Signal changes when second probe is attached to op-amp - Page 1
Forum section: electronics / beginners

author: TNb
author_rank: Regular Contributor
timestamp: March 22, 2017, 05:23:20 am
content: I have very basic circuit on a breadboard, it is op-amp buffer that takes signal from frequency generator and buffers it, no gain whatsoever.But for some reason when I attach only 1 probe to the output - I have the spikes on the edges of square wave and it is not just spikes that you see on uncompensated probe, no, probes are compensated and I doublechecked it. The spikes go all the way to rails, i.e. to 0 and 3.3 V.The interesting thing is that when I attach the second probe spikes go away and everything is bang on perfect, even though the only thing I changed is hooked another probe.I am well within specs, tested under 10kHz on 10MHz op amp (MCP6024).Here is when I have only output connected:And here I also connect input of op amp to second channel on the scope:I tried:1) To switch probes around - same behavior2) To put 1M resistor to ground on the input of op amp(thinking that probe impedance makes a difference).3) I've put decoupling cap on power pins(forgot) - no change.Have no other ideas actually what is wrong.Maybe somebody have an idea?Also additional information - with sine wave amplitude jumps around when sweeping through frequency and sawtooth waveform is a little round.All signals on input look perfectly normal and likewise - if I connect second scope probe everything goes fine.

author: TNb
author_rank: Regular Contributor
timestamp: March 22, 2017, 05:37:56 am
content: I guess I should draw a circuit what I have before op amp because this may be the problem, though I am quite sure it is OK, tested it for hour on various freq/amplitude without a problem.A short explanation: input signal (from bench generator) is attenuated 10 times and biased to 1.65 V, then it goes to op amp. Diodes are fast switching for extra op amp input protection, capacitors are variable caps to compensate for parasitic capacitance in diodes(~20pF). Joints are only where they are drawn to be, sorry for sloppy quick drawing.

author: forrestc
author_rank: Unknown
timestamp: March 22, 2017, 05:48:17 am
content: Quote from: TNb on March 22, 2017, 05:23:20 amWhere are you putting the probes? That's a bit unclear.Remember there's some caps in those probes as well...Without knowing exactly your grounding situation, one thing which does come to mind is that if your scope is isolated from everything else (i.e. the sig gen and power supply signals are ungrounded) is that you effectively have a signal path between the two probes. If these are 10X probes, then it's going to be equivalent to a 20M resistor... that is, a path from one probe, through the 10x and scope impedance back out the other probe's impedance and resistance into the other probe. Not sure if that would be enough.The other thought is that this is a probe grounding issue. Where are the ground probes attached to the circuit? I've had all sorts of oddities related to the ground leads - back when I was less experienced I often would only connect the ground lead from one probe, and then wonder why my signals were not as clean as they should be.IT might be helpful to post a couple of pictures with the probes in place.

author: TNb
author_rank: Regular Contributor
timestamp: March 22, 2017, 06:02:04 am
content: well, red trace on scope is from red circle on the circuit and yellow trace is from yellow circleI am sure that both scope and frequency generator are grounded, i.e. not isolated. Power supply is not grounded, but I have never had any issues with that.

author: rstofer
author_rank: Super Contributor
timestamp: March 22, 2017, 06:02:18 am
content: That first image is of a differentiator circuit. Somewhere you have series capacitance. Are you using DC coupling? I'm not sure that AC coupling would be a problem but it does have a series capacitor.http://www.electronics-tutorials.ws/opamp/opamp_7.html

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: March 22, 2017, 07:12:52 am
content: A complete schematic, not just the input, could be usefull, as well as showing your breadboard layout.I'm also not quite sure where the input in your schematic is connected. Is it connected to the parallel R1 and C2?If I interpret the schematic correctly, this is the AC equivalent circuit:In that case, the schematic you show has a pole and a zero.If fzerois much lower than fpole, then can to first order approximate the behavior well past the zero but not close to the pole with this schematic (ie, the C2is now a short):In other words, you are not going to attenuate the high frequencies.Even though you might be testing with a 10kHz signal, that is a square wave, so it a lot of high frequency content at those transitions. So while your circuit may attenuate the low frequency signals by a factor of 10, it will not attenuate the high frequency signals. The amplifier now tries to follow this.Why does it change when you connect the probe? My first guess would be that you adding the probe introduces a enought stray capacitance to bring the pole close enough to your zero that it compensates enough. if you look closely, it would seem that you still have a good amount of overshoot in your signal.What amount of capacitance do you have on the output?There could also be some issues with reflections of the signals but I don't think that will be such an issue at <10MHz frequencies.

author: nugglix
author_rank: Regular Contributor
timestamp: March 22, 2017, 07:47:45 am
content: Sorry, but as far as I can trace the OPs diagram R1 and C2 are in series and in parallel to C1 and R1.There is no connection from C2 upper side to signal.I can't see anything that would make R1 & C2 be in parallel.This looks really weird.Suggestion to the OP:1. Scrap it.2. Draw a complete and proper schematic.3. Build result from 2. on breadboard and see again.PS: Dear OP, next time use the screenshot function of the scope.

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: March 22, 2017, 07:54:32 am
content: Quote from: nugglix on March 22, 2017, 07:47:45 amI went of off this statement:Quote from: TNb on March 22, 2017, 05:37:56 amand there is clearly a connection from C2 to R1, and C2 is not connected to the 3.3V line, according to his schematic. It would also be the only point where I can understand an input coming from, esp. when you include his statement of a factor 10 attenuation. Otherwise, they are just connected on one end to the input of the opamp, and the other end is connected but just floating around. I do agree that the OP needs a better schematic, because this seems confusing (and has no input labeled which doesn't exactly help).

author: nugglix
author_rank: Regular Contributor
timestamp: March 22, 2017, 08:12:12 am
content: This is really hard to read.In your diagram R2 and R3 are in parallel, which is not given in the OPs schematic -- I think.They're in series from 3.3V to GND.V1 -> R3 -> R2 -> GND^|AA -> C2 -> R1 -> A ==> no clue what that should be.Edit:I give up, this is more guessing than anything else.

author: Zero999
author_rank: Super Contributor
timestamp: March 22, 2017, 09:27:49 am
content: Quote from: nugglix on March 22, 2017, 08:12:12 amIt's a potential divider with biasing, based on a circuit I posted in another thread, with the values scaled to the nearest E24 values.https://www.eevblog.com/forum/beginners/proper-biasing-to-virtual-ground-with-tle2426/msg1164997/#msg1164997Quote from: TNb on March 22, 2017, 05:23:20 amWhat were the probe settings? x1, x10?A lot of op-amps behave erratically when driving a capacitive load so the probe capacitance might be a factor here.What's the maximum frequency of interest (bear in mind that if the signal isn't a sinewave the bandwidth needs to be much higher due to the harmonic content)?

author: TheUnnamedNewbie
author_rank: Super Contributor
timestamp: March 22, 2017, 09:46:09 am
content: Quote from: nugglix on March 22, 2017, 08:12:12 amThey are in series for DC, but in AC analysis you use superposition. The DC source is then shorted to ground, and thus thd resistors are in parallel.I assumed the input would be on the node where r1 and c2 are connected together, since that would make sense in the OP's description. (10x attenuation and dc bias)

author: Zero999
author_rank: Super Contributor
timestamp: March 22, 2017, 12:31:52 pm
content: Quote from: TNb on March 22, 2017, 05:37:56 am20pF? Is that for each diode, if so, it's huge. If it's for both diodes added together, then it's not so bad. What diodes are you using?I'd recommend a Schottky diode, such as the BAT54, because it has a forward voltage of only 300mV which will limit the input voltage below the maximum specified on the op-amp data sheet. It has a capacitance of 10pF.http://assets.nexperia.com/documents/data-sheet/BAT54_SER.pdfThe capacitance of both of the diodes and the op-amp's common mode input are effectively in parallel at AC. One of the capacitors (C1 in your schematic) is making matters worse, rather than better, because it adds to the parasitic capacitance of the diodes and op-amp.Only one capacitor is required across R3 and it should be1/9of the value of the parasitic capacitance of the op-amp + diodes, forming a 10:1 capacitive divider, in parallel with the resistive divider.If the op-amp has a common mode capacitance of 6pF and the capacitance of both diodes added together is 20pF, the total parasitic capacitance is 26pF, so the compensation capacitor, in parallel with R1 should be 26pF/9pF = 2.889pF, with 3pF being the nearest standard value.http://ww1.microchip.com/downloads/en/DeviceDoc/21685d.pdfYou could have also chosen better resistor values, see attached.What bandwidth do you need?With the component values shown above, the bandwidth, without the compensation capacitor will be 51kHz. Adding the 3pF compensation capacitor, across R3, will increase the upper cut-off frequency to above the op-amp's gain bandwidth product.FC= 1/(2pi*RC)WhereR = R1|R2|R3 = (R1-1+R2-1+R3-1)-1C = parasitic capacitance = 26pF

author: TNb
author_rank: Regular Contributor
timestamp: March 22, 2017, 06:19:56 pm
content: Sorry for my terrible schematics, didn't have much time in the morning.Here is better one:Quote from: Hero999 on March 22, 2017, 12:31:52 pmYes, I done research and chosen BAT41 for that, quite similar to BAT54, so I doubt I have problem with diodes.20pF is the capacitance of variable caps, so I adjust them to the value that yields good square wave without overshoots, prior to op-amp I can get very clean square wave even at 1.5MHz.Quote from: Hero999 on March 22, 2017, 12:31:52 pmI would like to do at least 1 MHz, since op amp has 10 MHz bandwidth product, so in theory I should be able to get 1 MHz easily (and I do, but only with second probe connected, haha).Hope new schematic drawing will ring a bell for someone, thanks for replies.I just came back home and gonna try to figure out myself too.

author: Zero999
author_rank: Super Contributor
timestamp: March 22, 2017, 07:31:38 pm
content: Quote from: TNb on March 22, 2017, 06:19:56 pmOh, I can see the need for the additional capacitor across the diode now. It's there because the diode's capacitance isn't known and can't be predicted using the data sheet. I still think two trimmer capacitors is overkill though. You could probably make C1 3.3pF or 4.7pF fixed, then you only need C2 to be adjustable.QuoteNote the following points:1) You actually have well in excess of 1.5MHz there, as it's a square wave with plenty of harmonics. Not a problem since your op-amp is good for 10MHz.2) Check the op-amp's slew rate specification isn't limiting the rise/fall times. Going by your images, I don't think it's a problem but it's always good to check.3) Your oscilloscope probe will add capacitance to the divider. This is why I think you see overshoot. You connect the 'scope to the op-amp's input, adjust the divider to see a perfect square wave. Now, when you move the probe to the output, the capacitance on the the input has been reduced below the optimal value, so it overshoots. This will also explain why the output looks good with a probe on both the input and output. You need to compensate your circuit without the 'scope probe connected to the divider. Connect the 'scope to the op-amp's output and nowhere else.

author: bson
author_rank: Unknown
timestamp: March 22, 2017, 08:01:34 pm
content: Another problem here is you tweak capacitances on the order of pF on a breadboard. The breadboard itself is on the same order.

author: StillTrying
author_rank: Super Contributor
timestamp: March 22, 2017, 11:39:12 pm
content: Are the scope probes on X10.It could easily be just the op amp not liking the probe's capacitance on it's output. Putting the other probe on it's input could be slowing it down just enough for it to cope with the probe on it's output.

author: Zero999
author_rank: Super Contributor
timestamp: March 23, 2017, 12:02:20 am
content: Quote from: StillTrying on March 22, 2017, 11:39:12 pmMy first reaction was the op-amp not liking the probe capacitance but it shouldn't be enough to cause any trouble with a decent op-amp. I'd also missed the fact that he had one probe on the input and the other on the output.It's certainly, the probe's capacitance adding to that of the the compensation capacitor. I bet if he just connected one probe to the output and trimmed for a good square eave, then added the probe to the input, the overshooting edges would reverse and become rounded.

author: StillTrying
author_rank: Super Contributor
timestamp: March 23, 2017, 01:57:19 am
content: I've read every post and I still don't know what the OP is trying to do.

author: Audioguru
author_rank: Super Contributor
timestamp: March 23, 2017, 03:05:49 am
content: The datasheet of the opamp shows a perfect squarewave when the output has a 60pF load, a proper supply bypassing discussion and a discussion about how to prevent "frequency peaking".A solderless breadboard with its stray capacitance all over the place? Guaranteed trouble. Use a proper pcb instead.

author: Zero999
author_rank: Super Contributor
timestamp: March 23, 2017, 09:28:26 am
content: Quote from: StillTrying on March 23, 2017, 01:57:19 amHe wants to scale -16.5V to 16.5V to 0V to 3.3V to measure with an ADC in a micro controller.Quote from: Audioguru on March 23, 2017, 03:05:49 amWhat makes you believe that's the cause of the problem here? I think you saw the words "solderless breadboard" and jumped to conclusions.The circuit works perfectly fine when the probe is on the op-amp's output.Another problem might be that the probe will be slightly resistive, creating a high pass filter at a much higher frequency.

author: Audioguru
author_rank: Super Contributor
timestamp: March 23, 2017, 04:32:36 pm
content: Like I said, the datasheet shows a perfect squarewave when the output has a 60pF capacitance to ground. The second probe and random solderless breadboard capacitance created 60pF for itIf you bump a wire on a solderless breadboard then it will be different and might even oscillate. Of course I mentioned the problems with solderless breadboards because many threads have problems with it.

author: Zero999
author_rank: Super Contributor
timestamp: March 23, 2017, 06:39:09 pm
content: Quote from: Audioguru on March 23, 2017, 04:32:36 pmNo. The circuit worked perfectly when the probe was connected to the op-amp's output. It only started to misbehave, when an additional probe was connected to the op-amp's input. He didn't try connecting two probes to the output because it would be pointless, since they'd both read the same signal.The 60pF given on the datasheet is the worst case scenario. I wouldn't be surprised if the op-amp can drive more than 60pF before it oscillates, although wouldn't deliberately connect it to over 60pF.Also look at the waveform on the 'scope again. There's no sign of any oscillation. There's some overshoot but no sign of any ringing, which you'd expect if the op-amp were oscillating. It's very similar to the waveform shown in the simulation I posted above.

author: StillTrying
author_rank: Super Contributor
timestamp: March 23, 2017, 07:09:38 pm
content: Quote from: Hero999 on March 23, 2017, 06:39:09 pmI took it the opposite way, a 2nd probe on the +ve input was slowing it's transition enough for itto workwith the 1st probe still on the output. But who knows!Looking closely at the spike's decay/recovery I think the op amp might be internally limiting/saturating, at 3.3V there's very little supply volts left for driving <100ns edges with a scope probe there.And there's this...4.7 Supply BypassWith this family of operational amplifiers, the power supply pin (VDD for single supply) should have a localbypass capacitor (i.e., 0.01 ?F to 0.1 ?F) within 2 mm for good, high-frequency performance. It also needs a bulk capacitor (i.e., 1 ?F or larger) within 100 mmOn a breadboard a MCP6024 needs the 0.1uF to be straddling the middle of the IC to get close to both of it's supply pins.Are the scope probes on X10? LOL