Help with Resistor for LED

I was seeing if anyone could help me check these calculations, since I'm not sure whether I'm doing them correctly. I already did a search and found this thread, which helped me a little.

For this flashlight: http://store.advancedmart.com/5wamesiledfl.html

There is a review at FlashlightReviews which speculates that it is using an 8mm 100 mA LED (http://www.powerleds.com/data/pro4.html). Another review measured the current draw at 91 mA, so that seems correct.

Since it's a direct drive flashlight, modifying it to use NiMH batteries means that it needs a resistor in series, right? So here are my calculations:

Since each NiMH battery starts at about 1.4V, 1.4V x 3 = 4.2V

LED operating values: 3.2 V @ 100 mA

4.2 - 3.2 = 1V drop

Using V = IR,
1V drop at 100 mA = 10 Ohm resistor

Does this seem correct? Any help would be greatly appreciated. Thanks in advance!

sounds right.... but wouldn't that 1.4 drop to 1.2 pretty fast? 4 Ohm?

3.6/3.2 10 ohm = 40mA
3.6/3.2 4 ohm = 100mA

i think...

Does that mean I should use a resistor with lower resistance then? Won't that cause the LED to be hit with a lot more current until the voltage of the NiMH drops to 1.2 V from 1.4V?

Is it safe to run a LED at currents above absolute maximum ratings?

litlmh said:

Since it's a direct drive flashlight, modifying it to use NiMH batteries means that it needs a resistor in series, right? So here are my calculations:

Since each NiMH battery starts at about 1.4V, 1.4V x 3 = 4.2V

LED operating values: 3.2 V @ 100 mA

4.2 - 3.2 = 1V drop

Using V = IR,
1V drop at 100 mA = 10 Ohm resistor

Does this seem correct? Any help would be greatly appreciated. Thanks in advance!

Click to expand...

If we are certain that the LED is operating on 3.2V @ 100ma (which is not overdriving it at that voltage or current level),then

V = IR
(1V) = (100ma)*R
R = 1/0.1 ohm = 10 ohm,

and I think it is correct.

Unless we are drianing the batteries high up to 500ma+, it shouldn't be a very severe drop from the beginning(the battery voltage sag), and I believe the discharge curve might look like one from the MJ leds(which is a flat very gradual slope)

#######

And if we were to assume the operating voltage for the 8mm to be remained around close to 3.2v with higher current, then if we were using lower value resistor(say 5 ohm):

V = IR
(1V) = I*(5 ohm)
I = 0.25A = 250ma.

It will most likely severely overdrive the LED and might even permanently damage it.

Should my calculation is indeed incorrect, please also feel free to correct.