n_i[+]</code> rst_n_i = '0' in the next cycle, repeating a non-zero number of repetitions</p>
<p><code>{not rst_n_i[*1]; rst_n_i[+]}[*1]</code> Matches for exactly one occurrence of the whole sequence.</p>
<p>So the only valid sequence is exactly that one you want:</p>
<pre><code>clk_i: -_-_-_-_....-_-_
rst_n_i: -_______....____
</code></pre>
<p>The assume directive has a subtle different meaning. It's useful to constrain the behavior of inputs, so that all properties only have to hold under these constraints.</p>
<p>From the book <strong>A Practical Introduction to PSL</strong>:</p>
<blockquote>
<p>assume: This directive instructs a verification tool to constrain the verifi-
cation so that a property holds. In other words, it instructs the verification
tool that any asserted properties need not hold on traces that do not obey
the assumptions.</p>
</blockquote>
| <p>I am trying to learn Formal Verification for VHDL design (using Property Specification Language). My module has an active-low reset input. I would like to drive tool to set the reset input to '0' for 1 clk and to '1' for all others.</p>
<p>I've tried the following code:</p>
<pre><code>vunit i_test(test(rtl))
{
-- set all declarations to run on clk
default clock is rising_edge(clk_i);
f_reset_0 : assume (rst_n_i = '0');
f_reset_1 : assume always (next rst_n_i); -- tried also without this line
f_post_reset_out_en : assert always {rst_n_i = '0'} |=> {out_en_o = '0'};
f_sum : assert always {rst_n_i = '1'} |=> {not(out_o = x"AE")};
}
</code></pre>
<p><a href="https://i.stack.imgur.com/vOnWr.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vOnWr.png" alt="enter image description here" /></a></p>
<p>What I am doing wrong?</p>
| Formal HDL verification with PSL | 2023-12-21T15:32:43.457 |
695034 | |transistors|amplifier|bjt| | <p>Output voltage swing simply means the range of values your output voltage can take. The reason it is negative in this amplifier is because in this amplifier the output voltage Vce decreases to a minimum value of 0.3V.</p>
| <p>What is meant by maximum allowable negative signal swing in the case of BJT amplifiers? What is the formula to find the same? What can be a practical example for the same?
<a href="https://i.stack.imgur.com/Zz4iZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Zz4iZ.png" alt="enter image description here" /></a></p>
| Maximum allowable negative signal swing | 2023-12-21T15:57:27.567 |
695039 | |i2c|c|microprocessor|beaglebone-black|real-time| | <p>I have found the issue with the given code after some tedious debugging; turns out the delays given in the library were not enough, and the library failed to clear the bit flags that indicate an error occurred. As a result, the first transaction would occur as normal, but when attempting to read the data from the RX FIFO, the AERR flag would be set since the FIFO was still empty when trying to read due to the delays not being enough. Then, when trying to perform the second read, the error flag is still set and the data read from the RX FIFO is actually the previous value. After adding a few more microseconds of delay between transactions and making sure to handle any errors that occur everything works well. The updated ReadBytes function from the library I've rewritten to:</p>
<pre><code>uint8_t pru_i2c_driver_ReadBytes(
const uint8_t i2cDevice,
const uint8_t address,
const uint8_t reg,
const uint8_t bytes,
uint8_t* buffer,
i2c_flags_t* flags
)
{
uint8_t tries = 0;
_init_section:
if(!pru_i2c_initialized[i2cDevice-1]) {
if(!pru_i2c_driver_Init(i2cDevice)) {
*flags |= I2C_FAILED_INIT;
if(tries++ < MAX_TRIES){
pru_i2c_driver_DelayMicros(1);
goto _init_section;
}
return 0;
}
}
tries = 0;
_write_init:
if (!pru_i2c_driver_WaitBB(i2cDevice))
{
*flags |= I2C_FAILED_WAIT_BB;
if(tries++ < MAX_TRIES){
pru_i2c_driver_DelayMicros(1);
goto _write_init;
}
return 0;
}
tries = 0;
CT_I2C[i2cDevice-1]->I2C_SA_bit.I2C_SA_SA = address; // 7 bit address
CT_I2C[i2cDevice-1]->I2C_CNT_bit.I2C_CNT_DCOUNT = 1; // 1 byte to transmit
CT_I2C[i2cDevice-1]->I2C_CON = 0x8601; // EN/MST/TRX/STT
pru_i2c_driver_DelayMicros(7);
_write_start:
if (!pru_i2c_driver_WaitXRDY(i2cDevice))
{
*flags |= I2C_FAILED_WAIT_XRDY;
if(tries++ < MAX_TRIES){
pru_i2c_driver_DelayMicros(1);
goto _write_start;
}
return 0;
}
// write register to read
CT_I2C[i2cDevice-1]->I2C_DATA = reg;
CT_I2C[i2cDevice-1]->I2C_IRQSTATUS_RAW_bit.I2C_IRQSTATUS_RAW_XRDY = 0b1;
// wait access to registers
if (!pru_i2c_driver_WaitARDY(i2cDevice))
{
*flags |= I2C_FAILED_WAIT_ARDY;
if(tries++ < MAX_TRIES){
pru_i2c_driver_DelayMicros(1);
goto _write_start;
}
return 0;
}
pru_i2c_driver_DelayMicros(5);
CT_I2C[i2cDevice-1]->I2C_IRQSTATUS_RAW_bit.I2C_IRQSTATUS_RAW_ARDY = 0b1;
if (CT_I2C[i2cDevice-1]->I2C_IRQSTATUS_RAW_bit.I2C_IRQSTATUS_RAW_AERR)
{
*flags |= I2C_AERR_WRITE;
if(tries++ < MAX_TRIES){
pru_i2c_driver_DelayMicros(1);
goto _write_start;
}
*flags |= I2C_AERR_WRITE_UNRESOLVED;
return 0;
}
if(CT_I2C[i2cDevice-1]->I2C_IRQSTATUS_RAW_bit.I2C_IRQSTATUS_RAW_NACK)
{
*flags |= I2C_NACK_DETECTED;
CT_I2C[i2cDevice-1]->I2C_IRQSTATUS_RAW_bit.I2C_IRQSTATUS_RAW_NACK = 0b1;
return 0;
}
tries = 0;
_read_init:
// read data
CT_I2C[i2cDevice-1]->I2C_CNT_bit.I2C_CNT_DCOUNT = bytes; // bytes to receive
CT_I2C[i2cDevice-1]->I2C_CON = 0x8403; // EN/MST/STP/STT
pru_i2c_driver_DelayMicros(20);
_read_start:
// wait data
if (!pru_i2c_driver_WaitRRDY(i2cDevice))
{
*flags |= I2C_FAILED_WAIT_RRDY;
if(tries++ < MAX_TRIES){
pru_i2c_driver_DelayMicros(1);
goto _read_start;
}
return 0;
}
pru_i2c_driver_DelayMicros(10);
int8_t count;
for (count = 0; count < bytes; count++)
{
_looptop:
// read byte
buffer[count] = CT_I2C[i2cDevice-1]->I2C_DATA;
pru_i2c_driver_DelayMicros(10);
if (CT_I2C[i2cDevice-1]->I2C_IRQSTATUS_RAW_bit.I2C_IRQSTATUS_RAW_AERR)
{
CT_I2C[i2cDevice-1]->I2C_IRQSTATUS_RAW_bit.I2C_IRQSTATUS_RAW_AERR = 0b1;
if(tries++ < MAX_TRIES){
pru_i2c_driver_DelayMicros(10);
goto _looptop;
}
return 0;
}
if (CT_I2C[i2cDevice-1]->I2C_IRQSTATUS_RAW_bit.I2C_IRQSTATUS_RAW_NACK)
{
*flags |= I2C_NACK_DETECTED;
CT_I2C[i2cDevice-1]->I2C_IRQSTATUS_RAW_bit.I2C_IRQSTATUS_RAW_NACK = 0b1;
return 0;
}
_wait_next_data:
// require next data
CT_I2C[i2cDevice-1]->I2C_IRQSTATUS_RAW_bit.I2C_IRQSTATUS_RAW_RRDY = 0b1;
// wait data
if (!pru_i2c_driver_WaitRRDY(i2cDevice))
{
*flags |= I2C_FAILED_WAIT_RRDY;
if(tries++ < MAX_TRIES){
pru_i2c_driver_DelayMicros(10);
goto _wait_next_data;
}
return 0;
}
tries = 0;
}
_transaction_end:
// wait for access ready
if (!pru_i2c_driver_WaitARDY(i2cDevice))
{
*flags |= I2C_FAILED_WAIT_ARDY;
if(tries++ < MAX_TRIES){
pru_i2c_driver_DelayMicros(10);
goto _transaction_end;
}
return 0;
}
_wait_bus_free:
// wait for bus free
// wait data
if (!pru_i2c_driver_WaitBF(i2cDevice))
{
*flags |= I2C_FAILED_WAIT_BF;
if(tries++ < MAX_TRIES){
pru_i2c_driver_DelayMicros(10);
goto _wait_bus_free;
}
return 0;
}
CT_I2C[i2cDevice-1]->I2C_IRQSTATUS_RAW_bit.I2C_IRQSTATUS_RAW_ARDY = 0b1;
CT_I2C[i2cDevice-1]->I2C_IRQSTATUS_RAW_bit.I2C_IRQSTATUS_RAW_XRDY = 0b1;
CT_I2C[i2cDevice-1]->I2C_IRQSTATUS_RAW_bit.I2C_IRQSTATUS_RAW_RRDY = 0b1;
return count;
}
</code></pre>
<p>Furthermore, the delay function was plain wrong. It did not delay the requested microseconds and was optimized away by the compiler as soon as optimizations were turned on. I fixed it by changing it to:</p>
<pre><code>void DelayMicros(uint8_t micros){
// Factor of 29 gives most accurate timings verified with logic analyzer
uint16_t cycles = ((uint16_t)micros * 29u);
uint16_t i = 0;
#pragma MUST_ITERATE(29, , 29)
#pragma UNROLL(1)
for (i = 0; i < cycles; i++)
{
__delay_cycles(1);
};
}
</code></pre>
| <p>I am using <a href="https://github.com/LinuxDroneLab/pru-i2c-lib" rel="nofollow noreferrer">this</a> library to sample some sensors over the I2C bus using the PRUs on the Beaglebone Black.</p>
<p>It seems as if only the first transaction works correctly, after which the library seems to only attempt to read from register address 0xFF, followed by the SCL line being pulled low (see attached image captures by logic analyzer). I have already opened an issue on their github, but am not expecting an answer there as the repository hasn't been updated in years and the author hasn't been active on GitHub since 2019.</p>
<p><a href="https://i.stack.imgur.com/qpsCj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/qpsCj.png" alt="Signal Analyzer Output" /></a></p>
<p>While trying to find the problem myself, I decided to write the register address and received data in the shared memory to inspect using prudebug. For this test, I am attempting to read registers 0x00 and 0x03. Interestingly, it writes the correct register addresses into memory (see image), but only the first returned value is correct (0xEA). The value written into the received data buffer in shared memory is 0xFF, which does not make sense. As can be seen from the logic analyzer, 0xFF is the address it attempts to read from.</p>
<p><a href="https://i.stack.imgur.com/JSSwE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/JSSwE.png" alt="Memory analysis" /></a></p>
<p>I don't know what I am doing wrong here, but any help would be appreciated! My main code is shown below, and the library functions can be found in the linked repository <a href="https://github.com/LinuxDroneLab/pru-i2c-lib/blob/master/pru_i2c_driver.c" rel="nofollow noreferrer">here</a>.</p>
<pre><code>uint8_t curr_data_idx = 0;
uint8_t tmp = 0;
uint8_t reg = 0x00;
volatile uint8_t * buf = ((volatile uint8_t*)(RESERVED_SMEM_ADDR + 32));
int main(void)
{
// enable OCP
CT_CFG.SYSCFG_bit.STANDBY_INIT = 0;
pru_i2c_driver_DelayMicros(500000);
uint8_t res = 0;
res = pru_i2c_driver_Init(1);
uint8_t saddr = 0x68;
pru_i2c_driver_ReadReg(1, saddr, reg, &tmp);
buf[curr_data_idx] = tmp;
pru_i2c_driver_DelayMicros(1000);
curr_data_idx++;
reg = 0x03;
pru_i2c_driver_ReadReg(1, saddr, reg, &tmp);
buf[curr_data_idx] = tmp;
pru_i2c_driver_DelayMicros(1000);
curr_data_idx++;
__halt();
return 0;
}
</code></pre>
<p>The Technical Reference Manual of the CPU and the I2C peripheral can be found <a href="https://www.ti.com/lit/ug/spruh73q/spruh73q.pdf" rel="nofollow noreferrer">here</a>, page 4583.</p>
| I2C peripheral fails after a single transaction | 2023-12-21T16:11:15.860 |
695048 | |pcb|usb|soldering|usb-device|usb-c| | <p>I believe from the blurry photo that pins of missing component L2 connect to ground and pins of missing component L1 connects to VBUS. These are the two widest tracks to the connector, so they must be the supply pins. The two narrower tracks are the USB data pair.</p>
<p>Feel free to measure yourself with a multimeter.</p>
<p>However, as this is a Type-C receptacle, there will be no voltage on connector unless a cable to another device is plugged in.</p>
<p>And since this is is Type-C, the voltage can be negotiated between the devices. You never told which device this is so the voltage it requests is unknown. Type-C currently supports up to 20V and 5A.</p>
<p>So just a warning that soldering blue LEDs to the VBUS and Ground pins of the connector may not be very useful as it always requires the cable to supply power, and you must know what voltages this device negotiates from the power source.</p>
| <p>I'm modifying my Nintendo controller, and I need to solder some stuff onto the <strong>usb type c</strong> <strong>female</strong> plug. However, I noticed there are <strong>five</strong> pins, and I have <strong>no idea</strong> what a single one is. Could you explain what each pin is?</p>
<p>Sorry it's blurry; this is my usb c female plug:</p>
<p><a href="https://i.stack.imgur.com/fkkET.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fkkET.jpg" alt="board" /></a></p>
| How do you wire a usb type c female connector? | 2023-12-21T17:00:20.760 |
695057 | |power|relay|power-electronics|components| | <h3>Component 1</h3>
<p>Component 1 might be a <a href="https://sinolec.co.uk/en/manual-reset-thermostat/1211842-thermal-cut-out-110c-nc.html" rel="nofollow noreferrer">Thermal Cut Out</a>, based upon a quick search of similar looking components. Is there any marking on the white body of component 1?</p>
<p>The black <em>button</em> could be a manual reset, to be used after an over temperature has occurred (and the oven has cooled down).</p>
<h3>Component 2</h3>
<p>Component 2 appears to have the manufacturers name of HUIDE. <a href="http://www.huidewk.com/en/showproducts.asp?id=2486" rel="nofollow noreferrer">Rising thermostatboa...</a> has the following picture which is taken from a different angle, but could be the <em>heat knob</em>:</p>
<p><a href="https://i.stack.imgur.com/V7lPO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/V7lPO.png" alt="enter image description here" /></a></p>
<p>From the above manufacturers website can't seem to find any data nor part numbers.</p>
| <p>I am disassemblying an oven to make a reflow oven out of it. I need help identifying two components. The simplified schematic of the system is as follows:</p>
<p><a href="https://i.stack.imgur.com/2UhbH.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/2UhbH.png" alt="oven schematics" /></a></p>
<p>Component 1 (picture below) is marked as Component ? in the schematics. Is this just a connection between the two wires or is there more to it?</p>
<p><a href="https://i.stack.imgur.com/tGw8D.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/tGw8D.jpg" alt="Component ? to identify" /></a></p>
<p>Component 2 is the heat knob (picture below). A google search didn' yield much. How does this component work? Does it set the heat "PWM style"?</p>
<p><a href="https://i.stack.imgur.com/t9cET.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t9cET.jpg" alt="Component 2 to identify" /></a></p>
<p>Thanks a lot for your help.</p>
| Identify Components in Oven | 2023-12-21T18:20:50.937 |
695064 | |operational-amplifier|differential| | <h2>discussion</h2>
<p>I believe you mentioned that you haven't covered Laplace in lectures, yet. This, plus your starting approach, makes me think you are working on things more like analog computers (old days.)</p>
<p>For notation convenience set <span class="math-container">\$D=\frac{\text{d}}{\text{d}t}\$</span>, so that <span class="math-container">\$D\$</span> is just an operator and then your equation becomes:</p>
<p><span class="math-container">$$\begin{align*}
\left[4D^3+10D^2+8D+6\right]v_{out}&=12v_{in}\tag{1}
\\\\
\left[D^3+\frac52D^2+2D+\frac32\right]v_{out}&=3v_{in}\tag{2}\label{td}
\\\\
\therefore \frac{v_{out}}{v_{in}}&=\frac3{D^3+\frac52D^2+2D+\frac32}\tag{3}\label{lap}
\end{align*}$$</span></p>
<p>I'll get to Eq. <span class="math-container">\$\ref{lap}\$</span>, later.</p>
<h2>analog computers</h2>
<p>For now, re-arrange Eq. <span class="math-container">\$\ref{td}\$</span> to get:</p>
<p><span class="math-container">$$\dddot{v_{out}}=3\left[v_{in}-\frac12v_{out}\right]-\frac52\ddot{v_{out}}-2\dot{v_{out}}\tag{4}\label{ac}$$</span></p>
<p>That's great because a standard RC integrator has <span class="math-container">\$v_o=-\frac1{R\,C}\int v_{i}\$</span>, which means such a stage can be used to develop each of the required terms found in Eq. <span class="math-container">\$\ref{ac}\$</span>.</p>
<p>Start by assuming <span class="math-container">\$\dddot{v_{out}}\$</span> exists. That can be integrated to <span class="math-container">\$-\ddot{v_{out}}\$</span>, then to <span class="math-container">\$\dot{v_{out}}\$</span>, and finally to <span class="math-container">\$-v_{out}\$</span>. Since <span class="math-container">\$v_{out}\$</span> is desired, flip the sign and assume to start with <span class="math-container">\$-\dddot{v_{out}}\$</span>, instead. Apply:</p>
<p><a href="https://i.stack.imgur.com/bRxsV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bRxsV.png" alt="enter image description here" /></a></p>
<p>Then to that, apply:</p>
<p><a href="https://i.stack.imgur.com/a6SJR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/a6SJR.png" alt="enter image description here" /></a></p>
<p>And now to get the final result apply:</p>
<p><a href="https://i.stack.imgur.com/B9pzb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B9pzb.png" alt="enter image description here" /></a></p>
<p>You now have the important terms for:</p>
<p><span class="math-container">$$\dddot{v_{out}}=-3\left[\vphantom{\frac52}v_{out}-\frac12v_{in}\right]-\left(+\frac52\ddot{v_{out}}\right)+\left(\vphantom{\frac52}-2\dot{v_{out}}\right)\tag{5}\label{acx}$$</span></p>
<p>All that can be handled now with a few more opamps to complete the picture (if I didn't miss anything -- look closely over this as I may have made a mistake in translating the image below -- but at least it points out the general approach even if I get a sign wrong below):</p>
<p><a href="https://i.stack.imgur.com/Y3u9u.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Y3u9u.png" alt="enter image description here" /></a></p>
<p>It's possible this is the approach required of you. But I'll move on from here.</p>
<h2>transfer function</h2>
<p>It's just a notation change to turn the transfer function shown by Eq. <span class="math-container">\$\ref{lap}\$</span> into Laplace form:</p>
<p><span class="math-container">$$\frac{V_{out}}{V_{in}}=2\cdot\frac{\frac32}{s^3+\frac52s^2+2s+\frac32}\tag{6}$$</span></p>
<p>This is recognizable as a low-pass, 3rd order filter with a DC gain of 2. There are several alternatives to turn this into a filter design. One such filter that could be applied is shown at <a href="https://electronics.stackexchange.com/a/694851">this answer</a>.</p>
<p>However, this is about having a straight-forward process so I'll take a simpler-to-apply approach that uses two opamps. That will help isolate the issues.</p>
<h2>design</h2>
<p>Given it is 3rd order, there will be a real root and a conjugate pair. The real root will be used for one opamp for a 1st order response and the conjugate pair will be used in my example for another opamp with a 2nd order response. The schematic will look like this:</p>
<p><img src="https://i.stack.imgur.com/b4zGQ.png" alt="schematic" /></p>
<p><sup><a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fb4zGQ.png">simulate this circuit</a> – Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a></sup></p>
<p>I'll be using SymPy/Python/Sagemath as my tools. But if you wish, you can refresh your memory on <a href="https://en.wikipedia.org/wiki/Cubic_equation" rel="nofollow noreferrer">cubics</a> and solve it exactly. Regardless, step 1 is to get the roots:</p>
<pre><code>for i in roots(s**3 + 5/2*s**2 + 2*s + 3/2,s):i.n()
-1.85809432949655
-0.320952835251724 - 0.839206763026694*I
-0.320952835251724 + 0.839206763026694*I
expand((s-(-0.320952835251724-0.839206763026694*I))*(s-(-0.320952835251724+0.839206763026694*I)))
s**2 + 0.641905670503448*s + 0.807278713565862
</code></pre>
<p>I'll center around <span class="math-container">\$10\:\text{k}\Omega\$</span> resistors for <span class="math-container">\$R_1\$</span>, <span class="math-container">\$R_2\$</span>, and <span class="math-container">\$R_3\$</span>.</p>
<p>Normally, it would be better to consider the capacitor values first and then pick odd-valued resistors, since they come in wider varieties. But this is schoolwork. So I'll set the resistors instead and let the capacitor values fall where they may. Feel free (as an education) to set the capacitor values to readily available values and work out the resistors. I'll leave that to you.</p>
<p>The transfer function, with gain of 2, for the 1st order stage is <span class="math-container">\$2\cdot\frac{\omega_{_0}}{s+\omega_{_0}}\$</span>. Given the real root above, <span class="math-container">\$\omega_{_0}=1.85809432949655\:\frac{\text{rad}}{\text{s}}\$</span>. So the capacitor for the 1st order stage (also the first stage) is:</p>
<p><span class="math-container">$$C_1=\frac{1}{\omega_{_0}\cdot R_1}=\frac{1}{1.85809432949655\:\frac{\text{rad}}{\text{s}}\,\cdot\,10\:\text{k}\Omega}\approx 53.8186\:\mu\text{F}$$</span>.</p>
<p>The transfer function, with gain of 1, for the 2nd order stage is <span class="math-container">\$\frac{\omega_{_0}^{\,2}}{s^2+\frac{\omega_{_0}}{Q}\,s+\omega_{_0}^{\,2}}\$</span>. Given the 2nd order denominator made from the roots above, <span class="math-container">\$\frac{\omega_{_0}}{Q}=0.641905670503448\$</span> and <span class="math-container">\$\omega_{_0}^{\,2}=0.807278713565862\:\frac{\text{rad}^2}{\text{s}^2}\$</span>. The feedback capacitor for the 2nd order stage is:</p>
<p><span class="math-container">$$C_2=\frac{1}{\frac{\omega_{_0}}{Q}\,\cdot\,\left(R_2\mid\mid R_3\right)}=\frac{1}{0.641905670503448\:\frac{\text{rad}}{\text{s}}\,\cdot\,\left(10\:\text{k}\Omega\mid\mid 10\:\text{k}\Omega\right)}\approx 311.5723\:\mu\text{F}$$</span>.</p>
<p>The final capacitor (grounded) for the 2nd order stage then falls out as:</p>
<p><span class="math-container">$$\begin{align*}C_3=\frac{1}{\omega_{_0}^{\,2}\,\cdot\,R_2\,\cdot\,R_3\,\cdot\,C_2}&=\frac{1}{0.807278713565862\:\frac{\text{rad}^2}{\text{s}^2}\,\cdot\,10\:\text{k}\Omega\,\cdot\,10\:\text{k}\Omega\,\cdot\,C_2}\\\\&\approx 39.7574\:\mu\text{F}\end{align*}$$</span></p>
<p>And that's it.</p>
<h2>validation</h2>
<p>I'll provide a schematic using LTspice that includes both the transfer function itself (no components or parts) and then the final schematic:</p>
<p><a href="https://i.stack.imgur.com/nlkTd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nlkTd.png" alt="enter image description here" /></a></p>
<p>This shows the ideal response (red) and the actual response using models for actual parts (green, and in this case a decompensated LT1127 opamp.) It's clear that the transfer functions match well up to a point.</p>
<p>The process above works reasonably well and uses fewer opamps and parts.</p>
<h2>time domain</h2>
<p>If we set:</p>
<p><span class="math-container">$$\begin{align*}
\alpha_{_0}&=-1.85809432949655\\
\beta_{_0}&=0.839206763026694\\
\alpha_{_1}&=-0.320952835251724
\end{align*}$$</span></p>
<p>then,</p>
<p><span class="math-container">$$\begin{align*}
\left[D^3+\frac52D^2+2D+\frac32\right]v_{out}&=3v_{in}
\\\\
\left[D-\alpha_{_1}\vphantom{\left(\right)^{\,2}}\right]\left[\left(D-\alpha_{_0}\right)^2+ \beta_{_0}^{\,2}\right]v_{out}&=3v_{in}
\end{align*}$$</span></p>
<p>Just to verify:</p>
<pre><code>collect(expand((s-(-1.85809432949655))*((s-(-0.320952835251724))**2+0.839206763026694**2)),s)
s**3 + 2.5*s**2 + 2.0*s + 1.5
</code></pre>
<p>That pans out. The only remaining problem is that this isn't homogeneous and we don't know what <span class="math-container">\$v_{in}\$</span> is.</p>
<p>If we happen to suggest, for example, that <span class="math-container">\$v_{in}=A\,\sin\left(\omega_{_0}\,t\right)\$</span> then the right side can be annihilated as follows:</p>
<p><span class="math-container">$$\begin{align*}
\left[D-\alpha_{_1}\vphantom{\left(\right)^{\,2}}\right]\left[\left(D-\alpha_{_0}\right)^2+ \beta_{_0}^{\,2}\right]v_{out}&=3v_{in}
\\\\
\left[\left(D-\alpha_{_2}\right)^2+ \omega_{_0}^{\,2}\right]\left[D-\alpha_{_1}\vphantom{\left(\right)^{\,2}}\right]\left[\left(D-\alpha_{_0}\right)^2+ \beta_{_0}^{\,2}\right]v_{out}&=\left[\left(D-\alpha_{_2}\right)^2+ \omega_{_0}^{\,2}\right]3v_{in}
\\\\
\left[\left(D-\alpha_{_2}\right)^2+ \omega_{_0}^{\,2}\right]\left[D-\alpha_{_1}\vphantom{\left(\right)^{\,2}}\right]\left[\left(D-\alpha_{_0}\right)^2+ \beta_{_0}^{\,2}\right]v_{out}&=0, \text{where }\alpha_{_2}=0
\end{align*}$$</span></p>
<p>and now it is homogeneous. It's also in a form we can use, as:</p>
<ul>
<li><p><span class="math-container">\$\left[D-\alpha\right]^m\$</span> annihilates any solution of the general form:</p>
<p><span class="math-container">\$\quad\quad f\left(t\right)=\sum_{k=0}^{m-1} A_k\,t^k\,e^{^{\alpha\,t}}\$</span>.</p>
</li>
<li><p><span class="math-container">\$\left[\left(D-\alpha\right)^2+\beta^2\right]^m\$</span> annihilates any solution of the general form:</p>
<p><span class="math-container">\$\quad\quad f\left(t\right)=\sum_{k=0}^{m-1} A_{2k}\,t^k\,e^{^{\alpha\,t}}\cos\left(\beta\,t\right) + A_{2k+1}\,t^k\,e^{^{\alpha\,t}}\sin\left(\beta\,t\right)\$</span></p>
</li>
</ul>
<p>So, the general form of the solution with unspecified constants is:</p>
<p><span class="math-container">$$v_{out}=A_{_0}\cos\left(\omega_{_0}\,t\right) + A_{_1}\sin\left(\omega_{_0}\,t\right)+A_{_2}\,\exp\left(\alpha_{_1}\,t\right)+\exp\left(\alpha_{_0}\,t\right)\left[A_{_3}\cos\left(\beta_{_0}\,t\right) + A_{_4}\sin\left(\beta_{_0}\,t\right)\right]$$</span></p>
<p>Some terms are just part of the general solution of the associated homogeneous equation and some are remaining terms expressing the particular solution to the nonhomogeneous equation with undetermined coefficients. But at least you are now in a position of working through a process to determine the coefficients.</p>
<p>For example, to start you might note that at <span class="math-container">\$t=0\$</span> you know that <span class="math-container">\$A_{_0}+A_{_2}+A_{_3}=0\$</span>. From the derivative, at <span class="math-container">\$t=0\$</span>, you also know that <span class="math-container">\$A=A_{_1}\cdot\omega_{_0}+A_{_2}\cdot\alpha_{_1}+A_{_3}\cdot\alpha_{_0}+A_{_4}\cdot\beta_{_0}\$</span>. And so on.</p>
<p>Just something to consider if you were looking to create a time domain solution.</p>
| <p><span class="math-container">$$ 6b(t)+\frac{8d}{dt}b(t)+\frac{10d^2}{dt^2}b(t)+\frac{4d^3}{dt^3}b(t)=12a(t) $$</span></p>
<p>I would like to implement this linear equation by using any type of op-amps. The solution will be the function b(t), so we can set the initial conditions to be zero. a(t) is the input.<br />
It was hard to grasp this differential equation so it was a challenge for me to apply this circuit. I will be glad for your help. You are free to use inductors and/or capacitors if you want. Thank you.</p>
<p>This is the circuit that I have tried to build but it only gives 0.
<a href="https://i.stack.imgur.com/3wrIF.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3wrIF.png" alt="enter image description here" /></a></p>
| Using Operation-Amplifier circuit to apply differential equation | 2023-12-21T18:49:08.323 |
695069 | |h-bridge|efficiency|full-bridge|class-d| | <p>Datasheet says: "Effective Output Capacitance, Energy Related" = 15pF</p>
<p>Coss (Cds+Cgd) capacitance is charged from the power supply every time the FET turns off, which stores some energy. When the FET turns on, it shorts the cap so the energy turns into heat.</p>
<p>Power loss: FCV² = 0.8W with your numbers, so a large part of losses are from charging and discharging FET drain capacitance.</p>
<p>This amp should be able to output 5A@400Vpeak so 1kW real watts into a 80 ohm load. 37% efficiency at 3W is pretty good. It will be lower once you factor in the rest of low voltage circuits, opamps, gate drive, etc.</p>
<p>Also... I think the FET turns off and goes into diode mode where the arrow is pointing, so maybe you could reduce dead time. A bit more zoom wouldn't hurt to check. That will only matter for efficiency at high current though.</p>
<p><a href="https://i.stack.imgur.com/P2d1W.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/P2d1W.png" alt="enter image description here" /></a></p>
| <p>I am simulating an H-bridge/class-D amplifier based on four GaN transistors with integrated gate drivers (<a href="https://navitassemi.com/wp-content/uploads/2022/06/NV6113-Datasheet-FINAL-05-19-22.pdf" rel="nofollow noreferrer">NV6113</a> from Navitas). I am driving the gates via PWM signals where the PWM frequency is about 330 kHz and the signal frequency is about 30 kHz.</p>
<p><a href="https://i.stack.imgur.com/hwFId.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hwFId.png" alt="enter image description here" /></a></p>
<p>Basically, the circuit works and I get a sinusoidal output voltage across the load (although the filtering can of course be improved).</p>
<p><a href="https://i.stack.imgur.com/Oed11.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Oed11.png" alt="enter image description here" /></a></p>
<p>However, I don't understand why the efficiency is so poor. An effective power of 3 W results across the load Rload. However, the circuit draws a power of about 13 W from the bridge voltage V_SUPPLY_FB. The power dissipation of the four transistors is around 2.5 W each.</p>
<p>I suspect the problem is that very high current peaks (10 A) occur in the drain current. Does anyone have any ideas on how the efficiency or the circuit could be improved?</p>
<p><a href="https://i.stack.imgur.com/gVp2s.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gVp2s.png" alt="enter image description here" /></a></p>
<p><strong>Update 1:</strong></p>
<p>I introduced about 18 ns of dead time. As before, an effective power of 3 W results across the load Rload. However, the circuit still draws an average power of about 8.1 W from the bridge voltage V_SUPPLY_FB (about 1.3 W average loss for each transistor). Thus, the efficiency improved from 23 % (without dead time) to 37 % (with dead time).</p>
<p>As the following signals (only shown for half-bridge) show there is still significant power loss in the transistors during switching. I noticed that the voltage on the first transistor only changes after the second transistor has been switched. Is this because the load has a relatively high impedance? Any ideas on how to further improve the efficiency?</p>
<p><a href="https://i.stack.imgur.com/uwEb5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/uwEb5.png" alt="enter image description here" /></a></p>
<p>Just for comparison the same signals <strong>without dead time</strong>:</p>
<p><a href="https://i.stack.imgur.com/UcRwA.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UcRwA.png" alt="enter image description here" /></a></p>
| Poor efficiency of an H-bridge | 2023-12-21T20:52:40.723 |
695076 | |adc| | <blockquote>
<p>Can I expect the number of levels applicable to the accelerometer to be proportional to the voltage of the accelerometer output, ie 4096/10*3.3? or how does it work?</p>
</blockquote>
<p>Short answer: Yes, provided that the output voltage of the sensor (the accelerometer in this case) is limited to between 0.0V & 3.3V.</p>
<p>In other words, you are losing about 67% of the available ADC range. One bit of the ADC output will represent 2.4mV (10V/4096). For an input voltage ranging from 0 to 3.3V the ADC output range will be 0 to 1,352; meaning that ADC output counts from 1,353 to 4,095 will not be used.</p>
| <p>I am using a ADXL335 accelerometer on a GY-61 board. Apparently the output is radiometric so the 0g measurement output is at half of the 3.3V supply voltage (1.65V), -3g is at 0v, and +3g is at 3.3V. I am connecting it to the analogue input of a STG-680 PLC that has 0-10v 12bit ADC so expecting 4096 levels. Can I expect the number of levels applicable to the accelerometer to be proportional to the voltage of the accelerometer output, ie 4096/10*3.3? or how does it work?</p>
| What is the resolution of an accelerometer output to my PLC ADC | 2023-12-21T22:08:16.847 |
695080 | |rf|transmitter|spark| | <p>About as small as you want. Consider the <a href="https://www.smecc.org/j_c__bose_-_radio_and_microwaves.htm" rel="nofollow noreferrer">pioneering microwave work by J.C. Bose</a>: the spark generator and antenna is only some cm across, the full pieces being mostly support.</p>
<p>Air ionization can act impressively quickly; spark gap risetime is mostly limited by spark length, hence we get risetimes of single to fractional ns in EMC work for example (EFT, ESD). With a controlled spark length and environment, and transmission line techniques, bandwidths of many GHz are possible. At some point, emission and ionization dynamics take over, and sparking (relaxation oscillator) gives way to a softer negative-resistance characteristic, and eventually stops altogether, at a cutoff frequency. I don't know the cutoff frequency typical of sparks in air, but evidently it's in the 10s of GHz!</p>
<p>(For comparison, a similar experiment conducted with a small neon lamp, finds relaxation behavior up to maybe 10s of kHz, and negative resistance in the low 100s kHz, with little or no negative impedance beyond there. The low pressure, modest dimensions, and characteristics of the gases used, cause fairly slow recombination. The high pressure of atmospheric air causes much faster recombination.)</p>
<p>I don't know if he was able to achieve any kind of CW emission; I haven't read the original work. If it was powered by electrostatic generator, it was probably impulsive regardless, though how long those impulses were (i.e. tone burst vs. impulse and ringdown) is still a valid question. Not that it would've been resolvable at the time, but we can easily* scope such signals these days.</p>
<p>*Given suitable budget that is, heh. But they are standard off-the-shelf equipment. Or a wideband detector can be used, with a somewhat lower bandwidth scope to read the envelope.</p>
<p>CW, or tone burst, is an interesting possibility, as there were negative-resistance <a href="https://en.wikipedia.org/wiki/Arc_converter" rel="nofollow noreferrer">arc type oscillator/amplifier systems</a> in use, briefly before vacuum tubes took over. It's conceivable that similar mechanisms could be employed at microwave frequencies, probably with some difficulty, and great precision (as you can't simply scale everything down proportionally, but parts that do, need to be proportionally more accurate).</p>
| <p>How small can a spark-gap transmitter be made? Wikipedia's got plenty of pictures ranging from "desk-sized" to "room-sized", but those are mostly either early, inefficient transmitters or exceptionally high-powered ones.</p>
<p>Would it be possible to make a portable backpack-sized or even hand-held one for short-range communication, or is there something about the spark-generation process that requires a larger minimum size?</p>
| How small can a spark-gap transmitter be made? | 2023-12-21T23:00:55.467 |
695094 | |transistors|amplifier| | <p>The differences between voltage and power amplifiers has been pretty well covered already so I'll focus on amplifier classes.</p>
<p>Classes A through C are based on conduction angle. Imagine a <span class="math-container">\$2~V_{pk-pk}\$</span> sine wave, the voltage will be:</p>
<ul>
<li><span class="math-container">\$~~~0\$</span> V @ <span class="math-container">\$0^{\circ}\$</span></li>
<li><span class="math-container">\$+1\$</span> V @ <span class="math-container">\$90^{\circ}\$</span></li>
<li><span class="math-container">\$~~~0\$</span> V @ <span class="math-container">\$180^{\circ}\$</span></li>
<li><span class="math-container">\$-1\$</span> V @ <span class="math-container">\$270^{\circ}\$</span></li>
<li><span class="math-container">\$~~~0\$</span> V @ <span class="math-container">\$360^{\circ}\$</span>
<a href="https://i.stack.imgur.com/nEPcB.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nEPcB.jpg" alt="Sine Wave" /></a></li>
</ul>
<p>In a Class A amplifier the active device (transistor, tube, etc.) will be biased to conduct for the entire <span class="math-container">\$360^{\circ}\$</span>.</p>
<p>In a Class B amplifier it will be biased to conduct for <span class="math-container">\$180^{\circ}\$</span>.</p>
<p>In Class C it will conduct for something less than <span class="math-container">\$180^{\circ}\$</span>.</p>
<p>And in Class AB it will conduct for something between <span class="math-container">\$180^{\circ}\$</span> and <span class="math-container">\$360^{\circ}\$</span>.</p>
<p>All of these can be done using a single active device, Class B and AB don't necessarily indicate a push-pull arrangement as some have suggested, it's just that they are usually used that way.</p>
<p>Class A is used when you want to minimize distortion. Since there's no point where the active device is in cut-off it avoids the most non-linear portion of the device's characteristic curves. Since it conducts all the time whether there's a signal or not it wastes power.</p>
<p>Class B is used when you want to reduce power dissipation in the active devices and improve efficiency. It is generally used with two devices in a push-pull or complementary symmetry arrangement so that one device amplifies from <span class="math-container">\$0^{\circ}\$</span> to <span class="math-container">\$180^{\circ}\$</span> and the second device amplifies from <span class="math-container">\$180^{\circ}\$</span> to <span class="math-container">\$360^{\circ}\$</span>. This way neither device is conducting when there is no signal, increasing efficiency. The trade off is that when conduction switches from one device to the other they go through their non-linear region creating what's called cross-over distortion.</p>
<p>Class AB is a trade-off between A and B. The active devices are biased slightly into conduction reducing the cross-over distortion seen in Class B while not conducting as much with no signal as a Class A would, increasing efficiency.</p>
<p>Class C is used in RF amplifiers, the active device only conducts for a small part of the waveform and depends on the flywheel effect of resonant circuits to fill in the missing part of the waveform.</p>
<p>Technically you could use any of these classes with any transistor circuit topology, CE, CB or CC, but some combinations are more useful than others.</p>
| <p>I am trying to learn more about the different types of amplifiers and got confused. I understand that there are 2 amplifiers: Voltage amplifier and a power amplifier. Do power amplifiers or voltage amplifiers or both amplifiers have class A, B, AB, and C. And is common emitter, common base, and common collector different ways to set up these amplifiers?</p>
| Amplifier Classes | 2023-12-22T02:31:47.600 |
695101 | |stm32|can|bus|peripheral|canopen| | <p>The bus was working between two MCP2515, but not with an STM32 involved.
The issue turned out to be a disabled HSE (external crystal) on the STM32.
CAN apparently only reliably works with an external crystal.</p>
| <p>I am using an STM32 to transmit data over CAN using CAN2. Currently, my code is only setting up the can peripheral and sending periodic messages.
My messages usually look like this on the receiving side:
<code>can0 002 [1] 00</code>, as recorded by candump.</p>
<p>Unfortunately, sometimes the stm32 spits out this: <code>can0 002 [2] 00 D0</code>. I am definetly not sending <code>D0</code>, yet it keeps coming back.
<code>01</code> becomes <code>01 C6</code> and<code>03 C6</code> sometimes.
<code>02</code> becomes <code>02 C6</code> and <code>05 EA</code>. The patterns in a failure are predictable.</p>
<p>Increasing the bitrate seems to reduce the chances of failure, but they still occur.</p>
<p>It is like the CAN peripheral is being interrupted mid-transmission, but is that even possible?</p>
<p>I measured the garbage data on the STM can_tx pin, so the transceiver/wiring should not be to blame.</p>
<p>Is there any possibility of interference with my CAN2 peripheral?
Is there a setting I am missing?</p>
| STM32 Can bus random data | 2023-12-22T05:20:57.817 |