ORIGINAL MESSAGE:
On Sat, 7 Nov 2009 17:28:31 +0100, Ulf Tjerneld <star@frizon.org> wrote:
> He explains that they
>serve a a return path to ground for surge currents and that the 100
>ohm resistor is there to act as a backup for the diodes. But he does
>not mention anything about how the metering is accomplished in this
>case; with a 100 ohm you would have maybe 25-35 volts over the
>resistor, so a meter would have to be shunted. And when shunted we
>would bring down the resistance to a low value again which in its turn
>would make it unnecessary with such a high value in the first place.
>At least according to my logic!
REPLY:
Your logic is correct. To have a 100 ohm "backup" for three large diodes is
silly. In my 50+ years of electronics I have never seen a silicon diode fail in
the open circuit mode. Invariably when they fail, they develop a dead short. To
have three fail open, well...... you would have a better chance of winning the
lottery ten times in a row.
That 100 ohm resistor will be the loneliest part in the whole amp., :-)
73, Bill W6WRT
_______________________________________________
Amps mailing list
Amps@contesting.com
http://lists.contesting.com/mailman/listinfo/amps
|