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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?

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For the first filter (high pass):

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For the second filter (low-pass):

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  • \$\begingroup\$ Have you looked at: en.wikipedia.org/wiki/Sallen%E2%80%93Key_topology ? \$\endgroup\$ – Bimpelrekkie Apr 14 '17 at 13:14
  • \$\begingroup\$ TI has an excellent document: ti.com/lit/an/sloa024b/sloa024b.pdf \$\endgroup\$ – Rohat Kılıç Apr 14 '17 at 13:16
  • \$\begingroup\$ Surely the passband gains of the filters is unity .So if they are cascaded to give bandpass you would get a gain of one. \$\endgroup\$ – Autistic Apr 14 '17 at 13:19
  • \$\begingroup\$ @Autistic I do not want a band pass. I want the gain of each of the filters. They are not cascading. \$\endgroup\$ – Carmen González Apr 14 '17 at 13:22
  • \$\begingroup\$ @RohatKılıç The filters in this document are slightly different from mine. Mine has no resistance at the negative input of AMPOP on Sallen-Key. \$\endgroup\$ – Carmen González Apr 14 '17 at 13:27
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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.

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  • \$\begingroup\$ 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 $$. \$\endgroup\$ – The Photon Apr 14 '17 at 16:01
  • \$\begingroup\$ @ThePhoton Done. \$\endgroup\$ – horta Apr 14 '17 at 20:06
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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.

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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.

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