# Simplifying capacitor input network

For a design I am working with a capacitive feedback network. I can't show you the actual network, but I intend this to be a more general question as it should be applicable to any input network.

Imagine an amplifier with a capacitive feedback network. These are generally as the follows: simulate this circuit – Schematic created using CircuitLab

These have a transfer function $$H(s)=\frac{C_{\text{F}}}{C_{\text{IN}}} \text{ for } s > 0 \\ H(s)=0 \text{ for } s = 0$$

Now, imagine there is a more complex input network, say: simulate this circuit

I would like to again remind everyone that I am not looking for a solution to this specific network, but a general solution for capacitive input networks.

My intuition tells me that if I can write the input network as an single equivalent input capacitor using thevenin's theorem, I can just plug that equation in the transfer function and know the behavior of the circuit as a whole, without having to start from the an equation for every node and solve this for the circuit (which can get very clumsy if you have a circuit with more than a handfull of nodes).

Is this correct?

• Yes it is correct but asking aquestion that only requires a yes or a no seems a bit trivial i.e. it would be embarassing to make an answer of it. – Andy aka Aug 27 '17 at 9:29
• I guess that makes sense. I'll perhaps extend it to better represent what I am struggeling with – Joren Vaes Aug 27 '17 at 9:31

## 1 Answer

This question has a very simple answer: Yes.

Because in practice the capacitors are not ideal, things may change a bit.

Because in practice the opamp won't be ideal, the shown circuit will probably not work. The input pin of the opamp will consume a very small current. Often nano-amps (1000th of a micro-amp!) or even pico-amps!

With only capacitors connected to the input pins, that will fully charge or discharge the capacitor and might end up with a circuit that is locked against a power supply rails....

• Good point about the power rail issue. I guess that is why I don't see these circuits used with single-ended amplifiers often, I didn't think about the leakage having that impact. – Joren Vaes Aug 28 '17 at 6:14