LED power-on indicator

Started by yunger, June 08, 2008, 06:31:38 PM

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yunger

does anyone know how to use a 3.5vdc 20ma LED light as a power-on indicator with a 120vac supply?
i tried just using a 10k resistor before the light and smoked the resistor....although the LED lit.

Seljer

wire a regular diode in parallel reverse as in http://www.muzique.com/news/?p=18 , I don't know how much LEDs can take in the opposite direction but i think 120V might kill it
and i think 10k should be enough as far as the resistor goes
and you might want to try half watt resistors as 1/4 watt ones might just not cut it

yunger

i'm not so concerned about the LED. it's the resistors that keep frying on me...even 1/2w.
the circuit is 120vac into a thermal switch that powers resistance wire like you would find in an electric blanket. maybe the heat from the heater wire is damaging my resistors.

dxm1

Hmm... Let's see. Power (in watts) is equal to the curent (in Amps) times the voltage (in Volts).  So, we're talking 120V times 0.020A. That comes out to... around... 2.4 watts. (!!!)  Maybe try a 3W 10K resistor - as well as the reversed rectifier diode (1N4001) to keep the PIV down in the LED...

yunger

oh, so that's what all that ohms law stuff is about!!!!!!!! i guess i should stick with my day job.
thank you both for the useful info.

R.G.

Be sure to put in the reverse diode. LEDs can be destroyed by reverse current with over a few volts. This is part of LEDs 101 for EEs.
R.G.

In response to the questions in the forum - PCB Layout for Musical Effects is available from The Book Patch. Search "PCB Layout" and it ought to appear.

yunger

should i use the 1n4003 since i'm dealing with higher voltages?

dxm1

Yes, a 4003 would be mo' betta.

Seljer

Quote from: dxm1 on June 08, 2008, 07:31:31 PM
Hmm... Let's see. Power (in watts) is equal to the curent (in Amps) times the voltage (in Volts).  So, we're talking 120V times 0.020A. That comes out to... around... 2.4 watts. (!!!)  Maybe try a 3W 10K resistor - as well as the reversed rectifier diode (1N4001) to keep the PIV down in the LED...

if we forget about the drop over the LED
(120 volts)^2 / 10kohms = 1.44 watts

hmm, yeah, a 1/2 watt wont cut it

scratch

you might want to check out the following discussion, on using a capacitor to drop voltage on the AC side ... I've never done it myself, but filed for future reference (like now).

http://cr4.globalspec.com/thread/13464/Capacitors-as-AC-Voltage-Drop-Components


Denis,
Nothing witty yet ...

new

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