17-06-2022, 05:37 PM
If it's a pentode or beam tetrode, measure the Ia (not the Ik which is greater by the screen current!) and also measure Vak.
Then optimum impedance for maximum power is approximately (Vak - 25) / Ia.
This is valid as long as the grid never needs to swing positive at the maximum signal current peaks.
For triodes it's rather harder, because when the anode voltage falls, it reduces its own current so the grid drive has to be much larger - and avoiding positive grid is not trivial. But a reasonable guide is Rl = 2 x ra.
(If it's a circuit with severely limited signal voltage, such that the valve is never driven to non-linearity, then you can apply the Maximum Power Transfer theorem and use Rl = ra. But that's a very rare scenario).
Then optimum impedance for maximum power is approximately (Vak - 25) / Ia.
This is valid as long as the grid never needs to swing positive at the maximum signal current peaks.
For triodes it's rather harder, because when the anode voltage falls, it reduces its own current so the grid drive has to be much larger - and avoiding positive grid is not trivial. But a reasonable guide is Rl = 2 x ra.
(If it's a circuit with severely limited signal voltage, such that the valve is never driven to non-linearity, then you can apply the Maximum Power Transfer theorem and use Rl = ra. But that's a very rare scenario).
