hi. i'm not sure if this is the right place to post this but i've been googling my question for about 2 hours and haven't found an answer!

I am working on an LED project for my car (12V supply)... and i calculated the resistors needed to be 470ohm. The resistor can be either 1/4W or 1/2W because the calculated resistor wattage needed is 0.16. I decided to get the 1/2W resistors from radio shack but found them to be too big as there is a limited amount of space to work around... therefore i am leaning toward some 1206 SMD resistors but all of the ones I found were 1/8W... from reading an article on google... it said SMD resistors dissapiate heat very well...

do u think the 1/8W SMD resistors hold up without blowing or overheating?? any other ideas?

 

If the SMT resistor is rated at 1/8 W ( 0.125 W), your calculated 0.16 W exceeds that value. As a rule of thumb we like to buy a resistor of twice the calculated dissipation. This is a derating factor to increase the life and reliability of components.

If you are a hobbiest, you can probably get away with the 0.125 W SMT resistor. I would not expect it to smoke. Build it, try it. However, if you are an engineer designing a reliable product for production, you need to derate the resistor.

A design engineer would also look at the highest voltage that would occurr during charging of the battery and recalculate the resistor dissipation based on this voltage.

If the ambient operating temperature is high enough, the resistor cannot dissipate the nominal rated value. There may be a derating graph in the data sheets for the resistor showing what it can dissipate versus temperature. In your case, the 2x rule of thumb is probably sufficient-- unless you are an aerospace engineer.

 

Additionally, you might want to check whether the 1/8W is specified with the resistor mounted on a board (what density?) on a minimum area of copper and your application conform to these conditions. Some manufacturers specify these to get better values in their datasheets.

 

OK, sounds like they may just be Current limit resistors for your LED's..?? Am I correct..?
If this is true, then why not just reduce the drive current and recalculate your values.
An example just off the top of my head, LED has approx 2 volts drop @ 20mA, The limit resistor will be 500ohms (10volts drop) = 0.2 watts resistor dissappation
IF you halve the drive current, the LED will still be reasonably bright, the limit resistor should now be 1K, and the dissappation has dropped to only 0.1 W.
Driving LED's is fairly non critical, and there can be some wide tolerances as long as parameters arn't exceeded.
Other applications can have fine tolerances however, so disregard this advice if this isn't how the resistors are being used, and perhaps supply a circuit....

 

There are several ways to come around this. If you have the space on your pcb, I would have split the resistor into, let's say 3. 3 resistors of 150 ohm in series makes 450 ohm total and a possible power dissipation of 3/8W or 0.375W this is well above the 2x value of 0.32W.

I'm a bit confused of your wattage.
A 1206 SMD resistor will normally take 1/4W, at least in my lists. A 0806 SMD resistor takes 1/8W. These are from KOA, RK73series.

If you use 2x 1206 resistors to split the power, you can use any equal combination of serial or prallel connection. 2 X 1k in parallel or 2x 220/250 in series gives the same result.

Whatever you do, keep in mind that you use resistors with equal values. This way you can add the wattage directly.

TOK