Capacitor charging

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

ken

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

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.

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.

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.

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

NO.

A resistor *resists* current.

You need to read a book about the basics of electricity.

Graham

(snip)

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?

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

Let's start with 'back emf' for example.

Graham

Go for it.

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 ?

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...

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

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

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.

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!?

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

It will if you have a 300 volt battery..

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