# doubling voltage on a capacitor

Discussion in 'Electronic Basics' started by lerameur, Jun 20, 2007.

1. ### lerameurGuest

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

Have Fun!
Rich

3. ### EeyoreGuest

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

Graham

4. ### lerameurGuest

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

k

5. ### John PopelishGuest

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

6. ### ChuckGuest

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

Tell us more.

Chuck

7. ### JasenGuest

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

8. ### ChuckGuest

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

9. ### JoeGuest

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

10. ### ChuckGuest

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

11. ### JoeGuest

<snip>

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

12. ### ChuckGuest

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

13. ### ChuckGuest

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

14. ### JoeGuest

Chuck,

You persist in your word-deluge, and now you have some dishonest editing.

Here is what the OP said:

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

15. ### ChuckGuest

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

16. ### JoeGuest

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