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# LED's, Resistors, and Potentiometers Answered

I"m having some trouble with my LED and pot.
My LED is a white 2.1v / 30mA
My power source is 2 AAA batteries in series at 3v
my pot says 500K on the back and three leads labeled 1, 2, 3
When my pot is at the off position, my LED is lit up perfect. but when I slightly turn the dial my LED starts to burn out and turns bright orange, and if I turn the dial further (Or let it sit for a minute) my LED turns bright red.
I know that's ruining the LED...
But what I'm dying to figure out is how do I know what resistors I would need to use with the pot, What's the math?

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## Discussions

All the information you need is in my answer :- Look at the specification of the LEDs you're using. The red one will be something like Vf = 1.8V, If = 20mA. Put these numbers into the calculator I've linked along with your supply voltage and it will come up with the value for the resistor (for example 180R if you're using a 5V supply). The circuit for each LED will be exactly the same as the calculator shows you, except you'll also put in a potentiometer in the circuit as well, like THIS. What voltage are you going to be running it from?

The 500k is way too much, and sounds like it may be a logerithmic pot, rather then a taper, and even if it is not so, small movement makes for large jumps in resistance. A 50 ohm to 1 K would serve you better.

Go to the calculator on line that Caitlinsdad mentions so you know the upper and lower range you should use. This way you can get closer and the turns will be more "gradual" then such a high resistance one.

look up LED calculator. The potentiometer is a variable resistor so the bottom end of the range of resistance value should be the max that the LED can tolerate - it is when the LED is the brightest. Increasing the resistance when you turn it up should dim the LED a bit.