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In books, I often see that they state that the op-amp has a virtual ground property when it has a high open-loop gain and in a negative feedback configuration.

First Question: The negative feedback part makes sense to me, but why is it necessary to have a high open loop gain to satisfy the virtual ground property?

Second Question: How does steady-state error from control theory tie into op-amps? Increasing open-loop gain improves the error? I'm trying to make a link between the two courses.

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    \$\begingroup\$ (1) What happens if the open loop gain is small, say 10? Can you express this in quantitative equation form? (It may help if you show us a specific schematic, too, because I think some other constraints that you didn't mention will show up when you do this.) Because if you can, I think you will have your answer. (2) Once you have (1), I think you have this, as well. \$\endgroup\$
    – jonk
    Apr 29 at 18:57

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Scenarios 1 and 2 below apply when the op-amp is either open-loop or closed-loop: -

  1. If the open-loop voltage gain is only 100 (for instance) then the voltage difference between the two op-amp inputs needs to be 10 mV to produce 1 volt at the output. 100 mV is needed to produce 10 volts at the output.
  2. If the open-loop voltage gain is 10,000 then the the voltage difference between the two op-amp inputs need only be 0.1 mV to produce 1 volt at the output. 1 mV is needed to produce 10 volts at the output.

In the second scenario, the difference between the op-amp inputs is mainly less than 1 mV and you would say that this is much more of a virtual earth/ground than the first scenario. And for many op-amps (at DC), the open-loop voltage gain is in the order of 100,000 to 1,000,000.

How does steady-state error from control theory tie into op-amps? Increasing open-loop gain improves the error? I'm trying to make a link between the two courses.

An op-amp is a control system and so it applies 100%.

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