- across projector LED: 2.8V
- across other LED: 2.7V
- across proposed joule thief location (see last post): 6.25V
- across 1Ohm resistor: 0.017V
- across 33Ohm resistor: 0.57V
- across photoresistor: 1MOhm
With the tape off (LEDs off):
- across proposed joule thief location: 1.1V
- across 33Ohm resistor: 0.6V
- across transistor (collector-emitter): 0.5V
- across photoresistor: 4kOhm, 0.43V
What this is telling me is that some sort of voltage regulation is happening. I think it has something to do with the zener diodes. The AC-DC converter looks a lot like typical capacitor (transformerless) DC power supply, but with zener diodes as part of the rectifier. Looking at the circuit diagram in last post, can anyone tell me how this power supply works?
I was close with the 5V guess. It's actually around 6.2V when the LEDs are on. But the problem is that this drops to 1.1V when the LEDs are off (light room). While that's good for power dissipation while off (estimating about 20mW using 33Ohm resistor current and 1.1V), it means that if I have a 6.25V joule thief hooked up to it in a light room, a lot of current will have to be dissipated by the 33Ohm resistor and the transistor, which is bad.
One (very non-elegant) work around would be to cut a trace, have a way to switch between JT or AC operation, have a transistor inline with the JT, and a tiny microcontroller with leads spliced across the photoresistor. In JT mode, the microcontroller would read the photoresistor and tell the JT transistor to turn on or off. In AC mode, the JT would be taken out of the circuit by a switch and it would operate as it was designed to.
Another thing I might be able to do is splice off of the photoresistor-base transistor connection and use that to turn a transistor in the JT circuit on or off directly.
I'll have to think about this some more. Figuring out how the power supply works would probably help.
-learn how the power supply works
-design JT circuit for 6.25V and ~20mA
-figure out efficient way to integrate JT circuit