schamp wrote:I was specifically thinking of the DC case, where the capacitor was (for example) serving as a bypass capacitor or as a power source.
I was wondering about this case, because I've been studying the solar engine circuits over on BEAM wiki, particularly the Miller Solar Engine (
http://www.beam-wiki.org/wiki/Miller_solar_engine), in which it looks like D1 is used not for its current direction limiting features, but for its voltage drop, so that the voltage that is presented to the voltage detector's GND (and that serves as the basis for its comparison) is higher than the GND voltage for the entire system.
In working with supercapacitors, they are usually easier to come by with smaller voltage ratings. I was wondering if you could use a diode in such a configuration to lower the potential voltage across the capacitor, to essentially increase the margin for the rated voltage. For example, if I had 2x 10F 2.5Vcapacitors, instead of running them in series (for 5F @ 5V), I could run them in parallel for 20F @ 2.5V, and used the diode (or some combination thereof) to let me use these in a system with a higher working voltage (say, 3.3 or 5V).
But if the diode would prevent the capacitor from discharging, then that wouldn't help much.
OH! You were talking about something else! OK, first of all, a bypass capacitor, (I assume this means a filtering capacitor intended to remove noise from Vcc) and a supply capacitor are never placed in series in a circuit, it's actually in parallel what you're talking about. Like this:
- parallel.cap.png (2.77 KiB) Viewed 2374 times
Here, the capacitor is acting as a battery, or a noise filter, like you've described, but strictly speaking it's in parallel with the rest of the circuit. Oh, I added R1 just because there's always a little trace resistance in any circuit. And because L.A.W. might object if I didn't.
The diode won't stop the cap from discharging in this case, because the charge is accumulating on the top, positive side of the capacitor. If you were to put the diode on the positive side, pointing down, then yes, I don't think the diode would allow the cap to discharge. But as is, the diode isn't really doing much.
As for increasing the voltage rating, via the diode drop, I guess it does that, but it simultaneously reduces the amount of voltage the capacitor "sees" to accomplish that. So a 2.5V diode and a 2.5V rated supercap can be used with Vcc = +5V GND = +0V, but you'll only be storing half the charge as if you'd not put in the diode.
Why does that matter? Well the formula for capacitance is C = Q*V. You store half the charge at the identical voltage, ergo the effective capacitance of your supercap drops to half the original capacitance. So you've accomplished precisely the same thing as putting two supercaps in series.
If you need the full capacitance
and twice the voltage rating,
there's no angle, no champagne room: You need four supercaps. Two pairs in series, and then the pairs wired up in parallel.