Jose said:
"suppressing the original 0.82 ohm/2W 'fuse' resistor leads me to think I am
reducing the circuit protection in case of a, let's say, shorted tube. Of
course, it
would also destroy the 10 ohm resistor... but would it take longer so will
damage anything else in the amp or power supply?. What about putting them in
series?. Best of both worlds"
Jose, the "glitch" resistor is designed not to fail under high current and
not be a fuse, since a fuse subjects the 3-500 final tubes to probable
destruction of the grids if the fuse works.
The original design was poor engineering. Since the grids are not protected
from overdrive, they will absorb the full 100 watts of drive if the plate
circuit opens. And, once the plate circuit opens when the fuse fails, the
output will drop to the input power and as I recall the plate current will
wiggle a bit from the cathode current flowing in the metering circuit.
Since there is still output and "plate" current, the operator will drive the
amp harder for awhile until he figures out what has gone wrong. By then,
the tubes might be fried.
So, unless the amplifier tube grids can dissipate the full power of the
driver, the fuse is a very bad idea. Perhaps some drivers will shut down
since the input swr will increase, but all of my modern rigs do not have swr
shutdown and operate full power into mismatches.
So, the best of both worlds is to eliminate the original flawed design and
replace it with a 10 or preferably a 20 ohm resistor of sufficient value in
wattage and design that it will never fail.
This will remove one of the weak links of the L7 design. It is a nice
amplifier and should give you lots of qso. The only other weak link on that
amplifier is the same weak link that all my amplifiers have - an operator
that is somewhat flawed in design and subject to periodic failures. This is
the perfect reason for purchasing a fully automatic amplifier. Even better
is a totally automated station that will eliminate the operator altogether.
73, Colin K7FM
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