Hey-o… if anyone’s still around this time of year, I’ve got a question for ya.
I want to attach 2 LED’s in series to a 12v power supply from my computer (EFY5366x). I just wanna make sure I have the the calculations for the resistor right (see diagram below).
Their total current draw is 60mA, and their total voltage drop is 3.6V. So
12v - 3.6v = 8.4v
And from V=IR, we get
R = 8.4v / (60/1000)A = 140 ohm.
Is that right, or do I not understand the problem at all? :ahh:
Also, I seem to remember that you’re supposed to put the resistor before the LEDs… is this right?
[Edit]Just realized the LED’s I linked to aren’t the 1.8v 30mA I put into the diagram… for sake of argument, let’s say I was using 1.8v 30mA LED’s…
First of all, 30ma seems a little high for a yellow LED. I’m used to 20ma for yellows, higher for blue and white, although it will work higher (and be brighter). Since it’s just your computer, it won’t matter if they burn out in 3 years rather then 10.
Second, since they are series, the total current draw for both is 30ma, not 60ma.
It doesn’t matter whether you put the resistor before or after the LEDs.
*Originally posted by Joe Ross *
**First of all, 30ma seems a little high for a yellow LED. I’m used to 20ma for yellows, higher for blue and white, although it will work higher (and be brighter). Since it’s just your computer, it won’t matter if they burn out in 3 years rather then 10.
**
Yeah, I realized after I posted it that the specs I listed were for a different LED - not the one I linked to.
*Originally posted by Joe Ross *
**Second, since they are series, the total current draw for both is 30ma, not 60ma.
**
3 weeks out of school and you start to mix up your physics already. whoops!
Anyways, fixing that mistake, that means my resister should be a 280ohm?
Super,
Here’s a couple of quick steps and some electrical info. LED’s can be considered to have 2 volt drops when forward biased and current in series circuits remains the same throughout. So you do not have 60 ma total. If they are high brightness LED you might have 30ma but 20ma is more the standard. So for calculations…
12v - 2(2v) = 8 volts (resistor drop)
so
R = 8/.030 = 266 ohms.
