hybridfree/eevblog-posts huggingface dataset:
Infrared Recievers - Page 1
Forum section: electronics / beginners

author: skillz21
author_rank: Frequent Contributor
timestamp: October 16, 2016, 01:40:25 am
content: Hi guys,A while back, i got a pack of IR transmitter LEDs and receiver LEDs (is that even what their called?). I know i can just use the... ones with a chip built in, but i want to know whatthesethings are used for.Thanks!PS.Sorry for the informality in my posts

author: Ammar
author_rank: Regular Contributor
timestamp: October 16, 2016, 03:30:15 am
content: How many pins do the receiver ones have?If it has two pins it is likely a phototransistor or photodiode.The ones with three pins contain a photodiode as well as a bunch of other magic (filters, amplifiers, demodulators) to ensure that the device receives signals that are modulated at a certain frequency, typically 10s of kHz. They will have a supply pin, a ground pin and a signal pin.

author: skillz21
author_rank: Frequent Contributor
timestamp: October 16, 2016, 03:56:46 am
content: two pins, clickHEREfor image.

author: Ammar
author_rank: Regular Contributor
timestamp: October 16, 2016, 04:01:07 am
content: I am guessing the dark one is the infrared LED although it could be the other way around.The clear one is likely a phototransistor. The longer lead is probably the emitter and the shorter one the collector. You can test it out by putting a 10k pull up resistor on the collector to a supply voltage, say 5 to 10V. Shine the infrared LED on the phototransistor and measure the collector voltage with a multimeter. Watch it drop to around 0.7V when the light is shining on it.You can also check to see if an infrared LED is actually glowing with a CMOS camera. Although filters have gotten better these days, so maybe not. My camera phone still shows glowing infrared LEDs as purple light.
INFRARED SENSOR - Page 1
Forum section: electronics / beginners

author: Abayomi
author_rank: Contributor
timestamp: August 22, 2016, 05:55:50 pm
content: I was building a remote circuit which IR sensor requires 5v but mistakenly I connected a 12v. I don't know if the sensor got spoilt so I decided to get a new IR sensor out of my old CD player . Unfortunately , I don't know the frequency of the new IR sensor be cause no number was written on it but the old IR sensor frequency was 38kHz because MYS1838 was written on it so I downloaded the data sheet.

author: Bukurat
author_rank: Regular Contributor
timestamp: August 23, 2016, 11:29:40 pm
content: Is there a question in there somewhere?

author: Abayomi
author_rank: Contributor
timestamp: August 24, 2016, 12:39:52 am
content: How do I know the frequency of the infrared sensor

author: Ammar
author_rank: Regular Contributor
timestamp: August 24, 2016, 01:17:49 am
content: It is probably 38KHz too. It is the standard. Good luck not frying the next one!

author: Alex Eisenhut
author_rank: Super Contributor
timestamp: August 24, 2016, 01:22:25 am
content: Quote from: Ammar on August 24, 2016, 01:17:49 amIt's "a" standard. There's other ones. My stupid split AC has a remote that uses some bloody thing my universal remote doesn't see.https://en.wikipedia.org/wiki/Consumer_IR"30 kHz, 33 kHz, 36 kHz, 38 kHz, 40 kHz, and 56 kHz."

author: Ammar
author_rank: Regular Contributor
timestamp: August 24, 2016, 02:01:21 am
content: Didn't know about the others. Good to know, thank you.

author: NexusKoolaid
author_rank: Contributor
timestamp: August 24, 2016, 01:57:47 pm
content: Quote from: Alex Eisenhut on August 24, 2016, 01:22:25 amSome of those units use some peculiar protocols. The GREE protocol I had to work with a few years back has an unusually long leading (non-repeating) sequence that's large enough to fool most universal remotes. If that wasn't bad enough, the sequence conveyed three different types of command information.

author: Luchik
author_rank: Newbie
timestamp: August 25, 2016, 09:18:15 am
content: Most IR Receivers I used (or found in my home appliances) were 38KHz. Some were 36KHz. conveniently enough, the 38KHz will usually work with 36KHz and 40KHz transmitters or LEDs, since the BPF is usually very wide as you can see in the following optic-preamplifier graph from Vishay:It is actually quite easy to test: you need LED connected to a simple uC I/O or PWM. Send a pulse of about 10mS (modulated) and check the component output.I believe that you have another problem: the new sensor may be intended for a different wavelength (example: 750nm IR), while the your old one (MYS1838) was 950nm, according to the datasheet.

author: bitslice
author_rank: Frequent Contributor
timestamp: August 25, 2016, 09:38:55 am
content: If you know the make of the CD player then you should be able to track down what frequency it uses.A lot have been documented in order to make universal remotes.
infrared sensors / detectors - Page 1
Forum section: electronics / beginners

author: ezzo
author_rank: Newbie
timestamp: February 13, 2016, 12:21:46 pm
content: Hi, I need help by someone who is very knowledgeable with infrared sensors / detectors etc. I have bought a sleep mask that can detect when a person is in REM sleep (dreaming) The sleep mask has a small circuit board inside with an infrared detector that detects eye movementthroughout the night.When the infrared detector picks up a lot of eyeball movement it then knows the person is dreaming and will flash 2 bright LED lights at the closed eyes allowing the dreamer to know when they are dreaming. The device is acurate and does what it says 'on the tin' Anyway, my issues are as follows. I am concerned that all night i have a electronic sensor constantly monitoring my eye for movement.The reason this is a concern is that it could damage the eye / long term problems etc. Im no expert but having infrared detector / device 1mm from the eye all night could be dangerous?This is a passage from their website ........... "The REM-Dreamer uses infrared sensors to detect when you are in REM (dreaming) sleep. At that point the REM-Dreamer gives you sound and light cues (beeps and flashing lights) to remind you that you are dreaming. Thus, external world stimuli are transferred to the world of your dreams, and it becomes easy to achieve lucidity"Please take a look at their website to get full details on this device, thanks youRemDreamer description -http://remdreamer.com/index.phpRem dreamer photos -http://remdreamer.com/rem-dreamer_pro.phpThank you Terry

author: Rager
author_rank: Contributor
timestamp: March 02, 2016, 03:43:36 pm
content: Hello Terry,I don't think that you need to worry about that. just think about the ammounts of infrared light your eye is receiving every day. This device runs on two coin cell batteries and to have a more or less long lifetime it will only pulse the IR LED for a really short time. The only thing I would worry about is the price ^^ That is quite a lot of money, for such a device

author: michaeliv
author_rank: Frequent Contributor
timestamp: March 02, 2016, 06:34:02 pm
content: Quote from: Rager on March 02, 2016, 03:43:36 pmIt's a small price to pay for "extending" your life by 1/3. Now you won't waste any time just laying around doing nothing in your dreams. You can continue that design on the power supply you've been working on or laying out that 8-layer analog board.Curious though ... why can't it just send out beeps and flashes anytime you're sleeping, why would it have to wait for REM ?
Infrared vs Hot Air for Rework Stations - Page 1
Forum section: electronics / beginners

author: requim
author_rank: Regular Contributor
timestamp: January 03, 2012, 07:01:48 pm
content: It seems as if the higher end rework stations all use infrared instead of hot air. What are the advantages/disadvantages of using infrared instead of hot air? I assume one advantage is that infrared can be focused more tightly. Anything else?

author: Unknown
author_rank: Unknown
timestamp: January 03, 2012, 09:22:38 pm
content: Not blowing small parts away appears to be a convenient advantage.

author: xygor
author_rank: Regular Contributor
timestamp: January 03, 2012, 10:17:59 pm
content: Infrared doesn't work too well on shiny things.It's more difficult to get accurate temperature with infrared.

author: requim
author_rank: Regular Contributor
timestamp: January 04, 2012, 12:30:35 am
content: Quote from: alm on January 03, 2012, 09:22:38 pmI would agree, however I've never had any problems with parts blowing away or shifting while using a heat gun.
Inguard/Outguard Power Supplies, What Are They? - Page 1
Forum section: electronics / beginners

author: jfriend
author_rank: Contributor
timestamp: June 15, 2023, 04:22:59 pm
content: The subject pretty much describes my question. I see Inguard and Outguard on schematics and even marked on some PCB boards. But I have yet to find an explanation what that means and what "creates" it. Can someone explain what this is all about?

author: TimFox
author_rank: Super Contributor
timestamp: June 15, 2023, 04:48:29 pm
content: Are these "guard" terminals, in the usual sense of a terminal at the same potential as a "delicate" terminal, allowing a guard shield around that terminal?https://protecequip.com/blog/what-is-the-purpose-of-a-guard-terminal/https://en.wikipedia.org/wiki/Driven_guardhttps://www.tek.com/en/support/faqs/what-difference-between-guarding-and-shielding"Guard" and "Ground" are often confused, but are quite different.

author: jfriend
author_rank: Contributor
timestamp: June 15, 2023, 08:10:18 pm
content: No, it's not guard. I know what that is. I've attached a couple photos of a power supply schematic and one from the index page of an HP 3458A repair manual. But it doesn't say what actually defines what "inguard" and "outguard" means, only what parts of the unit are "in" and "out". I've seen "Inguard" mentioned here on eevblog several times, but still nothing technical that describes how it is created.

author: MathWizard
author_rank: Super Contributor
timestamp: June 17, 2023, 04:57:10 pm
content: Do they mean the fuses and protections ? And maybe the jack is sturdy ?
injection transformer "termination" - Page 1
Forum section: electronics / beginners

author: exe
author_rank: Unknown
timestamp: February 11, 2020, 05:44:56 pm
content: Hi!I want to reproduce this experiment: . I have two questions about this schematic:  (I attach copy of the image for the sake of convenience).1) What's the role of R4? Does it need to match R3?2) Why R3 is 10 ohms and not something else? 10 ohms looks quite low to me. I usually see 50 ohm, is it to match siggen impedance?PS on a second thought, may be R4 and R5 are a resistor divider, but still I'm not sure it is needed because transformer makes R3 "visible" on the primary winding. So, R3 and R4 can be replaced with a single R3 with 5 ohm resistance.References:https://wiki.analog.com/university/courses/electronics/electronics-lab-loop-gainhttps://wiki.analog.com/_detail/university/courses/electronics/amlg_f3.png

author: fcb
author_rank: Super Contributor
timestamp: February 11, 2020, 06:20:52 pm
content: This has the same ground on both sides? Can you show how this is inserted into the FB loop of your DUT/system? Think there is some context missing that will help.Reason you might have the resistors (R3/R4) either side is to reduce the effect (and phase shift) that the transformer introduces to the local FB. Try and imagine what it would look like without them. Same reason that the 100ohm (R5) is many times bigger than R4, it reduces the effect that load variation will have on the operation of the circuit.Ignore that - just had a look at your linked sources. The OPAMP is the DUT.You could reduce the 100R if your signal generator has the drive; 100R will have a protective function to your signal generator.R3 is the part you are injecting across, it is low so that you don't affect normal operation of the circuit - Dr. Ray Riddley has an excellent set of videos that show this.
Inline Resistor for LC Filter Power Measurements - Page 1
Forum section: electronics / beginners

author: wardenclyffe
author_rank: Contributor
timestamp: February 13, 2013, 11:55:12 pm
content: Hey guys,I stumbled across this article when I was trying to figure out how to measure component power -->http://www.anandtech.com/show/6529/busting-the-x86-power-myth-indepth-clover-trail-power-analysis.Basically, they seem to suggest that you can measure the power consumption of different components by inserting a small value resistor inline with their input LC filters. My guess is that you would have some sort of switching regulator providing a voltage to an output LC filter that would smooth it into DC for the component in question. They say that by measuring the voltage drop across the resistor, you can figure out how much current is flowing into the component and thus how much power it is consuming.I tried this method myself, but both my multimeter and oscilloscope seemed to say the voltage differential was 0 across the resistor (resistor value was 5mohm). I thought about it and I was wondering if the series inductor is going to make it such that the RMS voltage is always 0 across the resistor. I tried switching the coupling on the oscilloscope but that didn't see to make any difference. What am I doing wrong? Let me know if you have any thoughts.Thanks!!
Input Attenuator - Page 1
Forum section: electronics / beginners

author: timj
author_rank: Newbie
timestamp: March 06, 2019, 11:03:39 am
content: Hello, I'm a beginner in electronics. I've been tasked with a practical that I've attached on here. To my understanding this is just a simple voltage divider. As it states, I need to modify the circuit such that the output is half the input AC signal. The audio generator that I have has an output of 600 ohm. Would I just include that in the voltage divider rule V_out = (R2/R1+R2) x V_in? And then change either R1 or R2 until I get the half the input signal?
Input Bias Current Errors In An Audio Amp - Page 1
Forum section: electronics / beginners

author: eev_carl
author_rank: Regular Contributor
timestamp: July 16, 2018, 01:40:02 pm
content: Hi,I built an inverting amp and was experimenting with a resistor from the positive input versus a wire to GND. For audio applications, is this resistor important? The op amp I'm using, LME49720NA, has an IB of 10nA.I was reading this article which uses an extra resistor in Figure 3:http://www.analog.com/media/en/training-seminars/tutorials/MT-038.pdf.Thanks,Carl

author: Wimberleytech
author_rank: Super Contributor
timestamp: July 16, 2018, 01:57:24 pm
content: Quote from: eev_carl on July 16, 2018, 01:40:02 pmIB can be as high as 72nA according to the spec. Without extra resistor in the + leg, this could result in 0.72mV offset (for a 10k resistor).Datasheet also indicates input offset current to be as high as 65nA, so for 10k resistor this could add 0.65mV offset voltage.Without the extra resistor, you might see 0.72mV input-referred offset.With the extra resistor, you might see 0.65mV input-referred offset.Appears to be a wash. I don't think you need it.

author: David Hess
author_rank: Super Contributor
timestamp: July 16, 2018, 07:27:10 pm
content: Ideally the resistor is not needed for audio. It adds noise and the amplifier can be configured for AC coupling so DC offsets are removed or minimized.

author: floobydust
author_rank: Super Contributor
timestamp: July 17, 2018, 04:31:26 pm
content: Ib causes big problems in audio use, as the output coupling capacitor's polarity has to be decided. Does the electrolytic cap (+) or (-) go to the op-amp's output?So sometimes a little intentional DC offset is desirable, from omitting the extra resistor.I've seen in Neve consoles the NE5534's age and output offset drift and change polarity, which aggravated output coupling capacitor failures.
Input Capacitor on cmoy headphone amp - Page 1
Forum section: electronics / beginners

author: lawfreak
author_rank: Newbie
timestamp: March 28, 2016, 02:15:40 am
content: Questions have been asked here before regarding this circuit.http://tangentsoft.net/audio/cmoy/misc/cmoy-tangent-sch.pdfThe 0.1 uF input capacitor seems to be far too high an impedance in the audio range to be an appropriate choice. Simply testing the input pre/post cap suggests significant attenuation of any audio input. I hope there is a reason for the choice of 0.1uf given the thoroughness of the site, but it seems counterproductive to attenuate the entire audio signal before amplification. To caveat this, I had trouble trying to observe any significant difference pre/post cap using a scope.Two questions now... Capacitors in the 10uF range seem better suited to the task, but are there other issues to consider? What drove the original design decision of 0.1uF capacitors?

author: daveshah
author_rank: Unknown
timestamp: March 28, 2016, 07:40:18 am
content: The difference will only be noticeable at low frequencies: in combination with the 100k input resistor a RC low-pass filter with a cutoff frequency around 16Hz is formed. This is unlikely to make much difference listening to actual music with actual headphones.I suspect 0.1uF was chosen because the original creator wanted to avoid electrolytics and 0.1uF is one of the largest readily available non-electrolytic and non-MLCC capacitor sizes.

author: ciccio
author_rank: Frequent Contributor
timestamp: March 28, 2016, 11:41:22 am
content: You can buy film capacitors in small sizes and low voltage up to 2.2 uF, but there is no need to increase the suggested value.On the other side, the power supply section is too simple... you can design a better virtual ground circuit.Best regards

author: danadak
author_rank: Super Contributor
timestamp: March 28, 2016, 11:57:47 am
content: Use a charge pump for the split rail solution.http://www.ti.com/lsds/ti/power-management/buck-boost-inverting-or-split-rail-products.pageOr do design as a single supply design.Regards, Dana.
Input Common-Mode Voltage Range - Page 1
Forum section: electronics / beginners

author: IanBerry
author_rank: Contributor
timestamp: April 14, 2017, 02:14:35 am
content: I've been playing with this op-amp lately because I want to make a super high-frequency SSTC driver (Not much reason other than to see what would happen). The issue is that the op-amp has been doing some weird things. Lots of distortion of a square wave, sometimes making it almost sinusoidal. I started to get frustrated because I was nowhere near the maximum frequency of the op-amp. I looked at the datasheet again and again until I found a little part that said:Input Common-Mode Voltage Range VCM MIN VEE - 0.2 MAX VCC - 2.25 VI was connecting the inputs to rail voltages. Do those numbers mean that the input voltage has to be 2.25 volts below the supply voltage or above 0.2 volts below the ground? I'll link the datasheet below.https://datasheets.maximintegrated.com/en/ds/MAX4012-MAX4020.pdf

author: digsys
author_rank: Unknown
timestamp: April 14, 2017, 02:33:30 am
content: Absolutely correct. Even with RRI (rail-rail-Inputs) and RRO (rail-rail-Output), you have to be careful. Sometimes they are rated with no load / ideal conditons.It may only be a few 10s-100s uV, can it is a trap. Instrument op-amps can go way above / below Supply.

author: Ian.M
author_rank: Super Contributor
timestamp: April 14, 2017, 02:43:06 am
content: Quote from: IanBerry on April 14, 2017, 02:14:35 amBoth. AND not OR. If either input goes below Vee-0.2V or above Vcc-2.25V bad things happen. The lower limit is right down to the ground (or negative) rail, but the upper limit is a PITA as it means you've got to prevent the inputs going closer to the positive rail than that, and you should design for a little more margin, say Vcc-2.5V, for safety.

author: Rerouter
author_rank: Super Contributor
timestamp: April 15, 2017, 07:00:56 am
content: On the flip side, you can find op amps with common mode input ranges far outside there voltage supply, which make things like bidirectional current monitoring much easier, but they are harder to find.
input current regulators - Page 1
Forum section: electronics / beginners

author: christos
author_rank: Regular Contributor
timestamp: November 03, 2014, 03:46:32 pm
content: first ,,this is a question for my motorcycle cuz the output wires from the coils that generates power are getting hot after changing the regulator. the coils are in good conditionif a battery needs more current from a regulator and the regulator can supply it,,does the input current from the regulator increase?or am i wrong?
INPUT DC Link Capacitor of Flyback Converter - Page 1
Forum section: electronics / beginners

author: khatus
author_rank: Regular Contributor
timestamp: August 20, 2018, 09:07:29 am
content: Hello guys , currently I am searching for documents to design a smps using flyback converter. I have found a document which uses the following formula to calculate the INPUT DC Link Capacitor of Flyback Converter. Here is the picture of their designAnd here is the formula they usedMy question is from where they derive the formula??

author: T3sl4co1l
author_rank: Super Contributor
timestamp: August 20, 2018, 08:25:43 pm
content: The diagram basically gives it away... do some geometry and prove it.What would you need, what are you missing, to do the proof? Inputs? Methods? How much can you infer from what you know about the circuit's operation, and the laws of circuits and components?I don't know offhand, but suspect that, the equation is approximate -- conduction angle factors into it, as does the discharge curve (which in general, is not going to be a linear ramp as in the diagram). In general, I think the true solutions will be transcendental -- that is, involving a relation of the form x = sin(x), which cannot be solved, only approximated. So they chose to assume conduction angle (letting you make your own error on that parameter), and the rest follows from a basic power balance and a constant current assumption. That's my guess.Tim
Input EMI filter design for two parallel DC-DC converters - Page 1
Forum section: electronics / beginners

author: Phil Smith
author_rank: Regular Contributor
timestamp: September 04, 2023, 02:36:31 pm
content: Hello everybody,I'm in the process of design a power supply for my simpit display. Basically, it is a 24inch Samsung monitor and raspberry pi driving it.Main power supply is +27VDC (nominal) which coming from external AC-DC converter via 5-7m cable. Grounding is done via simulator's metal chassis, therefore there is no negative wire.An EMI filter is being designed using this series of videos:https://www.youtube.com/playlist?list=PLE3Pq-hBtiMS-1gFecr9FcBjf_p4joRZPPlease find attached the block diagram of the supply.I got couple of questions please regarding the design of device's input bulk capacitor size and EMI filter.1) How to calculate C_bulk in this configuration?2) Is it correct to assume, that total input impedance of the power supply Z_in for EMI Filter will be sum of parallel input impedances of each DC-DC converter Z_in = (Z1_in * Z2_in)/(Z1_in + Z2_in)?3) In my case input ground is chassis, so is it correct to connect common mode caps 0.5*C_cm from each line to ground node before common mode choke as shown on my diagram?Any help is much appreciated.Kind regards, Phil.
Input filter capacitor for USB - Page 1
Forum section: electronics / beginners

author: Mp3
author_rank: Frequent Contributor
timestamp: May 07, 2020, 09:54:48 pm
content: My goal is to have the 5v from a USB cable go through an input filter capacitor and diode into an isolated DC-DC converter which will be the source for some cleaner 5v for a separate PCB.My thought was USB 2.0 spec says to put a capacitor with a maximum of 10µF in parallel with the 5v, but that's not the filter, and I don't know how to figure the value for the filter cap? The DC-DC isolation converter has a minimum draw of 25mA and my circuit draws something like 200mA directly off the computer USB.I realized i don't know how to calculate the value of the filter capacitor i want as i don't know the frequency of the input 5VDC. I thought frequency of DC is 0 so i am a little confused...

author: pigrew
author_rank: Frequent Contributor
timestamp: May 07, 2020, 10:44:52 pm
content: Quote from: Mp3 on May 07, 2020, 09:54:48 pmBeware noise from your DC-DC converter... it could be worse than what you're starting with.Also, USB _does_ allow >10uF if it is not in-circuit when the USB cable is first connected. There exist IC like NCP330 that will do the slow startup for you, you can put 4.7uF on the input side, and 22 uF, or more, on the output.However, you won't get great filtering with only a capacitor. The diode won't filter much at all (high capacitance across it). You'll want things like ferrite beads, common-mode chokes, and maybe ESD transient protection. Also a capacitor (class Y?) from input to output of the SMPS (across the isolation) can help in many cases.Knowing the type of noise you're filtering out is very important (what frequency).Please post a schematic of what you propose, for a better analysis.Quote from: Mp3 on May 07, 2020, 09:54:48 pmDC is freq=0, but that's not what you want to filter out. You might be wanting to filter out switching noise from the USB power source (100 to 700 kHz?) or maybe RF interference from nearby radio stations.
Input for a filter - Page 1
Forum section: electronics / beginners

author: prophoss
author_rank: Contributor
timestamp: April 24, 2022, 09:41:31 pm
content: I have been wanting to create an FIR filter using some code like C or python. I have found plenty of code but have been wondering where or how can you get the input for the filter? What would be the simplest way to get some input data to apply that FIR filter to? If this doesn't make sense let me know and I can try and explain again.

author: Manul
author_rank: Super Contributor
timestamp: April 24, 2022, 10:40:42 pm
content: 1. Reading samples from wave (uncompressed) audio files.2. Writing a function to generate array of samples of various waves. Sine wave can be made with, you guessed it, sin(x) function. You can add in harmonics if you want. You can also produce square wave, triangle. Store the samples in an array and feed your filter.

author: prophoss
author_rank: Contributor
timestamp: April 24, 2022, 11:02:25 pm
content: That may well fit the bill. I did not know about python's sin(x). Thank you for that.edit:Was able to create a file with the sine wave using for loop from 0 through 25. That can be adjusted to any amount really. Now I will have something to play with. Now I need tp look into adding the harmonics.Thanks, Manul
Input Impedance of a Basic Common-Emitter Amplifier - Page 1
Forum section: electronics / beginners

author: beeboopbeep
author_rank: Contributor
timestamp: August 31, 2017, 01:46:37 am
content: Hi everyone,I was reviewing analog circuits and was studying common-emitter amplifier from Art of Electronics (3rd edition) but I got confused analyzing a circuit. The book states that when looking at the input impedance of a common-emitter amplifier as shown below, it would be R1 || R2 || (B*R_E) where B is the current gain of the NPN.However, why wouldn't input see the R_C resistance? The "Ohm model" of the NPN transistor is as shown above, so it makes sense to me that at the input node (i.e., base of the NPN), it would see R_C resistance as part of the input resistance. Is it not included because you don't include any impedance at the output node of the circuit?

author: Ratch
author_rank: Regular Contributor
timestamp: August 31, 2017, 02:06:32 am
content: In the active mode of operation shown, the collector-base (CB) is reversed biased. This causes the the CB junction to present a very high resistance to the base input circuit. Since this resistance will not appreciably affect the input impedance, it is ignored.Ratch

author: beeboopbeep
author_rank: Contributor
timestamp: August 31, 2017, 02:47:30 am
content: Quote from: Ratch on August 31, 2017, 02:06:32 amThat makes sense! Thank you for the answer.

author: danadak
author_rank: Super Contributor
timestamp: August 31, 2017, 10:38:40 am
content: Here is a decent treatment of the question -http://www.electronics-tutorials.ws/amplifier/input-impedance-of-an-amplifier.htmlRegards, Dana.

author: b_force
author_rank: Super Contributor
timestamp: August 31, 2017, 12:14:37 pm
content: Maybe a more general answer would be helpful to give.In general, if looking at AC signals, all DC/GND nodes can be seen as the same (shorted).This can be really a trap for people when biasing opamps or other circuits on Vcc/2 (or any other bias voltage for that matter)
Input impedance oscilloscope  - Page 1
Forum section: electronics / beginners

author: alin_im
author_rank: Regular Contributor
timestamp: June 03, 2015, 10:49:07 am
content: Hy guys,I have a have an oscilloscope that has the imput impedance of 1M, do you think i can change the input impedance to 50 ohm if i put a 50 ohm resistor in parallel with the input, i also what to try with 75 ohm and 600 ohm. Do you think i could get some problems with this idea or it is perfectly fine.

author: Dago
author_rank: Frequent Contributor
timestamp: June 03, 2015, 11:24:27 am
content: Quote from: alin_im on June 03, 2015, 10:49:07 amYou can buy a "pass-through terminator" which has an integrated 50 ohm termination. Or a simple T-adapter with a terminator in one leg works fine.Do notice that your oscilloscope probe/cable is not matched to 50 ohm however. So you need to connect your circuit/device to the scope with a regular 50 ohm coax.

author: tggzzz
author_rank: Super Contributor
timestamp: June 03, 2015, 02:30:11 pm
content: Quote from: alin_im on June 03, 2015, 10:49:07 amNo, you don't. You probably have 1Mohm in parallel with 20pF - calculate the impedance of 20pF at, say, 100MHz and you will see the 1Mohm is completely irrelevant.QuoteIt works well, and a few scopes already have a 50ohm termination option built into them.A real advantage of a 50ohm input is that it allows "low" impedance Z0 probes to be used - which have ahigherimpedance than a traditional 1Mohm *10 probe, and have an upper frequency limit that is 10* higher than 1Mohm probes.Seehttps://entertaininghacks.wordpress.com/library-2/scope-probe-reference-material/for more information.
Input impedance selector - explanation needed - Page 1
Forum section: electronics / beginners

author: Tomer
author_rank: Regular Contributor
timestamp: March 03, 2019, 10:23:30 pm
content: Hello All, and thank you for your help.In Pro Audio pre-amps, there is usually an input impedance selector. A good example can be a classic pre-amp and compressor from UA model 6176.https://www.sweetwater.com/store/detail/6176--universal-audio-6176The 6176 offers an impedance range of 50Ohm to 2.2MOhm, and apply the first 3 options to balanced inputs and the last 2 option to an unbalanced inputs.Appreciate if you can explained how the impedance selector is working and what circuitry is behind such selector.Any referral to reading materials will be great.Thank you for your help, helping me to understand the topic.Tom

author: ataradov
author_rank: Super Contributor
timestamp: March 03, 2019, 10:52:32 pm
content: On a basic level it simply connects a resistor with a specified value parallel to the input. Assuming that input amplifier itself has an infinite impedance. If not, then the resistor is calculated so that total impedance matches what's on the label.In real life it may be a bit more complex that this, there may be some frequency compensation circuitry as well.

author: Audioguru
author_rank: Super Contributor
timestamp: March 04, 2019, 01:25:08 am
content: The very old vacuum tubes circuit claims to produce "vintage coloration" instead of high fidelity. The variable input impedance is part of that.An electric guitar can sound muffled or bright by changing the input impedance of its preamp. An old moving magnet phono cartridge is the same.

author: Tomer
author_rank: Regular Contributor
timestamp: March 04, 2019, 09:47:38 pm
content: Thank you ataradov.Can you explain why the connection needs to be in parallel for this case ?In addition, doesn't the input device see the load as a complete unit (the impedance sum, I am lacking a better explanation here) and not only the first resistor ?Thank you,Tom

author: ataradov
author_rank: Super Contributor
timestamp: March 04, 2019, 10:30:09 pm
content: Because that's what impedance is if we only consider the real (resistance) part of it.As I said, a resistor of that value is only valid if impedance of the amplifier following the input is large (close to infinite). This way your total input impedance of the device is mostly represented by the input resistor, which is adjusted by the knob.
Input Impedence - Page 1
Forum section: electronics / beginners

author: jj5
author_rank: Unknown
timestamp: February 24, 2023, 09:45:55 pm
content: The manual for my scope says that its input impedance is 1 MΩ. I hooked channel 1 on my scopeup to my digital multimeterand the measured resistance was 0.7767 MΩ.Does that mean that the other 0.2233 MΩ is reactance?

author: bdunham7
author_rank: Super Contributor
timestamp: February 24, 2023, 10:14:34 pm
content: Hmmm, no, the 1M should be the DC resistance. Can you try changing the vertical scale for CH1 to at least 5V/div?

author: Bud
author_rank: Super Contributor
timestamp: February 24, 2023, 10:18:57 pm
content: One Chinese made gizmo is measured using another Cinese made gizmo. Nothing can go wrong.You should start with checking accuracy of your Owon.

author: jj5
author_rank: Unknown
timestamp: February 24, 2023, 10:23:54 pm
content: Quote from: bdunham7 on February 24, 2023, 10:14:34 pmAh, so the problem seems to be that the BNC to banana adapter that I was using had a 1.5 MΩ resistance across it. I used a different cable (BNC to banana) and the measured resistance is now 1 MΩ as expected. Thanks.

author: shapirus
author_rank: Super Contributor
timestamp: February 24, 2023, 10:33:31 pm
content: Quote from: Bud on February 24, 2023, 10:18:57 pmthese chinese gizmos tend to be surprisingly accurate and rarely the place to start troubleshooting at these days.

author: BillyO
author_rank: Super Contributor
timestamp: February 25, 2023, 12:13:46 am
content: Quote from: shapirus on February 24, 2023, 10:33:31 pmIndeed.I have two trombones here, both from China. One is $200 and is what you'd expect. The other is $6000 and is a full out professional instrument of the finest quality. Seems painting everything from China with the same brush is a little silly.

author: coppercone2
author_rank: Super Contributor
timestamp: February 25, 2023, 12:26:53 am
content: that adapter is shit and you should throw it away. You can try cleaning it but be careful dipping those in iso, I had them warp from even 50% iso dip.I just measured 3, one from 1950 or 60, and 2 from the last 10 years. They all read open circuit.SO it does seem like a chinese gizmo was the problem... at least I feel better for buying pomona from a supplier .

author: jj5
author_rank: Unknown
timestamp: February 25, 2023, 12:30:26 am
content: Quote from: coppercone2 on February 25, 2023, 12:26:53 amYeah, I binned it.

author: jj5
author_rank: Unknown
timestamp: February 25, 2023, 12:49:31 am
content: For those bringing China into it, let's not forgetHow it Works.

author: BeBuLamar
author_rank: Super Contributor
timestamp: February 25, 2023, 08:48:15 pm
content: Quote from: BillyO on February 25, 2023, 12:13:46 amYes I agree with you. Chinese made products ranging from the best to the worst so a Chinese made gizmo doesn't mean anything.
Input protection and power feed - Page 1
Forum section: electronics / beginners

author: rrinker
author_rank: Super Contributor
timestamp: June 26, 2019, 06:13:23 pm
content: There's nothing really magic or critical about my circuit, it is just a micro with some pushbutton inputs, driving a couple of RC servos and relays. It's been a while, but I did post the original circuit for critique quite some time ago, and implemented the suggestions in my schematic. As it stands, it actually works just fine built on a breadboard - WITHOUT any of the attached power protection and bypass caps, other than the ones directly on the leads of the micro. The main difference is, on the breadboard, it is running off my lab power supply. "In the wild" I plan to use a higher voltage power bus, and tap off for each instance of the circuit using one of those small LM2596 modules. I've chosen this approach because the area that will have these circuits (a large model railway) is already too large to use just one master power supply, the voltage drop even with fairly heavy wire would be too much, so rather than try to distribute 5 volts, I feel it would be better to distribute a higher voltage that I can tap off for multiple uses. Plus if I use say 12 volts, and at the furthest distance from a given power supply, the voltage has fallen to 11 volts, I can still get a solid 5.5 or so from the buck converter.The reason I did what I did here - the MOSFET is for reverse protection. Do i need it? Well, I'm the only one using and wiring this circuit. It's to protect myself from being an idiot, mostly - no one else will be using this. I then have the capacitor to smooth out transients caused by the servo motors, which are fairly high current draw (relative to the rest of the circuit - worst case I have gotten is a bit over 1 amp total, with the servo stalled, which will quickly burn it out). So there is the tap off for the servo power lead. The diode is to prevent the surge to the servos from draining the bypass on the micro side and causing a voltage drop.Option 1 - what I originally designed for, was to put the buck converter off board, and since the servos are rated for up to 6 volts, set it a bit above 5 volts to compensate for the diode drop. Option 1A would be to make a footprint and put the buck on the main board - just needs 4 pin headers to mount it - and put the MOSFET in front of it to protect the buck.Option 2 would be to use 2 buck converters, either on or off the board, one supplying the servos and one supplying the rest.An additional thought - should I also use the servo power as the source for the relays? There are 4 relays, transistor driver with a suppression diode on each one.As I said, on a breadboard it works without any of this stuff, but I'd rather follow best design practices where practical. And I want to get it right, this will be the most complex PCB I have done and want to try to make it right. I need to build dozens of this circuit. And I have another that accompanies this, but I want to get one finished before starting the next.
Input protection for a charge amplifier (OpAmp) - Page 1
Forum section: electronics / beginners

author: JohannesS
author_rank: Newbie
timestamp: March 28, 2023, 05:45:18 pm
content: I want to use an OpAmp in a simple charge amplifier configuration to amplify the signal from a piezo element. Im only interested in signals below 1 V (and <1 kHz), but accidentally hitting the piezo might generate tens of volts with a very fast rise-time, so I need to protect the inputs. Can someone give me some feedback on my implementation?- R58 and R59 limit the current- The inputs of the OpAmp should both be at GND (or virtual GND), in case the OpAmp can't keep up D14 and D15 start conducting at 0.7 V and clamp the voltage- R57 and C38 form a lowpass filter with 1 kHz cutoff- D9-D12 clamp the voltage to +/-1.4 V, so the ADC (which runs on +/- 1.8 V) doesn't get damaged

author: Infraviolet
author_rank: Super Contributor
timestamp: March 28, 2023, 07:57:37 pm
content: Zener diode pointing from the ground rail to the op amp's non-grounded input may help, the diode will only reverse conduct if the voltage on the relevant input gets higher than the zener voltage above ground. You can get zeners with whichever reverse voltage you think appropriate. You could already seem to have a regular diode (D13)ready to conduct if the non-grounded op amp input gets taken below 1 diode drop (0.7V ish) below ground. If your desired input is 1V then D14 may be aproblem actually, as it will conduct if the non-grounded op amp input gets >0.7V above ground. A zener pointing the opposite way in the same location, as per the details at start of this post, would let you select a better voltage to start conducting at.Whether the ADC needs protection from the op amp's output will depend on what rails the op-amp is powered from and what voltages the ADC is rated for.

author: JohannesS
author_rank: Newbie
timestamp: March 29, 2023, 05:09:33 pm
content: I think I'll power the OpAmp with the same +/-1.8 V as the ADC to prevent issues, but my thinking was that a high voltage spike with high frequencies would couple through (is that the correct nomenclature?) C39 into the ADCs input, if the protection diodes before the OpAmp don't start conducting fast enough.Is that a warranted concern?Regarding the zener diode, my question was a bit unprecise. I'm fine with a +/- 0.7V input range, it doesn't even need to be 1V...
Input Sensitivity \L25D Amp - Page 1
Forum section: electronics / beginners

author: christos
author_rank: Regular Contributor
timestamp: August 19, 2016, 03:41:27 pm
content: Hello All!!.I just bought 2 mono amps L25D By LJM,it says its 200-250w at 8 ohm load(MONO) but doesn't sound like a 200-250w amp, its like a 50w ampIt says Input Sencitivity 1,6V(using as input a laptop 1v)is that the maximum or the minimum Input?

author: singapol
author_rank: Frequent Contributor
timestamp: August 19, 2016, 08:33:02 pm
content: Quote from: christos on August 19, 2016, 03:41:27 pmNormally it means 1.6V input to get maximum power whatever that means. look at the power transistor specs or datasheet. You have to supply a suitable supply voltage to get the current and power that you want. What does this amp manual say? Eg. if it's 50V then it's unlikely you can come close to 200W.

author: christos
author_rank: Regular Contributor
timestamp: August 19, 2016, 10:13:46 pm
content: it says 70v+- DCmin 60v max 80vI used a +- 55v dchttp://prntscr.com/c7x9v8here's the schematic

author: singapol
author_rank: Frequent Contributor
timestamp: August 20, 2016, 07:16:44 am
content: Quote from: christos on August 19, 2016, 10:13:46 pmSince it says 60V min. but you used 55V. Do check CP1 and C2 voltage rating before you try increasing voltage, you may damage your amp. Also check IRFB4020n datasheet.http://www.infineon.com/dgdl/irfb4020pbf.pdf?fileId=5546d462533600a4015356158ffd1e05
Input voltage below supply voltage but within datasheet input common-mode voltag - Page 1
Forum section: electronics / beginners

author: Bigman
author_rank: Regular Contributor
timestamp: April 01, 2015, 09:28:10 am
content: Hi,attached a screenshot about the Input common-mode voltage range statement from the OPA347 datasheet.(source: Burr-Brown/Texas Instruments, OPA347, November 2000 - revised July 2007).It says: (V-) - 0.2V and (V+) + 0.2VI am a niewbie and based on my theoretical knowledge, this would mean: I can go EVEN BEYOND the supply voltages and the OpAmp-Output would still behave fineFor example: for the attached schematic of a simple inverting amplifier, input signals down to -0.2V are fine? What's the catch?Thanks and cheers

author: ruffy91
author_rank: Regular Contributor
timestamp: April 01, 2015, 09:32:39 am
content: Is the screenshot from the electrical characteristics page or the absolute maximum ratings page? If the latter it just means that the opamp won't self destruct, not that it behaves correctly.

author: Bigman
author_rank: Regular Contributor
timestamp: April 01, 2015, 09:37:06 am
content: The screenshot is from the electrical characteristic page.I made a further screenshot. On the bottom, you see the Voltage common-input mode voltage range.(source: Burr-Brown/Texas Instruments, OPA347, November 2000 - revised July 2007)

author: JohnnyBerg
author_rank: Frequent Contributor
timestamp: April 01, 2015, 10:01:36 am
content: Yes, it will behave fine. It is not uncommon that inputs can go beyond the supply rails. It depends on the construction of the input stage.Compare as a example a LM358 and a TL072:- the LM358 has transistor inputs, and can operate very close to the negative rail, but is bad at the positive rail.- the TL072 had FET inputs, and can operate very close to the positive rail, but is bad at the negative rail.
input voltage detection circuit - Page 1
Forum section: electronics / beginners

author: hsn93
author_rank: Regular Contributor
timestamp: January 30, 2018, 08:19:40 am
content: hello ,im trying to understand voltage detection circuit here:so whats the benefit of this "input voltage detection" circuit ? what does it detect exactly ?as much as i can understand .. the input voltage will close the N-channel fet (Q102) which will short P-channel fet (Q101) to GND which will close it also to deliver (input voltage - voltage drop on Q101) ??

author: helius
author_rank: Super Contributor
timestamp: January 30, 2018, 08:51:23 am
content: It works together with the "power multiplexer" to isolate EXT_P5V0 when power is not being supplied, and if it is, prevents power being drawn from VCC_EDBG or VCC_TARGET. You can think of it as a relay switch that closes only as long as a positive voltage is supplied on its input.

author: hsn93
author_rank: Regular Contributor
timestamp: January 31, 2018, 06:45:06 am
content: thank you ,so to understand, this circuit is there because there is 3 power inputs right ?also, wouldn't the power pass the Q101 because of the diode it has ?or its ok because if there is voltage anyway the mosfet will be on so the diode doesnt matter ..

author: agehall
author_rank: Frequent Contributor
timestamp: January 31, 2018, 06:52:52 am
content: Yes, power will pass thru the diode of Q101 when it is applied, but only for a brief period of time. Once the power is applied, Q101 will be turned on via R103 and all power will flow thru the MOSFET itself.

author: helius
author_rank: Super Contributor
timestamp: January 31, 2018, 07:05:32 am
content: No, the body diodes do not pass any current when the MOSFETS are off.You need to draw a flowchart to see how the circuit operates.Step 1. Is a positive voltage (w.r.t. ground) available on EXT_P5V0?If Yes, then: Q100 turns on; Q102 turns on; Q101 turns on; EXT_P5V0 goes to U101./EN, which will disable U101 if it is a logic high (~>2.5V). Stop.If No, then: Q100 turns off; Q102 turns off; Q101 turns off; R107 pulls U101./EN down to ground, which activates U101. Go to step 2.Step 2. Is VCC_EDBG_USB_P5V0 above the threshold of U101.VSNS?If Yes, then U101.IN1 opens and passes to U101.OUT. Stop.If No, then U101.IN2 opens and passes to U101.OUT. Stop.For extra credit, describe what happens if EXT_P5V0 is between 0 and 2.5V.

author: T3sl4co1l
author_rank: Super Contributor
timestamp: January 31, 2018, 08:05:40 am
content: Quote from: helius on January 31, 2018, 07:05:32 amWhat happens when the left side or right side is in the 0-2V range, indeed!To OP:This looks to be a subcircuit which wasa. Poorly implemented following some intention,b. Poorly intended following its implementation, orc. Poorly documented / commented following either of the above.Or something altogether worse, who knows. In short, it doesn't seem to do what's claimed, and you certainly shouldn't take it at its word.Tim