# How does the capacitor push up the voltage in a diode clamp?

I am having a hard time understanding how a capacitor is able to charge to a higher voltage level than the peak of the input source, like in a clamp circuit. I've been reading articles on clamp circuits, like this one, but I am still confused as to how a capacitor can charge up to a voltage beyond what the voltage source is giving it. I.e. a peak detector makes sense to me, but the clamping circuit does not.

I have simulated the clamp circuit with the falstad simulator

The simulation didn't help that much; I did notice that the voltage on the right side of the capacitor bottoms out when the diode conducts, but I'm still confused about this.

• It doesn't really give you more voltage- compare the peak-to-peak input vs. output. – Spehro Pefhany Oct 13 '16 at 7:50

Assume we start with $t=0$ and a positive-going sine wave for the voltage source. Also assume that $V_C=0\:\textrm{V}$ at $t=0$.