Quote:
Originally Posted by johnwmclean /img/forum/go_quote.gif
I need a little help figuring out the value of a resistor I need to run in series from my psu to my latched Bulgin switch.
The psu (twisted pair LCBPS) is 15V dc and the LED in the Bulgin switch is 12V dc. I tried a couple of online calculation sites, I still don’t get it, any help would be most appreciated.
Cheers
John

That's because you need the current to know how to solve the threevariable equation represented by Ohm's law. You can't solve for one variable when there are two.
Most LED's are rated for a maximum of 20ma. So, if we plug in 0.02 with your voltages, you'd have: 15V12V = 0.02 x R, R = 3V/0.02 = 150 ohms.
Personally, I try to size LED resistors so that they're no more than half that current rating  there are other variables that either make this calc too much trouble or are unaccountable. (Technically, you'd include some calculated resistance of the LED in the total series resistance and then correct accordingly. Also, there may be variances on your voltage that are not presently known.) If you exceed the LED's current limit by much for an extended period  they'll burn out. Yet, brightness is diminished by very little, even at half current rating.
So, if we say 300 ohms, that would result in I = 3V/300 = 0.01A (10ma). Note that you should also check the Power equation to insure you have the proper rating for the resistor. P = (I^2) * R = 0.0001 * 300 = 0.03W. So, a 1/8th W resistor would be plenty. It's also interesting to see that if we used the original values, we'd end up with 0.06W, which is close to onehalf of 0.125W (1/8th W). We typically size resistors for twice the power rating for a good safety factor, so it would be right on the edge of bumping up to a 1/4W resistor.
Hope that helps.
P.S. I didn't see anything on the switch's data sheet that would be of much help, unfortunately.