# Charging a Capacitor with High Frequency DC voltage

by gedfire
Tags: capacitor, charging, frequency, voltage
PF Gold
P: 10,219
 Quote by BruceW I read somewhere online that it is good practice to only use half the voltage rating of the capacitor. So if your peak input voltage is 500v, then why not just send half of it somewhere else? Or make less in the first place?
It would all depend upon the details of the way the capacitor is rated. I agree it would be best to chose one that is rated a bit higher - as a matter of 'conservative principle' but you could find yourself paying more money than you need if you went for twice the voltage rating. Other factors also could become relevant, though. The RMS current can also be relevant to how much stress a capacitor is having to take.
HW Helper
P: 2,985
 Quote by Drakkith Sophie, since this is pulsed DC, should some sort of filter be used to smooth it out? Would that help at all?
I assumed this is exactly what he is using the capacitor to do. (simply to smooth out the pulsed dc a bit).
HW Helper
P: 2,985
 Quote by sophiecentaur Other factors also could become relevant, though. The RMS current can also be relevant to how much stress a capacitor is having to take.
yeah. I am too much in the habit of thinking of circuit components to always act in the 'ideal' way. I'd guess kHz frequency is nice enough though.
PF Gold
P: 10,430
 Quote by BruceW I assumed this is exactly what he is using the capacitor to do. (simply to smooth out the pulsed dc a bit).
I don't think so. From his description and linked videos I think he wants to use the capacitor as an actual power source once charged. I don't know why, as a battery would seem to be a much better choice, but it's not my project.
 HW Helper P: 2,985 ok. i think you are right. I am trying to get my head around it. Ok, so supposing he does charge the capacitor, then discharges it through an incandescent bulb, it will take of the order of RC (about 1 second) for the first capacitor he mentioned to discharge. (assuming ideal resistance). I guess this should be long enough for him to see the flash of light. I am not so sure about how he is going to charge it using pulsed dc though. pulsed dc will vary from zero voltage to the peak voltage, right? So if he lets the capacitor charge, then disconnects it from the pulsed dc, then the voltage 'stored' in the capacitor could be anything from zero volts to the peak voltage...
 Sci Advisor PF Gold P: 10,219 The nearest thing that I can think of to this is a camera flash circuit. But the load is of a specific type - well fitted to using the energy stored in a capacitor. Anything that requires a particular operating voltage would just not be suited. Or is this "load" something slightly 'mischevious'?
PF Gold
P: 10,430
 Quote by BruceW I am not so sure about how he is going to charge it using pulsed dc though. pulsed dc will vary from zero voltage to the peak voltage, right? So if he lets the capacitor charge, then disconnects it from the pulsed dc, then the voltage 'stored' in the capacitor could be anything from zero volts to the peak voltage...
It would be the average or RMS voltage or something like that I believe.
P: 609
 Quote by Drakkith It would be the average or RMS voltage or something like that I believe.
If you model it as an ideal capacitor in series with a voltage source and some internal resistance then I would expect it to be an exponentially decaying weighted average. The time constant on the exponential decay would be determined by the capacitor rating in farads divided by the power supply resistance in ohms. In the limiting case of an ideal voltage source with negligible internal resistance, BruceW has it right -- the final charge is an instantaneous sample of the final supply voltage.

This model means that the charging current would end up being AC, not pulsed DC.

If you were to use a diode or similar as a "ratchet" to enforce one-way charging current then the equilibrium state would be at peak voltage. No averaging required, RMS or otherwise.
PF Gold
P: 10,430
 Quote by jbriggs444 If you model it as an ideal capacitor in series with a voltage source and some internal resistance then I would expect it to be an exponentially decaying weighted average. The time constant on the exponential decay would be determined by the capacitor rating in farads divided by the power supply resistance in ohms. In the limiting case of an ideal voltage source with negligible internal resistance, BruceW has it right -- the final charge is an instantaneous sample of the final supply voltage. This model means that the charging current would end up being AC, not pulsed DC. If you were to use a diode or similar as a "ratchet" to enforce one-way charging current then the equilibrium state would be at peak voltage. No averaging required, RMS or otherwise.
You're saying that without some kind of resistance in the circuit the capacitor charges and discharges so fast that it follows the pulsed DC almost exactly?
 Sci Advisor PF Gold P: 10,219 We have taken 27 posts, just talking around how a normal rectifier circuit with a reservoir capacitor works. Apart from the relatively high voltage and the 'non-mains' frequency involved, is there any difference from what we find in every conventional power supply?
P: 609
 Quote by Drakkith You're saying that without some kind of resistance in the circuit the capacitor charges and discharges so fast that it follows the pulsed DC almost exactly?
Yes. This does also suppose zero inductance.
PF Gold
P: 10,430
 Quote by sophiecentaur We have taken 27 posts, just talking around how a normal rectifier circuit with a reservoir capacitor works. Apart from the relatively high voltage and the 'non-mains' frequency involved, is there any difference from what we find in every conventional power supply?
Some of us like to take the "scenic" route!
HW Helper
P: 2,985
 Quote by sophiecentaur We have taken 27 posts, just talking around how a normal rectifier circuit with a reservoir capacitor works. Apart from the relatively high voltage and the 'non-mains' frequency involved, is there any difference from what we find in every conventional power supply?
no, as Drakkith was saying, the OP'er is not using the capacitor to smooth out the pulsed dc. He wants to charge up the capacitor, then discharge it separately. So the capacitor is not being used as a reservoir capacitor (which is what I initially assumed, too).
 Mentor P: 15,674 Place the capacitor in series with a diode to keep the capacitor from discharging. If you cannot get a voltage regulator then place that in parallel to a zener to set the max voltage. Place that in series with a resistor to limit peak current. The capacitor will basically charge like a normal RC circuit, but the time constant will have to be scaled up by a factor of the duty cycle of the source.
PF Gold
P: 10,219
 Quote by Drakkith Some of us like to take the "scenic" route!
HAHA

 Quote by BruceW no, as Drakkith was saying, the OP'er is not using the capacitor to smooth out the pulsed dc. He wants to charge up the capacitor, then discharge it separately. So the capacitor is not being used as a reservoir capacitor (which is what I initially assumed, too).
So now it's like a camera flash? Also very much a 'known art'.
 HW Helper P: 2,985 yeah, the tricky bit is thinking of what the pulsed dc would do. A camera flash uses normal dc to charge the capacitor I would guess. But for the pulsed dc, even though the voltage is always in the same 'direction' across the capacitor, it would generally increase and decrease. If the pulsed dc had very low frequency compared to the time constant, the voltage across the capacitor would equal the instantaneous voltage of the pulsed dc at all times (since the capacitor can 'catch' up, before the pulsed dc voltage can change to another value). So in this limit, the voltage stored in the capacitor will be any value of the instantaneous voltage of the source. And in the limit of very high source frequency compared to the time constant, I would intuitively think that the voltage stored in the capacitor would tend to some non-zero value which is approximately constant with time. I haven't done the calculation though. jbriggs and dalespam have the interesting idea of using a diode to ensure that current only travels in one direction. This seems like it might work... I'm guessing it means that a lot of the power from the pulsed dc is going to be 'wasted' in resistance of the diode. But maybe this is the best way.
PF Gold
P: 10,219
 Quote by BruceW yeah, the tricky bit is thinking of what the pulsed dc would do. A camera flash uses normal dc to charge the capacitor I would guess. But for the pulsed dc, even though the voltage is always in the same 'direction' across the capacitor, it would generally increase and decrease. If the pulsed dc had very low frequency compared to the time constant, the voltage across the capacitor would equal the instantaneous voltage of the pulsed dc at all times (since the capacitor can 'catch' up, before the pulsed dc voltage can change to another value). So in this limit, the voltage stored in the capacitor will be any value of the instantaneous voltage of the source. And in the limit of very high source frequency compared to the time constant, I would intuitively think that the voltage stored in the capacitor would tend to some non-zero value which is approximately constant with time. I haven't done the calculation though. jbriggs and dalespam have the interesting idea of using a diode to ensure that current only travels in one direction. This seems like it might work... I'm guessing it means that a lot of the power from the pulsed dc is going to be 'wasted' in resistance of the diode. But maybe this is the best way.
That whistle you hear is the inverter - which produces pulses, surely, after rectification. How would th voltage on the capacitor decrease except through the load?
Until the capacitor is loaded, the time constant will be very long.
The forward voltage of a diode is hardly relevant to power dissipation in this sort of circuit.

I don't understand why there is so much arm waving on this thread. Put the values of the components into a simulator (if the sums are 'too hard') and see what emerges. Of course, the whole thing depends totally on the values of the critical components- like the capacitor, the operating frequency. If the OP can supply them then it can all be solved with standard tools.

Perhaps this should be on the Electrical Engineering Forum.

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