Why does this high pass filter/rc differentiator behave like this?

Question

Take a look at this circuit below:

If I'm correct, the circuit above is an RC differentiator/high pass filter. However, I do not understand its behavior.

The circuit produces an output like so: (assuming CLK is a square wave)

However, I don't see how this happens?

For instance, here is what I think would happen

1. CLK starts low, capacitor has no charge, no current goes through resistor and out is 0v
2. CLK goes high, charges capacitor, but out is still 0v and no current still goes through the resistor. The capacitor then blocks current from CLK since it is fully charged and there exists no potential difference.

Of course, I'm wrong because the behavior in the simulator does not match what I think should happen. But I still have no idea how the circuit's behavior is found out.

So why does this circuit behave like this? Thanks.

Answer 1

If I'm correct, the circuit above is an RC differentiator/high pass filter

It is a high-pass filter. It's not an ideal differentiator.

CLK goes high, charges capacitor,

Yes, but it charges the capacitor eventually.

but out is still 0v and no current still goes through the resistor. The capacitor then blocks current from CLK since it is fully charged and there exists no potential difference.

This is approximately correct, asymptotically, after charging is complete. This corresponds to the flat part of the output curve after each spike dies down.

However, all of this presumes that the capacitor was charged to reach the necessary charge across it to make Vout = 0. In order to actually charge it, a current must flow, as predicted by the branch constituent equation \$I = C \frac{dV}{dt}\$. Each time the input voltage changes, the capacitor charges and the charging current dies down exponentially.

By KCL at the output node, that current must go through the resistor (there is nowhere else for it to go), and that leads to a proportional voltage due to Ohm's law across the resistor.