Re: Additional info

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In Reply to: Additional info posted by SLM on April 26, 2002 at 02:02:54:

If I understand correctly, in your existing design, the voltage for the plate of the driver tube is obtained from the voltage drop across the cathode resistor of the output tube.
If this is the case, then the only thing you will need to be concerned about is how much AC drive voltage you need on the grid of the output tube to obtain the output you're after.
Assuming 200 Volts available and 150 Volts on the plate of the driver, you have 50 volts to play with, less a minimum of 5 Volts compliance for the C4S, so your AC drive will be limited to a peak of 45 Volts. Paul and VS, please chip in here if I have gone astray!
If this is adequate for your purposes, then the only other thing to be aware of is the 2mA additional current that you will need to bias the LEDs on the C4S. In other words, if you want 35mA for the tube, the C4S will require 37mA, 35mA for the tube and 2mA for the LED bias. I mention this so that you may figure out how that extra 2mA is going to affect the voltage drop across the cathode resistor of the driver tube.
Hope this helps,
Jim

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Topic - Voltage drop across c4s - SLM 15:19:41 04/25/02 (7)

Thanks all you're a great help !!! - SLM 15:13:35 04/26/02 (0)

Re: Voltage drop across c4s - VoltSecond 21:19:27 04/25/02 (0)

Re: Voltage drop across c4s - SpongeIt 20:18:43 04/25/02 (0)

Re: Voltage drop across c4s - Jim Carlon 18:53:09 04/25/02 (3)

Additional info - SLM 02:02:54 04/26/02 (2)

Re: Additional info - Jim Carlon 11:58:14 04/26/02 (0)

Re: Additional info - AnandR 06:35:18 04/26/02 (0)

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Re: Additional info - Jim Carlon 11:58:14 04/26/02 (0)

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