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Open loop gain comes from the internal circuit of op-amp which is very high, but in closed loop configuration the gain is limited by the feedback resistance. So my question is:

  1. What happens to open loop gain which should be there, because we haven't disturbed the internal circuit of op-amp after closing the loop around?
  2. If open loop gain still exists, then how can the two different voltages exist on same output node (one due to open loop and other due to closed loop)?
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  • \$\begingroup\$ how can the two different voltages exist on same output node -- Because you change the behavior of the circuit when you add the feedback! If you want to measure the open-loop gain, you can only do that in a circuit without feedback (unless there's a trick I don't know about), and if you want to measure the closed-loop gain you must have the feedback in place (I'm sure there isn't a trick for that). \$\endgroup\$
    – zwol
    Oct 4, 2017 at 20:43

3 Answers 3

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The open loop gain does not change.

The feedback drives the difference between the + and - terminals to a very small value such that the small value times the open loop gain is the correct output voltage for the closed-loop topology.

enter image description here

In the picture above, the closed loop gain is -R2/R1. So the feedback causes the voltage across RIN to be whatever value is needed to give that closed loop gain when multiplied by the large gain "A".

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Let me put some of what you said in a different order:

What happens to open loop gain which should be there, because we haven't disturbed the internal circuit of op-amp after closing the loop around

in closed loop configuration the gain is limited by the feedback resistance.

The op-amp is still providing the same gain, but the input differential voltage is very small (in the ideal case we say it goes to 0).

One way to look at this is that what's happening isn't that the feedback is reducing the gain, but that the feedback is driving one of the inputs (in the simplest circuits, the inverting input) to be (nearly) equal to the other input. Because of that, even the large gain of the amplifier doesn't produce a very large output voltage.

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Open loop gain comes from the internal circuit of op-amp which is very high, but in closed loop configuration the gain is limited by the feedback resistance.

OK, so far.

  1. What happens to open loop gain which should be there, because we haven't disturbed the internal circuit of op-amp after closing the loop around?

The op-amp gain is still the same. The output will be the open loop gain times the difference between the inputs. It's just that the feedback makes this difference very small.

  1. If open loop gain still exists, then how can the two different voltages exist on same output node (one due to open loop and other due to closed loop)?
  • The closed loop gain is the gain of the complete amplifier circuit.
  • The open loop gain is the gain of the op-amp itself.

schematic

simulate this circuit – Schematic created using CircuitLab

Figure 1. A non-inverting op-amp amplifier with a gain of 2.

If this op-amp has an open loop gain of 1,000,000 then the voltage at Vin- will be 0.999999 V (0.000,001 V below the non-inverting input).

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