[Herald1360]

Just for the record, in case it's not clear, in the calculations above, V=voltage, I=current and R=resistance.

(I just wondered for the first time "Why I?".... apparently it came from M Ampère from the phrase "intensité du courant".)


Anyway, just to be clear, you can measure the heater current by measuring the voltage across R39 and dividing by its resistance. (Should be 190 ohms, but check with the set off and disconnected from the supply). The supply paths for the rectifier feed and heater feed are split and only heater current flows through R39.


In round figures, you need to lose about another 25V in the heater chain series resistors which at the 100mA heater current will need 250 ohms dissipating 2.5W. Simplest fix is to add a 220 ohm 5 or 6 watt wirewouind resistor in series with R39. That'll be "near enough" in engineering terms!


The higher HT will not be a problem unless the voltage rating of any of the capacitors subjected to HT related voltages is exceeded. If you do need to reduce it, extra resistance in series with R38 can be used. Value would need some experiment, but start around 100 ohms with a 2 or 3 watt rating.