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Hi everyone,

I've been seeking the answer to this question for a long time. I've looked at posts like...

https://www.physicsforums.com/threads/op-amp-transient-state.855512/

and

https://www.physicsforums.com/threa...edback-in-an-op-amp-work-conceptually.584881/

...but I am struggling with one concept. Say we apply a 1V step input, with an Aol of 1000, and a feedback factor of 0.5 (so ideal closed loop gain of 2), how does the op-amp "know" to stop rising at Vout = 1.996V?

The finite gain equation says Vout = Aol(V+ - V-), but in this case if V+ = 1V, and V- = 0.5(Vout), then Vout would have to be 1.996V in order for it to simultanteously be 2V. ???? Since the input differential is dependent on the output, what is this equation even saying?

The op-amp does not "know" you want 2V on the output, per say...so what's stopping it from rising to 1.997V, and producing 1.5V on the output?

Here's how I've been thinking about it:

At first there's a large input voltage differential, (1V-0V), so the op-amp "tries" to reach a very high output voltage (1000V) but is physically limited by its slew rate. As the output rises as quickly as it can, it subsequently reduces the input voltage differential and thus it "tries" to reach lower and lower voltages. In this case, when the voltage reaches 1.99V, it is "trying" to reach 5V and at 1.996V, it is "trying" to reach 2V.

I'm just not sure how the op-amp says "this is a good place to stop!" when reacting to that step input.

Thanks a lot.

Tim

I've been seeking the answer to this question for a long time. I've looked at posts like...

https://www.physicsforums.com/threads/op-amp-transient-state.855512/

and

https://www.physicsforums.com/threa...edback-in-an-op-amp-work-conceptually.584881/

...but I am struggling with one concept. Say we apply a 1V step input, with an Aol of 1000, and a feedback factor of 0.5 (so ideal closed loop gain of 2), how does the op-amp "know" to stop rising at Vout = 1.996V?

The finite gain equation says Vout = Aol(V+ - V-), but in this case if V+ = 1V, and V- = 0.5(Vout), then Vout would have to be 1.996V in order for it to simultanteously be 2V. ???? Since the input differential is dependent on the output, what is this equation even saying?

The op-amp does not "know" you want 2V on the output, per say...so what's stopping it from rising to 1.997V, and producing 1.5V on the output?

Here's how I've been thinking about it:

At first there's a large input voltage differential, (1V-0V), so the op-amp "tries" to reach a very high output voltage (1000V) but is physically limited by its slew rate. As the output rises as quickly as it can, it subsequently reduces the input voltage differential and thus it "tries" to reach lower and lower voltages. In this case, when the voltage reaches 1.99V, it is "trying" to reach 5V and at 1.996V, it is "trying" to reach 2V.

I'm just not sure how the op-amp says "this is a good place to stop!" when reacting to that step input.

Thanks a lot.

Tim

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