# 1/8 w resistor for 12v?

Hi, I've been trying to add a LED light for my glove box, since it didn't come with any.

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

This is the old (and good) trick of using a resistor, placed in series with a battery and a LED, to limit the current to the LED. The assumptions that go into this trick are simply that the battery voltage is constant, and the voltage drop across the LED is also constant (over a wide range of current). So there is a constant voltage drop across that resistor, and the current is basically determined by the size of the resistor you choose. This current is just the voltage drop across the resistor divided by R.

Vbat = 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.

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Thank you for your detailed explanation! I understand this much more now.

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

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750 ohms is good too. That would actually draw less current than 600 ohms, and the power dissipation per resistor is less too.

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.

;-)

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Thanks, I went out to buy some 1/2w 680ohms resistors to try out, and I feel like the LED is dimmer now. Is it me, or is the 1/2w resistor doing it? thanks

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I dunno. It should be approximately the same current, and brightness.

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The resistor depends on the LED. We need to know the rated current and forward voltage, to advise you further.

You can calculate what you need from that information.

R= (12 - forward voltage) / rated current

and Power rating = rated current ^2 x R

Steve

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The voltage isn't important its the amount of current the LED draws. You place a resistor in series with an LED to help limit the current from the power source. So that resistor needs to be able to handle the amount of power coming into it. You can try it with 4 or 5 150 Ohm 1/8W resistors and see what happens. If that isn't enough resistance then you will blow the LED. If the resistors can't handle the current then you will smoke some resistors. If you want to be safe then wire several of your 150 Ohm resistors into a series parallel arrangement. If you need 750 Ohms then wire up 4 groups of LEDs in series to create the 750 Ohms needed. Then wire the 4 groups together in parallel so each group is sharing the load. Then you have basically made a 750 Ohm 1/2W resistor.

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