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I have read a lot of articles about how to determine the correct feedback compensation capacitors capacitance in transimpedance amplifier circuits, but they "all" use the photodiode as an example.

When using the photo diode as an example when calculating the feedback capacitors value and the cut-off frequency the photo diodes shunt capacitance is used. I understand that the electrical equivalent model of the photo diode is often modeled with a shunt capacitance and that this is often in the few hundred to thousands of picofarad range.

My question is:

If I replace the photo diode with a circuit like the one shown below, where the range of the capacitor value is about 50nF to 50pF, how much would the equivalent shunt capacitance be?

Is it zero?

schematic

simulate this circuit – Schematic created using CircuitLab

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In general, assuming an inverting amplifier with an op-amp, and the network you show being in series with a zero-impedance voltage source (several assumptions), then you can simply use the same network as the feedback, with capacitance divided and resistance multiplied by the desired gain (invert the sign, of course).

So if you want a gain of -10, you could use a feedback network of 10K in series with (10K parallel with 100pF).

Edit: This is what I mean.. if you run the simulation on the schematic, you'll see that the circuit has a gain (20dB) that is flat from 1Hz to over 100kHz, down about 3dB at a bit over 300kHz, which is about what you'd expect since that op-amp has a gain-bandwidth product of 3MHz, and the gain is (minus) 10.

The impedance of the feedback network is 10 times the impedance of the input network at all frequencies.

schematic

simulate this circuit – Schematic created using CircuitLab

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  • \$\begingroup\$ I have no idea what you are saying :P I can reformulate my question like this: If a voltage source is connected on the left side of the circuit shown in the question, what is the shunt capacitance seen from the right side of the circuit? \$\endgroup\$ May 27, 2014 at 23:32
  • \$\begingroup\$ It's not a pure capacitance, so that question doesn't really have a direct answer. \$\endgroup\$ May 27, 2014 at 23:40

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