LED indicator help

[fikri]

Hi !

I have 12V 3A tranny to drive a relay switching system plus the LED indicator. I can easily connect the 12V to power up the relay, but to power up the LED indicator, i have to lower the voltages. I use 100 ohm (in series with the LED) to lower it down to around 2V. Now the problem is, the Resistor will get too hot and it kills my LED one by one ! 
Am i doing this right ? or is there something that i should know more ?

Thanks !   

[burnt fingers]

OHMs law will help you here.  You should read up on it but in the meantime check out the bottom of this page.  I think it will clarify for you.

http://www.phantasmechanics.com/fcghost3.html

Scott

[R.G.]

An LED is a diode with roughly 2V drop in the forward conduction direction, which is how they're used. The LED will always have a 2V drop in the forward direction.

The LED current is then determined by the difference between the supply voltage (12V; is that AC or DC?? Did you rectify and filter, or are you putting AC on the LED?).

If you are using 12Vdc made from the 12Vac on the transformer, then you are putting 10V across the resistor/LED, and the 100 ohm resistor is letting through 10V/100ohm = 100ma. LEDs are usually rated for 20ma max, so the LED is quite possibly being damaged by the overcurrents. The resistor is dissipating 10V*0.1A, or 1W. A 1W resistor there will get hot enough to burn your finger.

By the way, you don't mention whether you're using the LED on AC or DC voltages. If AC, you need to connect a rectifier diode such as one of the 1N4001... 1N4007 series in anti-parallel so that when the voltage reverses, the LED is not reverse-conducting. This wears them out quickly. In addition, a "12Vac" supply to the LED makes the overcurrent worse. The peak of a 12Vac sine waveform is 12*1.414 = 16.9V. The peak LED current is then 16.9-2/100= 149ma. The resistor current is higher and the dissipation bigger too.

What to do?
1. Use a bigger resistor!! Put a 1K in there - or up to about 4.7K if the LED is still bright enough. The LED gets brighter up to about 20ma, but don't go over 20ma unless you have done the detailed design on the LED to know you are not killing it.
2. If you do not rectify/filter the 12Vac, use a silicon rectifier diode in antiparallel (A to K, K to A) to prevent LED reverse conduction.

[fikri]

Thanks burnt fingers ! (what happened exactly to your fingers ?  ). I will keep the articles for my further reference. And Thanks alot R.G ! Your explanation is very clear and straight forward (as always  )

[burnt fingers]

Fikiri,

Nothing serious really happend to my fingers. When I started doing this though, I kept burning my fingertips with the soldering iron because I wasn't paying attantion to what I was doing.  ( I've burnt my fair share of wire, componants, and even left the soldering iron on the wet sponge until there was fire).  Anyway, i just picket burnt fingers for my user name.

Scott