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Re: Voltage doubler (fwd)
---------- Forwarded message ----------
Date: Thu, 15 Apr 1999 19:15:23 -0700
From: Ed Phillips <evp-at-pacbell-dot-net>
Reply-To: ed-at-alumni.caltech.edu
To: Tesla List <tesla-at-pupman-dot-com>
Subject: Re: Voltage doubler
Tesla List wrote:
>
> Original Poster: David Trimmell <davidt-at-pond-dot-net>
>
> Hello, I was wondering if anyone on the list, who has a good understanding
> of what is actually going on within a voltage doubling circuit, could
> explain it to me. I am, at this time, using 3 uFd capacitance in my doubler
> to feed the plates on my 833A tube coil, and can range from 0-2890 VAC into
> the doubler (safely, keeping the tubes happy). I am very interested in
> finding the optimum capacitance value for a specific power and voltage
> value. I understand the principle of the halfwave voltage doubler circuit,
> but would like to know if the voltage is truly doubled, and how I can find
> the best value for the capacitance? I suspect that the higher the value for
> C, then the greater power available, but I need to find that optimum value ;-)
>
> Thanks for any assistance,
>
> David Trimmell
>
> PS. I am using the doubling circuit that Dave Sharpe and John Freau have
> used on their tube coils, i.e., cathode grounded on the diode.
Your assumptions about "the bigger C the better" are correct. There
are design nomograms for half and full-wave doublers in many
publications, but as a crude rule of thumb, if
2 pi f R C
is greater than about 50, the average output voltage will be about 180%
of the peak input voltage. R is the load resistance (Vload/Iload), f is
the supply frequency, and C is the capacitance in FARADS. If the same
factor is about 100, the average output voltage will be about 190% of
the peak input voltage. I can't give you the ripple without digging out
references, but assume that won't be important in your application.
Ed