upgrading light source for a Foucault Tester

Am building a newer / better Foucault tester for telescope mirrors. Will be using a COM-08285 super bright green LED for the light source. Using a couple double “A” batteries for 3 V power source. If my math is correct, I need a 166.6 ohm resistor to limit current. I might want a rheostat to control light intensity. Pretty simple project…

Is my math correct so far? Any suggestions for a rheostat? Buy a COM-10969 resistor assortment to come as close as I can to the 166.6 ohm resistor, which I’ll bet doesn’t exist??

Any advice / information / correction welcome!!

Hi Davey.

I’m guessing you’re calculations were based on a current of 18 ohms?

If so, there’s a little bit you’re not taking into account for.

LEDs have a forward voltage drop that you need to add to your calculation as well and that can be found in the data sheet. The forward voltage eats into your power supply and changed the values you need for calculating the resistance needed.

Now, the data sheet for COM-08285 says the LED has a forward voltage of between 3.1 and 3.3 volts. We will just call that 3.2 volts since it’s in the middle. So what that means is you need at least 3.2 volts to light the LED.

Now, if you’re using ohms law to calculate the resistor needed, you’d take your power supply voltage, subtract the LEDs forward voltage and then use the result to calculate resistance. So, in this case, we have a power supply voltage of 3V, and a forward voltage higher than what we have available. The LED shouldn’t light even without a resistor since 3 volts minus 3.2 volts is a negative voltage.

However, it turns out the datasheet is wrong! I measured one of these LEDs and the forward voltage seems to be about 2.1 volts.

So, with a little math, we take our power supply voltage of 3 volts, subtract 2.1 volts from that and get 0.9 volts that we have to work with.

Generally you want to run LEDs between 15 and 20mA to get the best amount of light without damaging the LED. If we shoot for 20mA using ohms law we get:

R = E / I
0.9V / 0.020A = 45Ω

That tells us the smallest size resistor we could use would be 45Ω. (Larger resistors will just reduce the current a bit)

Now 45Ω isn’t a common resistor value, but you could probably find one at one of the larger parts distributors. The next closest common value resistor would be 47Ω and that would work just fine for your requirements. We’re out of stock at the moment, but our [resistor kit does have 47Ω resistors in it. If you need something right away, two [100Ω resistors wired in parallel would give you 50Ω and that would work just fine too.

Now, if you wanted to dim the LED with a rheostat, that could be done as well by adding a rheostat in series with the resistor and LED. The existing resistor keeps current down to a maximum of 20mA and adding resistance with the rheostat will just dim the LED further. Just make sure your rheostat can deal with the amount of power your circuit is dissipating, which in this case would be about 20ish mW.](Resistor 100 Ohm 1/4 Watt PTH - 20 pack (Thick Leads) - PRT-14493 - SparkFun Electronics)](https://www.sparkfun.com/products/10969)

Thank you for the information; thank you for the education!! I appreciate it.

By the way; could you recommend a rheostat, or potentiometer that would work well in this application??

[COM-09939 would probably work.](Rotary Potentiometer - 100k Ohm, Logarithmic (Panel Mount) - COM-14624 - SparkFun Electronics)

Thank you for the assistance!! Now to use up the gift certificate from Christmas to make an order…