Using 12VDC spotlight charger on LED's?

I have an old rechargable spotlight that I never use, so I figured I would use it for parts.  Could I use the charger 120VAC/12VDC to power some LED's, like if I wanted to make an under the counter light?
If so, would it be better to run the LEDs parallel or in series?  I was reading an instructable on beginning LED's and saw that series will divide the 12 volts to each LED (4 LEDs at 3 volts each), and running parallel will send the entire 12 volts to each LED.  Would I use resistors even if I ran the LED's in a series?    If I ran the LED'S parallel, would the follow crude schematic be correct? 

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You can't run them in parallel.

Its not good to run them without a resistor, or some current control. Better have three LEDS in each string, with a 150 Ohm resistor on the string.  You can have lots of strings.

pennsteve (author)  steveastrouk7 years ago
Was that a yes to my new schematic?  I seem to be havig problems loading another photo, so I deleted my reply. sorry
You've got the right idea.
What colour of LEDs are you using ? 

pennsteve (author)  steveastrouk7 years ago
I went to radio shack and got 2 white and 2 red.  Not sure what I am using yet.  I know that from now on I think I'll send to Honk Kong for the LED's, unless I can find them cheaper around here, which I doubt. lol
pennsteve (author)  pennsteve7 years ago
Ok, another question.  If I run a string of three 3Volt, do I use 9 volts (3 LEDS at 3 volts each) in the calculation to find the resistor I need, or just 3 volts?  
You ADD the "forward" voltages of your LEDs together - so, say you have two that drop 1.2V and 1 that drops 3 v, you have a total of 2.4+3 or 5.4V in your LED string.

Take the supply voltage (ideally at least a couple of  volts more than the forward drop - so here, I'd like to see a supply of at least 7V. 

Take the ideal current through the LEDS, lets say for the sake of example, 20mA and then calculate (supply volts - diode drop)/ ideal current - this gives you the series resistor you need. 

( 7-5.4)/0.02 = 80 Ohms.