Gain of Inverting Active Bandpass Filter

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}\$$.

• I added the rest of your "update" to this answer, which will then be the best answer (and I upvoted it), since as you show the gain is better approximated as R2/(2*R1). (which will appear once peer reviewed). Commented Sep 26, 2018 at 4:18

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.

• Thanks for the reply. Your explanation seems to assume that bandwidth $| (R_1 C_1)^{-1} -(R_2 C_2)^{-1} |$ should be sufficiently large for the approximation to hold. The link talks of Narrowband (low bandwidth) bandpass filters.
– MUB
Commented Sep 26, 2018 at 3:17
• Yes, it must be sufficiently large or the approximation be just that- an approximation. The gain would approach that as the two cut-offs become further and further apart. You can plot the asymptotic slopes on a log plot to see how far off in dB the real answer would be (using -6 dB/octave for the RC roll-offs involved). Commented Sep 26, 2018 at 3:20
• But citing the link "If the value of quality factor is greater than ten then the pass band is narrow and bandwidth of the pass band is also less. This band pass filter is called as Narrow Band Pass Filter."
– MUB
Commented Sep 26, 2018 at 3:24
• Each cutoff is -3 dB from the maximum gain point, which is only asymptotically approached. I agree with you that the result is an approximation. Consider the extreme case when fc1 = fc2; at this point the impedances of the capacitors equal the impedances of the resistors and therefore the series impedance is sqrt(2) larger and the parallel impedance is sqrt(2) smaller. So in this extreme worst case the gain would be off by a factor of 2 of -6 dB. (and basically summing the two -3 dB cutoffs). So the gain is -R2/R1 with up to -6 dB less depending on bandwidth. Commented Sep 26, 2018 at 3:39
• I am not complaining if it being an approximation. I understand your last comment and agree with it, but your explanation in the answer is for LARGE bandwidth filters, which is not the case when $f_{c,1} \approx f_{c,2}$. And with your recent explanation: $\sqrt{2} \times \sqrt{2}$ should give us a $R_2/(2 R1)$ ?
– MUB
Commented Sep 26, 2018 at 3:42

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.

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