Still searching for an explanation why an anode-grid arc might blow the
cathode zener, something might have come up from an off-list exchange
with K5GW (reproduced here with permission).
>
>I have lost many cathode zener bias diodes in my 2m 8877 amp over the years
>(none since installing the 50ohm plate resistor). I always assumed that when
>the plate arced to ground and by way of grounded grid, perhaps plate to grid,
>that the HV power supply b plus was almost instantly placed at ground
>potential. If this is the case, the B minus lead must also nearly instantly
>rise to a negative value close to the full HV dc voltage less whatever amount
>has been lost during the nearly instant period of time. This being the case,
>the cathode is now negative several thousand volts compared to the still
>grounded grid. Looks to me like a large current would flow in this case,
>possibly damaging either/both cathode and grid. I know enough current flows to
>blow the smithereens out of the grid meter!
>
>Am I overlooking something obvious here? 73 de Gerald, K5GW
No - and maybe you've just provided the final clue!!
As you say (and EIMAC confirm) the arc pulls the B+ line very close to
chassis potential and tries to push the B- line a long way negative. If
you sketch out a simplified circuit, it goes something like this:
1. From B+ rail
2. Through surge limiting resistor (if used)
3. Arc to grid/chassis (about 50V according to EIMAC)
4. From chassis "upwards" through grid current meter
5. Through anode current meter
6. To B- rail.
The bias zener is connected to the junction of the two meters. The surge
current is trying to drive this point a long way negative (with respect
to chassis). If it succeeds, it will also drive the tube cathode
negative, and then the cathode-grid part of the tube will conduct like a
diode. Then part of the surge current will be flowing through the grid,
cathode and zener - maybe enough to blow the zener and cause the much
publicized damage to the grid.
[Note to the reflector: this is a good case for a small GIF file to make
the circuit more clear. If you want, and with the lsit-owner's
permission, I could post one.]
Part of the solution is the surge limiting resistor in the B+ lead. It's
very significant that you haven't lost a cathode zener since you added
this resistor.
The other part of the solution is a pair of enormous diodes across the
meters, capable of handling the full surge current. There is a circuit
on AG6K's web pages.
I use a packaged 25A bridge rectifier, connected like this:
"+" terminal to chassis.
Both "AC" teminals connected together, and connected to negative of G1
meter, positive of anode meter. This is also where the bias zener
connects in.
"-" terminal of rectifier to negative of anode current meter.
This puts two big diodes in parallel with each meter. If the normal
voltage drop across each meter is less than 0.5V, the diodes will have
no effect on the calibration.
When there is an arc, the surge current stretches the voltage drop
across the diodes to maybe a few volts each, but the diodes hold it
there until a fuse blows. The meters can usually handle that kind of
mild overload, and come up smiling.
In most published circuits you see wimpy 1N400x 1A diodes "protecting"
the meters. Well, maybe their surge current rating is high enough, and
maybe they would fail short-circuit - but nobody will guarantee that.
It's gambling with several $$$ worth of meters and other parts, and
possibly the tube as well. For a few pennies more it's worth using
really big diodes that you KNOW will handle the whole surge.
You can also use a 1N400x connected the reverse way for good luck - but
I can't see how a surge could ever go "backwards" in the normal GG
circuit.
73 from Ian G3SEK Editor, 'The VHF/UHF DX Book'
'In Practice' columnist for RadCom (RSGB)
http://www.ifwtech.demon.co.uk/g3sek
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