Re: Tube regulator for removing ripple . . .

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

Original poster: "Jim Lux by way of Terry Fritz <teslalist-at-qwest-dot-net>" <jimlux-at-earthlink-dot-net>

At 07:42 AM 4/2/2003 -0700, you wrote:
>Original poster: "davep by way of Terry Fritz <teslalist-at-qwest-dot-net>" 
><davep-at-quik-dot-com>
>
>
>>I'm currently building a 4kV DC supply for my VTTC which will be used for a
>>high-quality plasma speaker.  The only problem is removing the ripple on 
>>the output of my full-wave rectifier to the point where the DC is very clean
>>(<10V ripple)
>
>         Such a regulator will 'eat' voltage and power.
>         Max possible efficiency of the regulator is
>         50%...

50% is only true if the regulator has to operate over the full voltage 
range, which it does not.  It's not strictly a Class B amplifier. Or, 
perhaps, you could consider it as 50% eff over the regulation range, which 
is a long way from 50% overall.

Simple numerical example:

1 Amp output current at 4000V (i.e. 4 kW)
100V ripple into the regulator, regulated down to zero.  To make life easy, 
we assume ripple has a triangular waveform (or any, similar, waveform made 
of straight lines).

We'll further assume that the regulator has a minimum voltage drop of 5 volts
The voltage drop across the regulator will range from 5 to 105V, with an 
average of 55V (the straightline aspect, above).  The current is constant, 
at 1 Amp.  Therefore, the average power dissipated in the regulator is 55W, 
or, about 1.3% of the output power of the supply. This is, for all 
practical purposes, down in the noise, along with the iron and copper 
losses in the transformer, the Vf losses in the diodes, etc.



>         (ponder how the regulator must do its work:
>         'wasting' the excess voltage across the tube...)
>
>         best
>         dwp
>
>...the net of a million lies...
>         Vernor Vinge
>There are Many Web Sites which Say Many Things.
>         -me
>
>