LED question.

[Davefx]

This is probably a dumb question, but all the same, here goes.  I can put 18v on an LED, can I not? I'm using an 8.2k limiting resistor with it, is that sufficient, or is it wise to run that high of voltage. I have only this 1 LED in my house and I don't wish to ruin it testing it. I know it's sad, but I'm not an overly stocked guy, and RS is closed.  

[smoguzbenjamin]

Ohm's law my friend! Just calculate what resistance you need at that voltage to get about 20 mA... so R=V/I is the formula you need, R = 18/0.02 = 900 Ohms. But to be safe, use the 8k2 resistor anyway. I've fried many LEDs by running them at their maximum current. I run LEDs of 9V with 10k resistors and they work fine that way. Note that with a 900 ohm resistor you're sucking 20mA of your power just driving the damn LED.

The current with 8k2 to settle my own curiosity: I = V/R : I = 18/8200 = 0.002 or 2mA.

I have Ohm's law on an A4 sized paper in front of me whenever I'm building!   Hope this helps!

[Davefx]

Thanks buddy..  Yes I DO know Ohm's law believe it or not!!  :lol:  For some reason, I wasn't sure if I wanted the 20ma in the equation, because I didn't want it to draw that much... Phooy... Sometimes I need that 2nd opinion...  Low self confidence, I guess.  Thanks again for the help..

[brett]

This is nit-picking, I know, but I can't help myself.  When calculating the current flow through an LED, you should allow for the voltage drop across the LED.  For example, at 9V, there's about 2V fall across the LED and 7V across the resistor, so the resistor for 20mA at 9V is 7/0.02 = 350 ohms, NOT 9/0.02 = 450 ohms.  I find that 2mA is a good compromise between battery life and brightness in most high-brightness LEDs (anything over 1000mcd). So at 9V, that's a 3.3k resistor and at 18V that's an 8.2k resistor (10k would be close enough too).

Sorry for being pedantic.

[Samuel]

Ha! Dave you asked *EXACTLY* the question I was pondering tonight! Right down to "not sure if the 20mA goes into the equation..."

This board is great.

[Davefx]

Yeah, Samuel.  I guess it's like Brett posted.  You have to know how much current is sufficient for decent brightness that won't drain too much current from your power supply, and then you can use ohms law and figure out the resistor you need.  I was just fearing that 18v was too high, now I know it's not .  Oh, and thanks Brett!

[smoguzbenjamin]

Forgot about voltage drop  :roll:  Stupid me. Thinking I'd give a smart reply and I give an incomplete answer... Oh well it doesn't really matter that much.  I always use a 50k pot in series with the minimum resistance for 20mA to get the brightness I want, then I measure the resistance and get a fixed resistor  of that value. Or close to it anyway  :wink: