Looking at the heater string, it looks like the two tubes adds up to 81V. One could avoid the power dropping resistor by using an unfiltered diode off the powerline to heat these heaters to their correct temperature. I'd use a separate diode connected between the grey wire feeding the bridge rectifier and its cathode to pin 12 of V1. Disconnect the dropping resistor. The extra diode would allow only half the AC waveform thru to the heaters. As heat is power, and power is volts squared over resistance, you'd want a heater string of 120V * (1/square root of 2) = 83V. Not 60V. The diode divides the delivered power by 2.
This would cut waste heat inside the cabinet. Today a diode would be cheaper than a power resistor.
02-22-2016, 04:50 PM
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