Re: calculating ripple

["Tesla list" <tesla-at-pupman-dot-com>]

Original poster: "Crow Leader by way of Terry Fritz <twftesla-at-qwest-dot-net>" <tesla-at-lists.symmetric-dot-net>

My actual circuit will be 14.4kVAC -at- 60Hz into a full wave bridge rectifier
which will charge a filter cap to run my coil. I'm assuming the following

14.4kVAC * 1.4 = 20kVDC
I wish to run my coil at upto 500 breaks per second using a cap of
approximately 0.05uF. The power needed would then be

J = 1/2 uF * (kV^2)
10Joules=  0.5 * 0.05 * 20 *20

so, to charge this cap at 500Hz you need

Watts = Joules/seconds

5000W=(500 * 10J)/1

and 5000W from 20kV power supply is

I = P/V
0.25 = 5E3/20E3

for what is about a 250mA draw from 20kV, true the load on the filter caps
is not purely resisitive, but we do know the total amount of energy being
drawn from the power supply so except for possibly high RMS currents between
the filter and tank cap the average draw should still be near 250mA

10% ripple sounds like a fun arbitrary number and comes out to 2kV. With the
below formula I find that a 1uF filter cap should allow for about 2150volts
of ripple, which is close to 10%.

Does all this math look ok, and are there any real world problems that may
make all this fall apart?
Lastly, where does al that other strange math I came across from for
calculating ripple come from?

KEN



> Original poster: "Jim Lux by way of Terry Fritz <twftesla-at-qwest-dot-net>"
<jimlux-at-earthlink-dot-net>
>
> Ignoring such things as source impedance and assuming constant current
> draw...
>
> The voltage drop after the peak will be I/C * delta T, where delta T is
1/2
> cycle time (8.6 mSec for 60 Hz)
>
> So, for 1 Amp, and 1000 uF, 1/(1000E-6) * 8.6E-3 = 1E3 * 8.6E-3 = 8.6
> Volts... Not too bad for a 1000 V power supply, terrible if it's a 5V
> supply...
>
> Since a lot of circuits are sensitive to percentage ripple, as opposed to
> absolute ripple, you can see why you need a much bigger capacitor for a
low
> voltage supply.  Fortunately, as capacitors get lower voltage, getting
more
> uF gets cheaper.