# Capacitor charging

Discussion in 'Electronic Basics' started by Ken, Feb 23, 2006.

1. ### KenGuest

Hi,

This is a newbi questions I suppose.
If I take a voltage source say a battery, I put a capacitor with rating of
300volts to it, the charge went up to the maximum charge of the battery
(12.4volts). I thought the charge would accumulate up to 300 volts. Why does
it go to 300 volt ?

thank you

Ken

2. ### KenGuest

AA ok good, Thank you
I guess I need to put a resistor in there to creat current.

ken

3. ### KenGuest

yes you are right, Dj said that I could get up to maximum value. I tried
addidn a resistor in the circuit , but still did not get more then the 12.4
volts, the battery voltage.
I guess there is no easy way to charge my capacitor to 300 volts from a 12v
battery.

ken

4. ### DJ DelorieGuest

No, that rating means if you put more than 300 volts across the
capacitor, it explodes (or fails in some other way .

If you put a constant CURRENT into the capacitor, it will continue
accumulating voltage until it reaches its max rating and fails.

You are not supplying a constant current, you're supplying a constant
VOLTAGE. The capacitor will charge to that voltage, and then the
current flow stops and the charge stays at that voltage.

5. ### John PopelishGuest

The voltage rating on a capacitor tells you the maximum voltage you
can safely apply to it and still have it function as a capacitor.

The relation between current and voltage for a capacitor is:

I=C*(dv/dt), where I is in amperes, C in farads, and dv/dt is the rate
of change of the voltage across the capacitor in volts per second.

When you first connect the capacitor across a voltage source, like a
battery, there is a large pulse of current, limited only by the
internal series resistance of the source and inside the capacitor.
But this large current pulse results in a high rate of change of
voltage and the capacitor quickly charges up till its voltage matches
that of the source. At that point, you have two equal voltages
bucking each other, and the current heads toward zero.

This is something like attaching a small air storage tank to a much
larger one. There is a brief blast of air through the connecting
hose, and then, as the small tank pressure approaches that of the
larger tank, the flow through the hose falls toward zero. You can not
get the small tank to reach a pressure higher than that inside the
larger tank that is filling it.

6. ### DJ DelorieGuest

A resistor will only *reduce* the current, not *create* it. Thus, the
capacitor will charge slower, but still stop charging when it's
voltage reaches that of the power source.

If, for example, you had a power supply capable of providing 500 V at
1 mA, you could charge the capacitor - slowly - up to 500 V (or until
it fails, that is). But a 12 v supply cannot charge it past 12v,
because at that point the voltage across your resistor is zero, and
zero volts means zero amps means charging stops.

7. ### Pooh BearGuest

You won't.

You can't get any more volts that way than there are to begin with without using
some stup-up circuitry.

What you're doing is a bit like complaining that diesel fuel poured into a can
marked gasoline still won't work in a gasoline engine.

Graham

8. ### Pooh BearGuest

NO.

A resistor *resists* current.

Graham

9. ### Guest

Sure there is. You just need the electronic equivalent of a pump. One
way works something like an ignition coil does. You apply 12 volts
across an inductor till the current ramps up to some value, then turn
that currnet off, very quickly. The inductor produces a large voltage
in an attempt to keep the current going. You pass that pulse through a
high voltage rectifier and dump it into your capacitor.

The flyback transformer in any TV or monitor is such an inductive
component. Got any junk monitors sitting around?

10. ### Pooh BearGuest

Do you seriously imagine the OP has even half a clue what you mean ?

Graham

Go for it.

12. ### KenGuest

Until now I was looking as voltage multipliers that includes diodes and
capacitors.
The idea of back emf is good. A quick search on google led me to some motors
running on back emf. Is there any truths on John Bedini's motors ?

13. ### KenGuest

I guess you tried it and didn't work.
he is saying I guess, that the back emf gives off voltage (and current) that
is greater then the one that came in. Seems to defies some laws here...

14. ### KenGuest

For what he is saying that his machine is doing. Not for what is really
going on

15. ### John PopelishGuest

The motors, themselves tell truth.
Bedini wouldn't understand that truth if it bit him on the ass.

16. ### DJ DelorieGuest

No, same total energy. Recall that inductors store energy as magnetic
flux. You can create high flux with low voltage and high current,
then use that flux to create high voltage and low current. The
wattage is the same, hence conservation of total energy.

This is very similar to the way a step-up transformer works, except
you're doing it with only one coil instead of two.

17. ### John PopelishGuest

Back EMF and any other kind of EMF is just another word for voltage.
What he doesn't want to understand is the the time-continuous
integration of the product of volts and amperes. Instantaneous volts
times instantaneous amperes is instantaneous power. The integration
of power over time is energy. He measures average or peak volts and
average or peak amperes, separately, and guesses that energy is being
created. He doesn't make the measurements necessary, nor does the
math on those measurements necessary to find out what the answer is.
If he did, it would ruin his day.
Ya think!?

18. ### Guest

Yes. We must never confuse one of those for the other. ;-)

19. ### ClintonGuest

It will if you have a 300 volt battery..

20. ### ArtGuest

Must be a "Flux Capacitor" running on banana juice?? Developing voltage from
quasar sources.