<<<One caution however: there is no protection for the meter if there
is an arc and
the current rises to a large value. The meter could be destroyed. The best way
to protect the meter is to use a pair of heavy duty diodes such as the 6A10,
connected in anti-parallel (back to back with reverse polarity) across the
meter. BUT!! Most meters of this type have a very low internal resistance,
typically less than .1 ohms. and the diodes will not begin to conduct in time to
protect the meter. The solution is to connect a resistor in series with the
meter and connect the diodes across both meter and resistor. The value of the
resistor is chosen so that it and the meter will show a drop of about .5 volts
when the current is about 120-150% of full scale. For example, if the meter is 1
amp full scale and has a resistance of .1 ohm, you would use a ..317 ohm
resistor in series for a total of .417 ohms. .417 ohms will develop a drop of .5
volts at 1.2 amps and the diodes will conduct, shunting any further current
around the meter. Of course, it is not likely you will fine a resistor of
exactly .317 ohms, but you should get as close as you can.>>>
I'm not sure of the meaning of "anti-parallel." do you mean two
diodes anode to cahode to anode to cathode from one end of the
resistor to the + side of the meter, or by "back to back" do you mean
diode cathode to anode to anode to cathode?
Also, what about the diodes must be known in order to choose a drop of
1/2 volt for the meter/resistor combination? It seems like there is
some information there that you left out. A conduction threshold
voltage? it looks like if you reverse the diodes you are dealing with
PIV which on a 6A10 is 1 kv isn't it? So I'm obviously confused about
some details.
<<<Me too. I don't think there is any wonderfully safe way, as all three
power supplies - HV, grid and screen - are only connected to chassis
through the G2+ connection. In particular it needs a fully floating HV
supply with a very well insulated negative side.>>>
What is a G2+ connection?
<<<Then the resistor forms the last line of defence to prevent
the B-minus rail from floating far away from ground.>>>
Isn't B- far removed from ground potential anyway, usually around -3kv
+- a few 100 v. in typical ham amp power supplies? there is clearly
something I'm missing here also.
73
Rob
K5UJ
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