How does negative feedback in an op-amp work conceptually.
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Mar7-12, 10:40 PM
Consider an op-amp in which there is a DC voltage of 100mV for example applied to the non-inverting (positive) terminal of the op-amp and the negative terminal of the op-amp is connected to the output directly. Also take the voltage supplied to power the op-amp, Vcc, to be 15V and the gain, A, to be 10000 for example.
Now from my understanding, excluding common-mode signals, the output voltage of the op-amp is given by: vo=A(vp-vn)
Initially vo=10000(0.1-0) = 1000V (theoretically) (But because Vcc is 15V the voltage will be clipped to 15V)
Then this is fed back to vn and so vn=15V.
vo=10000(0.1-15) = -149000V (again it will be clipped but to -15V this time).
Now, vo=10000(0.1+15) = 151000V (clipped to 15V)
So I don't understand how this works as a voltage follower, to me it seems the output is constantly oscillating between positive and negative saturation? Any help on what is going on would be great.
I understand from mathematics derivation that you argue that because Vo/A = 0 if A is large then vp=vn=vo so we have unity gain. But then that doesn't exactly make sense because if negative feedback isn't applied vp!=vn always. But from a practical standpoint I don't understand how unity gain is achieved (I don't like just accepting things).
As the output voltage (and hence the - input voltage) passes through the + input voltage, what happens...?