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Re: A 20 kV DC tank supply (more on 3 phase rectifiers)

Tesla List wrote:
> 
> Original Poster: "Marco Denicolai" <Marco.Denicolai-at-tellabs.fi>
> 
> I would like to built a 20 kV DC supply for my big coil. These are the
> tentative specs:
> 
> - 20 kV DC output
> - primary capacitor 0.1 uF, charged at 20kV --> 20 Joules
> - maximum DC deviation: 1%
> - maximum BPS: 800 Hz
> 
> If I use a 3-phase 14 kV pole pig and 6 diodes I get a ripple of 50*6 = 300
> Hz. Voltage varies between 14 kV * 1.41 = 19.8 kV and
> 14 kV * 1.41 * 0.866 = 17.1 kV.
> 
> I need a flywheel capacitor to provide, roughly, enough energy for 3
> primary capacitor chargings (at BPS = 800 Hz, I get almost 3 bangs between
> each ripple top), that is 20 * 3 = 60 Joules allowing the top voltage to



Here are some "cookbook" numbers for a 3phase bridge rectifier feeding a
large choke input filter.

Ripple with no filter: 4.2% (rms ripple/output voltage)
Ripple frequency = 6 * line frequency

RMS current per rectifier: .578*Avg DC output current
Peak reverse volts = 1.05 * output DC voltage
Output Voltage = 2.33 * Line to neutral RMS AC voltage
Output Voltage = 1.35 * Line to Line RMS AC Voltage
Secondary VA = 1.05 * DC Watts out


Choke input filter design

Choke selection:
L min = (K/fs)* Rload(min)
fs = source frequency (50 or 60 Hz, as appropriate)
K = 0.06 for single phase, 0.0017 for fullwave three phase

For your application, 20kV -at- 1 Amp, Rload = 20K

Lmin = .0017/50 * 20E3 = 0.68 Henry (quite reasonable size)


The LC product must be bigger than a minimum required to get ripple
factor low enough.

r = (.0079/*L*C) * (60/fs)^2

L in henries, C in uF, for 3 phase full wave (6 pulses/cycle)
For ripple voltage factor of 0.01 (i.e. 1%), and figuring a 1 H filter
inductor...

.01= 1.44 * (.0079/ (1H*C) =>>   C = 1.13 uF

Certainly a reasonable value for a filter cap.

You can add a second LC section to reduce the ripple even more:

Ripple after/Ripple before = 1e6/((2*pi*fripple)^2*L*C)
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