Hi guys, I have an LED indicator light which is supposed to let me know if power is getting to a certain part of my circuit. That part of the circuit is getting ~173 VAC. I have only a 110 VAC LED indicator. I figure I need to put a resistor in series with the LED, but I'm really not sure what value, or how to do the analysis. The LED indicator I have is designed for 110 VAC, but I'm not sure how. Its in a small case so it may have some circuitry like a diode and/or capacitor. Could you let me know if an LED in series is the best way to drop the voltage for the LED (I'm really limited here since I can't change the circuit or the LED directly)? Also how would one do the analysis be hand to determine how much voltage will be dropped across the resistor (assuming I know how much current the LED takes, which I don't). Thanks.
Use a capacitor to have a voltage drop. Choose the value as per Xc=1/2*pi*f*C depending on the voltage drop. Don't forget to check the capacitor voltage aroud 1.5 times the AC voltage
Generally a LED using high voltage (ie 110VAC) MUST have a diode across it, as they have a really low PIV. After that condition is met a low value capacitor and resistor is needed. Truth to tell, I've never heard of a 110V LED indicators, I suspect it is something else (there are lots of gadgets that can do this better than an LED, and I'm not thinking of a neon).
Any LED will light up with DC and AC, here is what you do: Now since your doing AC you'll need to find the rms voltage: 0.707 X 173 = about 123V. So imagine if you had a lightbulb (LED would be fried, just example) with 100V DC and 100VAC P-P. Will the lightbulb be more brilliant with the DC or AC? The DC! Because DC is a steady voltage while AC varies. So the rms is to take that AC and basically see how bright the bulb would light compared to DC. So we have 123 rms V. Now for the LED: You could actually use any LED but you need to know the Forward voltage drop along with which current this happens at. Most LED's typically have a forward voltage drop of 3.2V with 20mA. So 123-3.2 = 128.8 or 129V. So R = E/I - 129V/20mA = 6,450Ω resistor. Now make sure you have a higher power resistor than a 1/4 of a watt because your using high power! Your LED is different so check your specs for the forward voltage drop and the current at which that happens.