Grokking Capacitors

[schamp]

Stupid expensive, exactly the reason.

Your explanation makes sense. Most datasheets I've looked at don't show I at V=0, so I didn't know what to expect.

So another, related question. Say we replace that diode with an LED with a (say) voltage drop of 1.5V. Before the cap is fully charged, I would expect the current to light up the LED, which would get dimmer and dimmer as the cap charges, finally going out when the current goes too low (i.e., to zero, when the cap is fully charged). Do I have that right?

[stinkbutt]

Yeah, that sounds about right. As the positive charge carriers accumulate on the positive end of the cap, negative charge carriers accumulate on the negative end of the cap, resulting in a brief current flow during charging that decays as the capacitor's voltage asymptotically approaches Vcc.

Note that positive charge carriers and negative charge carriers are just abstractions. The type of charge accumulated on the capacitor (either electrons or electron "holes") is largely irrelevant - They act like positive and negative charge carriers; Beyond that we don't care how the sausage is made.

[pstemari]

It also depends on how leaky the cap is.

Hooking a cap in series with a diode is an interesting thought experiment, but in practice the diode blocks AC and the cap blocks DC, leaving you with nothing.

The current through the diode is exponential with voltage, and is also temperature dependent. The "voltage drop" is really the knee in the curve. There's some leakage when the diode is reverse-biased, and the diode collects charge like a cap until it gets to the zener point, where it starts conducting again. That happens when you've pulled all the electrons out of the N side, leaving behind immobile positive charge in the crystal lattice (the nuclei of the dopant), and all the "holes" out of the P-side, leaving extra electrons bound in the valance.

[stinkbutt]

It also depends on how leaky the cap is.

Hooking a cap in series with a diode is an interesting thought experiment, but in practice the diode blocks AC and the cap blocks DC, leaving you with nothing.

No, he's not talking about putting the cap in series. He's talking about putting it in parallel, connected to ground on the other end. Only he's looking to reduce the voltage the cap "sees" by taking advantage of the diode drop, because he doesn't want to buy four Supercaps.

The circuit he intends is like this:

[ImaginaryAxis]

How is the capacitor in parallel with the diode?

[stinkbutt]

It's not, and that's not what I said.