Replacement LED

[cpemma]

Originally posted by DreadPirate
OK, I think I got this figured out, thanks for the tips and links. What I need to do is remove the existing LED and check its resistance.

An LED is a diode, diodes don't obey Ohms Law. The forward voltage (nominally 2.1V) changes by only a fraction of a volt for currents from under 1mA to over 30mA.

Your maths looks OK for finding the resistor, but another thing about LEDs is they'll glow quite adequate for an indicator on 2mA or less. But there's bound to be a suitable resistor already in the circuit. Easy way to check is to fix a 1k across where the LED was and measure the voltage drop across the 1k. Knowing there's 20V altogether you can get the existing resistor value.

With round LEDs there's usually a flat on the rim indicating the cathode lead (which connects towards ground); handy to know if the leads have been cropped.

[glenncal]

"If there is no current flowing, there will be no voltage drop".

Most voltmeters have a high 'impedance' (formerly called "Ohms per Volt") so they don't load the circuit under test. So--if you measure a 20v source in series with an LED current limiting resistor, but without the LED in the circuit, you will still read 20v. Without the LED, it's an open circuit. The resistor will not have any significant current flowing, therefore (by Ohms Law) it will drop no voltage.

Likely there is already a limiting resistor in the circuit, so adding another resistor would lower the current to the new LED.