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707534

|switches|esp32|esp8266|npn|

<p>Ex 1 will not work, it won't do anything.</p> <p>Ex 2 should work.</p> <p>The only problem is, you need to know which pin on the motherboard is ground and which is the button input that should be connected to ground to activate it.</p> <p>Assuming one of the button pins even is a ground. One of them should be, but there is no official standard that defines how the button pins should work, except that there are two pins and pushing the button shorts these two pins together.</p>

<p><strong>Hi all!</strong></p> <p>I'am trying to use an <strong>ESP12S</strong> to turn a desktop computer on/off remotely, using a NPN <strong>2N2222 transistor</strong>, while also trying to keep the physical power button working.</p> <p><strong>Much like this project:</strong></p> <blockquote> <p><a href="https://www.ajfriesen.com/pc-switch" rel="nofollow noreferrer">https://www.ajfriesen.com/pc-switch</a></p> </blockquote> <p>Now my background is Computer Science, I had a bit of electronic classes in my Uni, so I've searched a little bit, and I was able to come up with two designs. Pardon if they feel too cumbersome...</p> <p>First one, a sketch on TinkerCad that simulates the physical pressing of button AND the output from the ESP. I add two buttons to be able simulate <strong>either being pressed</strong> should trigger the <strong>LED ON</strong>. <a href="https://www.tinkercad.com/things/2rcvq1ql6S4-wol-debug/editel?returnTo=%2Fdashboard&amp;sharecode=Nn6cBOiARjH6eFKsJ0Fil7gZPo4UCg-UkIfwJLs4kKE" rel="nofollow noreferrer">link here</a></p> <p>With the success test from the above design ( <strong>LED turned ON by pressing either switch</strong> ) I've tried to <strong>design a PCB on EasyEDA</strong> to actually print it and put it to use.</p> <p><a href="https://i.stack.imgur.com/Jw18s.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Jw18s.png" alt="easyEDA" /></a></p> <p>Q1: NPN 2N2222 transistor R2: 200 Ohm resistor H2: Power switch connection on motherboard H1: Physical power button</p> <p>My question is, which is the correct way of using the NPN transistor?</p> <ul> <li>EX.1</li> <li>EX.2</li> </ul> <p>Is there a better way of using the NPN transistor for this particular case?</p> <p>Thank you in advance :)</p>

ESP12 to switch computer power

2024-03-26T22:51:02.640

707538

|pcb-design|altium|ddr|autorouter|circuitmaker|

<p>Short answer: You don't.</p> <p>Long answer: You don't do this in Circuit Maker at all and you even don't let Altium's autorouter work in AD (all versions including the most current AD24). The state of Altium's autorouter is abysmal.</p>

<p>I need to do autorouting in Altium CircuitMaker. I have only found information how to do autorouting in Altium Designer.</p> <p>What I need to know is to set up rules and enviroment for autorouting with the same length. This image shows a DDR3-memory and that requries the same length for all tracks.</p> <p>Sure, I can drag every track by hand....but it would take a month.</p> <p><a href="https://i.stack.imgur.com/bz543.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/bz543.png" alt="enter image description here" /></a></p>

How can autorouting be done in Altium CircuitMaker?

2024-03-26T23:50:44.017

707547

|power-supply|switch-mode-power-supply|tl431|

<p>In a TL431-based regulator, the REF pin of the TL431 will be kept at the IC's internal reference voltage: ~2.5V for TL431/432 or ~1.24V for TL<strong>V</strong>431. The former is more common.</p> <blockquote> <p>Is there something stopping the SMPS outputting 3V or 50V, if the TL431 commands it so (apart from the voltage ratings of the output caps and diodes)? Is the transformer/coil optimized for just a narrow optimum output voltage?</p> </blockquote> <p>Without knowing anything about the circuit, it's safe to say that the theoretical lower bound for adjustment is the REF voltage. For your example, you can't expect an output voltage lower than 2.5V. In practice, at such low voltages, the ripple (percentage-wise) can be really high, so this could be a limiting factor. For example, if the design is optimised for 1% ripple at 19V then you may end up with, say, 5% ripple (or maybe even higher) if you adjust the output to 5V.</p> <p>As for the upper bound, it all depends on the rest of the circuit. Without knowing the internals we can't give a satisfactory enough but there are some limiting factors:</p> <ul> <li><p><strong>Duty Cycle Limitation:</strong> Most laptop chargers are isolated flyback converters where the duty cycle, which determines the power delivery and the output voltage, is somewhat internally limited. Assuming there's no active power factor correction (APFC) circuit inside, we can expect the transformer to be designed for a considerably wide input voltage range (e.g. European: 180..264 Vac or universal: 85..264 Vac). Now if you try to adjust the output to a relatively high voltage (e.g. 50V) then you may not get the desired output when the input voltage is relatively lower due to the aforementioned duty cycle limitation.</p> </li> <li><p><strong>Component Ratings &amp; De-ratings:</strong> You are ignoring the exceeding of capacitor &amp; diode ratings but higher output voltage means higher reflected output to the primary (assuming for a flyback) and this will increase the stress of the primary-side components. Although the voltages are clamped, there's still a risk of having higher stress which may result in failure. If the design is a two-switch flyback then this may not be a problem but it's unlikely to see a two-switch design due to the aggressive market and cost targets.</p> </li> </ul> <p><strong>NOTE:</strong> The placement of the adjustment pot is important. As you might already be aware, the TL431 takes the output voltage sample from a resistive divider. If you put the pot to the high side then you'll change the dynamics of the circuit as the high side of the divider network appears in the transfer function. So, you may end up with an unstable system by adjusting the output from a pot put on the high side. The pot should be a part of the low side of the divider.</p> <blockquote> <p>Also, if the rating was originally 19V@4A=76W, how does that rating change in practice as I go lower in voltage, can I consider 76W to be my guiding max number, or the 4A to be a maximum regardless of voltage (I suppose overheating the rectifying diode would be my first concern with higher amp flow at a lower voltage?)?</p> </blockquote> <p>Again, it's difficult to answer without knowing the internals but we are sure that the transformer and the output section (PCB/layout) are designed/optimised for 4 Amps output current. Lowering the output voltage doesn't mean the charger can deliver a higher output current. But if you adjust the output to a higher voltage, the internal power limitation may not allow you to draw 4 Amps. So, at 10V you can possibly draw 40W but at 30V you may not draw 120W.</p> <blockquote> <p>Is a 24V power brick for example almost identical to a 12V one, with just different resistors on the TL431, or are there more design considerations to adapt to a different voltage?</p> </blockquote> <p>The transformer design is different, even if we can expect the rest to be the same. If the same transformer is used for 12V and 24V output, assuming the input voltage is the same, the losses will quite possibly be higher for 24V even at the same power level.</p>

<p>So here I have a standard 19V noname laptop power brick rated at say 4A output. Inside is a typical SMPS regulated by a TL431 and an optocoupler.</p> <p>I added in a potentiometer to the reference resistors on the TL431 so am able to vary the output voltage (say 10-19V now). I am curious what the limiting factor is regarding the possible voltage window that can be achieved. Is there something stopping the SMPS outputting 3V or 50V, if the TL431 commands it so (apart from the voltage ratings of the output caps and diodes)? Is the transformer/coil optimized for just a narrow optimum output voltage?</p> <p>Also, if the rating was originally 19V@4A=76W, how does that rating change in practice as I go lower in voltage, can I consider 76W to be my guiding max number, or the 4A to be a maximum regardless of voltage (I suppose overheating the rectifying diode would be my first concern with higher amp flow at a lower voltage?)? Is a 24V power brick for example almost identical to a 12V one, with just different resistors on the TL431, or are there more design considerations to adapt to a different voltage?</p> <p>Cheers!</p>

Modified output voltage range of laptop power brick? (TL431)

2024-03-27T02:54:07.183

707575

|transformer|ground|

<p>That's quite possibly connected to the <strong>shielding winding</strong> which is usually an <strong>unshorted</strong> copper foil or a few turns of magnet wire located between the primary and the secondy. The wire connection then goes to the mains GND / PE (protective earth).</p>

<p>I am rewinding a transformer that got fried. It is a single phase, reducing transformer.</p> <p>On the primary we have mains, and the secondary is tapped at different points. See figure for details.</p> <p>Unfortunately the windings had been destroyed at several points, and I could not figure out what the GND pin was connected to.</p> <p>The ground pin was physically located on the primary side of the transformer. That doesn't mean that it is electrically connected to the primary, though.</p> <p>When unwinding the coils, first there was the secondary. Next came the GND wire, but it had only a short length of wire, which had been severed from wherever it connected to. Finally there was the primary, with a bit missing at the beginning.</p> <p>Any idea of where to connect this GND pin? <a href="https://i.stack.imgur.com/BkdOj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BkdOj.png" alt="TransformerDiagram" /></a> <a href="https://i.stack.imgur.com/Dmc67.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Dmc67.jpg" alt="Picture" /></a></p>

Where to connect transformer ground?

2024-03-27T12:40:08.897

707595

|dc-dc-converter|

<p>The peak voltage seen at the drain of a MOSFET in a Ćuk converter is approximated as the power supply voltage plus the magnitude of the output voltage plus one diode drop. For instance: -</p> <p><a href="https://i.stack.imgur.com/9znD5.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9znD5.png" alt="enter image description here" /></a></p> <p>So, my supply is 12 volt and the magnitude of the output voltage is 5.887 volts. This adds up to 17.887 volts. Add on the diode forward drop of 0.757 volts and you get 18.644 volts. This is 0.28 volts lower than the actual peak drain voltage of 18.924 volts.</p> <p>Always choose a MOSFET rating that as about 1.5 times the expected peak value to cover uncertainties and mild circuit abuse. If you do this then the professors approximation is good enough.</p> <p>And, always simulate this type of circuit to double check things.</p>

<p>I know voltage stress occurs during 1-D (switch off).</p> <p>So voltage stress is going to be Vc1/(1-D).</p> <p>But my professor said voltage stress is Vin + Vout because of charge balance law, but I didn't understand that.</p> <p><a href="https://i.stack.imgur.com/xPrXe.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xPrXe.png" alt="enter image description here" /></a></p>

What is Ćuk converter's switch voltage stress?

2024-03-27T14:40:43.763

707601

|power-supply|circuit-analysis|measurement|voltage-measurement|

<p>I'm guessing your data logger is connected at the load. If that's the case, you are probably seeing the resistive drop of the leads from the power supply to the load based on the current you're drawing.</p> <p>Some power supplies allow you to mitigate that by running additional sense lines from the load back to the supply. (I don't see any connections for sense leads on your supply, unless they're on the back.) This allows the supply to accurately measure the load voltage (since the sense lines conduct very little current) and compensate its regulation to adjust the voltage at the load rather than at the supply.</p> <p>Without such a feedback mechanism, the only place the supply has to measure its output is at the supply terminals.</p>

<p>I have one of <a href="https://rads.stackoverflow.com/amzn/click/com/B09MVSJHY9" rel="nofollow noreferrer" rel="nofollow noreferrer">these power supplies</a>:</p> <p><a href="https://i.stack.imgur.com/1iX72m.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1iX72m.jpg" alt="enter image description here" /></a></p> <p>I also have an <a href="https://tandd.com/product/mcr4v/" rel="nofollow noreferrer">MCR-4V data logger</a>:</p> <p><a href="https://i.stack.imgur.com/MvToEm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/MvToEm.png" alt="enter image description here" /></a></p> <p>If I connect wires directly from the power supply to the data logger (red to red, black to black), then the voltage output on the power supply closely matches the measured voltage, e.g. a 10.00V shows consumption of 0.000A (i.e. &lt; 0.0005A, very low resistance), and measured voltage is about 9.95V.</p> <p>Now I also have a homemade hot plate. Hooking this up directly to the power supply, at a constant 10.00V it draws about 2.000A (hence 20W total).</p> <p>However if I now connect my measurement wires to measure the voltage between the black and red terminals, the device no longer reads 9.95V but rather 9.70V.</p> <p>This is well outside the stated error margins of the measurement device (it claims ~0.3%, or 30 mV).</p> <p>I'm thinking the most likely culprit is that at a higher load, the power supply's display is lying. Is that indeed the likeliest possibility? What other sources of error can I look for? I would tend to trust the measuring device as it's much more expensive and also appears to be high quality. But what I trust least of all is my circuit building skills!</p> <h1><strong>Update</strong></h1> <ol> <li><p>If I connect the datalogger directly at the power supply terminals, I get the 9.96V reading.</p> </li> <li><p>If I connect the datalogger to the end of the alligator clips coming out of the power supply leads, I get the same ~9.96V reading:</p> </li> </ol> <p><a href="https://i.stack.imgur.com/cQlY9l.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/cQlY9l.jpg" alt="enter image description here" /></a></p> <p>To me this indicates the extremely low resistance of the circuit up to this point.</p> <ol start="3"> <li>Now if I connect my hot plate wires <strong>later on the alligator clip</strong>, keeping all else the same, the reading on the data logger now drops to 9.688V:</li> </ol> <p><a href="https://i.stack.imgur.com/EBH6ml.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/EBH6ml.jpg" alt="enter image description here" /></a></p> <ol start="4"> <li>I get this same ~9.67V reading if I try the following connector options too:</li> </ol> <p>a. Join the data logger and hot plate wire, clip in the middle:</p> <p><a href="https://i.stack.imgur.com/QIA6d.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QIA6d.png" alt="enter image description here" /></a></p> <p>b. Clip the alligator clip to the end of the hot plate wire, and wrap the data logger wire a bit before:</p> <p><a href="https://i.stack.imgur.com/QQk6D.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/QQk6D.png" alt="enter image description here" /></a></p> <p>What puzzles me is why the sole act of hooking up the hot plate wires is changing the voltage here. The resistance of the measuring circuit is very small. As I understand it, this forms a parallel circuit -- one going to the data logger, and another going to the hot plate -- where each branch should have the same voltage as the whole circuit, namely 9.96V as measured at the power supply leads.</p>

Why might my circuit's measured voltage drop when I add more load to it?

2024-03-27T15:15:38.797

707605

|circuit-analysis|differential|dac|lvds|

<p>Yes, this should work fine.</p> <p>As noted in the comments, the input has a built in 100R termination. It is in fact this termination that makes the circuit work. This has a range of 85R to 135R, which may have some impact on the common mode and differential voltage range, but should still be in spec at both extremes.</p> <p>With the 100R in series with the other two resistors, you get a voltage of 300mV from P to N, with a common mode voltage of approximately 1.2V. Both are well within the specifications for an LVDS input, and will provide a suitably strong 1 or 0 (depending on which resistor you connect to the pullup).</p> <p><a href="https://i.stack.imgur.com/ecxirm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ecxirm.jpg" alt="Circuit simulation showing resistors and internal termination" /></a></p> <p>With the voltages in spec, there is no danger of damaging the inputs to the chip - it's no different from feeding the receiver with a continuous stream of zero bits from a normal LVDS transmitter.</p> <p>Note that each input bias network will sink about 3mA of current, but that is no different from what you would get with a normal driver, and can't be improved on because the internal termination prevents using higher value resistors to provide the bias.</p>

<p>I have a chip with 14-bit LVDS input data lines (DATA0P/N to DATA13P/N) on the chip DAC3174. I don't have enough space left to route all these data lines, so I chose to set the two least-significant bits (LSBs) to permanently HIGH or LOW. I know that LVDS has common-mode voltage of 1.2V, V_OL of 1V and V_OH of 1.4V (so a peak-to-peak differential voltage of 400mV) just as it's shown here:</p> <p><a href="https://i.stack.imgur.com/LUjvd.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LUjvd.png" alt="enter image description here" /></a></p> <p>I thought about this resistor network:</p> <p><a href="https://i.stack.imgur.com/z6BNa.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/z6BNa.png" alt="enter image description here" /></a></p> <p>On the left side the voltage is 1.26V at the input of the chip, so the pin would see a permenant HIGH. On the left side the the P/N are reversed, while the voltage stays the same, P is at VSS and N is at VCC, so the pin would see a permenant LOW.</p> <p>Here's also the table of the DAC3174 chip about the DATA lines: <a href="https://i.stack.imgur.com/aLoNh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/aLoNh.png" alt="enter image description here" /></a></p> <p>Will this work as intended? I will eventually use zero ohm resistors to switch between both cases. Please suggest a better solution, if this network isn't optimal for this purpose.</p>

Can I set LVDS permanently to HIGH or LOW with this simple circuit?

2024-03-27T15:53:37.563

707612

|circuit-analysis|zener|solenoid|vintage|

<p>It's there because the coil has <em>inductance</em>. When the transistor tries to turn off the inductance fights against the dying of the current by changing voltage in the direction that tends to maintain the current as much as possible. That means it goes from about +40V to 0V and continues to go negative until something stops it. That something could be the transistor breaking down (which is very hard on the transistor and could certainly affect reliability) or it could be something across the coil or the transistor conducting. Or it could just resonate with the coil distributed capacitance and ring for a while if the transistor has a high enough breakdown voltage (it might be hundreds of volts though).</p> <p>If we didn't give a toss about speed we would just slap a (normally reverse-biased) diode across the coil and the Vce of that transistor would be limited to maybe -41V, since the coil voltage would only go to one diode drop below ground. Easy and cheap. But that's a bad solution if we want the maximum speed since the current will continue for longer that if we allow the voltage across the transistor to get higher, perhaps to -62V by using Zener diode.</p> <p>That's harder on the transistor, but not as hard as letting the transistor absorb the energy that now goes into the Zener diode. The total energy in the coil is <span class="math-container">\$ L I^2/2\$</span>, and the initial current will be the coil current when on. Some of the energy goes into the coil resistance, some into the Zener diode, and some goes into the transistor as it is trying to switch off.</p> <p>The transistor will have a specification called SOA (Safe Operating Area) that indicates the allowable current and voltage that transistor can reliably handle.</p>

<p>I am trying to understand a vintage (1970s) impact printer circuit. Each of the pins in the printer are driven by a small individual coil driver circuit, that was originally specified to look like this:</p> <p><a href="https://i.stack.imgur.com/0u8ce.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/0u8ce.png" alt="enter image description here" /></a></p> <p><em>(Detail from a schematic <a href="https://deramp.com/swtpc.com/PR_40/PR40_Schematic.pdf" rel="nofollow noreferrer">here.</a> )</em></p> <p>I understand how this works since <a href="https://electronics.stackexchange.com/questions/706049/why-is-this-zener-in-series-at-the-base-of-transistor-and-has-something-failed">I asked a question about it earlier</a> and got some great answers, but the company that sold the printer issued an addendum to the circuit with an update, where a 62V Zener diode (1N4759) is added from the collector to the emitter:</p> <p><a href="https://i.stack.imgur.com/hF8om.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/hF8om.png" alt="enter image description here" /></a></p> <p>(SWTPC PR-40 addendum detail.)</p> <p>What is the purpose of this additional diode?</p> <p>I'm assuming it's to protect something (the coil?), since it was an after-the-fact addition to the circuit design, but I'm having a hard time intuiting what's happening here.</p> <p>I don't think the Zener will do anything in normal operation-- the 40VDC rail is below the 62V breakdown voltage so it won't bypass the transistor. Is this a protection for if the 40V rail somehow manages to jump up to &gt; 62V? Is something else going on?</p>

Why is this Zener across the collector and emitter of this transistor driving a solenoid coil?

2024-03-27T16:53:15.447

707630

|pcb-design|layout|pcb-layers|trace|clearance|

<p>Your issues with the LED lines bring up two issues. One is the interruption of the return path for any signals crossing the gap. The return currents have to find their way around the gap, creating a loop antenna where you wanted a transmission line.</p> <p>The second has to do with the current flow in your ground plane. Gaps in the ground plane can act as slot antennas.</p> <p>If it's absolutely impossible to implement these on the top/bottom, another approach would be to run them around the edge of the board instead of through the middle.</p> <p>Both of these concerns are also present for the slots you've created with lines of vias spaced so closely that there is no ground plane connection. If it's possible to stagger those so the pad clearance areas don't overlap, you're better off. Another approach is to reduce the pad size to zero on the planes they're not connecting to, which makes the pad clearances smaller.</p> <p>As far as the RS232 and I2C lines, 0.25mm spacing has always worked for me. These signals are slow enough that they don't need much in the way of precise tuning; you primarily need to ensure the edges are monotonic. Most signal integrity issues I've seen on I2C are due to improper selection of pullup resistors.</p>

<p>I have almost finished my first 4-layer PCB design using Signal(Top)-Ground-Power-Signal(Bottom) stack-up and have some questions before sending it to production. Due to the very tight space on the top and bottom planes, I had to pass RGB LED PWM signal traces on the top signal layer through the power plane to the bottom layer across the 5V power supply ground rail, where the filtering with decoupling capacitors is made. These three orange colored traces can be seen in Figure 1, where they switch to blue colored bottom layer through vias. In Figure 1 only bottom and power layers as well as top and bottom component margins are visible. Since these traces split the part of the power plane, which is a reference for the bottom 5V DC supply ground rail, does this cause EMI issues in my board? How should I mitigate the effects of it? Should I use jumpers and 0 Ohm resistors?</p> <p><a href="https://i.stack.imgur.com/ix5qb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ix5qb.png" alt="enter image description here" /></a></p> <p>My second question is about UART and I2C signal trace clearances. I have used 0.25 mm wide traces for UART signals. Is it safe to keep a clearance of 0.25 mm between each trace in order to avoid signal coupling and integrity issues?</p> <p><a href="https://i.stack.imgur.com/nqZrj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/nqZrj.png" alt="enter image description here" /></a></p> <p>Thank you...</p>

LED PWM traces splitting reference power plane adjacent to bottom 5V DC supply ground zone on 4-layer PCB

2024-03-27T19:19:44.130

707633

|resistors|opto-isolator|triac|

<p>R4 and R6 are connected directly to L, so when you switch on the gate through the opto coupler, that voltage will be applied and many current will flow until next zero voltage cross. See modification of these resistors connection. In that way, once the triac is switched on, almost no power will applied to R4 or R6 (only the voltage drop of triac). Only a short pulse what is needed, it is recommended a small capacitor and a 360 Ohm resistor to provide enough peak current (see MOC3021 datasheet). <a href="https://i.stack.imgur.com/FXfeZ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FXfeZ.png" alt="enter image description here" /></a></p>

<p>let me clarify first than I'm no expert in electronics, so, I could be doing something really stupid here.</p> <p>I want to replace the control board of my kitchen extractor motor with a custom version that can be controlled via ESPHome. I designed the full circuit, but I started implementing it by bits.</p> <p>So this is my circuit: <a href="https://i.stack.imgur.com/6l9V0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6l9V0.png" alt="enter image description here" /></a></p> <p>I assembled only the Optocoupler+TRIAC controlling part of the circuit, which is the one I'm more inexperienced in as AC and Mains are involved.</p> <p>The motor is controlled by two different triacs, but I just built one to try it out before going for the whole circuit. Soldered everything on a perfboard for security reasons. Then connected a NodeMCU to the MOC3021. <code>GPIO D2 -&gt; 220 resistor -&gt; Pin 1</code> and <code>GND to pin 2</code>. Double/triple checked for miswirings/shorts and everything is OK.</p> <p>With the test assembly done, I connected everything to mains and the &quot;LOAD&quot; accordingly (I didn't connected the motor, but a LED lamp) in case something went wrong.</p> <p>Right out the gate I noticed something &quot;weird&quot; as the LED lamp barely lit even if the triac was off. But I guessed it would be because the lamp was connected to mains and the &quot;Neutral&quot; was connected to the MT2 which is also connected to the snubber circuit that allowed enough current to bearly light the lamp. So far so good..</p> <p>Then I proceeded to tell the microcontroller to drive HIGH the MOC's GPIO.. and the lamp lit up fully for just about a second and something went &quot;click&quot; and the lamp returned to the barely lit state. No more reaction to GPIO input.</p> <p>Inspecting everything was normal, except R4 wasn't conducting anymore. It fried. I guessed &quot;ok, 1/4w resistor isn't gonna cut it&quot;.. So I went with a 1/2W resistor and tried again. Same scenario. Only this time the resistor died more spectacularly, some sparks flew by and the perfboard got a nice &quot;tan&quot;. I still don't know if the triac or the MOC are still alive, but I have brand new ones to try again.. I just wanted to be sure that's not a design flaw.</p> <p>I've been googling around and saw just one place where they suggested a 5W resistor for R4, everywhere else even say that 1/4w resistor should do it... So, I'm really confused here. Any hints on what's wrong?</p>

MOC3021M + JST16A-600B TRIAC Circuit problem: Resistor blowing up

2024-03-27T19:53:20.347

707652

|batteries|current|load|

<p>Ability to supply current from a battery is limited by internal impedance of the battery. The impedance value is dependent on manufacturing technology and physical battery size.</p> <p><a href="https://i.stack.imgur.com/mmqJj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/mmqJj.png" alt="enter image description here" /></a></p> <p>The first-order approximation of internal impedance is easily measurable with an ordinary DMM and a load resistor. Please start by visiting some <a href="https://learn.sparkfun.com/tutorials/measuring-internal-resistance-of-batteries/internal-resistance" rel="nofollow noreferrer">introductory places like Sparkfun</a> for explanations and examples.</p>

<p>So I know an average PP3 9-volt battery has about <a href="https://en.wikipedia.org/wiki/Nine-volt_battery#History" rel="nofollow noreferrer">500mAh of capacity</a>. If I wanted to, for example, draw 3.5 amps, the battery would last about 8 or 9 minutes, though that would be assuming the battery is even capable of outputting 3.5 amps, which is what I'm worried about. I've looked up several datasheets including ones from Duracell and Energizer and none of them mention maximum current.</p> <p>For the project I'm making, I have several options for parts that would have different loads on the circuit (rough calculations):</p> <ul> <li>26.45V @ 2.27A</li> <li>16.64V @ 3.54A</li> <li>10.56V @ 3.76A</li> <li>6.56V @ 8.90A</li> </ul> <p>I want to use common batteries (like 9-volt) and stack them in series to get the needed voltage, I also would rather go a little over the needed voltage than under. So the 26.45V option is good because I can use 3 9-volts to get 27V, but the 10.56V isn't because one battery is too little and two is way too much.</p> <p>All the batteries I've found are either common size, or car batteries that can output 1000 amps, I'm struggling to find something in between, or even know if the common ones can handle it.</p> <p>With all that being said, how can I find out for a given battery what current it can handle, and what type of battery/how many would be best for my project given the several options for parts and loads.</p>

How much current can I pull out of a battery

2024-03-27T22:12:19.380

707662

|identification|varistor|

<p>The exact family / product line may be obsolete, but Panasonic (Matsushita) still make varistors, ERZ-VA7D271 for example. Any 270V (175VAC) MOV of equal or greater diameter will suffice; there is very little distinction between product lines, they all use a common formulation.</p> <p>Mind that other components may be damaged. Smoke residue may be conductive or prone to tracking, and should be cleaned with alcohol, or scraping if necessary.</p>

<p>I have been trying to figure out what this part is called and what I can replace it with (see pics).</p> <p>It seems to be called a capacitor, varistor, or thermistor when I do an image search. It's black and says K271u 81 on the front. It's from a 1980s Panasonic Massage Lounger, Matsushita Electric Works, Ltd.</p> <p>The chair worked 2 years ago, but then we updated the wiring in the walls and the chair shorted out after turning it on. This part burned part of the circuit board. We are just now looking to fix it.</p> <p><a href="https://i.stack.imgur.com/V9HDR.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/V9HDR.jpg" alt="K271U 81" /></a></p> <p><a href="https://i.stack.imgur.com/avCpc.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/avCpc.jpg" alt="enter image description here" /></a></p>

What is this part called and what can I replace it with?

2024-03-28T00:10:25.153

707664

|pcb|usb|connector|usb-c|

<p>Usual solution to your problem of adapting a USB-A captive cable (as on many headsets) to a USB-C port is to buy a simple adapter like this one: <a href="https://i.stack.imgur.com/Tfedo.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Tfedo.png" alt="enter image description here" /></a></p> <p>No modification of any sort is needed.</p>

<p>I’ve recently undertaken a project to adapt my Razer Kraken X USB headphones from a USB-A to a USB-C connection. Originally, these headphones featured a male USB-A cable designed for PC connection. I modified them by removing the USB-A cable and soldering a male USB-C terminal from an OTG adapter. The modification works well; however, I’m struggling to find high-quality, soft cables like the USB-C cables used for iPhones. Specifically, I needeed a cable with a female USB-C on one end and a male USB-C on the other (didn't found).</p> <p>My question is: does anyone know of a PCB or adapter that would allow me to attach a female USB-C connection directly? (I have found only USB-C to USB A unidirectional adapters, and don't known how to recognize USB A male to USB C Female adapter)The goal is to use the headphones and be able to disconnect the cable from the earpiece on the headband.</p> <p>I’ve tried several PCBs from Amazon, but none have met my requirements:</p> <ol> <li><a href="https://rads.stackoverflow.com/amzn/click/com/B077S1G4HX" rel="nofollow noreferrer" rel="nofollow noreferrer">chenyang/Type-C USB-C to USB 2.0 OTG Adapter</a>: Seems to be unidirectional. Not working</li> <li><a href="https://rads.stackoverflow.com/amzn/click/com/B0C6M4797B" rel="nofollow noreferrer" rel="nofollow noreferrer">YoungSelly/Connectors usb C/Tunderbold compatible </a>: Not even working</li> <li><a href="https://rads.stackoverflow.com/amzn/click/com/B09WCSF8FC" rel="nofollow noreferrer" rel="nofollow noreferrer">ANMBEST/USB 3.1 type C Conector 24 Pin</a>: USB C To USB A unidirectional</li> <li><a href="https://rads.stackoverflow.com/amzn/click/com/B0CB395L99" rel="nofollow noreferrer" rel="nofollow noreferrer">Cermant/USB 3.1 Type C 16pin for data transfer</a>: USB C To USB A unidirectional</li> <li><a href="https://rads.stackoverflow.com/amzn/click/com/B0B55XVMKP" rel="nofollow noreferrer" rel="nofollow noreferrer">ANMBEST/USB 3.1 type C Conector 24 Pin</a>: USB C To USB A unidirectional</li> <li><a href="https://rads.stackoverflow.com/amzn/click/com/B0BW8ZDSFV" rel="nofollow noreferrer" rel="nofollow noreferrer">Henrety/Adapter USB C to USB, OTG converter Thunderbolt 4 typeC, adapter USB C male to USB 3.0 female</a>: It worked, but only with USB C Male output</li> <li><a href="https://rads.stackoverflow.com/amzn/click/com/B07KCL8WZK" rel="nofollow noreferrer" rel="nofollow noreferrer">Satechi/OTG type A male to type C female</a>: It works but not in the direction I want.</li> </ol> <p>Any recommendations or guidance would be greatly appreciated. Thank you in advance for your help!</p> <p>Some image here: <a href="https://i.stack.imgur.com/AvI1A.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AvI1A.jpg" alt="project so far" /></a></p>

Seeking USB-C Female PCB Connector for Razer Kraken X Headphones Adaptation

2024-03-28T00:48:45.250

707682

|tester|parameters|

<p><span class="math-container">\$I_{ceo}\$</span> (note that it's o, not 0) is the <strong>c</strong>ollector-to-<strong>e</strong>mitter current when the base is <strong>o</strong>pen. This is one measurement of the transistor's leakage current.</p> <p><span class="math-container">\$I_{ces}\$</span> is the <strong>c</strong>ollector-to-<strong>e</strong>mitter current when the base is <strong>s</strong>horted to emitter. This is another leakage characteristic.</p> <p><span class="math-container">\$β\$</span> (β, not B) is the transistor's current gain, <span class="math-container">\$\frac{I_c}{I_b}\$</span>. It's also called <span class="math-container">\$h_{FE}\$</span>.</p> <p>As for why β is so much lower than the datasheet number, I would guess because the tester isn't designed to measure Darlington pairs like this, and can't interpret such a high number. It's possible some digits got cut off, and the number is actually 13400 or something. Or it's possible that, at the low test current this device uses (who knows what it is, but it's probably no more than a few μA), the β actually <em>is</em> 134--the datasheet number is specified at 10 A of collector current, far more than this thing can probably provide.</p> <p>For completeness' sake, the other parameter listed, <span class="math-container">\$U_f\$</span> is the forward voltage of, most likely, the base-emitter junction. Or junctions, as it's a Darlington. This is usually written as <span class="math-container">\$V_f\$</span> in my part of the world, but <span class="math-container">\$U_f\$</span> is common in Europe and Asia. Since there's apparently a 70 Ω pull-down resistor in there (according to the datasheet you linked), I don't think the number it gives you is likely to be that useful.</p> <p>The tester seems to have also erroneously identified an emitter-collector diode; the datasheet makes no mention of this, and while there <em>are</em> BJTs that have one, it's pretty uncommon--you see that more often with IGBTs (and there's the inherent diode in a MOSFET, of course, but that's not added intentionally).</p>

<p>I have one of these cheapo GM328A transistor testers &amp; I'm diagnosing Darlington transistor with it. It's printing these parameters <span class="math-container">\$ICE0\$</span>, <span class="math-container">\$ICEs\$</span> &amp; <span class="math-container">\$B\$</span>, which I'm not sure what they mean:</p> <p><a href="https://i.stack.imgur.com/RYdt8.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RYdt8.jpg" alt="enter image description here" /></a></p> <p>I have a suspicion <span class="math-container">\$ICE0\$</span>, <span class="math-container">\$ICEs\$</span> are just CE leakage currents when <span class="math-container">\$V_B = 0\$</span>. Then there's <span class="math-container">\$B\$</span>. Is that just <span class="math-container">\$B_f\$</span>?? Because according to the <a href="https://www.semicon.sanken-ele.co.jp/sk_content/2sd2560_ds_en.pdf" rel="nofollow noreferrer">datasheet</a>, it's a minimum of 5000.</p>

Meaning of \$ICE0\$, \$ICEs\$ & \$B\$ parameters from this common transistor tester

2024-03-28T04:26:39.273

707695

|mosfet|switch-mode-power-supply|mosfet-driver|

<p>I think it's better to understand the pulse width and rise and fall times first. The following is a good representation nicked from <a href="https://academy.berkeleynucleonics.com/courses/664435/lectures/11838681" rel="nofollow noreferrer">here</a>:</p> <p><a href="https://i.stack.imgur.com/xNWvq.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/xNWvq.png" alt="enter image description here" /></a></p> <p>Apart from the minimum ON time, some chips also mention rise and fall times, but for specific capacitive loads which represent a random MOSFET's input capacitance.</p> <p>This is not the case for the minimum ON time (or minimum controllable ON time) because this is limited by the gate driver itself.</p> <p>The thing is, the chip generates a pulse and its duration (the dime difference between 50% of the rising and falling edges) is measured, probably right at the chip's pin but at the MOSFET's gate it can get shorter due to the trace resistance, externally put gate stopper resistance, MOSFET's internal gate resistance which comes from the manufacturing process of the MOSFET (e.g. bonding, semiconductor structure), and the MOSFET's input capacitance as all of these play a role on effective rise and fall times which determine the effective pulse width:</p> <p><a href="https://i.stack.imgur.com/jZILG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jZILG.png" alt="enter image description here" /></a></p> <p>In the image above, I tried to show how the rise time can increase due to the external factors such as effective series resistance. So the effective pulse width is shorter than that at the pin.</p> <p>This can't be under the chip's control. The chip guarantees a minimum on time for specific conditions (e.g. frequency, operation mode, etc) then the rest is up to the designer and the application.</p>

<p>I would like to better understand this topic: <a href="https://electronics.stackexchange.com/questions/646712/minimum-on-time-for-a-buck-converter/646724#646724">Minimum ON time for a buck converter</a></p> <p>One of the answers seems to say that the minimum on-time depends on the selected MOSFET. According to the datasheet that I can find on internet, there is no information about the minimum on time. Here is an <a href="https://www.ti.com/lit/ds/symlink/lm5177.pdf?ts=1711551018487&amp;ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FLM5177%253FkeyMatch%253DLM5177%2526tisearch%253Dsearch-everything%2526usecase%253DGPN" rel="nofollow noreferrer">example</a> :</p> <p><a href="https://i.stack.imgur.com/Hefbj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hefbj.png" alt="enter image description here" /></a></p> <p>Here is another <a href="https://www.ti.com/lit/ds/symlink/lm5123-q1.pdf?ts=1711535345756&amp;ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FLM5123-Q1" rel="nofollow noreferrer">example</a> :</p> <p><a href="https://i.stack.imgur.com/imSME.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/imSME.png" alt="enter image description here" /></a></p> <p>Does the minimum on-time depend on the MOSFET or not? It seems that it does not depend on the MOSFET, but I agree on the fact that if we took a bigger MOSFET, it will take much more time to switch ON, so it seems to have an effect on the minimum on time and on the output ripple. How can the manufacturer give minimum on-time without specifying the conditions on the MOSFET?</p> <p>When we look at what is the maximum input voltage? The formula given in the datasheet does not depend on the size of the MOSFET, so it appears that the size of the MOSFET has no effect on the minimum on time. Here is an example to get Vin_max from this <a href="https://www.analog.com/media/en/technical-documentation/data-sheets/ltc7804.pdf" rel="nofollow noreferrer">datasheet</a>.</p> <p><a href="https://i.stack.imgur.com/01Dq6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/01Dq6.png" alt="enter image description here" /></a></p>

Minimum on-time does not depend on MOSFET

2024-03-28T08:40:29.977

707698

|ltspice|sepic|

<blockquote> <p><em>I am unsure where I have gone wrong</em></p> </blockquote> <p>The problem is that you are not correctly adjusting the duty cycle when the stepping occurs. For a 15 volt DC input I get 25 volts on the output with a duty of 0.628: -</p> <p><a href="https://i.stack.imgur.com/OJgjS.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OJgjS.png" alt="enter image description here" /></a></p> <p>And, if I raised the input voltage to 35 volts I'd get 25 volts out when the duty cycle was 0.412: -</p> <p><a href="https://i.stack.imgur.com/03xDK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/03xDK.png" alt="enter image description here" /></a></p> <p>Hopefully, you can see that this works and learn how to correctly adjust duty cycle. I also expect that my duty cycles are a little different to yours because I didn't set my maximum timestep to anything special.</p> <p>However, what you might want to do is close the loop and let the duty cycle find its own level for the output voltage of 25 volts irrespective of the input voltage: -</p> <p><a href="https://i.stack.imgur.com/e95J0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/e95J0.png" alt="enter image description here" /></a></p> <p>I haven't particularly tried to set the control loop be fast. I've just bludgeoned in some values that I knew would result in a soft start and steady control.</p>

<p>I am designing a SEPIC that meets the following criteria:</p> <p>➢ Vin = 15 V-35 V<br /> ➢ Vout = 25 V<br /> ➢ Output voltage ripple 1% peak to peak<br /> ➢ Switching frequency 100 kHz<br /> ➢ Load 10 W-50 W</p> <p>Calculations have been done for each component with the circuit as follows:</p> <p><a href="https://i.stack.imgur.com/FMmcm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/FMmcm.png" alt="enter image description here" /></a></p> <p>The simulation for the output voltage is as follows, showing the Vin(min) of 15 V in blue, 25V in red, and max of 35 V in cyan.</p> <p>However, both inputs are supposed to have a voltage output of 25 V ± 1% ripple (pk-pk). I am unsure where I have gone wrong and am also not confident about the usefulness of my step parameter in achieving anything. It is not mentioned in the brief that a step parameter even needs to be used.</p> <p><a href="https://i.stack.imgur.com/w6CSO.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w6CSO.png" alt="enter image description here" /></a></p> <p>Other potentially relevant info:</p> <p>Duty cycle D_1 for 15 V input = 0.63<br /> Duty cyle D_1 for 35 V input = 0.42</p> <p>So, D_1 adjustable from 0.42 - 0.63.</p> <p>Resistance was taken from calculation using highest load, 50 W.</p> <p>T_on = 5.83 μs.</p> <p>I can provide other specific values/calculations I used as required.</p>

LTspice: SEPIC simulated output voltage does not match known output voltage

2024-03-28T08:44:29.563

707700

|pcb-design|connector|humidity|power-connector|

<p>For Q1)</p> <p>No, moisture will not cause a &quot;short circuit&quot;. Short circuit means usually a direct connection between terminals. Moisture causes corrosion of the contacts and contaminants in the moisture between the contacts make the water to conduct only a small amount of current, further causing electrolysis to corrode the terminals.</p> <p>For Q2)</p> <p>It is unknown if the connector is rated to be used in conditions that allow moisture to develop on it. Even if you cap the socket to avoid splashing water, the moisture in warm air can still condensate on cold connector under the cap. To prevent moisture from doing any harm to your connector, buy a connector intended to handle the environment you need.</p> <p>However, do note that when the charger is unplugged, the connector directly exposes the battery terminals.</p> <p>Someone might accidentally plug something metallic into the connector contacts and do cause a short circuit. Connecting the battery terminals directly to the connector may not be a great idea. Perhaps you can design in some additional level of protection like a fuse or current limiting or diode or something else.</p>

<p>I have an inquiry about a design with a 3-pin jack connector, the PJ-065A. <a href="https://www.mouser.es/datasheet/2/670/pj_065a-1778510.pdf" rel="nofollow noreferrer">Datasheet of this connector</a></p> <p><a href="https://i.stack.imgur.com/X5zFz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X5zFz.jpg" alt="PJ-065A" /></a></p> <p>I want to use this power jack connector for a design.</p> <p>I have selected this component because of its 3 terminals. My design comprises a Li-Ion battery, a circuitry to power (PCB with several components connected), and a charger for the battery. The battery and the PCB with the other components are enclosed in a box, and the charger is placed outside the box.</p> <p>The battery is 2447-3036-20 Ansmann 14,8V 3500mAh <a href="https://www.blubattery.com/home/li-ion-battery-packs/2447-3036-20-ansmann-batteria-ricaricabile-li-ion-14-8v-3500mah-type-4s1p-con-ptc-e-safety-board-certificata-un-38-3-e-iec-62133-2-2017.2.7.946.gp.10254.uw" rel="nofollow noreferrer">link</a>. The jack could be a 24V and max current would be 2A for this circuit. The charger I intend to use is a a MASCOT 2542 LI <a href="https://www.mascot.no/catalog/medical/2542li/c-24/c-1323/p-174" rel="nofollow noreferrer">link</a></p> <p>The PJ-065A will be the connector used for the charger. The 3 pins and working mode of this connector allow us to have a design where, when the charger is connected, the battery and the circuit are disconnected, and the charger is only powering the battery. In this scenario, when the charger is plugged, the battery is being charged but the circuit won't work. When the charger is not plugged, the battery and the circuit are connected again, and the battery powers the PCB and its components. We want it to work this way.</p> <p>Thus, we connect the pin 1 of the PJ-065A to the + of the battery and the + of the circuit. Then, the pin 2 is connected to the - of the battery, and finally the pin 3 is connected to the - of the circuit</p> <p><a href="https://i.stack.imgur.com/UradC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/UradC.jpg" alt="Circuit with and without charger plugged" /></a></p> <p>Now, my questions in this design are:</p> <p>The battery and the circuit are enclosed in a box, and the charger connector we selected (PJ-065A) will be panel mounted to that box, and connected as I explained in the last paragraph to the circuit and the battery. When there is no charger plugged, the battery is powering the circuit, so there is voltage and current running there. In this scenario, and because the connector is exposed to the exterior, there could be some moisture that appears inside the connector, which would cause a short circuit, right?</p> <p>Question1: is this assumption right? Would this moisture create a short circuit?</p> <p>Question2: in that case, how shall we prevent this? I know there are some caps we can place outside the connector, but would they be effective in my case?</p> <p>Question3: is there anything from this design you would change?</p> <p>Thank you dearly in advance!</p>

Power jack connector and moisture problems

2024-03-28T08:56:29.597

707709

|resistors|dc|surface-mount|repair|

<p>I am 99% sure that if you slap down the least expensive/most popular HC49/U 16MHz crystal and an appropriate size NP0 22pF capacitor it will work fine.</p> <p>There is nothing that requires extreme precision in timing in the GRBL controller and many use the Nano which does not even have a crystal (ceramic resonator only). So worrying about exact match of load capacitors is a waste of time and effort.</p>

<p>this is from a laser GRBL generic from China, 12v and 5v.</p> <p>The neighbor SMD appears identical to the one that was cracked. The multimeter measured 1.0 mega and milliohms, I couldn't get a reading.</p> <p>Do you know what is the piece and a good place to salvage, how should I proceed?</p> <p><a href="https://i.stack.imgur.com/WlSQ7.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WlSQ7.jpg" alt="enter image description here" /></a> <a href="https://i.stack.imgur.com/THLNb.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/THLNb.jpg" alt="enter image description here" /></a></p>

I obliterated a ceramic SMD, how to find out what piece it is?

2024-03-28T10:29:31.183

707711

|power-supply|switch-mode-power-supply|texas-instruments|ripple|

<blockquote> <p><em>no matter how many filter capacitors I add, the issue persists.</em></p> </blockquote> <p>This sort of problem usually comes down to your oscilloscope probes picking up magnetically induced voltages from the switching currents. Try shorting your probe out to the ground wire crocodile clip and see what sort of signal you get in the vicinity of your circuit board.</p> <p>Without making any electrical connection you will probably see the same level of ripple voltage (induced by the changing magnetic field).</p> <p>There are recommended ways to probe these types of circuit and those ways definitely <strong>do not</strong> include using a crocodile clip ground lead.</p>

<p>I'm designing a board that uses a Texas Instruments TLV62569DBVR <a href="https://www.lcsc.com/datasheet/lcsc_datasheet_1809251629_Texas-Instruments-TLV62569DBVR_C141836.pdf" rel="nofollow noreferrer">(datasheet here)</a> to convert 5V input from a USB power supply to 3.3 volts to be used to power an ESP32-WROOM-32D-N4. I am having lots of issues with rippling voltage and no matter how many filter capacitors I add, the issue persists.</p> <p>My Current PCB layout is as follows: <a href="https://i.stack.imgur.com/zd7ie.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zd7ie.png" alt="Current PCB layout. Traces have net labels if you zoom in." /></a> <a href="https://i.stack.imgur.com/RTJtw.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RTJtw.png" alt="bottom side carries 5V and 3V" /></a></p> <p>All capacitors are: CL21A226MAQNNNE. Have also tried the 10 uF version of this capacitor with the same results</p> <p>L1: <a href="https://www.lcsc.com/datasheet/lcsc_datasheet_2108131730_SXN-Shun-Xiang-Nuo-Elec-SMMS0530-2R2M_C181723.pdf" rel="nofollow noreferrer">SMMS0530-2R2M</a></p> <p>R16: 100K ohms</p> <p>R17: 400K ohms</p> <p>R20: 50K ohms</p> <p>U1: TLV62569DBVR</p> <p>I suspect it is either because the sense trace is way too big (Net FB) or that it is getting interference from the inductor. Any input is welcome!</p> <p><a href="https://i.stack.imgur.com/q3nSw.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q3nSw.jpg" alt="rippling voltage" /></a></p>

Switching power supply has high ripple voltage no matter how many capacitors are added

2024-03-28T10:39:09.360

707715

|identification|surface-mount|

<p>It is a <a href="https://www.infineon.com/dgdl/Infineon-IRLML2803-DataSheet-v01_01-EN.pdf?fileId=5546d462533600a4015356682aff260f" rel="nofollow noreferrer">IRLML2803</a> N-channel MOSFET from Infineon. <a href="https://i.stack.imgur.com/AXh2f.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AXh2f.png" alt="enter image description here" /></a></p> <p>Picture from <a href="https://assets.lcsc.com/images/lcsc/900x900/20180914_Infineon-Technologies-IRLML2803TRPBF_C2590_front.jpg" rel="nofollow noreferrer">LCSC</a>:</p> <p><a href="https://i.stack.imgur.com/AtLXt.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AtLXt.png" alt="enter image description here" /></a></p>

<p>I usually manage to find myself, but with this one I am stuck completely.</p> <p><a href="https://i.stack.imgur.com/S3DSG.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S3DSG.png" alt="enter image description here" /></a></p> <p><a href="https://i.stack.imgur.com/Wuakn.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Wuakn.jpg" alt="enter image description here" /></a></p> <p>I suspect it must be some kind of NPN transistor or FET. Most probably part of the marking is manufacturing site, but I also failed to find anything suitable using this pattern. Note dashes under and beyond the letters, maybe they mean something...</p> <p>Maybe you can explain the marking system here and share the hints.</p>

Identify 3 terminal SMD device, probably NPN transistor or FET, marked BKZE7

2024-03-28T10:56:27.717

707717

|power-supply|analog|ldo|linear-regulator|

<p>It's a NMOS LDO:</p> <p><a href="https://i.stack.imgur.com/TlEc9.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TlEc9.png" alt="enter image description here" /></a></p> <p>Let's look at the curves. Dropout voltage (between VIN and VOUT) can be very low as the left graph shows. But the NMOS gate needs to be biased to a voltage above VOUT, which is provided via BIAS pin. So you need sufficient voltage between BIAS and OUT: up to 1.2V, as the right graph shows.</p> <p><a href="https://i.stack.imgur.com/lNGKQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lNGKQ.png" alt="enter image description here" /></a></p> <p>The curves are only typical, and the table states <strong>BIAS should be 1.6V above OUT</strong>, to ensure it works in all cases.</p> <p>So if you connect VIN and VBIAS, this LDO becomes a HDO (High dropout regulator) and the dropout voltage is indicated by the plot on the right. If you want to get 1.2V from 3.3V, then it's absolutely fine. If you want to get 2.5V from 3.3V, then nope.</p> <p>To get 3.3V from 5V, that's 1.7V difference so it will work. If the 5V is actually 5V +/-5% then the lower bound of that (4.75V) is too low.</p> <p>Thermal resistance is quite high (it's a tiny chip) so when operating at high dropout, power dissipation needs to be taken into account.</p> <p>Also note that PSRR and dropout voltage have a special relationship. When the regulator is in dropout (when it operates at its minimum input voltage to still maintain the output voltage), the MOSFET acts like a resistor: it has no PSRR at all. This is why, while they brag about &quot;low dropout&quot; and &quot;ultra high PSRR&quot; you do not get both at the same time. And if you look at the fine print, the PSRR curves are measured at 0.5V dropout, far from the minimum. It's still quite good. But if you operate it at above 1.6V dropout as recommended, you will get the datasheet PSRR or better.</p>

<p>I am using <a href="https://www.sg-micro.de/uploads/soft/20221113/1668325365.pdf" rel="nofollow noreferrer">SGM2052 LDO</a> in my design. Using its fixed voltage version.In the datasheet you can see that it has a pin called Vbias.</p> <p>The description of Vbias is given below.</p> <p><a href="https://i.stack.imgur.com/zJB6M.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/zJB6M.png" alt="enter image description here" /></a></p> <p>Below is the typical application circuit.</p> <p><a href="https://i.stack.imgur.com/52juj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/52juj.png" alt="enter image description here" /></a></p> <p>What I understood from the datasheet is we need to connect a separate voltage at BIAS pin.</p> <p><strong>My question is can I connect Vbias to Vin as shown below?.</strong></p> <p><a href="https://i.stack.imgur.com/CnALL.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CnALL.png" alt="enter image description here" /></a></p>

SGM2052 LDO Vbias pin connection

2024-03-28T11:35:22.973

707724

|capacitor|multimeter|

<p>Any 2 conductors separated by an insulator will exhibit some amount of capacitance. In the case of a typical multimeter, the connections to the multimeter are metal and separated by air and plastic. Thus there will be a small capacitance between them, usually a few pF. When you connect the leads the capacitance will increase since you now have several feet of copper wire. If you separate the leads, the capacitance will decrease. If you wrap the leads together, the capacitance will increase. This is normal behavior for almost all measuring equipment. In most cases, it is small enough to ignore. For oscilloscopes, however, the input capacitance has a significant effect and is dealt with by adding a compensating adjustment to scope probes.</p>

<p>Why does a multimeter show some capacitance when its terminal are not connected to any capacitor?</p> <p>My multimeter is showing 30pF. Does this imply that my multimeter is not calibrated properly or it is due to humidity in the atmosphere or something else?</p>

Capacitance on floating terminals of a multimeter

2024-03-28T12:41:53.287

707747

|sensor|piezo|teensy|

<p>No, it does not look right.</p> <p>You say you have a Teensy 4.0.</p> <p>It does not have a 5V MCU, but a 3.3V MCU, so the analog reference and analog inputs are up to 3.3V only. 5V will damage it.</p> <p>Also the analog voltage to MCU has no defined voltage. You might read any value depending on what the DC bias happens to settle due to leakage currents etc.</p> <p>You might want to add maybe a 1 megaohm resistance over the piezo element. Then both sides of the piezo will idle at 1.65V half-supply when using a 3.3V reference.</p> <p>The 10 ohm resistors are also quite absurdly low values, wasting few hundred milliamps into heat. Maybe use 1k or 10k resistors, and possibly a bypass capacitor to lower the AC impedance.</p> <p>Also, as this circuit uses high impedance signals with, Schottky diodes may not be very suitable. They may have microamps of reverse bias leakage current which will bias your high impedance analog input. Standard diodes may be more suitable, but have higher forward voltage. You might want to put a series resistor of about 1kohm between diodes and MCU analog input.</p> <p>If the source impedance is too high for the ADC input, you can always put an op-amp as a buffer to drive the high impedance piezo signal into MCU analog pin.</p>

<p>I'm making a musical instrument, where you pluck &quot;teeth&quot; which have piezo disc sensors attached to them, and I want to get the output of the piezo into a microcontroller so I can interpret it and trigger musical notes:</p> <p><a href="https://i.stack.imgur.com/t7Bpv.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/t7Bpv.jpg" alt="enter image description here" /></a></p> <p>I'm using a Teensy 4.0, which has 14 ADCs, and I'll have 13 parallel piezo circuits. I'm a software engineer, so once I get the signal into the microcontroller I think I know what to do, but the electrical engineering I'm less sure about.</p> <p>Here's the idea I have for the circuit, though I've drawn it with only one piezo for simplicity:</p> <p><a href="https://i.stack.imgur.com/B4tmM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B4tmM.png" alt="circuit diagram" /></a></p> <p>The idea is that the Teensy's ADC reads from 0V to whatever you put on the V_REF pin, so I'll send 5V to the V_REF and the use R1 and R2 to make a voltage divider to bias the piezo up by 2.5V. I've measured the piezo output, and they're just about in the right range, but if you hit it quite hard it can make very large (positive or negative voltages) which I think I need to protect the microcontroller from. So I've put a pair of diodes (D1, D2) on the other side of the piezo so that if the voltage goes above 5V or below 0V they send the excess out.</p> <p>Before I build this and risk wrecking my Teensy if I got it wrong, does this look right? And are Schottky diodes a good choice for D1 and D2?</p> <p>EDIT 2024-03-29: here's an updated schematic taking into account the feedback from Justme:</p> <p><a href="https://i.stack.imgur.com/iTf5S.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iTf5S.png" alt="enter image description here" /></a></p> <p>I've lowered the voltage reference to 3.3V, marked the R1 and R2 voltage divider resistors 10k, and added a 1M resistor in parallel with the piezo.</p> <p>I'll try this circuit out feeding an oscilloscope before I try it with the Teensy.</p> <p>EDIT 2024-03-31: I've made a new circuit to take into account that the diodes are not perfect. Now there's a margin, both positive and negative:</p> <p><a href="https://i.stack.imgur.com/Q63Mp.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Q63Mp.png" alt="enter image description here" /></a></p> <p>I also switched it from an external voltage reference to the microcontrollers 3.3 volt pin.</p> <p>After testing with the oscilloscope I connected it to the microcontroller, and I'm getting good results! Here is the signal as observed by the ADC:</p> <p><a href="https://i.stack.imgur.com/71ypv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/71ypv.png" alt="enter image description here" /></a></p>

Is this a good circuit for connecting a piezo disc sensor to a microcontroller?

2024-03-28T17:09:24.020

707753

|operational-amplifier|circuit-analysis|circuit-design|speakers|sine|

<p>A LM386 is designed to drive an 8-ohm loudspeaker - it is a good solution.<br></p> <p>But let's modify your original circuit to give more output...<br> A 3.3V peak-to-peak sine wave driving an 8 ohm load requires over 0.4 amps: a 2N3904 is too feeble for this current and would overheat in any case. You'd need a substantially more robust transistor. Your class-A amplifier would also run hot, requiring a heat sink on that transistor.<br></p> <p>The 1.65V DC offset on that 0-to-3.3V input waveform also complicates biasing. By capacitively-coupling that waveform into a GND-referenced resistor, the sine wave will swing symmetrically +1.65 -to- -1.65 V. Now you can ground-reference the 8-ohm load resistor to those +6/-6 DC supplies.<br></p> <p>It will be far easier to add a PNP drive transistor that can pull output down, while an NPN drive transistor pulls up (on alternate half-cycles). These transistors will still run warm, but having two shares the load in this class-B arrangement.<br> I have not shown high-frequency bypass capacitors on the DC supply going into the opamp...each supply should be bypassed to GND to warn-off potential oscillations.</p> <p><img src="https://i.stack.imgur.com/siYme.png" alt="schematic" /></p> <p>^{<a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fsiYme.png">simulate this circuit</a> &ndash; Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a>}</p>

<p>I am trying to build a voltage follower op-amp circuit to hear a sine wave but I can't get it to work. I was using the LM386 but then found out that it has a gain of 20 by default so I can't use it. I am inputting a sine wave however when it gets to a capacitor it becomes heavily attenuated. What op amp can I use for this? How do I design this circuit? I've tried this circuit it didn't lead anywhere<a href="https://i.stack.imgur.com/Q1AJE.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Q1AJE.png" alt="enter image description here" /></a></p>

Building a unity-gain op amp with speaker?

2024-03-28T18:00:32.293

707757

|thermal|heatsink|pad|

<p>That foam thermal interface material is meant to be compressed to give reasonable thermal transfer between non-plane surfaces either at a slight angle or to accommodate variances in height between the source(s) - multiple chips to a single heatsink or a single chip to an uneven surface, or both - foam is mostly air so in its uncompressed state it's pretty low thermal conductivity. Apparently it's good enough in this application that they decided to roll with it to, as you said, increase the surface area. It's probably cheaper than a heatsink, doesn't require any mounting hardware, won't cause problems if it falls off, and they might have just had a bunch lying around.</p>

<p>I found this thermal pad on a controller board for SATA to USB enclosure. It's stuck on the SATA to USB bridge ICs. You can see 2 ICs to the left of it. There was an identical thermal pad adhered which I removed to inspect.</p> <p><a href="https://i.stack.imgur.com/z1m1R.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/z1m1R.jpg" alt="enter image description here" /></a></p> <p>I've never seen a thermal pad without a heatsink before. Typically a thermal pad transfers heat into a heatsink. I'm trying to understand what its purpose is here. There's nothing that comes in contact with it opposite from the IC. I have a feeling it acts as a thermal mass itself, similar to a piece of copper without fins. I guess it still increases the surface area of the ICs by quite a bit given its size. Is this an inappropriate use of a thermal pad or is it perhaps a clever one?</p> <p>Thank you!</p>

What's the purpose of a large thermal pad without a heatsink?

2024-03-28T18:59:57.903

707761

|soldering|

<p>If the solder is from a Chinese company and it forms a sludge when &quot;melted&quot;, it is likely counterfeit.</p> <p>Tin is significantly more expensive than lead -- 15 to 20 times more expensive. Fake 60/40 solder has a higher percentage of lead than the advertised 40%.</p> <p>The melting point of genuine 60/40 solder is <span class="math-container">\$188^{\circ}\$</span>. At that temperature it melts completely and shouldn't be pasty. A soldering iron of sufficient size at <span class="math-container">\$200^{\circ}\$</span> should melt solder. If the soldering iron is not large enough, the solder should melt completely with an iron at some higher temperature, such as <span class="math-container">\$250^{\circ}\$</span> or <span class="math-container">\$300^{\circ}\$</span>. But at whatever temperature is required to melt the 60/40 solder, it should melt completely. You should not need a iron temperature of <span class="math-container">\$350^{\circ}\$</span> unless the iron is small, or the solder diameter is large.</p> <p>Counterfeit solder will be pasty. Part of it will melt, but part of it will remain solid, because the ratio of tin to lead is incorrect.</p> <p><a href="https://www.youtube.com/watch?v=8cCja0FsYUs" rel="nofollow noreferrer">https://www.youtube.com/watch?v=8cCja0FsYUs</a></p> <p><a href="https://www.youtube.com/watch?v=5Ku7I3hA3AA" rel="nofollow noreferrer">https://www.youtube.com/watch?v=5Ku7I3hA3AA</a></p>

<p>I am not an expert at soldering but I am good enough for simple tasks. I am trying to solder some wires to a PCB.</p> <p>If I use some 'old' solder (Radio Shack, 60/40 Rosin-core) I get a result that is acceptable to me. Good flow of solder, shiny result, good connection.</p> <p>If I use a 'new' solder (63/37 Rosin Core, Flux 1.8%) I do not get anything good at all. The solder gets pasty, does not flow, does not stick.</p> <p>I am using the same soldering iron, the same temperature (370C) and the same technique. The 'new' solder is thinner (0.8mm) than the old one.</p> <p>What can I do to correct this? Thanks.</p>

Solder is getting 'pasty' when heated

2024-03-28T19:24:07.560

707765

|logic-gates|

<p>The chip accepts inputs up to 5.5 even without supply.</p> <p>It is specifically allowed to use it for e.g. level conversion of 5V logic to lower logic voltage levels.</p> <p>The output will be high impedance, but it can't exceed VCC so basically you can't e.g. tie the output with a pull-up to another supply.</p> <p>No smoke will come out. Chip won't be powered from input.</p>

<p>I'm using the <a href="https://www.ti.com/lit/ds/symlink/sn74ahc1g32.pdf" rel="nofollow noreferrer">SN74AHC1G32DBVRE4</a> in an application where the inputs to the gate may be energized, but the VCC input to the chip may be disconnected (i.e. floating, open-circuit). What, if anything, can I say about the state of the output in this condition? Is it high-impedance? If at least one of the inputs is at a voltage &gt; V_ih does the output rise to the same voltage? Does it let out the smoke?</p>

Output of OR-gate when inputs are active but gate is not powered

2024-03-28T20:00:07.197

707775

|circuit-analysis|circuit-design|power-electronics|simulation|proteus|

<p><a href="https://i.stack.imgur.com/eX393.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/eX393.png" alt="enter image description here" /></a></p> <p>The load was just too high in value and, back-emfs from the flux in the transformer are masking the true power output voltage waveform. From comments: -</p> <blockquote> <p>Try using a resistor for the load and, not one of value 100 kohm. Maybe 1 kohm – Andy aka</p> </blockquote> <p><em>&gt; @Andy aka Thanks, it works now – Serkan Kaya</em></p>

<p>I simulated this circuit but got an unexpected signal. The result appears on the graph. <a href="https://i.stack.imgur.com/dYvLR.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/dYvLR.png" alt="enter image description here" /></a> But I expected this result:</p> <p><a href="https://i.stack.imgur.com/Hjdad.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Hjdad.png" alt="enter image description here" /></a></p> <p>These pictures show my transformer and lamp settings: <a href="https://i.stack.imgur.com/fWStM.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/fWStM.png" alt="enter image description here" /></a></p> <p><a href="https://i.stack.imgur.com/VE4cm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/VE4cm.png" alt="enter image description here" /></a></p> <p>The schematic and oscilloscope picture are taken from this link: <a href="https://www.youtube.com/watch?app=desktop&amp;v=urKi9OLDdCA" rel="nofollow noreferrer">https://www.youtube.com/watch?app=desktop&amp;v=urKi9OLDdCA</a></p> <p>I mean, I don't understand. What's my mistake?</p>

12V to 220V modified sine wave inverter simulation

2024-03-28T21:40:23.973

707784

|power|transformer|motor|cables|

<p>Electrically it's possible.</p> <p>I'm assuming the transformer outputs (and motor inputs) are not grounded, and that the cable is thick enough for the currents that will flow.</p>

<p>Is it possible to independently run two motors using a shared common conductor?</p> <p>The arrangement is:</p> <ul> <li>there are two motors, each has an external transformer: from mains power down to 24V (I think it is 24 VAC, not 24 VDC)</li> <li>the input to each motor only requires a two core cable (the black and pink inputs in the diagram below)</li> <li>I have an existing 3 core cable between the location of the transformers and the location of the two motors. The 3-core is 2.5mm2, so I think it can carry 20A.</li> <li>the distance between the transformers and motors is about 10-15m</li> <li>the motors are for moving a pair of sliding gates, so they need to be operated at the same time, but independently.</li> <li>the motors consume 50-70W</li> <li>the duty cycle is very low: the motors may run for 30 seconds at a time, and a few times per day.</li> </ul> <p>Can I use the existing 3 core cable to power the two motors, by sharing the common wire (the pink lines in the diagram below)?</p> <p><img src="https://i.stack.imgur.com/0CSIH.jpg" alt="attachment" /></p>

Can I run two motors using a shared common conductor?

2024-03-28T23:41:07.373

707804

|arduino|i2c|display|diagram|oled|

<h4>Connections</h4> <p>D0 is the clock line and D1 is the data line, which is true for both SPI and I²C.</p> <p>Connections to the Arduino Uno (using hardware SPI):</p> <ul> <li>D0 =&gt; SCK</li> <li>D1 =&gt; MOSI</li> <li>CS =&gt; any IO pin</li> <li>DC =&gt; any IO pin</li> <li>RST =&gt; any IO pin</li> </ul> <p>The SMD resistors should have 0 Ohm values. To change from SPI to I²C:</p> <ul> <li>Unsolder R3</li> <li>Solder R8 and R1</li> </ul> <p>You do not need to use zero ohm resistors, just connect the two pads.</p> <p><em>As a sanity check that the unit is actually working, test the OLED with the current SPI configuration first. This will ensure that the unit is actually working before you start soldering away</em>.</p> <h4>Library</h4> <p>You can use the same library. For example, if you use <code>U8GLIB_SSD1306_128X64</code>, just uncomment the relevant line:</p> <pre class="lang-none prettyprint-override"><code>//U8GLIB_SSD1306_128X64 u8g(13, 11, 10, 9); // SW SPI Com: SCK = 13, MOSI = 11, CS = 10, A0 = 9 //U8GLIB_SSD1306_128X64 u8g(4, 5, 6, 7); // SW SPI Com: SCK = 4, MOSI = 5, CS = 6, A0 = 7 (new white HalTec OLED) //U8GLIB_SSD1306_128X64 u8g(10, 9); // HW SPI Com: CS = 10, A0 = 9 (Hardware Pins are SCK = 13 and MOSI = 11) //U8GLIB_SSD1306_128X64 u8g(U8G_I2C_OPT_NONE|U8G_I2C_OPT_DEV_0); // I2C / TWI //U8GLIB_SSD1306_128X64 u8g(U8G_I2C_OPT_DEV_0|U8G_I2C_OPT_NO_ACK|U8G_I2C_OPT_FAST); // Fast I2C / TWI //U8GLIB_SSD1306_128X64 u8g(U8G_I2C_OPT_NO_ACK); // Display which does not send AC </code></pre> <h4>Further reading</h4> <ul> <li><a href="https://forum.arduino.cc/t/running-a-7-pin-oled-with-4-pins-i2c/469007" rel="nofollow noreferrer">Running a 7 pin OLED with 4 pins (I2C)</a></li> </ul>

<p>I just bought the <a href="https://quartzcomponents.com/products/0-96-monochrome-oled-display-ssd1306?_pos=1&amp;_sid=99264004d&amp;_ss=r" rel="nofollow noreferrer">0.96” Monochrome OLED Display (SSD1306) </a> and it claims to support both SPI and IIC/I2C with <em><strong>7 pins</strong></em>.</p> <p>I want to use it with an Arduino Uno R3.</p> <p><a href="https://i.stack.imgur.com/oFgAA.jpg" rel="nofollow noreferrer" title="OLED front"><img src="https://i.stack.imgur.com/oFgAA.jpg" alt="OLED front" title="OLED front" /></a></p> <p><a href="https://i.stack.imgur.com/wGXdc.jpg" rel="nofollow noreferrer" title="OLED rear"><img src="https://i.stack.imgur.com/wGXdc.jpg" alt="OLED rear" title="OLED rear" /></a></p> <p>How do I go about doing this and ensure I don't <em>fry</em> it?</p> <p>Please give me a circuit diagram and tell me which pin to connect where and which pins to ignore to make it function with an Arduino Uno R3 using IIC mode.</p> <p>Also do I use the same library I would use for a <em><strong>4 pins</strong></em> OLED display??</p>

Using 7pin OLED 128x64 with iic

2024-03-29T04:10:57.840

707818

|switches|contact|conductive|

<p>It’s silicone grease. The reason they put it in the switch is for lubrication. Moving parts tend to need lubrication to reduce wear and make them easier to move.</p> <p>That the grease is not conductive has little if any affect on the contact resistance, there is enough pressure to get a good metal to metal contact. In fact you don’t want it to be conductive as that would cause leakage currents between the contacts.</p>

<p><a href="https://i.stack.imgur.com/S0Kom.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/S0Kom.jpg" alt="enter image description here" /></a> <a href="https://i.stack.imgur.com/DGVHz.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/DGVHz.jpg" alt="enter image description here" /></a></p> <p>Look at the picture of my teardown of the SP10T I ordered quite a while ago from DigiKey, <a href="https://www.ckswitches.com/media/1349/arotary.pdf" rel="nofollow noreferrer">https://www.ckswitches.com/media/1349/arotary.pdf</a>. Opening it, it's got grease on the contacts. Much more on the base side too, which is the second picture (a red circle points out one of the contacts that has a dab of the grease).</p> <p>Of course, despite the clear color, it'd be crazy if the grease was a dielectric grease. That would hinder electrical conduction. It'd had to be conductive grease. But, scooping a bit of the grease &amp; putting it on the tip of my multimeter's probes &amp; only have the grease touch, I found out it is an insulator. It IS dielectric grease.</p> <p>It's rated for 2.5A... Shouldn't the idea be maximizing contact to keep relatively high currents?</p> <p>So what gives? Why would the manufacturer put insulating lubricant on the contacts?</p>

Seemingly dielectric grease in my rotary switch

2024-03-29T08:49:21.813

707821

|voltage|level-translation|high-impedance|

<p>There exist families of 74xx style logic ICs that have exactly that behaviour. They are the 74HCT, 74ACT and 74LVC families (thanks to Hearth for reminding me of LVC). You start by selecting which logic function you wish to implement, such as NAND, AND, OR etc, and you acquire the appropriate IC:</p> <ul> <li><a href="https://www.diodes.com/assets/Datasheets/74HCT00.pdf" rel="nofollow noreferrer">74HCT00</a> or <a href="https://www.onsemi.com/download/data-sheet/pdf/74act00-d.pdf" rel="nofollow noreferrer">74ACT00</a>: 4 × 2-input AND gates</li> <li><a href="https://assets.nexperia.com/documents/data-sheet/74HC_HCT1G00.pdf" rel="nofollow noreferrer">74HCT1G00</a> or <a href="https://assets.nexperia.com/documents/data-sheet/74LVC1G00.pdf" rel="nofollow noreferrer">74LVC1G00</a>: 1 × 2-input AND gate</li> <li><a href="https://www.diodes.com/assets/Datasheets/74HCT08.pdf" rel="nofollow noreferrer">74HCT08</a> or <a href="https://www.ti.com/lit/ds/symlink/sn74act08.pdf" rel="nofollow noreferrer">74ACT08</a> or : 4 × 2-input AND gates</li> <li><a href="https://assets.nexperia.com/documents/data-sheet/74HC_HCT1G08.pdf" rel="nofollow noreferrer">74HCT1G08</a> or <a href="https://www.diodes.com/assets/Datasheets/74LVC1G08.pdf" rel="nofollow noreferrer">74LVC1G08</a>: 1 × 2-input AND gate</li> <li><a href="https://www.ti.com/lit/ds/symlink/cd74hct132.pdf" rel="nofollow noreferrer">74HCT132</a> or <a href="https://www.farnell.com/datasheets/2032042.pdf" rel="nofollow noreferrer">74ACT132</a>: 4 × 2-input NAND gates with schmitt trigger inputs</li> <li><a href="https://www.ti.com/lit/ds/symlink/sn74lvc1g132.pdf" rel="nofollow noreferrer">74LVC1G132</a>: 1 × 2-input NAND gate with schmitt trigger inputs</li> <li><a href="https://www.diodes.com/assets/Datasheets/74LVC1G17.pdf" rel="nofollow noreferrer">74LVC1G17</a> 1 × non-inverting buffer with schmitt trigger input</li> </ul> <p>and so on.</p> <p>The 74ACT family is faster than 74HCT, and may be more suitable for 25MHz signals, but I have had success using basic 74HCT gates at similar frequencies.</p> <p>These devices have TTL compatible <em>inputs</em>, meaning that an input above +2.0V will be treated as logic high, and an input below 0.8V will be considered low, which also happens to be compatible with standard 3.3V logic levels. Having MOSFET gates as inputs, their input impedance is huge, and shouldn't influence the signal of interest much, if at all. Be aware, though, that at 25MHz <em>any</em> capacitive load, even the few picofarads of these gates, will slow signal slew rate slightly.</p> <p>If frequency can drop to a few kilohertz, I would recommend using a schmitt trigger input device, which will have better immunity to false triggering due to slow slewing of the input signal.</p> <p>The output will be a function of the device's power supply, so if you supply it with +5V, then it will output either 0V or +5V.</p> <p>In terms of efficiency, the static supply current for these devices will be a few microamps (check the datasheet), but that will naturally increase with switching frequency. Don't forget to tie unused inputs to ground or +5V, to prevent parasitic oscillations, and increased power consumption.</p> <p>Most of the above gates have multiple inputs, but you can tie an input permanently high (or low, if you are using an OR gate) to have the device ignore it:</p> <p><img src="https://i.stack.imgur.com/cMfiO.png" alt="schematic" /></p> <p>^{<a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fcMfiO.png">simulate this circuit</a> &ndash; Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a>}</p> <p>Here's an implementation using a single 74HCT1G08:</p> <p><img src="https://i.stack.imgur.com/LXpzi.png" alt="schematic" /></p> <p>^{<a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fLXpzi.png">simulate this circuit</a>}</p> <p>Don't be tempted to omit C1. At 25MHz you absolutely must have supply decoupling close to the IC's power pins. And I repeat, if you use a package with multiple gates, like the 74ACT08, never leave unused gate inputs unconnected.</p>

<p>I'm looking for a component that can translate voltage levels from 3.3 to 5.0 V (LVTTL to TTL). This component will be placed in order to spy on a signal, it must therefore have a high input impedance so as not to disturb the signal.</p> <p>Does anyone know of a component or circuit capable of achieving this, knowing that the bandwidth can vary from a few kHz to 20 or 25 MHz?</p> <p>I am of course looking for the most energy efficient solution possible.</p>

How to change the levels of a logic signal

2024-03-29T09:04:10.320

707826

|component-selection|magnetron|

<p>You need the transformer from a matching micowave oven, as it needs a defined air gap in the core to function as expected. The silicon diode needs to be at least a 4kV 1A type, or higher depending on the anode voltage of the magnetron.</p> <p>It's not worth the effort. At best, you don't electrocute and neither microwave yourself or people you like.</p>

<p>I have a 1500 W magnetron. What diode and transformer can I use for it?</p>

Selection of diode and transformer for a 1500 W magnetron

2024-03-29T09:24:57.440

707842

|power-supply|voltage-regulator|low-power|battery-operated|ldo|

<p>Essentially, you have the choice between switching regulators and linear regulators with LDOs being just a flavor of linear regulators.</p> <p>If efficiency is a concern, you maybe want to use a switching regulator. Also switching regulators usually cover a wider input voltage range than linear regulators. With the right topology, the input voltage can even drop below Vcc.</p> <p>Linear regulators on the other hand are usually simpler to use and less noisy. But the input voltage always needs to be above Vcc. LDOs can have their input a little closer to Vcc than conventional linear regulators. But LDOs can also have a tendency to become unstable under certain circumstances.</p> <p>Considering your circuit is drawing 2 A and you need Vcc stable &quot;until the batteries are at least 80% depleted&quot; a switching regulator appears to be the more favorable option.</p> <p>If you used a linear regulator, the input voltage just being 1 V above Vcc would cause 2 Watts of power dissipation inside the regulator at 2 A. That would heat up the linear regulator quite a bit.</p>

<p>I'm making a portable device (well not really portable, but intended to be set up away from mains power) powered by AA batteries. I need a stable <strong>VCC</strong> as I have displays that I don't want the brightness to change on (I'm also considering using <strong>VCC</strong> as a reference, see <a href="https://electronics.stackexchange.com/questions/707840/why-use-a-voltage-reference-chip-rather-than-a-known-vcc-and-resistor">this question</a>). I need the <strong>VCC</strong> to remain stable as the batteries drain, atleast to a certain extent (say until the batteries are atleast 80% depleted).</p> <p>I'm thinking I'll use LDO(s) to provide power, as none of the circuits should consume that much power so I won't need anything more bulky. Is there any better solution for this kind of device, or is it fine to use LDO(s)?</p>

What's the best way to make sure I have a known VCC on a battery powered device?

2024-03-29T12:15:04.380

707843

|inductor|buck|switching-regulator|ripple|

<p>The spikes that you're seeing on the scope trace aren't actually present across the 47µF output capacitor. Instead, what's happening here is that the magnetic field produced by the circuit couples into the wire loop formed by the scope tip and its ground spring. This is why you're still seeing it even with the probe tip shorted to ground. It's essentially a (very bad) transformer that lets the scope see some of the voltage across the inductor, as well as the associated switching transients.</p> <p><strong>Your circuit is functioning perfectly fine, there's nothing at all wrong with it</strong>. Your PCB layout also looks to be quite good as the input and output current loops are pretty much as small as they can be.</p> <p>If you want to get rid of the magnetic interference, you could consider wrapping the circuit in an electromagnetic shield, such as mu-metal foil. I doubt that this will be necessary, though.</p>

<p><strong>SEE EDITS BELOW</strong></p> <p>I'm running into some problems with my Buck converter design. During the switching period the inductors magnetic field seems to collapse, which generates some large voltage peaks. That's what I think is happening at least.</p> <p>Some general specs of the converter are: (schematic attached as image below)</p> <ul> <li>AP64500</li> <li>Input Capacitor 2x10uF (one of them is in another part of the schematic)</li> <li>Output Capacitor 1x47uF</li> <li>Inductor 3.3uH</li> <li>Switching Frequency 500kHz</li> <li>Input voltage ~14.7V (lead acid battery), output 3.1V</li> <li>Specs are more or less copied from the recommended components, on page 14 of the datasheet <a href="https://www.diodes.com/assets/Datasheets/AP64500.pdf" rel="nofollow noreferrer">https://www.diodes.com/assets/Datasheets/AP64500.pdf</a></li> </ul> <p>As you can see in the image I have attached, there are some rather large peaks during the switching period, which I think originate from the Inductors magnetic field collapsing. The thing is, I don't really know what could be the cause for this. I've tried multiple inductors, since the original one I had selected had a rather low saturation current. But all of them had the same problem</p> <p>The inductors I've tried thus far are:</p> <ul> <li>VLS3010CX-3R3M-1 (3.3uH, Isat 1A, DCR 0.083)(this was the original one, which had a too low saturation current I think)</li> <li>IHLP1212BZER3R3M11 (3.3uH, Isat 3.3A, DCR 0.056)</li> <li>IHLP1212BZEV3R3MA1 (3.3uH, Isat 4.5A, DCR 0.090)</li> <li>74438336033 (3.3uH, Isat 2.85A, DCR 0.099)</li> </ul> <p>Now I wouldn't really know what could be causing these peaks. At first I thought the saturation current was the issue, but that doesn't seem to be the case. So now I don't really know where these peaks could come from, or what I could do to fix them. If anyone would know what could be the cause of these peaks, or what I could do to prevent them, that would be nice!</p> <p>Thanks!</p> <p><a href="https://i.stack.imgur.com/LGAI6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LGAI6.png" alt="enter image description here" /></a> <a href="https://i.stack.imgur.com/pGvFW.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pGvFW.png" alt="enter image description here" /></a></p> <p><strong>EDIT</strong></p> <p>As requested in the comments, here's a picture of where I'm measuring. I also forgot to mention that I adde some dummy load resistors which pull around 200mA, just for testing purposes. My actual load will also be around this much. There's a big ground plane on the bottom layer, and the inductor and output capacitor are connected using big copper planes</p> <p><a href="https://i.sstatic.net/7IY1IJeK.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/7IY1IJeK.jpg" alt="enter image description here" /></a></p> <p><strong>EDIT 2</strong></p> <p>Picture of the top and bottom of the PCB layout, as well as a picture of the scope screen with the ground spring shorted to the probe tip.</p> <p><a href="https://i.sstatic.net/rUptfHlk.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/rUptfHlk.png" alt="enter image description here" /></a><a href="https://i.stack.imgur.com/XKrNX.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/XKrNX.png" alt="enter image description here" /></a> <a href="https://i.stack.imgur.com/vSDLu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vSDLu.png" alt="enter image description here" /></a></p> <p><strong>EDIT 3</strong></p> <p>I did some more measuring and found that the peaks were at 500kHz, which is the programmed switching frequency. I assumed one full ripple would be at 500kHz, but it seems I was wrong about that. I also measured across some of the dummy load resistors (Blue waveform) and did indeed find that the peaks were significantly smaller as opposed to measuring on the converter output cap, which makes sense considering Jonathan S.'s answer.</p> <p><a href="https://i.stack.imgur.com/ItXEf.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ItXEf.png" alt="enter image description here" /></a>]<a href="https://i.sstatic.net/XImIh9Cc.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/XImIh9Cc.jpg" alt="enter image description here" /></a></p>

Inductor magnetic field collapses during switching?

2024-03-29T12:19:58.640

707849

|raspberry-pi|spi|flash|

<h2>Potential code issue with multiple pages to be written</h2> <p>The code structure uses a loop where each iteration:</p> <ul> <li>Issues a <code>Write Enable</code> command.</li> <li>Issues a <code>Page Program</code> command.</li> </ul> <p>The problem with this is that <code>Page Program</code> command can take some time to complete, before the flash is ready for a further program or erase command. Before issuing a program or erase command, a <code>Read Status Register-1</code> command should be used and the <code>WRITE IN PROGRESS</code> (BUSY) status bit checked to confirm the flash is idle before sending the next program or erase command. From the W25Q32JV <a href="https://www.winbond.com/hq/support/documentation/downloadV2022.jsp?__locale=en&amp;xmlPath=/support/resources/.content/item/DA00-W25Q32JV_1.html&amp;level=1" rel="nofollow noreferrer">datasheet</a>:</p> <blockquote> <p><strong>Erase/Write In Progress (BUSY) – Status Only</strong></p> <p>BUSY is a read only bit in the status register (S0) that is set to a 1 state when the device is executing a Page Program, Quad Page Program, Sector Erase, Block Erase, Chip Erase, Write Status Register or Erase/Program Security Register instruction. During this time the device will ignore further instructions except for the Read Status Register and Erase/Program Suspend instruction (see tW, tPP, tSE, tBE, and tCE in AC Characteristics). When the program, erase or write status/security register instruction has completed, the BUSY bit will be cleared to a 0 state indicating the device is ready for further instructions.</p> </blockquote> <hr /> <h2>Other debugging suggestions</h2> <blockquote> <p>When i use a Flash read command later, all the data read just appear as boxes in the terminal.</p> </blockquote> <p>The other <a href="https://electronics.stackexchange.com/a/707854/101063">answer</a> already notes the flash must be erased before it can be programmed - since erasing changes bits from <code>0</code> to <code>1</code> and programming can only changes bits from a <code>1</code> to <code>0</code>.</p> <p>Some other suggestions for debugging:</p> <ol> <li>Have any tests been performed to check basic communication between the Raspberry PI and W25Q32JV? E.g. send the following commands and check if the flash returns the expected byte values described in the datasheet: <ul> <li>Manufacturer/Device ID (90h)</li> <li>JEDEC ID (9Fh)</li> </ul> </li> <li>After sending the <code>Write Enable</code> command, issue a <code>Read Status Register-1</code> command and check if the <code>WRITE ENABLE LATCH</code> (WEL) bit is set to confirm if the write enable command was accepted.</li> </ol>

<p>My querry is, I am trying to write data from a file stored in by Raspberry PI to an external flash(W25Q32JV). The size of the file is greater than 256 bytes. As Winbond allows writing a max of only 256 bytes in 1 command, My idea of implementing a multiple page write was by first reading the entire data into an array at first. Then determine the number of pages to be written based on the size of the a fileread. Then run a for loop for &quot;number of pages&quot; times.. with each time the sequence being</p> <ol> <li>Write enable, ea</li> <li>Write hex command and address</li> <li>Write the required data (max of 256).</li> </ol> <p>I feel I am making some error when I am writing the data from the file to the array.. Am i missing out on the logic somewhere here.</p> <p>Edit: Adding a more specific question: When I run this piece of code, the data doesnt get transferred to the Flash. When i use a Flash read command later, all the data read just appear as boxes in the terminal.</p>

Multiple Page Write of Winbond W25Q32JV without polling Status registers

2024-03-29T13:05:23.203

707867

|feedback|stability|negative-feedback|loop-gain|

<p>The Python code, the handwritten notes, and the body of the question say that <span class="math-container">$$G=105$$</span>. But Simulink diagram says that <span class="math-container">$$G=105 \cdot \frac{1}{s}$$</span>.</p> <p>Moreover, the Python actually <strong>does not implement either of the above situation</strong>. It appears to implement <span class="math-container">$$ x_n = vin_n - B \cdot vout_{n-1}\\ vout_{n} = x_n \cdot G = G \cdot vin_n - G \cdot B \cdot vout_{n-1}\\ $$</span> where <code>n</code> represents <em>sample number</em> or <em>(discrete) time variable</em>.</p> <p><code>Tf = G/(1+GB)</code> only if it was implemented as <span class="math-container">$$ x_n = vin_n - B \cdot vout_{\color{red}{n}}\\ vout_{n} = x_n \cdot G = G \cdot vin_n - G \cdot B \cdot vout_{\color{red}{n}}\\ \implies \left(1 + G \cdot B\right) vout_{n} = G \cdot vin_{n}\\ \implies \frac{vout_n}{vin_n} = \frac{G}{1 + G \cdot B} $$</span></p> <p>i.e. <em>simultaneous equations</em>.</p>

<p>I have been reading about stability and feedback systems, and a very silly issue has appeared. We can describe a negative feedback system as what can be seen in the following picture:</p> <p><a href="https://i.stack.imgur.com/RDsem.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/RDsem.png" alt="negative feedback system" /></a></p> <p>Image source: <a href="https://www.electronics-tutorials.ws/systems/negative-feedback.html" rel="nofollow noreferrer">Electronics Tutorials - Negative Feedback Systems</a></p> <p>So here we will have that Vout = (Vin-B*Vout)*G</p> <p>We can find the transfer function as Vout/Vin = G/(1+B*G)</p> <p>The issue is... I know for a fact that a system with negative feedback should be stable, regardless of AB&gt;1 or AB&lt;1. But I've tried to simulate that in Python, and my signal oscillates and tends to infinity. The loop is running correctly, meaning if you perform the calculations by hand it gives those results.</p> <p><a href="https://i.stack.imgur.com/lLEex.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/lLEex.png" alt="enter image description here" /></a></p> <p>For the Python simulation, I assumed an input of vin = 1, an initial state of vout=0, and G=105 and B = 0.01, which is over unity. At this point I tried to do some calculations to find how can I make the system not oscillate (and thus avoid positive feedback), but I obtained a result saying that GB&lt;1, which I know for a fact isn't a requirement. I'll leave the calculations:</p> <p><a href="https://i.stack.imgur.com/whekJ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/whekJ.png" alt="enter image description here" /></a></p> <p>(This also includes the image, copied from <a href="https://www.electronics-tutorials.ws/systems/negative-feedback.html" rel="nofollow noreferrer">Electronics Tutorials - Negative Feedback Systems</a>)</p> <p>I know I'm making a mistake somewhere, because if a perform the simulation in Matlab Simulink I get the results that I expect to obtain. For example, in the following image, you have first a system with positive feedback with GB&gt;1 (which should be stable), secondly a positive feedback system with AB&gt;1 (which should be unstable), and finally a positive feedback system with GB&lt;1 (which should be stable). The results are attached and are exactly what I expected.</p> <p><a href="https://i.stack.imgur.com/HxCkU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/HxCkU.png" alt="enter image description here" /></a></p> <p><a href="https://i.stack.imgur.com/jK5bx.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/jK5bx.png" alt="enter image description here" /></a></p> <p>The question remains... Where is my mistake? Why isn't my Python script correct?</p>

About stability in system with loop gain AB>1 in negative feedback

2024-03-29T15:53:08.853

707872

|circuit-design|pcb-design|potentiometer|

<p>The reason such 'gang' potentiometers exist is to be able to handle multiple signals with one manual control. Prime example for a 2-gang pot: a stereo volume control.</p> <p>If you short the two pots together as you propose, you will get:</p> <ul> <li>overall (pin 1 to pin 3) resistance will be halved</li> <li>wiper (pin 2 to 1/3) resistance will be an average of each wiper, halved</li> </ul> <p>So if your pots are a linear 10k and you have the wiper at midpoint, shorting them as you propose will be:</p> <ul> <li>5k, pins 1-3</li> <li>2.5k pins 2 to pins 1</li> <li>2.5k pins 2 to pins 3</li> </ul> <p>A more complex model would account for any offset between them, but you get the basic point: you've turned a 10k gang pot into a single 5k one.</p> <p>Nothing 'bad' will happen if you short them, but there isn't a particular advantage of doing so either. If you only need one channel it's a waste of money to use a gang pot.</p>

<p>I have a 6 Pin Potentiometer. I know that the 3 pins are separately a potentiometer and the other three pins are a part of other potentiometer. My question is what will happen if I externally short the pins, i.e., 1-to-1, 2-to-2 and 3-to-3 and then use 1-3 as input and 2 as output? Will the circuit be short or is it just going to work fine? <a href="https://i.stack.imgur.com/ibzY7.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ibzY7.png" alt="enter image description here" /></a></p> <p>Reference: vu2nan (<a href="https://electronics.stackexchange.com/users/193409/vu2nan">https://electronics.stackexchange.com/users/193409/vu2nan</a>), How to wire up 6 pin potentiometers, URL (version: 2020-09-12): <a href="https://electronics.stackexchange.com/q/521166">https://electronics.stackexchange.com/q/521166</a></p>

6 Pin Potentiometer

2024-03-29T16:49:36.883

707878

|power-supply|power|boost|buck-boost|

<p>Loss in a linear 90v to 70V dropper at 21 mA = 20V drop x 21 mA ~+ 400 mW.</p> <p>You say have enough power from your converter to do this.<br /> It's entirely your choice whether to do it.<br /> It's low cost and simple compared to alternatives and reliable if designed properly.</p> <p>The loss is about 8% of your total energy use [(20x21)/(41x90 + 70x21) ] .<br /> Low cost, simple, reliable, modest energy loss - sounds reasonable.</p>

<p>I have a VFD matrix display that needs both 90V and 70V to power on. I already designed a boost converter that will give me the 90V.</p> <p>I was wondering if I can just put a voltage divider at the output to get the 70V.</p> <p>I need 41ma and 21mA for the 90V and 70V inputs respectively.</p>

Can I supply two different voltages with the same boost converter?

2024-03-29T17:27:20.443

707887

|schematics|review|

<p>An old <a href="https://electronics.stackexchange.com/questions/308878/how-were-schematics-drawn-before-cad">post</a> on this forum by Jonk outlines good practices for drawing schematics. <br>Your second schematic is OK, meaning it is readable with some difficulty, however, it needs some cleanup.</p> <p>Some thoughts on what I see:</p> <ul> <li><p>Schematic symbols and reference designators can be found in IEEE Std 315. It's best to use the standards. If your region has a different standard for drafting, use it.</p> </li> <li><p>When depicting transistors, use the proper symbol instead of a box. When others look at your schematic proper symbols are instantly recognizable. Don't make viewers do extra work parsing symbols. Modern schematic tools allow homogeneous and hetrogeneous parts.</p> </li> <li><p>The common symbol has &quot;GND&quot; along with it. This extraneous text adds to clutter. There are times to show labels with common symbols if you have multiple commons. Also, this looks like Altium with the non-standard common symbol with four horizontal lines (it should have three horizontal lines). For that reason, I use a triangle symbol in Altium and hide the &quot;GND&quot; text. The first schematic has ground symbols pointing upwards which is bad practice. You fixed that in the second schematic.</p> </li> <li><p>Q5 &amp; Q6 has crowding with junction dots touching Q5 &amp; Q6 which looks like there should be a pin on Q5 &amp; Q6. This adds confusion. Also the PyroContinuityn nets split the resistor symbols (R8 &amp; R9) from the labels. There is no need to do this as you have ample space to route the net around the resistors.</p> </li> <li><p>Be consistent in your labeling. Most of your symbols have the reference designation over the value (good). R8 &amp; R9, and other parts, have the value over the reference designation. The reference designation and vale should line up. C15, C16, &amp; Q6 are good examples of being sloppy. A schematic is also a work of art.</p> </li> <li><p>No need to include the omega symbol for the resistor values. It clutters up the schematic.</p> </li> <li><p>My personal preference is to have reference information horizontal. LED1 &amp; LED2 are examples of vertical text which I don't like.</p> </li> <li><p>My personal preference is to use upper case text except for design notes. Easier to read when printed. I also use a mono-spaced font with no ambiguities (i.e., 1 l I are distinct). However, Altium doesn't handle non-default fonts well when using the built-in PDF printer (solution is to use a third party PDF printer).</p> </li> <li><p>My personal preference is to use as few pages as possible. Having a one page schematic as you do is wonderful. On more complicated schematics I'll group functionality per page. Example: core schematic on page 1, power supplies on page 2. Unfortunately, Altium schematic editor gets very slow when there are lots of parts/pins on a single page. My cure is to work on small parts of the design on a separate worksheet page, then move (copy/paste) the section to the final page. Other approaches to the program slowness are to have a 20+ page schematic which makes it hard to follow.</p> </li> <li><p>Be careful about color selection. Some colors don't have good contrast when printed.</p> </li> <li><p>Check for connection errors like CN1-1 going to +VBAT. It looks like you are missing a junction dot. The design rules may show this as an error when you transfer the design to PCB.</p> </li> </ul>

<p>After reading all of your feedback on my flight controller schematic I posted two days ago in <em><a href="https://electronics.stackexchange.com/q/707649/101063">Model rocket flight controller schematic review</a></em>, I redid the entire schematic and tried my best to make it readable and easy to understand, so that others can more effective look over it and check for errors/mistakes.</p> <p>The main goal was to make connections clearer, and not so split.</p> <p>Is my second version better, in terms of readability?</p> <p>And what are things to improve? (mainly on the &quot;aesthetic&quot; side; not for the functional side of things)</p> <p>This is the first version:</p> <p><a href="https://i.stack.imgur.com/3oPZq.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/3oPZq.jpg" alt="Enter image description here" /></a></p> <p>And this is the new, second version:</p> <p><a href="https://i.stack.imgur.com/6Ptyu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6Ptyu.png" alt="Enter image description here" /></a></p>

How can I design a readable and understandable schematic?

2024-03-29T18:53:47.257

707924

|delay|

<p>Your circuit won't switch the load cleanly on or off, with such a slow-slewing battery voltage. The load's voltage will change faster than battery voltage, due to the transistor's gain, but still not as fast as you need. Also, near the &quot;switching&quot; threshold, any slower fluctuations in battery potential (due to changing loads) will appear across the load too, heavily amplified.</p> <p>For this kind of application you really need need hysteresis, a difference between &quot;on&quot; and &quot;off&quot; thresholds. That's usually implemented using positive feedback, which also has the benefit of causing switching to be fast and &quot;digital&quot;, never occupying a state between fully on or off. While this can be achieved with a couple of transistors, by far the best solution will employ either a microcontroller, or comparators. Microcontrollers require programming, and assuming you don't want that hassle, here's a comparator solution, using the two comparators inside a single 8-pin DIP package <a href="https://www.ti.com/lit/ds/symlink/lm393.pdf" rel="nofollow noreferrer">LM393</a>:</p> <p><img src="https://i.stack.imgur.com/nteUL.png" alt="schematic" /></p> <p>^{<a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fnteUL.png">simulate this circuit</a> &ndash; Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a>}</p> <p>R4 and R5 provide positive feedback for about 250mV of hysteresis between the upper and lower switching thresholds, which are +12.25V and +12.0V respectively. That is, supply voltage must rise above +12.25V for the load to switch on, and then fall below +12.0V for it to switch off. This prevents oscillations when supply voltage passes very slowly through the thresholds, and produces a fast and emphatic on/off for the load. Because of this, C2 and C3 are necessary to help keep the supply from fluctuating under sudden load changes. Install C2 as close as possible to M1's source and where the load's ground connects to this circuit. Connect C3 right across the LM393 power supply pins, as close as possible to those pins.</p> <p>When CMP1 detects crossing of the thresholds, and its output A changes state, capacitor C1 charges or discharges via R7 and R8, producing a slow rise or fall of potential at B. CMP2 goes high or low as B passes through about half of the supply potential, resulting in its output changing state several seconds after CMP1's output transition. That delay is adjustable between about 3s and 15s, by adjusting potentiometer R7.</p> <p>Here are some waveforms from a simulation. The first is input potential <span class="math-container">\$V_P\$</span> at node P, and the second shows potential <span class="math-container">\$V_A\$</span> at node A (blue), and output (load voltage) <span class="math-container">\$V_{OUT}\$</span> (orange):</p> <p><a href="https://i.stack.imgur.com/PVja2.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/PVja2.png" alt="enter image description here" /></a> <a href="https://i.stack.imgur.com/Fcnmg.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Fcnmg.png" alt="enter image description here" /></a></p> <p>The green markers show the slight difference between the upper and lower switching thresholds.</p>

<p>I have a set of security cameras powered by a solar powered battery bank. (4S5P LiFePO4 batteries charged by a 100W solar panel).</p> <p>As the voltage varies depending on charge level and how sunny it has been, the batteries can sometimes be drained over night before the solar panel starts charging them again. The batteries also have a BMS that will cut them out below a certain voltage.</p> <p>The cameras draw a higher current during startup than during normal operation, so powering all at once when the batteries are at a low charge level will cause voltage drops.</p> <p>To maximize battery life and to keep some of the cameras on longer than others during low battery conditions, and also to ensure that they don't all power on at once, I want to add a simple voltage dependent delay-on circuit that power some of them on a short while after the voltage goes above 12V. If the voltage drops below 12V, I want the cameras with this circuit to lose power after a while (not instantly). The duration of the on/off delay is not critical, as long as it is in the range of a second or a few. The 'prioritized' cameras will not have this circuit and will stay powered until the battery bank's BMS cuts the power due to too low voltage in the batteries.</p> <p>I also want to keep these delay-on circuits as simple as possible.</p> <p>My initial thinking is something like the circuit below: if the voltage goes above 12V the D1 zener will conduct and charge C1. When C1 is charged, M1 will conduct and power the load (here illustrated by a light bulb). If the voltage drops below 12V, R2 will discharge C1 and M1 will stop conducting. C1 can vary in size for each camera so that the on-delay is different for each camera.</p> <p>Is my thinking correct / will this work as intended?</p> <p>Any suggestions on changes or even a completely different solution are much appreciated!</p> <p><img src="https://i.stack.imgur.com/NcGcj.png" alt="schematic" /></p> <p>^{<a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fNcGcj.png">simulate this circuit</a> &ndash; Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a>}</p> <p><strong>Update:</strong> in front of each load, I also have a boost-buck regulator. Those are el-cheapo things from China. Unfortunately, I don't have a datasheet for them, but there is an input capacitor @ 470uF and a similar sized output capacitor. I have those regulators to ensure the cameras are fed with the 12-13V they expect and to not fry their built-in regulators with over- or undervoltage. The new switch circuit will sit in front of the regulator. I added this detail later in case the inrush current to the regulator's capacitors could be an issue.</p> <p><a href="https://i.stack.imgur.com/rwAdO.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rwAdO.jpg" alt="voltage regulator that is part of the load" /></a></p>

Simple voltage dependent on/off-delay switch circuit

2024-03-30T04:08:07.760

707926

|converter|

<p>The converter you drew is a boost converter, not a buck converter. I don't know why you shorted the inductor in the second phase. In that phase, the inductor is not shorted, and the voltage across the switch is <span class="math-container">\$V_O\$</span> (assuming an ideal diode with zero voltage drop). <a href="https://i.stack.imgur.com/6eLmU.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6eLmU.png" alt="enter image description here" /></a> Image taken from Wikipedia: Boost converter.</p>

<p><a href="https://i.stack.imgur.com/azQcJ.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/azQcJ.jpg" alt="enter image description here" /></a></p> <p>I know the answer is Vs=Vo, by simulation. BUT I don't know why. In steady state inductor is gonna be short. And Vs is parallel with Vin and Vout. Can you tell me what is wrong?</p>

Boost converter's voltage stress

2024-03-30T06:18:50.777

707929

|ttl|

<p>It seems to be a simple inverter.</p> <p>Assuming no other connections, resistors and supply as shown, we can simplify the circuit as so:</p> <p><img src="https://i.stack.imgur.com/GfZG9.png" alt="schematic" /></p> <p>^{<a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fGfZG9.png">simulate this circuit</a> &ndash; Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a>}</p> <p>which has a DC transfer function from V1 (A) to D,</p> <p><a href="https://i.stack.imgur.com/TsmSC.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TsmSC.png" alt="enter image description here" /></a></p> <p>give or take the setting of your R13, here set all the way towards C.</p> <p>It's more or less an emitter follower-buffered negative-voltage (PNP type) RTL inverter, with level shift at the input.</p> <p>The Darlington follower (Q2 into Q3) doesn't seem very useful, but without knowledge of application and ratings, that's neither here nor there.</p>

<p>There is a circuit based on TTL( transistor transistor logic I think) . The signals at points A B C D are described. I am new with this TTL, could you help me decompose this united circuit into the logic components which it consists from? Thanks.</p> <p><a href="https://i.stack.imgur.com/CXaqu.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/CXaqu.png" alt="enter image description here" /></a></p> <p><a href="https://i.stack.imgur.com/pYP7n.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/pYP7n.png" alt="enter image description here" /></a></p>

recognising patterns in TTL logic analog circuit

2024-03-30T08:10:04.923

707930

|led|multiplexer|led-driver|display|7segmentdisplay|

<p>Yes, you can drive the additional digits as they are both common-cathode.</p> <p>The AS1115 seems particularly well-suited since (according to the datasheet big print) it allows brightness to be set on a digit-by-digit basis. This is important because the typical luminous intensity at a given average current is not well matched between the two displays.</p>

<p>The title's a bit convaluted, so I'll ellaborate.</p> <p>I'm driving a 4-digit 7-segment display with a multiplexing controller, the kind where it sets one display at a time and relies on persistence of vision to make it look like they're all on. It's the AS1115, you can see its datasheet <a href="https://au.mouser.com/datasheet/2/588/AS1115_DS000206_1_00-1512924.pdf" rel="nofollow noreferrer">here</a>.</p> <p>Anyway, my device also has a few single-digit 7-segment displays. Since the controller can control 8 digit displays and 4 of the digits aren't in use, I was wondering if it was possible to connect these seperate displays to the same controller, to save money and power?</p> <p>Sidenote: Here's my <a href="https://au.mouser.com/datasheet/2/239/C5723HR-1141844.pdf" rel="nofollow noreferrer">4-digit display</a> and here's my <a href="https://www.we-online.com/components/products/datasheet/157142S12703.pdf" rel="nofollow noreferrer">single-digit display</a>.</p>

Can single digit 7-segment displays use free digits from a multi-digit multiplexer?

2024-03-30T08:21:18.680

707932

|atmega328p|clock-speed|internal-resistance|

<p>The details you need are defined in the waveform image right above the table you pasted.</p> <p>Yes, High Time is same as &quot;On Time&quot; usually but your definition is way off. High Time is the time that signal spends above the logic threshold definition for high logic level. In this case, it must be higher than Vih1, which is either 0.8×Vcc or 0.7×Vcc, depending on supply voltage Vcc range.</p> <p>The internal resistance will be approximately infinite so it is undefined. It is a CMOS logic level input. If you want to know capacitive loading or leakage current, they are defined in the electrical specs.</p>

<p>While looking for the rise time and fall time of external clock in ATMega328p, I came across the term &quot;High time&quot;. Is it same as &quot;ON time&quot; i.e. time taken to reach 10% of amplitude? I have attached snippet of the datasheet which I'm referring to <a href="https://i.stack.imgur.com/q066y.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/q066y.png" alt="ATMega328p external clock drive" /></a></p> <p>Also can someone help how much internal resistance is seen on the pin of the MCU? I'm using Arduino Uno here.</p>

High time in External Clock drive

2024-03-30T08:30:24.393

707939

|adc|dc|

<p>As such, the block is rather useless, if the signal source is an ideal voltage supply like in your simulator.</p> <p>Maybe the analog input is disconnectable or the signal source is simply a resistor to some supply, like NTC temperature sensor. So the 10k pull-down is necessary.</p> <p>Maybe there is a high impedance signal source, so the capacitor is required for the ADC to sample it properly without voltage sagging.</p> <p>Maybe there are long wires and the capacitor and the 100 ohm series resitor is to protect from ESD.</p> <p>Maybe someone just draw the circuit like that because someone else had also drawn the cicuit like that without knowing why they should draw it like that. It's called cargo-culting.</p>

<p>I was looking at a schematic block, for an VIn for an ADC and I saw this :</p> <p><a href="https://i.stack.imgur.com/TXYO3.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/TXYO3.png" alt="enter image description here" /></a></p> <p>What is the possible scope for this block?</p>

What is the scope of this circuit?

2024-03-30T09:55:27.787

707944

|arduino|voltage-regulator|

<p>In addition to other answers...</p> <p>All switching power supplies are a concern for EMC. Significant care needs to be taken to avoid your PCB or other components becoming a radio broadcasting at the switching frequency.</p> <p>Switching power supplies also inherently produce some ripple on the output voltage. This makes them unsuitable for higher-precision analogue circuits on their own. Best practise if you care about noise is normally to follow the switcher with a linear power supply, which drops the voltage further but gives a very much cleaner supply.</p>

<p>I am referring to <a href="https://www.haoyuelectronics.com/Attachment/XL6009/XL6009-DC-DC-Converter-Datasheet.pdf" rel="nofollow noreferrer">this boost converter.</a></p> <p>I want to know the disadvantage of this. It isn't possible that this could just generate free voltage, there would be some disadvantage. I actually need this to power some servo motors because I have a supply of 2 volts and want to multiply it to about 5-6 volts, and control using an Arduino. Is it possible, or is it too good to be true?</p> <p>EDIT: <a href="https://robu.in/product/xl6009-dc-dc-step-up-converter-performance-ultra-lm2577-booster-circuit-board/" rel="nofollow noreferrer">This is the full PCB.</a></p>

What is the disadvantage of using a boost converter?

2024-03-30T10:32:45.260

707959

|switches|logic-gates|

<p>The buttons pull the lines low, and this is using inverse logic. If either of the button inputs is high, then that indicates not pressed, and thus not resetting. Thus it should only pull reset low if neither of the inputs is high.</p> <p>Or in other words, if gnd is true and 3.3v is false, then by de morgans law, the 'or' gate is actually a &quot;low input and&quot; gate.</p>

<p>I don't understand the schematics for this rebooting system based on pressing momentary switches. They are the schematics of Flipper Zero. (I am trying to look some schematics online to start understanding them. I am new to electronics.)</p> <p><a href="https://i.stack.imgur.com/OniPz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OniPz.png" alt="SCHEMATICS OF THE BUTTONS" /></a></p> <p>The Flipper should reboot if BOTH the &quot;Back&quot; and &quot;Left&quot; switches are pressed for 5 seconds. But there is an OR gate and not an AND. Also it should hard reboot if the &quot;Back&quot; switch is pressed for 30 sec. But, what if I press the &quot;Left&quot; switch for 30 sec? The circuit is similar / somehow identical. source : <a href="https://docs.flipper.net/development/hardware/schematic" rel="nofollow noreferrer">https://docs.flipper.net/development/hardware/schematic</a></p>

Circuit for rebooting by pressing momentary switches

2024-03-30T14:35:39.400

707960

|adc|high-voltage|voltage-measurement|

<blockquote> <p>So what is the best way to proceed? Perhaps with a simple voltage divider.</p> </blockquote> <p>The &quot;best&quot; way depends on a few factors: budget, accuracy, resolution, ...</p> <hr /> <p>As Andy mentioned in his answer, you can use an HV probe with a scope and fetch the scope data later (e.g. in CSV format) but the bandwidth and the sampling rate should be high enough for an accurate representation. Otherwise, the pulse shape you'll see on the scope may get distorted, which means less accuracy.</p> <p>I don't know what the pulse is going to look like but if, for example, that 1-ms pulse has a rise time of 10 nanoseconds (the datasheet doesn't mention anything, though) then the probe's bandwidth should be <em>higher than</em> 35 MHz to get an accurate representation of the pulse (NOTE: Ignored the scope requirements as even the cheapest ones today have 100 MHz of BW and 1 GSps of sampling rate).</p> <hr /> <p>The simplest option might look like building a 1000:1 voltage divider and feeding it to a 3V3 ADC: Throw a few 1206-case resistors (their voltage rating is about 200V so 20 of them would be enough), buffer if needed, and done. Layout is very important, always keep your and the circuit's safety in mind (high voltage can &quot;jump&quot;, 1 kV/millimetre is a good starting point for clearance).</p> <p>The accuracy and the speed of the ADC might be a limiting factor here. Depending on the pulse shape and your accuracy requirements, you may need an ADC with a high sampling rate.</p> <p>Now being a bit pessimistic, if the rise time of the signal is presumably 10 nanoseconds and you want an accurate representation then the sampling rate should be <em>at least</em> 70 MHz (Nyquist). If each sample is 8-bit long, you'll need to store 70 kB of data.</p> <p>Today's popular cheap MCUs go beyond 1 MSps (megasamples per second) with their ADCs. With multi-channel sampling, you can increase the effective sampling rate (e.g. 3-ch interlaced on an STM32F4 will give you 2.4 MSps x 3 = 7.2 MSps). That may give you enough representation.</p> <hr /> <p>Personally, I'd probe the signal with a scope and fetch the data for further processing.</p>

<p>For research lab work, we use a high-voltage generator to create one pulse with a low intensity (&lt; 4 mA).</p> <p>The voltage range is variable from 100 V to 3000 V and the FWHM of the pulse is ~1 ms. The pulse is delivered to different chemical and biological solution with high conductivity value; this implies that the equivalent Rload is low and varying.</p> <p><a href="https://i.stack.imgur.com/LwVoh.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/LwVoh.png" alt="enter image description here" /></a></p> <p>When I delivered the pulse I would like to measure the pulse profile (digitize with an ADC). So what is the best way to proceed? Perhaps with a simple voltage divider? The HV module is an <a href="https://www.xppower.com/product/F-Series" rel="nofollow noreferrer">XP power F Series</a>.</p>

Measurement of high voltage

2024-03-30T15:09:09.523

707975

|circuit-analysis|integrated-circuit|stability|ota|

<p>You need to include the whole (loop) in the analysis. Analyzing the OTA by itself doesn't inform much about the whole system.</p> <p>While the transistor loads the OTA, the frequency response of the transistor (and the effects of its load) also affect the whole loop.</p> <p>Present that the transistor and S load (RE) are all part of a larger OTA. It has a reference (the OTA's +), and a feedback node (S of FET to OTA's -). You can break the loop there and analyze the loop characteristics.</p> <p>In the application, the FET's drain is the 'useful' output, but (if the transistor is reasonable), there is little interaction from the load on the D to the signals at the S (some D-S capacitance, r0 of the FET, and possibly some Miller effects of CDG if the load is high impedance). to be more precise, you should perform the analysis with the correct load on the D.</p>

<p>I am designing an OTA for a current buffer (images from <a href="https://link.springer.com/book/10.1007/b135984" rel="nofollow noreferrer">Analog Design Essentials</a>):</p> <p><a href="https://i.stack.imgur.com/9EXK3m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/9EXK3m.png" alt="enter image description here" /></a></p> <p>The OTA is just used to provide a high gain, to regulate the voltage on RE. My question is, how to test the stability of an OTA used in this application?</p> <p><a href="https://i.stack.imgur.com/4rc20.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/4rc20.png" alt="enter image description here" /></a></p> <p>The classic method we learned from classes is to configure the OTA as a source follower, and add a load capacitor representing the next stage at the output. In my case, the load is the gate of a transistor, which may be several hundreds of fF. This makes the gain bandwidth product (GBW) of the OTA very large (GBW<span class="math-container">\$=\frac{g_m}{2 \pi C_L} \$</span>), and makes it hard to achieve a large phase margin. In my case, the phase margin is only 13 if I connect nothing as a load, but increases to 80 if I add a 1pF load.</p>

How to test the stability of an OTA in this case?

2024-03-30T18:26:46.310

707976

|current|safety|maximum-ratings|

<p>Don't do this.</p> <p>The contacts could weld together, defeating the emergency stop functionality. The switch would probably overheat somewhere (remember that doubling the current quadruples resistive heating), which could at worst cause a fire or could cause the switch to fail in a less-spectacular way. I imagine liability issues in the result of an accident would fall squarely on your shoulders. Use the switch to control a relay; they're cheap.</p>

<p>what would be the effects of an emergency stop button rated significantly lower than the system's current draw?</p> <p>Say for example, an e-stop rated for 10A DC in a system pulling 20A DC. Would the e-stop still function as designed to?</p>

Undercurrent E-stop

2024-03-30T18:46:10.857

707983

|lithium-ion|current-limiting|bms|

<blockquote> <p>it still pumps ~14 A,</p> </blockquote> <p>No it doesn't. A battery doesn't &quot;pump&quot; current. The <em>load</em> sets the current (at a given voltage). If you want the load to draw less current, you must change the voltage supplied to it. The BMS cannot change the voltage: it's either on or off. Instead you must add a device between the battery (and its BMS) and the load, that reduces the voltage to limit the current. That device is a &quot;motor controller&quot;. For a permanent magnet DC motor, a motor controller is often referred to as an ESC (Electronic Speed Control).</p> <p><img src="https://i.stack.imgur.com/T8xiC.png" alt="schematic" /></p> <p>^{<a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fT8xiC.png">simulate this circuit</a> &ndash; Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a>}</p>

<p>Beginner in electronics looking for advice. I got a 6S BMS board from AliExpress rated at 25 A. I need to reduce current to 5-6 A. The board goes in a mobile vacuum cleaner, its motor is max. 120 W.</p> <p>I desoldered 4 MOSFETs but it still pumps ~14 A, with is way too much for the motor. What is the best way to reduce the current to 5 A?</p> <p><a href="https://i.stack.imgur.com/vGeBC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/vGeBC.jpg" alt="enter image description here" /></a></p> <p><a href="https://i.stack.imgur.com/OTDmm.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/OTDmm.jpg" alt="enter image description here" /></a></p>

Reduce the current from a Li-ion battery with 6S 18650 cells and BMS to a motor

2024-03-30T20:01:22.903

708002

|binary|truth-table|decimal|combinatorial-logic|

<p>Design of a combinational circuit with 3 inputs, x, y, z, and 3 outputs, A, B, C. When binary input is 0, 1, 2, or 3, the binary output is twice the input. When binary input is 4, 5, 6, or 7, binary output is half the input. How can 5 and 7, the outputs of which are 2.5 and 3.5, respectively, be represented in the truth table? Here is the truth table that I have filled out as much as possible.</p> <p><a href="https://i.stack.imgur.com/Xmqsv.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/Xmqsv.png" alt="enter image description here" /></a></p>

Combinational Circuit: How can 2.5 and 3.5 be represented in a 3-bit truth table?

2024-03-31T03:39:29.613

708005

|transistors|relay|opto-isolator|

<p>The optoisolator is not doing much as shown since you've tied the grounds together. You could just drive the transistor base directly with something like a 240Ω resistor.</p> <p>You should also have a diode across the relay coil. Something like a 1N4148 is fine.</p> <p>If you have an galvanically isolated supply just for the relays then the isolator makes more sense.</p> <hr /> <p>Your edited schematic looks fine. Let's do some calculations of the resistor values using the Vishay <a href="https://www.vishay.com/docs/83725/4n25.pdf" rel="nofollow noreferrer">4N25</a> datasheet:</p> <p>You have 220Ω and let's guess that the output will be typically about 3.1V when high and the LED voltage about 1.1V, so the LED current will be nominally about 9mA. The 4N25 has 20% minimum CTR so we can get as little as 1.8mA from the phototransistor, but typically 4.5mA (which will be limited by the base resistor).</p> <p>A 5V 100mA relay will be more like a 20mA 24V relay. We would like base current of about 1/10 to 1/20 of that, or about 1-2mA, so the base drive should be just adequate</p> <p>If you're doing a design that will operate over a wide temperature range and that should have a very long life you might want to use an optocoupler with a higher CTR and from a top-tier vendor to minimize LED aging such as a Sharp or Toshiba PC817B or C.</p> <p>Base resistor could be something like 3.9kΩ or 4.7kΩ.</p> <p>Vishay does not specify the maximum high temperature dark current, but at 25°C it's 50nA maximum (5nA typical), leading to a maximum collector current of maybe 10uA and typically perhaps 1uA. That will double about every 10°C so if you anticipate high temperature operation then an additional base-emitter resistor of perhaps 5-10kΩ will help deal with that leakage. If the top end of your ambient temperature range is higher than about 50-60°C you might want to make additional adjustments. In particular, using a better optoisolator (higher minimum CTR) will allow more margin.</p>

<p><a href="https://i.stack.imgur.com/iLp42.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/iLp42.png" alt="enter image description here" /></a></p> <p>Was look for a review to make sure I'm not doing something stupid. Looking at common relay modules on Amazon, it seems that something like this is going on. Would this work to control a relay from a 3.3V logic pin a la raspberry pi? I picked 4N25 and 2N2222 as components since I have them on hand.</p> <p>The opto-coupler can't drive the relay directly because it would be too much current on the collector right? Assuming the relay wants something like 60-100 mA of current.</p> <p>Here is an updated schematic using a different power supply for the relay. Not sure if the 1k base resistor is right though. <a href="https://i.stack.imgur.com/5uWd0.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/5uWd0.png" alt="enter image description here" /></a></p>

Controlling a relay with a raspberry pi pin 3.3 V

2024-03-31T04:19:25.723

708011

|dac|buffer|stm32f103|

<p>This is the simplest BJT solution I can think of, that won't sound horrible:</p> <p><img src="https://i.stack.imgur.com/T3RWK.png" alt="schematic" /></p> <p>^{<a href="/plugins/schematics?image=http%3a%2f%2fi.stack.imgur.com%2fT3RWK.png">simulate this circuit</a> &ndash; Schematic created using <a href="https://www.circuitlab.com/" rel="nofollow">CircuitLab</a>}</p> <p>It won't sound great either, and it's not efficient at all, but you'll get a few tens of milliwatts of power into the speaker.</p>

<p>I'm trying to connect a speaker to an STM32. The purpose is to create sound using the DAC. The speaker I'm talking about is 0.5 W and 8 Ω. Tell me if I'm right: based on <a href="https://www.calculator.net/ohms-law-calculator.html?v=3.3&amp;vunit=volt&amp;c=&amp;cunit=ampere&amp;r=&amp;runit=ohm&amp;p=0.5&amp;punit=watt&amp;x=Calculate" rel="nofollow noreferrer">this</a> the current will be 0.15 A while the max. current through an I/O pin of the STM32 (sink/source) is 25 mA.</p> <p><a href="https://i.stack.imgur.com/glSkz.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/glSkz.png" alt="enter image description here" /></a></p> <p>So, it would better to buffer it. Am I right? Main questions:</p> <ol> <li>What about a BJT buffer, e.g. something like <a href="https://www.changpuak.ch/electronics/BJT_Buffer_Amplifier.php" rel="nofollow noreferrer">this</a> or <a href="https://guitarscience.net/calcs/cc.htm" rel="nofollow noreferrer">this</a>?</li> <li>Should I put in a resistor of around 13 Ω in series and a diode in parallel with the speaker?</li> </ol> <p><strong>Edit 1</strong>: As it's mentioned in the <a href="https://www.st.com/resource/en/datasheet/stm32f103ze.pdf" rel="nofollow noreferrer">datasheet of STM32F103</a>, there is a buffer for DAC but I'm not sure if it could tolerate the current consumption of my speaker.</p> <p><a href="https://i.stack.imgur.com/gQdMV.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/gQdMV.png" alt="enter image description here" /></a></p>

What kind of BJT buffer should I use to connect a speaker to an STM32?

2024-03-31T06:12:12.510

708027

|switches|identification|surface-mount|rotary|

<p>From a google image search for &quot;rotary quadrature encoder&quot; found an in-stock Panasonic <a href="https://www.mouser.co.uk/ProductDetail/Panasonic/EVQ-VVD00203B?qs=tfZGHB2PWd3U%2F4kR6m1kxw%3D%3D" rel="nofollow noreferrer">EVQ-VVD00203B</a> on a distributor website:</p> <p><a href="https://i.stack.imgur.com/GC8SQ.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/GC8SQ.png" alt="enter image description here" /></a></p> <p>The <a href="https://www.mouser.co.uk/datasheet/2/315/ATC0000CE22-531565.pdf" rel="nofollow noreferrer">datasheet</a> has the following picture which looks to match the <em>shape</em> shown in the pictures in the question:</p> <p><a href="https://i.stack.imgur.com/X8Axb.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/X8Axb.png" alt="enter image description here" /></a></p> <p>And as @thebusybee suggested in a comment this is a quadrature encoder, based upon the following from the datasheet:</p> <p><a href="https://i.stack.imgur.com/KWRr4.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/KWRr4.png" alt="enter image description here" /></a></p> <p>If the encoder on the PCB could be measured, that would help to see if a match for the above Panasonic encoder. If not, could serve as an example of a range to check for a match.</p>

<p>I've tried all kinds of rotary encoders and rotary switches, but none seem to match this one exactly.</p> <p>It is from an old car radio (many radios have them). When rotated, the current flows through one of the three bottom &quot;pins&quot;. Inside there are three equally spaced &quot;forks&quot;. Can you guys help me identify it so I can get a new one?</p> <p>Maybe it is an SPDT rotary switch, but I haven't been able to find any similar components online.</p> <p><a href="https://i.stack.imgur.com/w2TON.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/w2TON.jpg" alt="enter image description here" /></a></p> <p><a href="https://i.stack.imgur.com/6dgD0.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/6dgD0.jpg" alt="enter image description here" /></a></p>

Need help identifying a rotary switch component

2024-03-31T09:15:20.123

708040

|pcb|production|gerber|stack-up|

<p>I have the following Gerber files from the manufacturer for order confirmation. As I inspected them, I have inferred that the order of the copper layers from top to bottom are <strong>tl</strong>, <strong>l3</strong>, <strong>l2</strong> and <strong>bl</strong>. But when I check the vias <strong>l2</strong> seems to be the ground layer. So the result Signal(Top) <strong>tl</strong>-PWR <strong>l3</strong>-GND <strong>l2</strong>-Signal(Bottom) <strong>bl</strong> seems to be in conflict with my PCB stack-up, which is Signal(Top)-GND-PWR-Signal(Bottom). Which sequence is correct?</p> <pre><code>bl bo bs drl gbp gbr gtp ko l2 l3 npth sk tl to ts vcut </code></pre>

Gerber files sequence

2024-03-31T11:46:25.707

708044

|simulation|system-verilog|testbench|

<p>The problem is in your <code>tt</code> module.</p> <p>You declared a handle to the <code>p4</code> object here, as expected:</p> <pre><code> eth_pkt p1,p2,p3,p4; </code></pre> <p>However, you tried to access the handle before you allocated memory to it:</p> <pre><code> p4.unpack(bitQ); </code></pre> <p>You need to construct the object using the <code>new</code> method:</p> <pre><code> p4 = new(); p4.unpack(bitQ); </code></pre> <p>This fixes the error.</p>

<p>I am practising the SystemVerilog class concepts. I got an error while running the code:</p> <pre><code>Fatal Error: RUNTIME_0029 code.sv (67): Null pointer access. </code></pre> <p>Here is the code:</p> <pre><code>typedef enum bit[1:0]{SMALL=2'b 00, MEDIUM=2'b 01, LARGE=2'b 11} pkt_type; class eth_pkt; //CLASS PROPERTIES protected bit[55:0] preamble; protected rand bit[47:0] sa; protected rand bit[7:0] len; protected rand byte data[$]; protected static int count; rand pkt_type ptype; //CLASS CONSTRAINTS constraint data_l{ data.size()==len;//data queue gets populated foreach(data[i]) data[i] inside{8'h 1F, 8'h 5F}; } constraint sa_value{ sa inside{[100:150]}; } constraint pkt_category{ len inside{[1:10]}; (ptype==SMALL)-&gt;(len inside{[1:3]}); (ptype==MEDIUM)-&gt;(len inside{[4:6]}); (ptype==LARGE)-&gt;(len inside{[7:10]}); } //CLASS METHODS function new(); preamble={28{2'b10}}; count++; endfunction function void print( string name=&quot;eth_pkt&quot;); $display(&quot;ptype=%s&quot;,ptype); $display(&quot;printing the fields of %s&quot;,name); $display(&quot;preamble=%0h&quot;,preamble); $display(&quot;sa=%0h&quot;,sa); $display(&quot;len=%0h&quot;,len); foreach(data[i]) $display(&quot;data[%0d]=%0h&quot;,i,data[i]); $display(&quot;count=%h&quot;,count); endfunction function void copy( output eth_pkt cpkt); cpkt=new(); cpkt.preamble=preamble;cpkt.sa=sa;cpkt.len=len; cpkt.data=data;cpkt.count=count; cpkt.ptype=ptype; endfunction function bit compare(eth_pkt cmp_pkt); if(cmp_pkt.preamble==preamble &amp;&amp; cmp_pkt.sa==sa &amp;&amp; cmp_pkt.len==len &amp;&amp; cmp_pkt.data==data &amp;&amp; cmp_pkt.count==count &amp;&amp; cmp_pkt.ptype==ptype) begin $display(&quot;pkts matched&quot;); return 1;end else begin $display(&quot;pkts don't matched&quot;); return 0;end endfunction function void pack(output byte qout[$]); qout={&gt;&gt; byte{preamble,sa,len,data,count}}; // pkt_type can't be packed as it's size isn't a multiple of byte endfunction function void unpack(input byte qin[$]); preamble={qin[0],qin[1],qin[2],qin[3],qin[4],qin[5],qin[6]}; sa={qin[7],qin[8],qin[9],qin[10],qin[11],qin[12]}; len={qin[13]}; for(int i=0;i&lt;len;i++) data[i]=qin[14+i]; //qin doesn't have pkt_type information as it couldn't be packed. we can get pkt_type with the help of length info which qin have if(len inside{[1:20]}) ptype=SMALL; if(len inside{[21:100]}) ptype=MEDIUM; if(len inside{[101:255]}) ptype=LARGE; endfunction endclass module tt(); eth_pkt p1,p2,p3,p4; byte bitQ[$]; initial begin p1=new(); assert(p1.randomize); p1.print(&quot;p1&quot;); p1.copy(p2); assert(p2.compare(p1)); p2.print(&quot;p2&quot;); p3=new(); assert(p3.randomize); p3.print(&quot;p3&quot;); p3.pack(bitQ); p4.unpack(bitQ); p3.print(&quot;p3&quot;); assert( p2.compare(p1)); end endmodule </code></pre> <p>Here is the code link <a href="https://www.edaplayground.com/x/9ZuG" rel="nofollow noreferrer">https://www.edaplayground.com/x/9ZuG</a> Someone please tell what mistake have I made.</p>

Null pointer access error in SystemVerilog

2024-03-31T13:11:17.847

708068

|capacitor|dc|

<blockquote> <p>I wonder what would be the graph of current be?</p> </blockquote> <p>Well, you say it's a &quot;pure capacitor connected with a pure DC source&quot; hence, the current will be infinite amps for an infinitesimally short period. Then the current is zero for evermore.</p> <p>That sounds similar to what you mentioned in your question: <a href="https://en.wikipedia.org/wiki/Dirac_delta_function" rel="noreferrer">the Dirac delta function</a></p>

<p>Here is the pure capacitor connected with a pure DC source. The moment switch is closed all the voltage of the source will appear at the capacitor plates causing the current to flow. I wonder what would be the graph of current be? Some says it's exponential others says it's a direc delta function. Can anyone please explain that to me that how current or graph of current will look like in this case?</p> <p><a href="https://i.stack.imgur.com/1Mb04.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/1Mb04.jpg" alt="enter image description here" /></a></p>

Capacitor connected with a DC source

2024-03-31T17:15:13.537

708070

|led|led-driver|

<p>Filtering the TI <a href="https://www.ti.com/power-management/led-drivers/products.html#480=1&amp;1129=current%20mode&amp;sort=1129;desc&amp;" rel="nofollow noreferrer">LED drivers</a> parametric selection for single channel and <em>current mode</em> devices to look for possible devices which don't use PWM.</p> <p>On possible candidate is the <a href="https://www.ti.com/product/LM36011" rel="nofollow noreferrer">LM36011 Inductorless LED flash driver</a>. The <a href="https://www.ti.com/lit/ds/symlink/lm36011.pdf" rel="nofollow noreferrer">datasheet</a> shows that it contains a <em>Torch Mode</em> described as:</p> <blockquote> <p>In torch mode, the LED current source provides 128 target current levels from 2.4 mA to 376 mA, set by the LED Torch Brightness Register (0x04 bits [6:0]). Torch mode is activated by the Enable Register (0x01), setting mode M1, M0 (bits [1:0]) to 10. Once the TORCH sequence is activated, the LED current source ramps up to the programmed torch current by stepping through all current steps until the programmed current is reached.</p> </blockquote> <p>Section <em>8.2.2.1 Thermal Performance</em> of the datasheet suggests this is a linear regulator, and therefore doesn't use PWM:</p> <blockquote> <p>The device power dissipation equals:</p> <p>P_DISS = ( V_IN - V_LED ) * I_LED</p> </blockquote> <p>While the LM36011 LED current is set by I2C, rather than an analog voltage from a DAC, it might be suitable for a LED with the current controlled by a microcontroller. Even if not directly suitable, indicates what sort of LED driver to look for.</p>

<p>I want to make a very smooth adjustment of the brightness of a single 350 mA 2.85 V LED without PWM. And I need to get low brightness values. I do not know how to control the LED properly. Constant current and a adjustable voltage such as &quot;DC dimming&quot; may be suitable. I also found an option on one transistor and want to connect a DAC to it. How to properly control the brightness of the LED without PWM so that it works longer? <a href="https://i.stack.imgur.com/rYKc6.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/rYKc6.png" alt="enter image description here" /></a></p> <p>3.1+ kHz pwm for flicker free <a href="https://i.stack.imgur.com/h0MyK.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h0MyK.png" alt="enter image description here" /></a> <a href="https://i.stack.imgur.com/h8w9m.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/h8w9m.png" alt="enter image description here" /></a></p>

How to make smooth LED brightness control without PWM?

2024-03-31T17:22:35.930

708088

|switches|ltspice|simulation|

<p>I just don't have your problems:</p> <p><a href="https://i.stack.imgur.com/SSAP1.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/SSAP1.png" alt="enter image description here" /></a></p> <p>But since you provide no additional circuit details and have chosen to show only a little part, it's difficult to say much more.</p> <p>The above is the <em><strong>complete</strong></em> circuit. And you can readily see that it behaves as expected. It shows Vhigh and Vlow. Yours does not. So all I have are your assurances. Not visible evidence. Similarly for your switch model. I have what you say. But I do not see what is actually being seen by LTspice. My example shows everything you need to see.</p> <p>Provide fix this situation by showing everything pertinent to the question and perhaps something useful comes of it.</p> <p>My netlist is as follows:</p> <pre><code>S1 B A P 0 MS1 V1 N001 0 PULSE(0 5 20u 1n 1n 5u 17u) V2 A 0 PULSE(0 18 0 50n 50n 15u 40u) AC 1 R1 B 0 1k A1 N001 0 0 0 0 0 P 0 BUF Vhigh=5 Vlow=0 .model MS1 SW(Ron=.1 Roff=100Meg Vt=2.75) .tran 0 300u 0 1n </code></pre>

<p>There is a circuit to control a pwm signal with a switch in Ltspice. But it does not switch of despite to off mode of controller voltage. Is there anyone to give me some suggestion. My and gate specifications -&gt; Vhigh=5 Vlow=0 My switch specifications -&gt; .model SW SW(Ron=1 Roff=1Meg Vt=2.75)</p> <p><a href="https://i.stack.imgur.com/WD9tj.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/WD9tj.png" alt="enter image description here" /></a></p>

I could not manage to use Switch Component in LTspice

2024-03-31T19:45:14.440

708094

|cpu|computer-architecture|mips|risc|

<p>Let say, just for talking purposes, that <strong>$1</strong>=150, <strong>$2</strong>=300, <strong>$3</strong>=540, and <strong>$4</strong>=1000. (Decimal values.) And let's say that the addresses for these three instructions are at locations 0, 1, and 2. And that there are NOPs after that.</p> <pre><code>CC1 Present address 0 to the instruction memory. &quot;ADD $1,$1,$2&quot; presented to IF/ID. CC2 Present address 1 to the instruction memory. &quot;ADD $1,$1,$3&quot; presented to IF/ID. Latch IF/ID, presenting $1 and $2 to REG file. REG outputs 150 and 300 to ID/EX. CC3 Present address 2 to the instruction memory. &quot;ADD $1,$1,$4&quot; presented to IF/ID. Latch IF/ID, presenting $1 and $3 to REG file. REG outputs 150 and 540 to ID/EX. Latch ID/EX, capturing prior 150 and 300. Because there is no forwarding active, captured ID/EX values of 150 and 300 are presented to ALU. ALU summed output of 450 is presented to EX/MEM as an address and 2nd mux to ALU output of 300 is presented to EX/MEM as data. CC4 Present address 3 to the instruction memory. &quot;NOP&quot; presented to IF/ID. Latch IF/ID, presenting $1 and $4 to REG file. REG outputs 150 and 1000 to ID/EX. Latch ID/EX, capturing prior 150 and 540. Latch EX/MEM, capturing address 450 and data 300. Because forwarding is active now, captured EX/MEM address value of 450 is selected by first input mux to the ALU while the captured ID/EX value of 540 is selected by the second input mux. ALU summed output of 990 is presented to EX/MEM. $1 and 450 presented to MEM/WB by EX/MEM. CC5 Present address 4 to the instruction memory. &quot;NOP&quot; presented to IF/ID. Latch IF/ID, presenting $0 and $0 to REG file. REG outputs 0 and 0 to ID/EX. Latch ID/EX, capturing prior 150 and 1000. Latch EX/MEM, capturing address 990 and data 540. Latch MEM/WB, capturing register $1 and address 450 (used as value.) Because forwarding is active now, captured EX/MEM value of 990 is selected by one of the input muxes to the ALU while the captured ID/EX value of 1000 is selected by the other input mux. ALU summed output of 1990 is presented to EX/MEM. REG file is presented prior register $1 and value 450 for writing. $1 and 990 presented to MEM/WB by EX/MEM. CC6 Present address 5 to the instruction memory. &quot;NOP&quot; presented to IF/ID. Latch IF/ID, presenting $0 and $0 to REG file. REG outputs 0 and 0 to ID/EX. Latch ID/EX, capturing prior 0 and 0. Latch EX/MEM, capturing address 1990 and data 1000. Latch MEM/WB, capturing register $1 and address 990 (used as value.) Because there is no forwarding active, captured ID/EX values of 0 and 0 are presented to ALU. ALU summed output of 0 is presented to EX/MEM. REG file is presented prior register $1 and value 990 for writing. $1 and 1990 is presented to MEM/WB by EX/MEM. CC7 Present address 6 to the instruction memory. &quot;NOP&quot; presented to IF/ID. Latch IF/ID, presenting $0 and $0 to REG file. REG outputs 0 and 0 to ID/EX. Latch ID/EX, capturing prior 0 and 0. Latch EX/MEM, capturing address 0 and data 0. Latch MEM/WB, capturing register $1 and address 1990 (used as value.) Because there is no forwarding active, captured ID/EX values of 0 and 0 are presented to ALU. ALU summed output of 0 is presented to EX/MEM. REG file is presented prior register $1 and value 1990 for writing. </code></pre> <p>At the beginning of CC8 the <strong>$1</strong> register will have had its final value saved into it. (Presented at the end of CC7 and latched into the REG file somewhere around the start of CC8.)</p> <p>Just read through the above and match up the details with your pictures. In CC4, where forwarding starts occurring, the EX/MEM value of 450 is pulled backwards via the forwarding units control signals to the input muxes that feed the ALU, so that the EX/MEM address value becomes one of the ALU inputs. That's why the red line shows as it does in your first diagram.</p> <p>I think it is just a matter of you being able to read the intend of the person drawing that diagram. When you get stuck on something like this, just work through the steps like I did, above. It becomes a little clearer when you get out a piece of paper and just write out the details.</p>

<p>I was reading Computer Architecture and Organization 5th edition by Patterson and Hennessy. In Chapter 4, section 4.7 on Data Hazards, I read the following excerpt regarding forwarding from the MEM Stage:</p> <blockquote> <p>As mentioned above, there is no hazard in the WB stage, because we assume that the register file supplies the correct result if the instruction in the ID stage reads the same register written by the instruction in the WB stage. Such a register file performs another form of forwarding, but it occurs within the register file. One complication is potential data hazards between the result of the instruction in the WB stage, the result of the instruction in the MEM stage, and the source operand of the instruction in the ALU stage. For example, when summing a vector of numbers in a single register, a sequence of instructions will all read and write to the same register:</p> </blockquote> <pre><code>add $1,$1,$2 add $1,$1,$3 add $1,$1,$4 </code></pre> <blockquote> <p>In this case, the result is forwarded from the MEM stage because the result in the MEM stage is the more recent result.</p> </blockquote> <p>That confused me, bringing up the following questions:</p> <ul> <li><p>When they say &quot;forwarded from [x] Stage&quot;, does it mean from the [x]/... Pipeline Register or the .../[x] Pipeline Register? In this case, do they mean that the result is forwarded from the EX/MEM stage or the MEM/WB stage?</p> </li> <li><p>If it were to be from the MEM/EX stage, wouldn't that mean we would be going &quot;backwards in time&quot; as in the picture below? <a href="https://i.stack.imgur.com/B2k5A.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/B2k5A.png" alt="enter image description here" /></a></p> </li> <li><p>Since we are going &quot;backwards in time&quot;, shouldn't the forwarding be like this below? <a href="https://i.stack.imgur.com/sSn4t.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/sSn4t.png" alt="enter image description here" /></a></p> </li> <li><p>This confusion may stem from the fact that I do not know which pipeline registers belong to which stage. That is, does every pipeline register that is between a stage belong to that stage or just the one from the right or the one from the left?</p> </li> </ul>

Why forward from MEM Stage in a sequence of add instructions that all contain the same register?

2024-03-31T20:36:03.153

708102

|capacitor|ac|inductor|

<p>The mechanical equivalent is an undamped spring and mass. Its resonant frequency is still defined by mass and the spring constant and, they have equivalent reactances just like any AC circuit. Their mechanical reactances do not depend on any external force just as electrical reactance does not depend on the externally applied voltage.</p>

<p>In a pure LC circuit that is left to oscillate at its natural frequency without any AC source, would there be any <span class="math-container">\$X_L\$</span> or <span class="math-container">\$X_C\$</span>? I don't think there is any reactance in such a circuit, because there isn't any AC source driving the the circuit. However, sometimes I think to myself, isn't <span class="math-container">\$X_L=\omega L\$</span> then why can't I say <span class="math-container">\$X_L=\omega_0L\$</span> in the considered case?</p>

Is there any capacitive and inductive reactances in an LC circuit without an AC source?

2024-03-31T23:07:57.580

708103

|connector|camera|ribbon-cable|

<p>That is a <a href="https://connectorbook.com/identification1.html?n=ffc_fpc_back_flip_sockets&amp;f=jI" rel="nofollow noreferrer">bilevel back-flip* FFC socket</a> (my site).</p> <p>(*) Funny name, ain't it?</p> <p>I count 51 circuits.</p> <p>You can find them here: <a href="https://www.digikey.com/short/vbq78jw5" rel="nofollow noreferrer">https://www.digikey.com/short/vbq78jw5</a></p> <ul> <li>MANUF. - SERIES</li> <li><a href="https://www.molex.com/molex/products/family/easyon_ffc_fpc_connectors" rel="nofollow noreferrer">Molex</a> - <a href="https://www.molex.com/pdm_docs/sd/5025982393_sd.pdf" rel="nofollow noreferrer">Easy-on 502598 504740 504754</a> (pdf) - <a href="https://www.digikey.com/short/b0230h37" rel="nofollow noreferrer">digikey</a></li> <li><a href="https://www.molex.com/molex/products/family/easyon_ffc_fpc_connectors" rel="nofollow noreferrer">Molex</a> - <a href="https://www.molex.com/pdm_docs/sd/5040703991_sd.pdf" rel="nofollow noreferrer">Easy-on 504070</a> (pdf) - <a href="https://www.digikey.com/short/nrfjfwdh" rel="nofollow noreferrer">digikey</a></li> <li>Hirose Electric - <a href="https://media.digikey.com/pdf/Data%20Sheets/Hirose%20PDFs/FH35_Series.pdf" rel="nofollow noreferrer">FH35</a> (pdf) - <a href="https://www.digikey.com/short/r387zb8f" rel="nofollow noreferrer">digikey</a></li> <li><a href="https://www.hirose.com/en/product/series/FH35C" rel="nofollow noreferrer">Hirose Electric</a> - <a href="https://www.hirose.com/en/product/document?clcode=&amp;productname=&amp;series=FH35C&amp;documenttype=Catalog&amp;lang=en&amp;documentid=D49358_en" rel="nofollow noreferrer">FH35C</a> (pdf) - <a href="https://www.digikey.com/short/9vf40hv0" rel="nofollow noreferrer">digikey</a></li> <li>Hirose Electric - <a href="https://www.mouser.com/datasheet/2/185/hiros09581_1-2269477.pdf" rel="nofollow noreferrer">FH39</a> (pdf) - <a href="https://www.digikey.com/short/z5w2v9m5" rel="nofollow noreferrer">digikey</a></li> <li>Hirose Electric - <a href="https://www.mouser.com/datasheet/2/185/hiros09581_1-2269477.pdf" rel="nofollow noreferrer">FH39J</a> (pdf) - <a href="https://www.digikey.com/short/5708fp7r" rel="nofollow noreferrer">digikey</a></li> <li>Hirose Electric - <a href="https://www.hirose.com/product/document?clcode=CL0580-2302-0-10&amp;productname=FH42-15S-0.3SHW(10)&amp;series=FH42&amp;documenttype=Catalog&amp;lang=en&amp;documentid=D49361_en" rel="nofollow noreferrer">FH42</a> (pdf) - <a href="https://www.digikey.com/short/db9dvzjj" rel="nofollow noreferrer">digikey</a></li> <li><a href="https://www.hirose.com/en/product/series/FH58" rel="nofollow noreferrer">Hirose Electric</a> - <a href="https://www.hirose.com/en/product/document?clcode=&amp;productname=&amp;series=FH58&amp;documenttype=Catalog&amp;lang=en&amp;documentid=en_FH58_FH58M_CAT" rel="nofollow noreferrer">FH58 (0.2 mm)</a> (pdf) - <a href="https://www.digikey.com/short/pjvt3p2d" rel="nofollow noreferrer">digikey</a></li> <li><a href="https://www.hirose.com/en/product/series/FH64MA" rel="nofollow noreferrer">Hirose Electric</a> - <a href="https://www.hirose.com/en/product/document?clcode=&amp;productname=&amp;series=FH64MA&amp;documenttype=Catalog&amp;lang=en&amp;documentid=D162160_en" rel="nofollow noreferrer">FH64MA</a> (pdf) - <a href="https://www.digikey.com/short/m3bjw53n" rel="nofollow noreferrer">digikey</a></li> <li><a href="https://www.jae.com/en/connectors/series/detail/id=64327&amp;type_code=T1030" rel="nofollow noreferrer">JAE</a> - <a href="https://www.mouser.com/pdfdocs/JAEFR02FPCDatasheet.PDF" rel="nofollow noreferrer">FR02</a> (pdf) - <a href="https://www.digikey.com/short/99tvq027" rel="nofollow noreferrer">digikey</a></li> <li>Omron - <a href="https://www.mouser.com/datasheet/2/307/XF2B_1110-37066.pdf" rel="nofollow noreferrer">XF2B</a> (pdf) - <a href="https://www.digikey.com/short/375r7cbz" rel="nofollow noreferrer">digikey</a></li> <li>Omron - <a href="https://omronfs.omron.com/en_US/ecb/products/pdf/en-xf3a.pdf" rel="nofollow noreferrer">XF3A</a> (pdf) - <a href="https://www.digikey.com/short/741prwq1" rel="nofollow noreferrer">digikey</a></li> <li>Omron - <a href="https://omronfs.omron.com/en_US/ecb/products/pdf/en-xf3b.pdf" rel="nofollow noreferrer">XF3B</a> (pdf) - <a href="https://www.digikey.com/short/7nm5qcnm" rel="nofollow noreferrer">digikey</a></li> <li>Omron - <a href="https://omronfs.omron.com/en_US/ecb/products/pdf/en-xf3c.pdf" rel="nofollow noreferrer">XF3C</a> (pdf) - <a href="https://www.digikey.com/short/thqdqdm" rel="nofollow noreferrer">digikey</a></li> <li>Omron - <a href="https://omronfs.omron.com/en_US/ecb/products/pdf/en-xf3z.pdf" rel="nofollow noreferrer">XF3Z</a> (pdf) - <a href="https://www.digikey.com/short/02mw0440" rel="nofollow noreferrer">digikey</a></li> <li>Panasonic - <a href="https://www.mouser.com/datasheet/2/315/con_eng_y2b-1518755.pdf" rel="nofollow noreferrer">Y2B</a> (pdf) - <a href="https://www.digikey.com/short/5chjp4d7" rel="nofollow noreferrer">digikey</a></li> <li><a href="https://na.industrial.panasonic.com/products/connectors/fpc-ffc-connectors/lineup/fpc-ffc/series/70457" rel="nofollow noreferrer">Panasonic</a> - <a href="https://www3.panasonic.biz/ac/cdn/e/control/connector/fpc-ffc/catalog/con_eng_y3b.pdf" rel="nofollow noreferrer">Y3B</a> (pdf) - <a href="https://www.digikey.com/short/c9b1rq3b" rel="nofollow noreferrer">digikey</a></li> <li>Panasonic - <a href="https://www.mouser.com/datasheet/2/315/con_eng_y3bl-1397630.pdf" rel="nofollow noreferrer">Y3BL</a> (pdf) - <a href="https://www.digikey.com/short/91v4v7pv" rel="nofollow noreferrer">digikey</a></li> </ul> <p>To narrow it down, you will need to <a href="https://forum.digikey.com/t/pitch-of-a-connector/172" rel="nofollow noreferrer">measure the exact pitch</a>. If you are unsure how to do that, measure the width of the cable (not at the tip with the notches) and we can do the rest. Also, whether the contacts are at the top or the bottom. And, you'll need to find one that locks onto the tabs on either side of the FPC (the &quot;cable&quot;).</p> <blockquote> <p>Before I buy something like an adapter</p> </blockquote> <p>Be aware that that might not work. An adapter (if you even find one, which is doubtful) adds capacitance, picks up noise, and causes reflections. The product may very well stop working.</p> <blockquote> <p>What ZIF type connector</p> </blockquote> <p>Know that ZIP Is not a type of connector. It is a feature that many different connectors have. For example, <a href="https://www.sparkfun.com/products/retired/9175" rel="nofollow noreferrer">this ZIF DIP socket</a> which doesn't look anything like an FFC socket. So, calling your socket a &quot;ZIF connector&quot; is not quite right.</p>

<p>I am modifying a Canon Powershot G7X camera so that I can couple the sensor to a tube lens and microscope objective. To do this conveniently I would like to extend the ribbon cable connected to the rear touch screen panel.</p> <p>The connector for this looks like a so-called “zero insertion force” connector however I am not too familiar with these. Before buying an extender cable and pcb, can anyone help me identify the connector? <a href="https://i.stack.imgur.com/7tF2e.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/7tF2e.jpg" alt="enter image description here" /></a></p> <p>Circled you can see the connector and ribbon cable. I have also taken two close-up shots. <a href="https://i.stack.imgur.com/BhqIS.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/BhqIS.jpg" alt="enter image description here" /></a> <a href="https://i.stack.imgur.com/j0fvC.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/j0fvC.jpg" alt="enter image description here" /></a></p> <p>To be honest I am not sure I am even counting the number of pins correctly. I thought it would be 26 because of the exposed pins near the connector, but looking at the ribbon cable there seem to be more than that, possibly with some sort of keying.</p> <p>Before I buy something like an adapter below I had better know what I am looking for first. <a href="https://i.stack.imgur.com/AsS34.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AsS34.jpg" alt="enter image description here" /></a></p>

What ZIF type connector is shown in this Canon Powershot G7X Camera?

2024-03-31T23:12:03.533