You have a circuit like this, with two caps in parallel voltage divider style, and you send a single square pulse into Vn, so it goes from 0v to 10V for a few ms then back to 0v.
In this case, is it correct to model the caps as a voltage divider?
So I would get something like this:
\$\frac{1}{C_1jw}\$ and \$\frac{1}{C_2jw}\$
Which would give
$$\frac{\frac{1}{C_2jw}}{\frac{1}{C_1jw} + \frac{1}{C_2jw}}$$
Which simplifies to just $$V_o = \frac{C_1 +C_2}{C_2} (V_n)$$
simulate this circuit – Schematic created using CircuitLab
Assuming all this is correct I now have my real question. What happens if I add a resistor in parallel with C2?
I'm thinking I now get an RC style response curve at Vo. It will start at 0v. Spike to \$\frac{C_1 +C_2}{C_2} (V_n)\$, then fade back to zero with a time constant of (C1+C2) * R.
When Vn drops back to zero, I think I'll get a negative spike at Vo, but I'm not really sure to what anymore, since \$V_o = \frac{C_1 +C_2}{C_2} (V_n)\$ when Vn is zero gives me zero at vo.
Am I on the right track here or am I doing this totally the wrong way?
Thanks!
Best Answer
It appears that with a square wave in, the output (OUT1, below) is:
$$V_o = \frac{C_2}{C_1 +C_2} (V_n)$$
The spikes From OUT2 do have a decay time, and if you want to find out what it is, the LTspice circuit list follows: