Originally Posted by JimDilks:
... I'm not sure I understand how the shorting of one resistor works, it sounds reasonable. I ran a jumper over to the output side of the resistors that are now wired in parallel. Everything works, but for how long?
Houston, we have a problem.
Some background. The higher the voltage the higher the power/heat. The lower the net heater resistance, the more power/heat is generated. In other words there are two things going on in that affect heat, (1) track voltage and (2) the net heater resistance.
As GGG notes, the Q2 transistor switches in the 2nd resistor when the track voltage reaches a threshold. So to use your measurement of 20 ohm resistors, at low track voltage the net heater is 20 ohms (only one resistor active); at high track voltage the heater is 20+20=40 ohms (both resistors active but in series). So the effect of the increased voltage is crudely demoted by the additional net resistance. That's about as a "simple" a regulator as you can make.
If you are seeing BOTH resistor "cherry-red" at low track voltage and/or you wired the two resistors in parallel, then those resistors are not long for the world. In other words you have an effective heater resistance of 20/2=10 ohms which does not change even when track voltage exceeds the simple regulator trip voltage. So as soon as you run the engine at running speeds with smoke on, those resistors will likely overheat and fail or the circuit board will further burn up.