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# 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.

You do not say what power supply voltage you are using.

See 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 maSelect as Best AnswerUndo Best Answer

Correct answer.

L

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What's your source voltage? Anyway, use a resistor and wire them in series.

V

_{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.

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Correction:

R≈16Ohms (I hit 9 instead of 5.5)

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Thanks for your reply. I'm still a bit confused however.

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 :)

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Sorry, misread the 3.5V. I get R= 0.05/0.35=0.15Ohms. However, this doesn't work for the way I've modeled it. The LED is hardly linear, and I was assuming it would be on in this linear model. With the voltage source being pretty much exactly the turn-on voltage, any linear assumption goes out the window. You probably could calculate the right resistance, but the complexity of the model would make it such that you'd be better off just trying different resistors until you get the right one. If you have a multimeter, you can narrow down on a resistor value in 30 seconds or less by just trying different values.

Remember:

In series R

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

_{T}= 1/R_{1}+ 1/R_{2}+...Select as Best AnswerUndo Best Answer