OK -- I'm still a bit confused here, so . . .
On Wed, 14 Oct 1998 jwelser@ccwf.cc.utexas.edu wrote:
> The HV voltage (PRV of the diode) is 14.5kV on a 19V2000
> and 12kV on a G05. The current drawn by the anode is 100uA on a
OK, so remembering my basic power supply knowledge, it's the PRV of the
diode that determines the output voltage.
Does this mean that the 19V2000 needs a DIFFERENT diode, to make the 14.5kv
instead of 12kV ? As in, SK7333 is actually wrong, since it crosses to H1812
instead of H1809 ?
> 19V2000. I don't have the number for the G05, but I'm sure it's
> similar. Thus, the average If of the diode must be at least that
> much. The HV can be within about +- 300V of its nominal value, and
> most of the replacements havea Vf of in the 50 - 100V range.
Mouser lists some general, high-voltage rectifiers. In the .35 amp range, we
have
PRV IFM (Surge A)
HVM14 14000 50
HVM16 16000 50
Would one of those do ? If I understand what you are saying the PRV value
has to be within +- 300V of 14.5kV ?
-Chris
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Chris Candreva -- chris@westnet.com -- (914) 967-7816
WestNet Internet Services of Westchester
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Received on Wed Oct 14 23:49:33 1998
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