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I have built an instrumentation amplifier out of two op-amps in a quad op amp (MCP6L04T). The input signal is coming from a low-impedanece source (~17k) and most of the information in that signal is below 10 Hz. Although I can see the outline of the desired signal, it is made of a bunch of peaks - as if the op amp's supply keeps getting cut off every time it reaches its intended output value. Why is it doing this? Both scope caps were taken at the output of the op-amp, before the filter, but they can also be seen after the filter.

voltage spikes outlining desired waveform instead of desired, smooth waveform

a zoomed-out shot of the offending, spiky waveform

The other two amplifiers in the package are being used as follows: one is a non inverting amplifier for the output of this amplifier, and the last is a peak detector for the output of that amplifier.

This is running off a fresh CR2032 battery. A previous design that used a dedicated instrumentation amplifier IC had the same input and output filters and power supply and worked flawlessly.

The op-amp has a 1 uF bypass cap right next to the pin, and a 10 uF cap further away. Both are ceramic. Looking at the supply pin with a scope revealed a fairly smooth supply - it doesn't jitter with these weird spikes.

FYI the paralleled resistors are there to increase the overall tolerance of each pair of resistors. It's giving me an extra ~3 dB of CMRR without the cost of 0.1% resistors.

The circuit: circuit schematic

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  • \$\begingroup\$ What are the input/output signal levels? And what does the signal look like in the first scope image, if you were to only look at a smaller section (zoom in)? \$\endgroup\$
    – KyranF
    Dec 9 '14 at 6:13
  • \$\begingroup\$ Have you tried / can you try supplying power from a benchtop power supply instead of from a coin cell battery? Also, I believe an 'instrumentation amplifier' conventionally requires three op-amps. \$\endgroup\$
    – vicatcu
    Dec 9 '14 at 6:37
  • \$\begingroup\$ What is the intended quiescent output voltage? If the inputs are not sufficiently biased above ground the signal will be clipped, ditto if the output goes too high or too low. \$\endgroup\$ Dec 9 '14 at 6:38
  • \$\begingroup\$ Input: 5uV-1mV Peak AC, 1-10 Hz. Output feeds into a separate circuit that the user uses to attenuate or amplify the output signal. If you zoom in to one of the spikes, you see a series of very sharp pulses, nearly identical and with a consistent frequency of ~20kHz:imgur.com/7JaYqfV and imgur.com/Na2oe3u show detail. It exhibits the same behavior when connected to a bench supply. The quiescent output voltage was intended to be near 0V - small compared to the output signal. I suspect you are right, @SpehroPefhany - the input signal is likely too small. \$\endgroup\$
    – DEED
    Dec 17 '14 at 2:29
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Without seeing the input signals, I'd say the circuit is working just fine. That is, it's working as it should given the topology. Unfortunately, it won't work as an instrumentation amplifier, or even as a differential amplifier. And I'm pretty sure a differential amplifier with gain is what you intended.

First, of course, is the fact that you're feeding an AC-coupled signal to a single-supply op amp, and this is guaranteed to clip all inputs which are less than the average. Are you sure you want to do that? Granted, it's a poor man's rectifier, but it's bad for the op amps.

Each amplifier has a nominal gain of 63. Let's assume the each has an input of 10 mV. Then the output of the lower amp will be 630 mV. If the upper amp were operating from a (very large) split supply, it would put out 630 mV - (63 * 620 mV), or -38.4 volts. This, obviously, is not going to happen. The problem you have is that your topology only works as a differential amplifier when the gain is 1.

There is a variant on this which supports gains >1

schematic

simulate this circuit – Schematic created using CircuitLab

Common mode range is limited.

Another problem you have is that your input AC coupling networks are not obviously matched. Identical AC signals can produce non-zero outputs due to differential phase shift. Combined with the clipping produced by your AC coupling circuit, I'd expect some very odd results.

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  • \$\begingroup\$ You caught a number of problems, @WhatRoughBeast, some of which were "intentional" in that they were the result of low cost prevailing over quality in this design. The single-supply op amp is indeed supposed to act as a (lame) rectifier. It looks like the root of my problem has something to do with the input of the amplifier, and I think I'll have to change the topology. I am still curious as to why it chops up the signal. \$\endgroup\$
    – DEED
    Dec 17 '14 at 3:27
  • \$\begingroup\$ What is going on is pretty simple, I think. You have a small common-mode AC signal at 2-3 kHz, and you expect the amplifier to reject it - after all, that's what the commercial instrumentation amp did, right? Well, no. For the circuit you've shown, a common mode signal will be amplified by approximately the product of the nominal gains of the two stages. That's what I was trying to point out. So you've got a tiny AC common-mode signal being amplified enormously. Where that comes from I don't know - that's for you to determine. \$\endgroup\$ Dec 17 '14 at 3:39

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