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I am referring to the approach taken in this video, at about 00:43. I know that the op-amp "tries" to balance the voltages at its both terminals but is it a completely correct approach to take V+ = V- when deriving the op-amp's closed-loop gain? As far as I know, in practice, V+ is never equal to V-. But from a theoretical point of view, would it always be correct to assume V+ = V- when calculating the op-amp's closed-loop gain?

Voyager
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Yes. V+ = V- in cases whenever an OP-AMP gives out stable output and negative feedback dominated cases. In the real world, they are not equal. But almost equal. The difference between V+ and V- is almost zero and the gain of the OP-AMP is huge. But for a theoretical purpose, we just take it as zero and assume the gain of the OP-AMP is infinity.

If you want a better answer, use the formula V0 = A.(V+ - V-). In the video, V+ = Vin and V- = R1.V0/(R1 + RF) and then derive V0.

user3219492
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  • My professor said that he had to take away some points from my score in an exam because I assumed V+ = V- when deriving the op-amp's closed-loop gain, even though I have correctly derived the expression. We both agree that in practice V+ is never equal to V- but from a scientific/theoretical point of view is it appropriate to assume V+ = V- in an exam to get full points? – Voyager Dec 03 '16 at 18:05
  • If the question specifies $A_{OL}$ then you deserve to lose points by assuming V+ = V-. – Spehro Pefhany Dec 03 '16 at 18:48
  • I believe there was a mention of the open-loop gain within the question text, so now I see why I deserved to lose points. Thank you for your clarification. – Voyager Dec 03 '16 at 21:00
  • It's always worth blasting through with infinite gain, if it's an exam make that assumption explicit, to do a rough cut. Then bring in the second order effects like finite gain to complicate and correct the first attempt. You can expect it to be close, if it isn't, then it's worth looking for a mistake. Check out the story about the Hubble mirror grinding, and the difference between the coarse corrector, and the 'precision' one. – Neil_UK Dec 03 '16 at 21:35