I finally looked at my discharge setup, into which I had put some research
(re: our recent discussion).
I used an 800K 25W resistor. The reason I used a wirewound resistor is that a
small carbon resistor can arc inside and fry, leaving you with an open circuit,
so when you think you're discharging, nothing is being drained.
V=IR, right?
Assuming a tube voltage of 10,000V:
V/R=I, 10000/800000 = .0125
So that's .0125A, or 12.5mA. But at 10000V:
P=IV, or P=(.0125)(10000)=125W
That's a whole lot of watts. But as we discharge that voltage very quickly,
over a wirewound resistor, that's not too bad.
I used a new screwdriver with a heavy handle and a 7" metal shank terminating
in a narrow flat tip. I used a 16 gauge lamp (I used both wires) cord with
heavy insulation, wrapping the wire around the screwdriver for a good
mechanical connection soldering the wire to the metal shank (I chose a
screwdriver with a frosted finish for better adherence) and then tied the cord
to the handle with a cable tie, and then used multiple layers of heavy
heatshrink over the solder joint to serve as a strain relief.
I put the resistor inline, with multiple levels of heat shrink over the ends,
wrapping the upper levels of heatshrink onto the resistor body to prevent
lead fatigue and failure. I then used two final layers of heatshrink on
top of the entire resistor portion of the cable, both to insulate the
resistor and provide additional strain relief.
Silicone-insulated high-voltage test lead wire would be a better choice, but I
did not have any at the time.
I would always measure the resistance of the discharge lead before and after
use, so as to be certain the cable did not fail. And remember that the
portion of the wire between the screwdriver and the resistor is at the anode
cap voltage potential during discharge, so avoid contact with it and inspect
it for damage. Treat it like an expensive test tool, and if the resistor or
wiring becomes bent or damaged, throw it away and make a new one.
-Chris
