amplifierfeedbackgainmathoperational-amplifier
I have a circuit very similar to the inverting band pass filter circuit here:
I was asked a question about the circuit's feedback transfer function. Is this the same as the transfer function for the whole circuit?
Which I calculated to be this (for the circuit above):
$$\frac{V_0}{V_i}=\frac{j \omega R_2 C_1}{1+j \omega R_2 C_2 + j \omega C_1 + \omega^2 C_1 C_2}$$
If you wish to redraw the circuit without the op-amp, try something like this:
simulate this circuit – Schematic created using CircuitLab
See the difference?
I can redraw the block diagram as follows:
Where $$H_3(s) = \frac{R_3}{R_2}$$ and $$H_4(s) = \frac{R_3}{R_1}$$
From this, $$\frac{V_o}{V_{in}} = \frac{-H_1(s)\times H_4(s)}{1+H_1(s)H_2(s)H_3(s)}$$
Best Answer
The feedback transfer function is not the same as that of the amplifier, because the overall circuit suppresses whatever is fed back. So roughly speaking, they are opposite. The transfer function in the negative feedback is a high pass filter, and therefore the amplifier is a low-pass filter. (The overall circuit is band-pass because of the high pass in front of it.)