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Here is what I know about the Ideal Op-Amp.

  • The open loop voltage gain is infinite
  • The output voltage is given by the following \$v_o = A(v_+ - v_-)\$
  • Only with a negative feedback loop is \$ v_+ = v_-\$

My query is with regards to the negative feedback loop as shown below: enter image description here

My queries are as follows:

  1. Clearly, \$v_o = A(v_+ - v_-)\$ should still apply and since \$ v_+ = v_-\$, shouldn't the output voltage \$ v_o = 0 \$ always?
  2. Since \$v_o = A(v_+ - v_-)\$ should still apply, is A still the open-loop voltage gain which for an ideal op-amp is infinity. Thus, would the output voltage always be infinity?
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  • \$\begingroup\$ google offsets in opamps \$\endgroup\$ – Transistor Mar 23 '18 at 2:04
  • \$\begingroup\$ v- will actually be very slightly lower than v+. In fact how much lower it is depends on the inverse of the op-amp's gain. As the gain goes to infinity the difference will go to zero. \$\endgroup\$ – immibis Mar 23 '18 at 2:47
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    \$\begingroup\$ \$A\$ is the open loop gain and it is infinity as you said and \$lim_{x\rightarrow\infty}x\times 0 \$ can be anthing \$\endgroup\$ – Curd Mar 23 '18 at 11:39
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When analyzing an op-amp circuit, we don't assume a priori that the differential input voltage is 0.

We assume that the gain of the amplifier is very large, and the input impedances are very large.

If we then set up a negative feedback circuit, we find that then in the limit as the op-amp gains goes to infinity, the differential input voltage will go to zero.

The differential input voltage would not go to zero if the output voltage were always zero. (And of course in a real op-amp the gain is not actually infinite and therefore the input voltage is not actually zero)

Main point: The input voltage goes to zero as a result of the gain being very high and the output voltage going to some non-zero value, not the other way around.

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  • \$\begingroup\$ Ah. That makes much more sense. One more thing. Why is it only when we have negative feedback circuit that the differential input voltage goes to zero? I understand why thats true but why isn't true for just open-loop op-amp since the gain is still infinite there. \$\endgroup\$ – AlfroJang80 Mar 23 '18 at 3:17
  • \$\begingroup\$ With open loop, the output would just be the gain times the input voltage (limited by the power supply voltage). With a closed negative feedback loop, a deviation of the input from zero drives the output in such a way that it drives the input back towards zero. \$\endgroup\$ – The Photon Mar 23 '18 at 4:43
  • \$\begingroup\$ Ah. I think I understand now. One last thing, so with an open loop op-amp circuit, the output voltage would be extremely high correct and would drive it to the supply voltages? The addition of the feedback controls this for this. Would that be correct? \$\endgroup\$ – AlfroJang80 Mar 23 '18 at 16:41
  • \$\begingroup\$ @Alfro, yes, that's basically correct. \$\endgroup\$ – The Photon Mar 23 '18 at 16:51
  • \$\begingroup\$ Hmm. I just came across input offset voltage defined as being a DC voltage that needs to be applied to the input to force the output to zero. That doesn't make sense to me. Why would we want to force the output to zero? Is this talking about the open-loop case? \$\endgroup\$ – AlfroJang80 Mar 23 '18 at 17:28
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Zero times infinity is indeterminate. It may be zero or infinity or something in between.

You should calculate the output voltage for a large open-loop gain and look at what happens to the output voltage as the gain approaches infinity.

You should find that it approaches -Vin*R2/R1

And you should find that the differential input voltage at the op-amp approaches zero, but for every (finite) value of gain, no matter how large, they will be a small, but non-zero, value for the differential input voltage.

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  • The open loop voltage gain is infinite
  • The output voltage is given by the following \$v_o=A(v_+\$−\$v_−\$)

Then A must be infinity and for a finite output voltage the two inputs have to be equal i.e. infinity x zero is finite.

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