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In an online course on op amps the instructor was analyzing a circuit using KVL and said that in the ideal behavior the resistance Rin between the two input terminals (not sure of their name but they are the ones at the left with + and -) is very big that no current passes and the voltage drop across the resistance is zero.

My question is why is that true? Shouldn't the very big value of the resistance compensate the very small value of the current and make a voltage drop?

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Edit:

This is the exact circuit she was analyzing but Rin is not shown here

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I think the instructor may have been confused. In most op-amp applications, the external circuit uses negative feedback to force an equal voltage on the two input pins. But if you use an op-amp in an atypical mode (e.g., if you use it as if it was a comparator) then it is very easy to get a significant voltage between the two input pins.

In the example circuit that you show, if $V_{in}$ goes up, then the op amp will try to drive $V_{out}$ to the positive supply voltage. But, as $V_{out}$ starts to rise, the voltage at the center of the $R_1$ / $R_2$ voltage divider also will rise until the voltage at the '-' input is equal to the new voltage at the '+' input. At that point, the circuit reaches a new equillibrium.

The similar thing happens if $V_{in}$ goes down.

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  • $\begingroup$ This is actually my fault. I am new to the topic and i mixed information from two videos without knowing they were different configurations. $\endgroup$
    – maged rifaat
    Apr 9, 2018 at 17:51

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