# 1/8 w resistor for 12v?

I bought one radio shack led with built-in 1/8w 680ohm resistors to see how it was made.

I read the 1/8w resistor wouldn't work and it would burn up? Using something like 1/2w would be better.

I tested it with a 9v battery and I didn't feel it getting hotter. Maybe just warm?

I got some LEDs off ebay many years ago and I have no idea what their specs are.

They are 3mm led bulbs.

I decided to make my own by using 5 1/8w 150ohms resistors. 150 ohms are the only resistors I have on hand.

With the combined resistance of 750 ohms, how much voltage is going to the LED?

Will I be fine with just using 4 150 ohms resistors?

Will is be dangerous to use the 1/8w resistors? I don't want a fire.

Will having more resistors also help spread out the heat among the resistors?

I drive 4 hours a day, so the light will only be on for that duration.

Thanks

active| newest | oldestVbat = Vres + Vled

Vres = Vbat - Vled

I = Vres/R = (Vbat - Vled)/R

But it would be helpful to have some actual numbers, so I am guessing the LEDs you are using have Vled=3.6V and Imax=20 mA = 0.020A. Numbers like that would be typical for those little 3mm diameter white LEDs. Here I am assuming your LEDs are white. LED forward voltage depends on color, and white is about 3.6V.

Also assuming Vbat is somewhere in the range: 14> Vbat >10 volts

Also asssuming you want R to be an integer multiple of 150 ohms, i.e you will make R as a series n 150 ohm resistors in series.

Based on those assumptions, I am going to recommend using either:

Plan A: 4 150 ohm resistorsand1 white LEDin series(R=600 ohms, Vled=3.6V, Imax=17.3mA, Imin=10.7mA)

or

Plan B:3 150 ohm resistorsand2 white LEDsin series(R=450, Vled=7.2V, Imax=15.1mA, Imin=6.2mA)

For a string of n identical resistors in series, the power dissipated by each resistor is the same. (Since they share the same I, and for each, P=I^2*R) For example if you have a series string of 3 resistors, then the power dissipated by each single resistor is (1/3) the power dissipated by all of them. Conversely the power rating of the string, as a whole, is 3 times the power rating of each individual resistor.

So the maximum power dissipation

per resistorforPlan A:

(1/4)*(10.4^2/600) = 0.045067 = 45 mW per resistor

Plan B:

(1/3)*(6.8^2/450) = 0.034252 = 34 mW per resistor

and both those numbers are less than 125 mW = (1/8)W

Note that maximum number is based on the highest expected battery voltage, Vbat=14.0V. That gives a voltage drop of 14.0-3.6 =10.4V across the R=600ohms for plan A, and a voltage drop of 14.0-7.2=6.8V across R=450ohms for plan B.

Also if you run out of white LEDs, the cheapest place to find more of them might not be RadioShack(r) or eBay. If you have a RadioShack in your town, you might also have a Dollar Store of some kind, and if you can find something to smash apart that contains 2 or 3 white LEDs, such as a cheap white-LED flashlight

http://www.dollartree.com/catalog/search.cmd?form_state=searchForm&keyword=873973

Then that might be the best, easiest, deal you can find on white LEDs, at the time of this writing.

I plan to go Plan A, would using 5 150 Ohms resistor would be over doing it?

I tested it in my car with 5 150 Ohm resistors and 1 led, it looked ok, if 4 is recommended, I can just cut off the 5th one.

Thanks

Imax = (14-3.6)/750 = 0.013867 = 14 mA

(1/5)*((14-3.6)^2/750) = 0.028843 = 29 mW per resistor

Really, if it works, why mess with it? It would be less work just to leave it like it is, and you know, declare victory.

;-)

You can calculate what you need from that information.

R= (12 - forward voltage) / rated current

and Power rating = rated current ^2 x R

Steve