Making a beacon indicator LED burn its brightest without burning out

I’d like to put an indicator LED onto my beacon, but I want it to burn its brightest. Yes, yes… the easy way would be to put it between the +battery and ground pins. I want it to put it in so its power is modulated, though… partly to learn more about the calculations involved in PWM signals, and partly for the cool strobe effect when moving the beacon quickly :wink: 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? Specifically, rather than using 7.2V (or whatever your battery is at) for your resistor calculation, can you use some kind of ‘perceived voltage’ that can be calculated when the frequency of the modulation is known?

I guess what I’m basically asking is does the LED act as if its power input is a modulated 7.2V minus drop-across-resistor (what it actually is), or does it act as if its power input is a constant ‘percieved-voltage’ minus drop-across-resistor. If the latter, can you safely ‘overclock’ the LED by using resistor with a lower resistance than you normally would use to make it appear as bright as it does when its power is unmodulated?

Here’s an excellent source for LED info:

Generally, when pulsing LED’s at low frequencies as you wish to do, i BELIEVE you need to treat them as you would simple DC. If you want a very bright source, use superbright LED’s - you can get them even at radioshack - superbrights are awesome.

Intensity is generally proportional to current through LED, but there is a limit as you know. You can hook them up directly to the battery and they will be VERY bright - for a millisecond before self destructing. I think of them like thermal fuses - more current, more heat and pop. BUT, your best bet is superbrights.

leds are spec’d with a constant current and a peak current. If you are pulsing them, the peak current is what you cant exceed, and then you have to be mindful of the duty cycle so the RMS (sort of like the averaged current) stays close to the constant current spec.

BTW - what beacon are you talking about?!

Speaking of blowing out LEDs…

Did you know that the infrared LEDs produce visible light when they have too much current?

I accidentally shorted one of the resistors while testing with a multimeter and fried one of our LEDs.

I am using a superbright, and I want to know how far I can push it.

Either beacon. What I want to know is how to figure out what the limit is when it is being pulsed so I can reach it. Like I said above, the easy solution is to just hook it up to the +battery and ground pins, but I want to start understanding how modulation applies to electronics. The LED is just an example I want to use in this case.

hooking it up to the battery directly will only blow it out in seconds, and you wont learn anything (expect to not do that :c)

LEDs work on current - the series resistor is there to limit the current.

when you pulse an LED rapidly, say at a 1000 Hz, then the heat at the juction dissapates a little between the ON part of the cycle, so you can drive it a little harder

but if you pulse it at, lets say, 1 Hz - the junction temp rises fast enough that it will burn out

you really need to get a hold of the spec sheet for the LEDs - I dont know if FIRST listed the supplier and part number for those - if they did you should be able to goto the suppliers website and get the data sheet.

It will tell you the max (do not exceed) current - and most LED manuactures data sheets explain what you need to know (the levels) when pulsing them instead of driving them steady.

BTW, if all you want to do is make it appear brighter, I dont think pulsing it with more current will do that - driving it at its rated steady current level will be as bright as it can get.

Did you know that the infrared LEDs produce visible light when they have too much current?

I accidentally shorted one of the resistors while testing with a multimeter and fried one of our LEDs.

As a matter of fact, I did know that. Wanna know how? One of our students hooked up the IR LED to a variable voltage supply to test it (we had a “magic mirror” IR detector). The conversation went about like this:

Me: You’re not applying the full voltage are you?
Him: No.
(Pause. I begin to see the LED light up.)
Me: You probably don’t want to add any more voltage, that doesn’t look good …
(BOOM! The LED exploded, shooting right past my face.)

Needless to say, I wasn’t too happy :frowning:

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^2R or .036^2150=.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.