doubling voltage on a capacitor

Hello,

I have a DC source (lets say 50v) connected to a capacitor with a switch in between. everytime the switch turns off and on again the capacitor voltage gets hit with 50v and adding it everytime. 50 then

100, then 150 atc.. how is it possible? what is the physics behind this ? thanks

ken

Reply to
lerameur
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Read some of these links:

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Have Fun! Rich

Reply to
Rich Grise

It's not *one* capacitor for starters is it ?

Graham

Reply to
Eeyore

It is one capacitor with no load. And it is DC , so voltage multipliers with Ac will not work.. i think

k
Reply to
lerameur

If you have accurately described the circuit components, their connections and the result, the physics is not anything previously seen in this universe.

Reply to
John Popelish

Sounds like maybe the capacitor's plates are moving away, decreasing capacitance and increasing voltage.

Tell us more.

Chuck

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Reply to
Chuck

What you intend is unclear, but it sounds a little like a Cockroft-Walton voltage multiplier or maybe a Marx (high voltage) generator

A circuit with a battery, switch and, (instead of a capacitor) an inductor can generate high voltages, automotive ignition systems were based on this idea. this circuit is called a flyback circuit.

Bye. Jasen

Reply to
Jasen

How about this.

If the DC source is a constant-current source, and the switch is closed only for short, repeatable durations, then the capacitor's voltage will increase by the same fixed amount, doubling each time the switch is closed, provided its self-discharge rate is negligible in the time between switch closures.

Since each "pulse" provides a fixed quantity of charge, and since V=Q/C, (delta V) = (delta Q)/C.

Obviously, this can't go on forever with a real capacitor and a real constant-current source. But it can easily go on for many orders of magnitude of voltage doubling.

Not "exactly" the way the problem is described, but close, and it is easily demonstrable within the laws of physics on this planet. ;-)

Chuck

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Reply to
Chuck

Unbelievable. "If the DC source is a constant-current source" is not magic. It would have to be a source that increases the voltage to the being-charged capacitor. So, a "constant currrent source" would be capable of charging any capacitor to voltages approaching infinity, at least until the capacitor broke down (exploded?) or until reality took over the constant current source, and the constant current source yelled "No mas!".

Too many people have given 1/2 vast explanations that have hindered my understanding in the past, so excuse my reaction to this latest "explanation".

--- Joe

Reply to
Joe

Hello Joe,

I assumed that

would have some meaning. Even the OP never said "forever".

I will admit one more time that real electronic components are not ideal: real capacitors will have some self-discharge, etc. The most accurate 1 MHz oscillator may never actually produce a 1 MHz output, etc.

With that qualification, you can easily design and construct a real circuit that will double the charge (and voltage) on a capacitor by some fixed amount every time you press a momentary contact switch for some constant time interval. For the physics, the magnitudes of the time interval and the voltage increase are irrelevant. If the OP were interested in absolute accuracies and limits on how high voltage could be increase, I think he would have said so.

I'm sorry to hear that your experiences with invalid explanations have made you hypercautious. But a better approach might be to raise specific questions about an explanation, rather than to reject it outright. If there is some aspect of what I suggested that is troubling you, I'd be glad to attempt a better explanation. Or perhaps someone else will jump in.

Good luck.

Chuck

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Reply to
Chuck

The OP said

"I have a DC source (lets say 50v)..."

Chuck said

"If the DC source is a constant-current source..."

Here, Chuck has already confused the issue by completely changing the setup given by the OP.

Chuck then goes on to say:

"Since each "pulse" provides a fixed quantity of charge, and since V=Q/C, (delta V) = (delta Q)/C."

Which further confuses the issue, because, with a constant-current source, no pulsing is necessary or relevant to the OP's original question and setup.

Stating that "Since each "pulse" provides a fixed quantity of charge, and since V=Q/C, (delta V) = (delta Q)/C." simply obfuscates that Chuck has replaced the constant 50 volt supply by one that is (automatically) ramping up its voltage, and thus is increasing the voltage on the capacitor.

The capacitor has little to do with the mechanism by which the constant-current source increases the supplied voltage, and so has little if any relevance to the OP's question.

Chuck then tells me:

"...invalid explanations have made you hypercautious..."

No, Chuck invalid explanations, especially wordy ones, cause a visceral reaction, not related to "caution".

The plain fact is that the OP most likely had some idea of a voltage doubler in mind via a switching arrangement of charging capacitor(s).

Re-casting the whole discussion into a constant-current source distorts it and certainly does not simplify anything relevant to the OP's question.

Chuck then tells me:

"...raise specific questions about an explanation..."

No question raised by me. Rather a comment -- the OP asked about apples

*, you told him about oranges **.

  • Constant voltage

** Constant current

--- Joe

Reply to
Joe

Hello again, Joe.

It just occurred to me that you might have thought my earlier post about moving the capacitor plates was one of the "1/2 vast" explanations. May I explain?

There are really only two ways to increase the voltage on a capacitor: increase the charge or decrease the capacitance. In the constant-current approach you commented on, the charge is increased when the button is pushed.

In my earlier post, which is more humorous than wrong, I imagined that the capacitor plates (or whatever) are mechanically linked, say by a ratchet, to the switch handle. Thus, every time the switch is activated, the capacitance is (first) reduced by a set amount. Every time the capacitance is halved, the voltage doubles. This is, of course, operated with a constant-voltage source.

The physics here is elementary, but technically, this stretches the rules set by the OP because after the first switch activation to place a charge on the capacitor, subsequent activations of the switch lever require no further connections to the constant voltage DC source. This could be handled mechanically, of course, in the switch design. The capacitor voltage would not change if it were connected to the source each time the button is pushed.

Does that help any?

Chuck

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Reply to
Chuck

Wow.

Okay, you're not hypercautious, Joe. But I do detect some basic confusion about electronics.

There is no way one can divine from "DC source" that the OP's intent is a constant voltage source. DC source only means not AC.

A constant current DC source is as much a DC source as a constant voltage source. Less familiar to some, maybe.

The fact is that the OP did not specify the kind of DC source. So I did.

But I'll go way out on a limb here and say that even with an imperfect constant voltage source (i.e., one with a "high enough" internal resistance) the circuit will function as I described, although with an error determined by how high the internal resistance is. What I'm getting at is that real DC sources constitute a continuum from ideal constant voltage to ideal constant current.

But Joe, even if I had blatantly violated one of the OP's rules, it wouldn't justify such hostility toward my post. There was no hidden plan in my post to mislead or trick anyone.

The exercise was most probably posted as a riddle intended to foster thought and discussion.

Try to be more calm and enjoy the discussion.

Chuck

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Reply to
Chuck

Chuck,

You persist ************** I have a DC source (lets say 50v) connected to a capacitor with a switch in between. everytime the switch turns off and on again the capacitor voltage gets hit with 50v and adding it everytime. 50 then

100, then 150 atc.. how is it possible? what is the physics behind this ? thanks

ken

**************

Chuck, Chuck, Chuck. You left off the "lets say 50v", so you could say that your constant-current setup is consistent with the OP's question.

Bullshit.

--- Joe

Reply to
Joe

That might be a valid point, Joe.

I think you're saying that a constant current source will not have an open-circuit voltage of "say 50v". That is true of an ideal constant-current source, for which the open-circuit voltage is infinite. There is a finite open-circuit voltage at the output of every real constant-current source. So there is NO inconsistency there.

Also, I didn't think there was any significance to the particular voltage the OP assumed. The physics ought to be the same for any voltage if no solid state devices are involved.

My interpretation of the problem focused on doubling the capacitor's voltage each time the switch was closed.

If you believe that the important part of the problem was to increase the capacitor's voltage above the source voltage, then that is fine. I did not do that with this suggestion. If you read my second two posts in this thread, you'll see I never claimed to do that. I did do it with my first post, however: the one in which the plates moved.

Hitting the books might be a better use of your spare time than continuing to vent your hostility on this newsgroup, Joe.

Thanks for taking the time to let me know my postings troubled you and for giving me a chance to explain my thinking.

Good luck.

Chuck

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Reply to
Chuck

In article , Chuck disgorged

You are a true bullshit artist, Chuck.

Please, I'm not "...vent(ing) ... hostility on this newgroup", Chuck.

Just trying to clear up the bullshit you persist in dropping.

--- Joe

Reply to
Joe

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