Charging a Capacitor with High Frequency DC voltage

[BruceW]

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).

 

[Dale]

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.

 

[sophiecentaur]

Some of us like to take the "scenic" route!

HAHA

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'.

 

[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.

 

[sophiecentaur]

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.