Gain of Inverting Active Bandpass Filter

Question

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.

Answer 1

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}\$.

Answer 2

Yes your reference helps make it clear how this condition can exist. It is important that the peak gain position in frequency be sufficiently above the cutoff frequency defined by R1 and C1; the cutoff is the frequency at which the capacitive reactance and the resistance are equal, and as we go higher from this point the capacitive reactance decreases such that the resistance would begin to dominate the net impedance and can then be assumed to simply be R1 (and the higher you go in frequency from this point, the more so this holds true as the capacitor approaches zero in impedance).

Similarly because of the parallel combination of R2 and C2, if the peak gain location in frequency is positioned sufficiently below the cutoff defined by these two components, then the resistance would similar dominate the net impedance in this case (and as you go lower in frequency from cut-off, the capacitor approaches an open circuit). Therefore, with sufficient separation in frequency between the lower and upper cut-off, the gain can indeed be approximated by simply -R2/R1 as in a standard inverting op-amp configuration with only these resistors.

Answer 3

For music-like broadband filters (20 Hz-20 kHz) over the entire passband:

$$R_1>>ZC_1$$ $$ZC_2>>R_2$$

It is not very professional, however, in these cases: $$R_1 + ZC_1 ≈ R_1$$ $$R_2 || ZC_2 ≈ R_2$$

This causes the gain to depend only on the relation between the resistances.

$$|Av|=R_2/R_1$$

For more selective filters this simplification cannot be done since ZC₁ and ZC₂ are closer in magnitude to R₁ and R₂ over the entire passband.

Another way of looking at this is to take into account that the filter center frequency and bandwidth govern the selectivity of the filter such that:

$$Q=\frac{fo}{BW}$$ Where FH=high freq and FL=low freq, center frequency is:

$$fo=\sqrt{FH\times FL}$$ $$BW = FH-FL$$

For wideband, the center frequency tends to zero (never 0), due to FL, and BW tends to FH for the same reason.

$$Q≈0$$

Therefore, during the pass band it is as if the capacitors did not exist and it was a simple inverting amplifier, in wide band filters:

$$Av=-R_2/R_1$$

In the equation that you determined, if you choose R=R₁=R₂, the narrowband gain equation is much simpler and more useful, since you can create a very selective filter (high Q value) and predict its attenuation (Av <1), and then add an independent amplification stage of its filter that compensates for the attenuation that occurred.

Even if it comes from using a voltage follower before, you can convert it into a non-inverse amplifier that, in addition to matching impedances, compensates the attenuation of your filter.

$$|Av|=\frac{C_1}{C_1+C_2}$$