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In Reply to: RE: question on ohm's law posted by vinnie2 on April 26, 2016 at 05:31:38
The suggestion (Ohm's law) not entirely good.
If you use tube, the first momentum to know tis tube's datasheet.
As you can see, 290V as the #26 tube's anode voltage is unrealistic (if the cathode is almost grounded).
The grid bias voltage (for example -9V) and the target anode current (6mA) specify exact (depends of the tube's condition) anode-cathode voltage (in this case 138V).
The anode resistor (as load) is minimum double of the tube's plate resistance (about 7k6), so 18-24k is enough.
If the anode resistor is 24k, the voltage over this resistor is 6mA*24k=144V
So, the calculated B+ is 138+144=282V
If your raw voltage is higher (400V), use Ohm's law for dropping resistor:
(400-282)/0.006= 19666.
Use 20k 2-5W resistor AND 22-47uF capacitor to shunt (AC grounded) 24K and 24k resistor common point.
Follow Ups:
Actually I am trying to use 110vdc on the plate, 400vdc-290vdc.
20K and 24k resistor common point.
Could you tap the voltage, a little further down the PSU rail? Maybe, where the voltage in closer to 300VDC? Maybe, in a bit more linear range for the tube's anode plate needs?
Don't think that is available, but I will check.
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