16-08-2021, 03:02 PM
From first principles.
With a diode dropper the power is switched on for half the time. So the heater or heater chain is run at half power. Power = V*I and V = I * R (ohms law) So for constant R, power is proportional to square of V. So we'll need to divide the incoming voltage by SQRT(2). Approx 1.414. Which is the same as multiplying by 0.707.
At these low voltages the diode drop will also be significant. So reckon on losing another 0.7V or so. You'll end up with about 8.6V on the valve heater. This is fine for a nominal 9V valve heater.
If you have a true RMS meter you can measure the voltage. Otherwise it's possible to calculate what an ordinary averaging meter would read but I really can't be bothered.
With a diode dropper the power is switched on for half the time. So the heater or heater chain is run at half power. Power = V*I and V = I * R (ohms law) So for constant R, power is proportional to square of V. So we'll need to divide the incoming voltage by SQRT(2). Approx 1.414. Which is the same as multiplying by 0.707.
At these low voltages the diode drop will also be significant. So reckon on losing another 0.7V or so. You'll end up with about 8.6V on the valve heater. This is fine for a nominal 9V valve heater.
If you have a true RMS meter you can measure the voltage. Otherwise it's possible to calculate what an ordinary averaging meter would read but I really can't be bothered.
www.borinsky.co.uk Jeffrey Borinsky www.becg.tv
