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Old 21st October 2012, 05:58 AM   #21
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The idea of using an LED in a cathode is to maintain bias voltage in no dependance to anode current. See LED characteristics.
LED is not a cure for heavy loads on B+. Well designed HV power supply is.
What you can do is to calculate what is ΔVbias in regard to expected ΔB+.
E.g.
B=+302V
Vbias = +2V
Ibias=10mA (one can take anode current change under the load onto account, i.e. ΔB influence on anode current)
ΔB +/- 5% or 30,2V (that is heavy I guess)
Rbias=(302V-2V)/10mA=30kΩ
Now,
30,2V/30kΩ=~1mA or 0,5mA around LED op point. Assume dynamic resistance of a LED is 5Ω
ΔVbias=1mA * 5Ω = 5mV
It is early in PL, so I may have made a mistake somwhere.
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Old 21st October 2012, 06:04 AM   #22
dgta is offline dgta  United States
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Yagoolar, you didn't understand his question. He wants to avoid the extra 6mA current draw necessary to put more current through the LED than the tube is passing.
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Old 21st October 2012, 06:09 AM   #23
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Quote:
Originally Posted by dgta View Post
Yagoolar, you didn't understand his question. He wants to avoid the extra 6mA current draw necessary to put more current through the LED than the tube is passing.
I was referring to post #19. The question was to use LV PSU and lower value resistor to feed LED. It is not THAT early

My intention was to ask question "what for" if that works well.
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