Baronvonrex420 said:

Let me start by "Yes, I know I'm supposed to have a current limiting resister

to each LED", but I don't. This project requires over 100 LED's and I need to

reduce drilling where I can.

I need to drive 5 and 6 3VDC, 20 ma LEDS in par. but I can't find a formula

for calculating the resister value. The life expectancy of this product is

less than 100 hours and current will be supplied with a 7812 regulated 12 VDC

current from a car battery.

I tried 33 ohm 1/2 watt, 100 ohm 1/4 watt, and tonight I tried a 10 ohm 1/2

watt resister. The LEDS work great but the resisters are hotter than heck!

What do I need to do to bring the resister temps down or calculate the actual

value I need? I need to encase this in resin and I can't afford the heat.

Any help would be appreciated! TIA!

Newbie

Okay, let's take the case of 6 LEDs. I don't know why you MUST put them in

parallel, which is particularly wasteful of power (meaning lots of heat made for

nothing), but let me suggest a better strategy.

Since your supply is 12 volts, and you have LEDs that run at 3 volts, why

not put three LEDs in series with a single resistor, and do this twice? Now,

you greatly simplify the picture and you don't drop 9 volts in the resistors.

Here is how you calculate your resistance and the power dissipation.

Three LEDs in series will drop 9 volts, leaving 3 volts to drop in the

resistor. You know that in series, the current consumption remains unchanged,

so 20 mA is the figure we use. Since R = E/I, we divide 3 volts by .02 amps and

get 150 ohms. Now, since P = EI, the resistor will have to dissipate (3 volts x

..02 amps) watts as heat. This is 0.06 watts. Almost any small resistor can

handle this with no trouble, even the little 1/8 watt ones.

You will put two of these assemblies in parallel, meaning that you will have

6 LEDs and 2 resistors for one of these assemblies. Total power turned to heat

is a mere 0.12 watts per pair of LED strings.

With your 100 LEDs on at once, if you were running all of them in parallel

and dropping 9 volts through your resistor, you would have been dissipating 18

watts as heat if and only if your resistor was (9 volts / 2 amps) or 4.5 ohms.

But it is very likely that you would also have had a few dim ones, and not only

that, 3/4 of the power from your battery would be heating up a resistor instead

of supplying illumination.

If you wanted to string them in threes (as I mentioned above), then fully

3/4 of the power would go to lighting and only 1/4 would be wasted.

Cheers!

Chip Shults

My robotics, space and CGI web page -

http://home.cfl.rr.com/aichip