# Electrical – Gain of Inverting Active Bandpass Filter

circuit analysisfrequency responsetransfer function

For the filter below

I calculate the transfer function
$$H(j \omega) = -\frac{R_2C_1 j \omega}{(1+C_1R_1 j \omega)(1+C_2R_2 j \omega)}$$ and thus calculate the gain $$A= \sqrt{ H(j \omega_o) H(- j \omega_o) } = \frac{R_2 C_1}{\left(R_1 C_1+R_2 C_2\right)}$$ where $$\\omega_o^{-1}=\sqrt{R_2 R_1 C_1 C_2}\$$.

Question: Please suggest as to why this is not the Voltage Gain, $$\ A:= \left|\frac{V_o}{V_i}\right| \$$ as many colleagues and sources over the internet claim the gain to be $$\ \frac{R_2}{R_1} \$$, for instance see here. Is there a way to derive this result of $$\ \frac{R_2}{R_1} \$$ by defining the Gain to be something else ?

EDIT: Here is one way I can justify the approximation to be $$\\frac{R_2}{2 R_1}\$$hold only for Narrowband Filters. For bandpass action we would need $$\\omega_1:=(R_1 C_1)^{-1} < \omega_2:=(R_2 C_2)^{-1}\$$. Now this allows us to write the gain as $$A=\frac{R_2}{R_1} \frac{\omega_1^{-1}}{\omega_2^{-1}+\omega_1^{-1}}.$$
This can also be expressed as $$A=\frac{R_2}{R_1}\frac{1}{1+\frac{\omega_1}{\omega_2}}.$$ Assuming the ratio $$\\frac{\omega_1}{\omega_2} \to 1 \$$ we get $$\A \to R_2/(2 R_1)\$$. This ratio approaching 1 signifies Narrowband Filters.

Here is one way I can justify the approximation to be $$\\frac{R_2}{2 R_1}\$$hold only for Narrowband Filters. For bandpass action we would need $$\\omega_1:=(R_1 C_1)^{-1} < \omega_2:=(R_2 C_2)^{-1}\$$. Now this allows us to write the gain as $$A=\frac{R_2}{R_1} \frac{\omega_1^{-1}}{\omega_2^{-1}+\omega_1^{-1}}.$$ This can also be expressed as $$A=\frac{R_2}{R_1}\frac{1}{1+\frac{\omega_1}{\omega_2}}.$$ Assuming the ratio $$\\frac{\omega_1}{\omega_2} \to 1 \$$ we get $$\A \to R_2/(2 R_1)\$$. This ratio approaching 1 signifies Narrowband Filters.
This answer is summarizing the discussion with Dan Boschen, for Wideband filters ($$\\frac{\omega_1}{\omega_2}\to 0\$$), we get the approximation of the gain to be $$\\frac{R_2}{R_1}\$$.