You are getting advice you do not need, Grant.
The .82 ohm resistor is not for glitch protection,
as some have assumed without looking at the L-4B
schematic. It is exactly for the purpose you mention
below...only to protect the diodes in the power supply
section. The advice about the glitch resistor is valid, but
it goes in the HV B+ lead where it exits the power supply
box. A glitch resistor is to limit the energy stored in the
filter capacitors to a few joules while the primary circuit
breakers/fuses have time to blow during a fault in the RF
deck. It must be robust enough to NOT BLOW UP during
an arc or fault in the amp. The value is selected to limit
the energy and the wattage is selected to handle the power
for the length of time required for the power supply to shut
itself down.
I only mentioned using a two watt resistor for the form for the
wire fuse to keep the cosmetics and appearance looking the
same as the original part.
Wow! A simple fix is turning into an epistle, as usual!!!
(((73)))
Phil, K5PC
>
> ? A 0.82-ohm, 2w glitch resistor is not good engineering practice. it
> can not possibly limit fault current to a safe value.
The .82 ohm resistor is not a glitch resistor. It is an "Aw-sxxx" resistor
to protect the diodes when one takes the lid off the power supply with
it plugged in.
Ok, I'm confused. Per the L4-B manual, "In the event of a short circuit,
resistor R-12 [the one we're talking about] ... will be destroyed. The resistor
protects the diodes in the power supply ...."
See above...
So this sounds to me like the intended function of the 0.82 ohm 2W, wire
wound, fire retardent resistor is to fail rapidly -- as in a fuse. It sounds
like
the wire in the resistor is simply there to act as a fusible link, much the same
way the Buss HV fuse is used in the KWS-1.
That is another topic for another time.
Grant
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