# How to reduce LED drain on battery?

I have a Cree Q5 LED, I established the typical forward voltage is 3.7V and the maximum current is 1000mA. I'm confused as to how I reduce the current, I want it to run at 350mA, I know this is possible, I'm just not sure on how to go about doing this. Can anyone point out how I do this.

Thank you.

active| newest | oldestSee the circuit for a current controlled driver that can work on your LED

from 5VDC to 35VDC. . . . . . . . . . . . . A

R

_{1}= 3.50 ohms gives an LED current of 357 maR

_{1}= 1.25 ohms gives an LED current of 1000 maV

_{source}-V_{LED}=V_{resistor}(Kirchhoff's Voltage Law)

Let's say you have a 9V source. The turn-on voltage of your LED is 3.5V meaning V

_{r}=5.5V.I

_{LED}=I_{resistor }(Kirchhoff's Current Law)

You want I

_{LED}to be 350mA.V

_{r}=I_{r}R(Ohms Law)

Solving for R you get R≈28Ohms for a 9V source. Do this same set of calculations to fit what you have.

R≈16Ohms (I hit 9 instead of 5.5)

1.) Where did you get 3.5V for the LED to turn on, why aren't you using the forward voltage of 3.7V?

2.) If my source voltage is 3.75 I get R = 6 Ohms at 350mA is that correct?

3.) Okay now pretend I wanted the LED to draw 1000mA am I correct in saying R = 2 Ohms?

4.) Assuming my calculation in question 2 was correct if I have 2 LED's and I want each to draw 350mA. I need two 6 Ohms resistors with the 1 resistor and 1 LED wired together, with the battery connect to them both in parallel. Is this correct?

Thank you, I really appreciate the help :)

Remember:

In series R

_{T}= R_{1}+ R_{2}+...In parallel 1/R

_{T}= 1/R_{1}+ 1/R_{2}+...