From: Jim Large <jml@nomos.com> Subject: Your LED Flashlight Recipe Date: Wed, 13 Jan 1999 19:06:23 -0500 Donald, I found your web page, http://www.cs.rutgers.edu/~watrous/ledlite.html, in a search engine while I was looking for something else, and it caught my eye because I've built several LED flashlights by the same method. The plan calls for a 150 ohm resistor to limit current through the LED to the recommended, 20 milliamps, but 150 ohms is too high. It will limit the current to about 7.6 mA, and the LED will not shine as brightly as it could. The plan fails to account for the forward voltage drop of the diode. If you build the circuit, and measure the voltages across each component, you should find 3V across the battery stack (as you would expect if the batteries are fresh), 1.85 V across the diode, and (3 - 1.85) or 1.15 V across the resistor. The 1.85 V drop across the LED is a characteristic of the diode, and remains nearly constant over the normal range of operating currents. You will probably find it listed on the back of the package that the LED came in. The current through each component is the same because they are wired in series. To compute it, we use Ohm's law, but note! The LED and the battery do not obey Ohm's law. The only component that obeys it is the resistor. Ohm's law says that I = E/R where I is the current through the resistor, E is the voltage across the resistor, and R is the resistance. Plugging in the numbers we get, I = 1.15/150 = 0.00767 A == 7.67 mA. I'm guessing that the value, 150, was chosen based on the erroneous assumption that there is no voltage drop across the LED, and that the resistor drops the full three volts of the battery stack. To get the recommended 20mA, you need to use a lower valued resistor. 57.5 ohms would be ideal, but you will probably need to substitute a different, more readily available value. A 47ohm (+/- 20%) resistor should give you somewhere between 20.4mA and 30.6mA. -- more than you need, but probably within the absolute limits of the LED. Even if the maximum current rating of the LED is 30mA (I forget whether it's 30 or 50), a 47 ohm resistor should be safe because there's another resistance in the circuit that I haven't accounted for. The batteries have internal resistance. I don't know how much, but it will be higher for smaller batteries. It might be significant if you're using AA or AAA size batteries. I'll leave the method of measuring battery resistance as an exercise for the reader (Hint: don't try simply hooking one up to an ohm meter). -- Jim Large P.S.: I bought a ready-made, two-AAA-cell, LED flashlight that died after two days. I took it apart, and found that there was no resistor at all! I guess the engineers thought the resistance of the batteries would be high enough so that most of the LEDs would survive, and I got one of ones that didn't.