# Gain of filter high-pass and gain of filter Sallen-Key

I can not calculate the gain of these two filters separately. I know the first filter is high pass and the second filter is a Sallen-Key, which is low pass. Could you give me a hand, please?

For the first filter (high pass):

For the second filter (low-pass):

• Have you looked at: en.wikipedia.org/wiki/Sallen%E2%80%93Key_topology ? Apr 14, 2017 at 13:14
• TI has an excellent document: ti.com/lit/an/sloa024b/sloa024b.pdf Apr 14, 2017 at 13:16
• Surely the passband gains of the filters is unity .So if they are cascaded to give bandpass you would get a gain of one. Apr 14, 2017 at 13:19
• @Autistic I do not want a band pass. I want the gain of each of the filters. They are not cascading. Apr 14, 2017 at 13:22
• @RohatKılıç The filters in this document are slightly different from mine. Mine has no resistance at the negative input of AMPOP on Sallen-Key. Apr 14, 2017 at 13:27

Gain is Vout/Vin. Using ideal op-amp theory and using impedances calculate Vout/Vin. Capacitances become 1/Cs.

For the high pass filter, you'd get something like this: $$I_2 = \frac{V_{mid}-V_{out}}{R_2}$$ $$I_1 = \frac{V_{in}-V_{mid}}{\frac{1}{C_1s}}$$
$$I_3 = \frac{V_{mid}-V_{out}}{\frac{1}{C_3s}}$$# Vout due to op-amp inputs being equal.
$$I_4 = \frac{V_{out}}{R_4}$$ $$I_1 = I_2 + I_3$$ $$I_3 = I_4$$ Using the info above, you should now be able to solve for Vout/Vin which is your gain. Repeat the same thing for the next filter.

Using I3=I4:
$$\frac{V_{out}}{R_4} = \frac{V_{mid}-V_{out}}{\frac{1}{C_3s}}$$ $$V_{out} = (V_{mid} - V_{out})R_4C_3s$$
$$V_{out}(1+R_4C_3s) = V_{mid}R_4C_3s$$ $$V_{out} = V_{mid}R_4C_3s/(1+R_4C_3s)$$ $$V_{mid} = V_{out}(1+R_4C_3s)/(R_4C_3s)$$

Using I1 = I2 +I3:
$$\frac{V_{in}-V_{mid}}{\frac{1}{C_1s}} = \frac{V_{mid}-V_{out}}{R_2} + \frac{V_{mid}-V_{out}}{\frac{1}{C_3s}}$$ $$V_{in}C_1s = V_{mid}(\frac{1}{R_2}+C_1s+C_3s) - V_{out}(C_3s +\frac{1}{R_2})$$ $$V_{mid} = \frac{V_{in}C_1s + V_{out}(C_3s+\frac{1}{R_2})}{\frac{1}{R_2}+C_1s+C_3s}$$

Combine top and bottom equations:
$$V_{out}\frac{1+R_4C_3s}{R_4C_3s} = \frac{V_{in}C_1s + V_{out}(C_3s+1/R_2)}{1/R_2+C_1s+C_3s}$$ $$V_{out}( \frac{1}{R_4C_3s} +1 -\frac{C_3s+1/R_2}{1/R_2+C_1s+C_3s} ) = V_{in}\frac{C_1s}{1/R_2+C_1s+C_3s}$$ $$Gain = \frac{V_{out}}{V_{in}} = \frac{ \frac{C_1s}{1/R_2+C_1s+C_3s} } { \frac{1}{R_4C_3s} +1 -\frac{C_3s+1/R_2}{1/R_2+C_1s+C_3s} }$$

$$\frac{\frac{C_1s(R_4C_3s)}{X}} {1+R_4C_3s-R_4C_3s\frac{C_3s+1/R_2}{X} }$$

$$\frac{ C_1s(R_4C_3s) }{ X+(R_4C_3s)X-R_4C_3s(C_3s+1/R_2) }$$

$$\frac{C_1sR_4C_3s }{ (1/R_2+C_1s+C_3s)(1+R_4C_3s)-R_4C_3s(C_3s+1/R_2) }$$ $$\frac{C_1sR_4C_3s }{ 1/R_2+C_1s+C_3s + R_4C_3s/R_2 + C_1sR_4C_3s+R_4C_3^2s^2-R_4C_3^2s^2-R_4C_3s/R_2 }$$ $$\frac{ C_1sR_4C_3s }{1/R_2+C_1s+C_3s + C_1sR_4C_3s }$$ $$\frac {R_4C_1C_3s^2}{R_4C_1C_3s^2 + (C_1+C_3)s + 1/R_2 }$$

I don't guarantee I didn't make a typo somewhere, but this should put you on the right track. Once you determine the basic currents/voltages, it's like any other circuit where it's just a lot of algebra.

• Please consider using Mathjax to make your answer more readable. Note EE.SE uses \$ to start and end inline Mathjax, rather than just $. Display equations still start and end with . Apr 14, 2017 at 16:01
• @ThePhoton Done. Apr 14, 2017 at 20:06

Each filter is a Sallen Key unity gain filter.

You can tell they're unity gain from the direct feedback between the op-amp output and the inverting input, which configures the op-amp for unity gain operation.

The top one is highpass. The bottom one is lowpass.

If you're calculating the DC gain of the filters... then that's pretty simple.

Capacitors are open-circuits (disconnected) at DC. From there, we see that we have only R1 + R2 into the "Plus" terminal of the OpAmp (so there's a voltage divider between the internal "input impedance"), and we have the output shorted to the negative feedback terminal.

So it should be something close to a unity-gain amplifier at DC.

For the high-pass filter, the capacitors are shorts at infinite frequency. Follow the logic again, and we see that the OpAmp is once again a unity-gain amplifier at infinite frequency.

In practice, this isn't the case because all OpAmps are bandwidth limited... but assuming "perfect theoretical components", that's how you do things.