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Before asking the question I would like to attach the image of the BGR regarding which I have some doubts:

enter image description here

From what I have read, due to the high gain of the op-amp, the two inputs of the op-amp, Va and Vb, are always at an equal potential such that the PTAT current flows across R1.

But, if the potential at the two inputs is equal, then shouldn't the output of the op-amp be 0 (or very close to it) as V0 = Av·(Vb-Va)?

Arnab Deb
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  • What has this question specifically got to do with band gap references or VLSI? – Andy aka Jul 07 '22 at 09:26
  • Well, this circuit as shown is a BGR circuit. I did understand the overall functioning of the circuit, but I just had a doubt regarding the op-amp. I am sorry I may have used some irrelevant tags as I am new to this platform – Arnab Deb Jul 07 '22 at 09:40
  • If you want to attract more answers, I would retitle your question to remove BGR because this isn't specifically about band gap references but about op-amp theory. You appear to be using the BGR as an example that uses an op-amp and that's only a tenuous connection as far as I can tell. – Andy aka Jul 07 '22 at 10:55
  • Yes sir, kindly retitle my question. I am learning from my mistakes and will put a better and more relevant title from next time – Arnab Deb Jul 07 '22 at 11:30
  • You should retitle your question. Given that I have made an answer, it's inappropriate for me to do this. Maybe something like "Op-amp output error due to non-infinite open-loop gain" – Andy aka Jul 07 '22 at 11:32
  • "Va and Vb, are always at an equal potential" -< this is only the case if your circuit implements negative feedback from the output of the opamp to its inverting input. – brhans Jul 07 '22 at 14:21

2 Answers2

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But, if the potential at the two inputs is equal, then shouldn't the output of the op-amp be 0 (or very close to it)

Yes, this is very true of any op-amp except ideal ones. It creates an error that sometimes can be annoyingly large. Consider an op-amp with an open-loop gain of 10,000 in a unity gain non-inverting circuit: -

enter image description here

Image from wiki.

If Vin is (say) 3 volts, we would expect the output to be 3 volts theoretically but, to get 3 volts at the output we need an input differential voltage of 3/10000 or 0.3 mV.

Because it's a unity gain amplifier, this means that the true output is not 3 volts but 2.9997 volts. That's at DC. At AC, the open-loop op-amp gain drops almost proportionately with frequency above the 3 dB point something like this: -

enter image description here

Image from here.

And, the impact of this is that the gain error gets larger with frequency. So, choose your op-amp carefully for your application and do an error analysis. Note also that for op-amp circuits with signal gain, you get an even bigger error.

For instance, at DC, if the open-loop gain is 10,000 and you require a circuit gain of (say) 10, if you fed 300 mV at the input (hoping to get 3 volts on the output), the actual output voltage would be 2.997 volts (previously 2.9997 volts for a unity gain configuration): -

enter image description here

Andy aka
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  • Thank you so much, sir. Now it is clear to me. The larger the gain less is the error, but the error will always be there since the infinite gain is practically impossible to achieve. Am I correct Sir? – Arnab Deb Jul 07 '22 at 11:33
  • You are correct @ArnabDeb – Andy aka Jul 07 '22 at 11:33
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    @ArnabDeb if we are done now, you should [take the two minute tour](https://electronics.stackexchange.com/tour) to understand what motivates folk to give answers for free. Pay particular attention to the bit about upvoting good answers and accepting the answer that worked for you. – Andy aka Jul 07 '22 at 12:52
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    "The larger the gain less is the error"......To be exact (and more clear to you) it should be mentioned that we are speaking about the OPEN-LOOP GAIN and about the error voltage when the loop is CLOSED. Hence, all this applies only when negative feedback is applied. – LvW Jul 07 '22 at 13:41
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    @lvw I have clearly stated that in my answer i.e. "Consider an op-amp with an open-loop gain of 10,000 in a unity gain non-inverting circuit" – Andy aka Jul 07 '22 at 13:43
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    Andy aka - I must admit that I have considered the values you have mentioned as a basis for the example you have presented. Of course, everything OK. On the other hand, I had the impression that it was not quite clear to the questioner that - IN GENERAL - the differential input for the opamp approaches zero only in case of negative feedback. Only THIS was my point. – LvW Jul 07 '22 at 16:52
  • Yes sir I was also talking about closed-loop configuration (negative feedback) as shown in the figure. – Arnab Deb Jul 08 '22 at 10:19
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    @ArnabDeb if we are done here you should formally accept an answer [like this](https://i.stack.imgur.com/mKrn6.png). – Andy aka Jul 08 '22 at 10:22
  • Sir also the error that is produced at the output in the above example (unity gain buffer), is it the input offset voltage of the op-amp? 3V at non-inverting terminal and 2.9997V at inverting terminal. – Arnab Deb Jul 08 '22 at 10:22
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    No, the input offset voltage is yet another error that needs to be accounted for and, is unrelated to the error produced by an otherwise ideal op-amp with non-infinite gain @ArnabDeb – Andy aka Jul 08 '22 at 10:25
  • Actually sir, in simulation to find out the offset voltage of an op-amp, the op-amp is configured as a unity gain buffer. The difference between the two terminals is then considered to be the offset voltage. That's why I have asked this question. – Arnab Deb Jul 08 '22 at 11:19
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From what I have read, due to the high gain of the op-amp, the two inputs of the op-amp, Va and Vb, are always at an equal potential such that the PTAT current flows across R1.

VA and Vb are, in general, only approximately equal, not exactly equal.

Math Keeps Me Busy
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