Adding LEDs to a 40v EHX pedal

[dano12]

I'm doing a re-house of an EHX Frequency Modulator into a smaller enclosure. I'm also adding status LEDs for the two switches. The problem is that the pedal has a 40v power supply and LEDs want a lot less than that. I've played around with some pots and resistors trying to find the correct value to lower the voltage to the LEDs, but at 40 volts it appears I'd have to have a *huge* resistor value.

Any ideas?

[cd]

Just use a big resistor.  It's not perfect but good enough for an LED.  I have used them in amps to drop 250V or more down to run FETs at 12V.  47k-100k ought to do it.  Just watch the power rating of the resistor.  If you're dropping say 35V with 10mA current draw that's 350mW, or around 1/3W.  Which means you have to use a 1/2W or higher resistor.

[dano12]

Just use a big resistor.  It's not perfect but good enough for an LED.  I have used them in amps to drop 250V or more down to run FETs at 12V.  47k-100k ought to do it.  Just watch the power rating of the resistor.  If you're dropping say 35V with 10mA current draw that's 350mW, or around 1/3W.  Which means you have to use a 1/2W or higher resistor.

That's what I was thinking but the values you suggest aren't getting me there.

The adaptor puts out around 49vDC according to my meter. I grabbed the biggest resistor I had which is 4.1 megaohms. I put it in series with the positive end of the power supply. My meter says I've dropped it only to 33 volts.

If I try a 10 meg ohm resistor, it drops to 24v.

Am I missing something (i.e. an alternative to simply putting the resistor in series?)

[Seljer]

The resistor is limiting the current, not the voltage. As the LED is a diode, not another resistor (your multimeter shows a lower voltage when you hooked up the 10Mohm resistor because of its own internal resistance).

series resistor = (voltage - forward voltage of the LED) / LED current(up to 10mA or so)

and watch the power rating like mentioned