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Originally Posted by SuperDanman
I'd like to put an indicator LED onto my beacon, but I want it to burn its brightest. The problem is, because the power for the LED is modulated, the LED is going to be perceived as dimmer than it would be unmodulated.
Lets say we have an LED with a 3.5V drop that takes 20ma. How could I calculate the resistor to make the LED appear the brightest?
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Dan,
Most LEDs have a 2 volt drop across them when turned on so you can use that as a rule of thumb. In looking at one of the high output Radio Shack Red LEDs, the spec was a 5000 mcd at 36 ma max current. So to find the voltage drop across a series resistor, subtract 2 volts from the power supply. Use Ohm's Law to solve for the resistor and you are done. So if we are working from the 7.2 volt battery 7.2-2=5.2 volts. 5.2 volts/.036 amps=144 ohms. The closest standard value is 150 ohms. P=I^2*R or .036^2*150=.19 Watts so a 1/4 watt resistor should be fine. If you are using the PWM signal fed to the IR LEDs than the relative brightness is a functon of the average current. For the 1mS pulse, the LED will appear about 1/10 it's full ON brightness and the 2 mS side will be about 1/5 it's full ON brightness. Since these pulses are occuring at a rate faster than your eye can detect you wll not likely see the switching (pulse) in the light output. If you wanted to cheat it a little, since the pulse repition rate is so slow, you can bump the resistor to a 120 ohm (standard value) for the 1 mS side and that might make both LED about the same brightness.
As a rule of thumb, the human eye stops seeing changes that occur faster than 24 times a second. That is why film is shot at 24 frames per second and TV has a frame rate near 30/sec.