> I am in the process of finishing up my 8877 amp, and the
> plans I am using
> show a 0.6 ohm 1 watter and a 50 ohm 50 watter in series.
> So in the event
> of a flashover (milliseconds ?) the 0.5 ohmer is supposed
> to blow open?
Use Ohm's law. We know that always works for heat.
Let's assume you have 10 ohms of impedance in the supply
between the filter caps stored energy and the anode system
with 4000 volts (this is pretty much typical). Without a
glitch resistor and an anode to grid flashover the surge
current could be 400 amperes.
Now you add a 1 ohm resistor and nothing else. If the
resistor doesn't short during a fault failure it would only
change current to 364 amperes. Hardly worthwhile.
Another problem is a small resistor when it fails arcs
across the body of the resistor of the gap between leads. It
continues to provide a low impedance path through the plasma
until the voltage drops enough to quench the arc. That's why
a small resistor for a fuse or surge limiting is a very poor
idea.
If you had a 50 ohm resistor (50 plus the 10 ESR) the surge
current would be 66.7 amperes. So long as the resistor
doesn't arc we can be reasonably sure current will be less
than 67 amperes and less. The peak dissipation in the 50 ohm
resistor would be 225 kW.
Now you have a .6 ohm resistor that we already know is
useless because of arc path length) in series with the 50
ohm. It is 1% of the value and has the same current, so it
would dissipate 1% of the power or 2.25kW. While that
resistor might fail, the arc path length is so short it
couldn't possibly limit current and open the circuit until
after the HV is virtually gone. All it does is add something
to be a nuisance later on. It doesn't really protect or fuse
anything, it's just something to replace later that really
never helps.
If you doubt the quench time, look closely at how HV fuses
are designed. Even with 250 volt fuses the design becomes
special. HV fuses are *very* long and normally have an
arc-quenching powder inside.
Why add a part that does nothing?
> can't see the value of the 50 watter. In case of
> extremely high plate
> current, such as hitting it with full drive and no load,
> say 2 amps of plate
> current flows and even then the big resistor only drops
> 100 volts, and heats
> up 200 watts. I wouldn't expect a wirewound power resistor
> to fail
> immediately under these circumstances, and I can't see how
> it is protecting
> anything.
> What am I missing here? Is the 50 watter the wrong type?
It's for fault limiting. For plate current protection you
need a real fuse.
73 Tom
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