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How to drop 9 volts to 3 volts at 32 ma Answered

Hello, I am a new B here. I would like to use a 9 volt battery to power lights for my RC car. The led lights operate at 3 volts 32 ma. What size resistor will I need to make the voltage drop. Will this also control the current so I do burn up the lights. The lights are somewhat expensive so would like to get it right the first time. Thanks for anyone's help.

papadewey221

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Jack A Lopez

4 weeks ago

This sounds like a job for Rob Arnold's LED wizard calculator, here:
http://led.linear1.org/led.wiz

I am naively assuming you are using two LEDs, each of which has a 3 volt forward drop.

If you want to, "get it right the first time," I suggest being conservative about the design current. Instead of the maximum, which is 32 mA, try about 1/5 of that, or around 6 mA.

For example, the inputs:

9.0 = Source voltage
3.0 = diode forward voltage
6.0 = diode forward current (mA)
2 = number of LEDs in your array

Return:
Solution 0: 2 x 1 array uses 2 LEDs exactly
+----|>|----|>|---/\/\/----+ R = 560 ohms

The wizard says: In solution 0:
each 560 ohm resistor dissipates 20.16 mW
the wizard thinks ¼W resistors are fine for your application
together, all resistors dissipate 20.16 mW
together, the diodes dissipate 36 mW
total power dissipated by the array is 56.16 mW