Generating bias with a cathode resistor is not fixed bias but self-bias (or auto-bias).
Neither your formula nor your circuit make sense.
An ammeter must be wired in series with the current, so that all current runs through the meter.
There might be a possibility to wire the meter in place of Rk.
But first measure the DC resistance of the meter and post the full original schematic.
Neither your formula nor your circuit make sense.
An ammeter must be wired in series with the current, so that all current runs through the meter.
There might be a possibility to wire the meter in place of Rk.
But first measure the DC resistance of the meter and post the full original schematic.
Most of the mA meter doesn't require any external resistor. They just need to connect in series with the cathode resistor of the power tube. Simple. No calculation is needed. The meter is with very low resistance. Therefore, it won't affect the bias of the power tube at all.
Johnny
Johnny
I just saw that the value of the cathode resistor is 750R.
The meter might have a DCR of a few Ohms.
So wiring it in series with Rk should work.
The meter might have a DCR of a few Ohms.
So wiring it in series with Rk should work.
I assume is the same as this
It is just an ammeter without a shunt inside made for cathode resistor bias , with advantages described in the video . If internal resistance is 1K6 and 1mA for full scale , then it would need max 1,6V . Then you calculate the cathode sensing resistor for fixed bias , surely 10ohm is not enough , but 20-30ohm or even more would not hurt the operation of the tube ( for direct heated tubes with high value cathode bias resistors ) . Put the meter in parallel with this resistor and that's it .
Last edited: