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Grokking Capacitors
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Re: Grokking Capacitors

by stinkbutt on Fri Dec 24, 2010 1:37 pm

Zener wrote:My issue with threads like this is they start with a beginner looking for useful knowledge, and then "physicists" or semantists start arguing how many angels fit on the head of a pin, and almost totally disregard the desire of the OP to learn something. Notice the OP has not been back.


Precisely.

LAW, you and tinsmith can go down all the physics rabbit-holes you like, but if I'm trying to actually give rwp42 an answer that's useful to him, I don't appreciate you hijacking the post. rwp42 doesn't care that I left out what - the internal resistance of the battery? Any more than he cares that I failed to account for the ESR and the leakage currents or the internal inductance of the capacitor. He just wants to know how capacitors work.

I don't know what the exact right perfect answer to rwp42's question was, but I'm quite certain that whatever that perfect answer is, it does not include any of these phrases:

"the voltage across a capacitor can change instantaneously if the current is infinite"
"power train analysis of a Class D, PWM H-bridge"
"Kirchhoff's Laws are an abstraction of field theory"
"complex impedance"
"displacement current produces magnetic field identical to "real" current"

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Re: Grokking Capacitors

by ImaginaryAxis on Fri Dec 24, 2010 10:14 pm

stinkbutt wrote:He just wants to know how capacitors work.


We all contributed from basics to more advanced topics. Next time explain it correctly.

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Re: Grokking Capacitors

by stinkbutt on Sat Dec 25, 2010 10:12 pm

Next time don't hijack the thread. And then follow it up by hijacking the post that attempts to right the thread.
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Re: Grokking Capacitors

by stinkbutt on Sat Dec 25, 2010 10:28 pm

It's Christmas, and I'm feeling charitable, so I've modified the circuit diagrams in my original post to reflect the internal resistance of the battery.
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Re: Grokking Capacitors

by tinsmith on Sun Dec 26, 2010 7:33 am

Wow, this thread got a bit antagonistic over the holidays.
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Re: Grokking Capacitors

by pstemari on Tue Dec 28, 2010 2:37 am

Any more of this and it will be slide rules at twenty paces :D

Kirchoff's current law, unlike the voltage law, is one of those things that depends on the details of the fine print. In particular, a capacitor isn't a node--wires are the nodes, and they're only an approximation of varying quality to a node.
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Re: Grokking Capacitors

by schamp on Tue Jan 04, 2011 5:21 pm

So...what would you expect to happen if you hooked up a diode in series after a capacitor, i.e., from the negative terminal of the capacitor to the anode, cathode to ground.

Like this:

Code: Select all | TOGGLE FULL SIZE
V+ ---||--->|---GND
     cap  diode


Would the capacitor charge and discharge as normal (except, charge only to the total voltage minus the forward drop of the diode)? Or would it be prevented from discharging, since the diode would prevent any reverse current?
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Re: Grokking Capacitors

by stinkbutt on Wed Jan 05, 2011 11:39 am

schamp wrote:So...what would you expect to happen if you hooked up a diode in series after a capacitor, i.e., from the negative terminal of the capacitor to the anode, cathode to ground.

Like this:

Code: Select all | TOGGLE FULL SIZE
V+ ---||--->|---GND
     cap  diode


Would the capacitor charge and discharge as normal (except, charge only to the total voltage minus the forward drop of the diode)? Or would it be prevented from discharging, since the diode would prevent any reverse current?


Well, as before there's DC and AC cases.

In the DC case, nothing. The circuit's basically open. In the steady state the capacitor is charged up on the positive side due to it's direct connection to V+, and then... crickets.

In the AC case, assuming V+ is some sinusoid, reasonably small with respect to the capacitance, then the signal gets transmitted through the capacitor to the diode, at which point the diode would allow the positive half of the signal to pass and would block the negative half.
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Re: Grokking Capacitors

by zener on Thu Jan 06, 2011 2:09 am

schamp wrote:So...what would you expect to happen if you hooked up a diode in series after a capacitor, i.e., from the negative terminal of the capacitor to the anode, cathode to ground.

Like this:

Code: Select all | TOGGLE FULL SIZE
V+ ---||--->|---GND
     cap  diode


Would the capacitor charge and discharge as normal (except, charge only to the total voltage minus the forward drop of the diode)? Or would it be prevented from discharging, since the diode would prevent any reverse current?


I think that is a really good question. I think that at first when DC voltage is applied, the cap charges up with the negative side at the Vf of the diode. However, as the current falls, so does the Vf of the diode. If we assume an ideal cap with zero leakage, then eventually I believe the Vf falls to zero. Yes? I could be wrong...

A discharge condition seems to present a paradox. You would seem to have a voltage present, but you cannot have any current, assuming an ideal diode. Hmmmm. But I am very tired...

I think this schamp is a troublemaker...

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Re: Grokking Capacitors

by schamp on Fri Jan 07, 2011 2:21 pm

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.
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Re: Grokking Capacitors

by stinkbutt on Fri Jan 07, 2011 4:08 pm

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
parallel.cap.png (2.77 KiB) Viewed 2322 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.
Last edited by stinkbutt on Fri Jan 07, 2011 4:14 pm, edited 3 times in total.
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Re: Grokking Capacitors

by stinkbutt on Fri Jan 07, 2011 4:10 pm

Also, keep in mind I'm not entirely convinced the diode trick would even work at all. Diode drops aren't really static, they're calculated based on an amount of current that they're passing at the time. A 0 amps, there might not be much drop to speak of at all, and in your circuit there's not going to be any current.

I'm not certain one way or the other about this bit. But my intuition says no, no drop. Let's see what everyone else sez.
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Re: Grokking Capacitors

by ImaginaryAxis on Sat Jan 08, 2011 12:40 am

stinkbutt wrote: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.


I knew I felt a disturbance in the Force. 8) My point was that the cap charges instantly, practically speaking, when shunted across a voltage source.

The example, as explained by Zener, is correct when using the exponential, PWL, or small signal model of a diode. The constant voltage drop or ideal diode models are another matter, but is not what will happen in real life. See Microelectronics by Sedra and Smith - excellent book.

The circuit itself does not operate as the example, because there is a current path.

Otherwise use the capacitor bank as suggested by SB.

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Re: Grokking Capacitors

by schamp on Sat Jan 08, 2011 8:11 pm

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.
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Re: Grokking Capacitors

by stinkbutt on Sat Jan 08, 2011 10:42 pm

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
nxp.ivsv.png (13.33 KiB) Viewed 3493 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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