Please sanity check this simple calculation...

I'm not very confident in my understanding of this stuff. I'm confident with the math, but am I considering everything?

I salvaged a LED strip from a laser printer. It has two parallel sets of four LEDs in series. Each set of LEDs is in series with a 5.1K resistor. (I read the color code and then verified with a multimeter.)

To produce a current of 20mA (ballpark what LEDs take), this circuit would require about 100V... right?

I'm thinking I'll probably solder in some different resistors in parallel with the old ones rather than play around with mains voltage...

Can you set out your calculation?

Four LEDs. White? Red? Green?

5k will take approximately 100V to get 20mA, plus the voltage drop across the LEDs. That seems a bit strange, though. How did you determine that the resistors are 5.1k? That's 2W in the resistors.

If all that is true, I'd definitely go with lower resistance. But we need to know how much voltage is dropped across the LEDs.

kjkrum:
To produce a current of 20mA (ballpark what LEDs take), this circuit would require about 100V... right?

It would also mean each resistor is dissipating 2W of heat.

ie. Your LED board would be a 16W heater, and that makes no sense at all.

polymorph:
Four LEDs. White? Red? Green?

Unknown. I think it's the "discharge lamp" described here: http://computer.howstuffworks.com/laser-printer3.htm

polymorph:
How did you determine that the resistors are 5.1k?

As I said, I read the color code and then verified with a meter.

fungus:
Your LED board would be a 16W heater, and that makes no sense at all.

Yeah, I thought that was weird because the power supply must provide some lower voltages for the logic boards. But doesn't the current through a LED vary with temperature? Maybe it was a cheap and easy way to make temperature have little effect on the current in the circuit. This came from a printer that was made in 1991. And the LED strip lived a couple inches away from a 150W heater (the fuser roller) so what'a another 16W, right?