High Power LED Strobe

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I am attempting to drive 3 High-power LED's in parallel and wish to strobe them. The issue I'm running into is in parallel, the LED's require 3v and 8.4A, however, the 555 timer requires higher input voltages and doesn't output enough amps. Secondly, I'm researching the use of an MCU but am very new to all this.

Question 1: Can I overdrive the 555 and it operate, or will it burn up?

Question 2: Will an MCU be the appropriate solution utilizing only a battery source of 3.7v and 12A? I only need it to strobe for about 10 seconds and then the circuit shut down.

Thank you

Best Answer

The 555 is not designed to drive large loads such as these, it will certainly "burn up."

MCU's are complex, but versatile. There is a steep learning curve...

It sounds like what you want to do is use the 555 just as a "pulse source", then have some other circuit element do all the power switching. Typically that is done with MOSFETs or other power-switching devices.

For your LED's, the important value is their pulsed current (amps) rating. If you don't know exactly what this value is, google the LED model number to find it's datasheet, the value should be in there. So say it is 2.8A. Another useful value is the Vf or "forward voltage drop" of the LED. Say this is 3.0v @2A. What we really want to know is what the drop is at 2.8A, often this is plotted in a chart.

So lets say we have an LED that should drop 3.2v at 2.8A. (Speculating numbers, replace with measured values.) But we have a 3.7v supply, so we can't just hook up a 3.2v LED to it directly because "overdriving it" will push far more than 2.8A through it and probably burn the LED up. So we need to drop 3.7v - 3.2v = 0.5v. So put a resistor in series with the LED. Ohm's Law: E=I*R. Solving for R, R=E/I. So 0.5v / 2.8A = 0.179 ohms. Round that to 0.180 (180 milli-Ohms.) That is the resistance needed.

Now resistors get hot, and that needs to be considered. Power (watts) = Volts * Amps. So for that resistor, it's dropping 0.5v * 2.8A = 1.4W. Now these are pulses, not a continuous duty cycle, so this can be de-rated. If the pulse duty is 50% (half the time on, half off) then the resistor would dissipate 0.7W. Now by convention, power-rating of resistors is always >2x the actual power. So for this 0.7W need, we'd spec a 2W resistor (minimum.) A 5W or even 10W will work just as well (and get less hot.)

Ok so you have a resistor for each LED. Then use a MOSFET to "turn on" each LED as shown in the above link. What kind of MOSFET? One that ideally has a logic-level input, so it can operate from the 555 output. And one "big" enough to switch at least 2x 2.8A. The lower it's "RdsON" value, the less heat it will dissipate, but there's a gotcha here. Most MOSFETS with very low RdsON values, have large gate capacitances, so when the 555 sends the signal to turn on, their response time is slowed due to this large capacitance. So it would be fun (and educational) to try several different ones.