On Wed, 28 Jul 1999 15:35:10 -0400 "Tom Rauch" <w8ji@contesting.com>
writes:
>> Incidently, I don't believe that recovery time is an issue in any
60 Hz supply.
>
>Hi Marv,
>
>Then what determines the voltage division at the moment of
>transition to cutoff?
>
Hi Tom,
Well, in a (un-equalized) series string the fastest (recovery) diode
will take the entire inverse voltage at that moment.
1N5408's exhibit a reverse recovery in the 10's of microseconds.
Let's assume for a worst case scenario that we have 9 realllllly slow...
100us parts and one really fast 1 us device.
In a 60 Hz application, with those aforementioned ten 1N5408's
connected in series to handle 3KV RMS with a capacitive output, at the
moment the diodes turn off with approximately 3000 volts on each side of
the string the dV/dT will be something on the order of 1.5V/uS.
So, even though the fastest diode, in this exaggerated example, will
support the entire inverse voltage for the first 100 uS., there will only
be 150 volts maximum to block during that interval.
After that time, all the diodes will return to their blocking state
and will share the reverse voltage... That part of the cycle was already
covered well, and at length, by John, K5PRO, a couple weeks ago.
73,
Marv WC6W
P.S. -- In my own construction over the years, I have employed unmatched
diodes with equalizing components, potted strings, factory matched
packaged die, or strings of avalanche diodes. I have never gone to the
effort of matching diodes myself in this application nor have I ever felt
lucky enough to try an unmatched string of as few as 2 diodes.
*
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