I have tried assuming OPA1 has negative feedback and then find the sign of OPA2 making assumptions on the latter. But the answer turns out to be wrong. How should I proceed with this question?
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2\$\begingroup\$ Information is missing that tells us what the circuit is meant to do i.e. is it a linear amplifier or, is it something else? I mean that you can assume A = negative input and get a linear amplifier or, you can assume A = positive input and get something else. and, it doesn't matter what that something else is because it's still valid as a circuit. \$\endgroup\$– Andy akaCommented Apr 6 at 16:18
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1\$\begingroup\$ This is clearly a homework question, so we shouldn't answer it directly! The point of the question is clearly to see whether you can recognise inverting-amplifier and non-inverting-amplifier op-amp configurations. Look for blocks of the circuit which do those things. We must not help you more than a hint in the right direction. \$\endgroup\$– GrahamCommented Apr 7 at 6:13
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3 Answers
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How should I proceed with this question?
Maybe we should try them in a simulator using: -
- Ideal op-amps with an open-loop gain of 1 million
- V1: 1 volt peak sinewave at 100 Hz
- V2: 1 volt peak sinewave at 200 Hz
If you then assume A is the non-inverting input you get this result: -
Then if you swap the op-amp input pins over on both devices you get this: -
In other words, exactly the same result. In fact if you swap any of the op-amp inputs over, it works the same. Cool circuit. Simulations courtesy of micro-cap 12.
My R2 was set at 8 kΩ mistakenly but, it makes no difference because when set to 2 kΩ although the output wave is slightly different, in either case of input polarity, the output remains the same.
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\$\begingroup\$ In fact, swapping the OpAmp terminals in most feedback circuits will cause a latch-up because of the high DC gain and positive feedback. \$\endgroup\$ Commented Apr 6 at 19:12
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\$\begingroup\$ @BenFM you are missing the overriding point that although one op-amp circuit looks like a Schmitt trigger (positive feedback), there is an overriding negative feedback that cancels it and makes it linear. \$\endgroup\$– Andy akaCommented Apr 6 at 19:30
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\$\begingroup\$ There is more to it than the simple argument here. OPA2 is embedded in two loops, and the overall "strength" of the negative feedback must overcome that of the positive one. This may be a coincidence with these resistor values, but a general statement cannot be made without the proper calculations. \$\endgroup\$ Commented Apr 6 at 19:52
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\$\begingroup\$ @BenFM you are missing the point of what the question is about. It's for the OP to solve and prove things mathematically rigorously if that is ultimately needed; that's not for me to do. My aim is helping the OP with the question by showing them that there are two viable linear solutions here and, in my comment under the question, I'm hinting that he hasn't provided enough details to get proper help. All your answer does is assume that you get overall linear performance with A as the inverting terminal. \$\endgroup\$– Andy akaCommented Apr 6 at 20:11
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2\$\begingroup\$ You've missed the point here. This is very clearly a homework question looking for them recognising op-amp circuit blocks. Simulating it is not the aim here. And we must not answer their homework for them! \$\endgroup\$– GrahamCommented Apr 7 at 6:15
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If A is input "-" of opamp ... then the DC behavior is "this" (?) ...
When V3 is greater than +/- 7 V, it locks saturated (?).
Otherwise, it looks "linear". But with a little "noise". The noise disappeared when R2 = ~ 6 kOhm.
Note that "slow" opamps were used in this circuit.
If the amplitudes are "great", some saturation occurs.
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Theoretically, OpAmps are mostly used in negative feedback. You must check the feedback loop sign and ensure it is negative. Positive feedback circuits using OpAmps, however, do exist. Your circuit looks like a signal-conditioning path, and assuming it needs negative feedback operation for stability, OPA1.A and OPA2.A are readily identified as the negative terminals.
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\$\begingroup\$ You should read the other answers. \$\endgroup\$– Andy akaCommented Apr 6 at 19:09
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1\$\begingroup\$ I did, and I found a lack of theoretical explanation. Simulations must VERIFY the theory, not propose their own. Thank you for your clarification. \$\endgroup\$ Commented Apr 6 at 19:10
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\$\begingroup\$ The question is this: How to determine the signs of the op amps in this circuit? <-- your question does not fully explore both possibilities to demonstrate that there is a high possibility that both versions are linear. I'm just saying this that's all. \$\endgroup\$– Andy akaCommented Apr 6 at 19:28