Re: Grokking Capacitors

Yes, stickbutt, that's exactly what I had in mind.

And I think I understand about how two caps in series is pretty much the same as the cap and the diode. In theory, I could use two cap-diode series things in parallel to get twice the capacitance out of the same number of caps than I would if I just had the caps in series together, right?

Any more insight about the voltage drop of a diode when there's no current passing through it? I know from looking at Schottky datasheets that the voltage drop often depends on the current. So, what voltage would be across the capacitor if it were fully charged and therefore passing no current? Would it not be the full (say) 5V from VCC to GND? In other words, would the diode fail to provide any voltage drop if there were no current running through it? If so, then it wouldn't achieve the intended purpose, the cap would exceed its voltage rating and (possibly) be destroyed.
schamp

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Re: Grokking Capacitors

Look on the datasheet for the diode in question. There'll be a curve that plots I vs. V that's going to be an exponential curve, but it probably won't start at V = 0. The point where it first starts going up? That's your zero-current voltage drop.

Still, I wouldn't do the diode-cap thing anyway. The voltage drop of your diodes isn't really a deterministic parameter. That is to say, I've never heard of a manufacturer actually guaranteeing a particular drop at zero current. Most datasheets that plot current vs. voltage (or voltage vs. current) seem to treat the current logarithmically, so they never ever get to zero. The one datasheet I was able to find from National Semiconductor for the 1N4148 (I looked at a couple of super-common diodes: The 1N4001, 1N914, 1N4148...) showed the zero-current case, but all the others did not, so use this information at your own risk:

nxp.ivsv.png (13.33 KiB) Viewed 3489 times

If you're willing to use current there are other options as well, from a Zener with a precise 2.5V breakdown, to an ordinary voltage divider. But really, the best answer here is four supercaps. I can only assume you don't want to use that option 'cuz it's stupid-expensive.
Last edited by stinkbutt on Sun Jan 09, 2011 1:55 am, edited 1 time in total.
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stinkbutt

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Re: Grokking Capacitors

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?
schamp

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Re: Grokking Capacitors

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.
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stinkbutt

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Re: Grokking Capacitors

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.
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pstemari

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Re: Grokking Capacitors

pstemari wrote: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:

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stinkbutt

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Re: Grokking Capacitors

How is the capacitor in parallel with the diode?

ImaginaryAxis

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Re: Grokking Capacitors

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

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