0
\$\begingroup\$

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

\$\endgroup\$
  • 1
    \$\begingroup\$ Why are you not feeding it into the PWM input of the LED driver? \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 3 '15 at 18:30
  • \$\begingroup\$ I did some quick checking and it appears the PWM's is an awesome idea, however, the specs don't align. Of the ones I found on Mouser, they either require too high a supply voltage or don't have a high enough output amperage. I like this idea... simple, does anyone know of one I can search for that would work... again I'm new at this and truly appreciate your input!! \$\endgroup\$ – JonH Jul 3 '15 at 18:52
  • \$\begingroup\$ Reconfiguring them to be in series may solve that. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 3 '15 at 18:57
  • \$\begingroup\$ Voltage gets too high for a compact power source. In series, it jumps to 9v, thus a source greater than 9v would be required and I'm not aware of a compact battery source at that level. \$\endgroup\$ – JonH Jul 3 '15 at 19:05
  • \$\begingroup\$ Inductive drivers can handle that, but may need a fairly large inductor to store the flux for it. \$\endgroup\$ – Ignacio Vazquez-Abrams Jul 3 '15 at 19:07
1
\$\begingroup\$

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.

\$\endgroup\$
  • \$\begingroup\$ very cool, thank you - I'm reseraching MOSFET now and will see what it looks like. I looked up the data sheet and each LED has a peak forward pulse current of 5.5A and a forward voltage drop of 3v. Based on the current parallel circuit design it requires a .44 ohm resistor (~0.5 ohms). This was calculated on the typical LED current of 2.8A, not the peak, so I may need to recalculate. \$\endgroup\$ – JonH Jul 3 '15 at 19:27
  • \$\begingroup\$ Based on the new calculation, the parallel circuit requires 16.5A, the peak pulse current of the battery source is 15A, so based on this, the circuit will require about a .2 ohm resistor, but if I interpret this correctly, then I should not be able to overdrive the LED's with this power source. Now to research the MOSFET. Please add where you think I may be going astray... thank you so much. \$\endgroup\$ – JonH Jul 3 '15 at 19:34
  • \$\begingroup\$ Things are sounding pretty good so far. There is only so much calculating that can be done; sometimes actual prototyping and testing the real thing and taking measurements needs to be done. I hope the project is working well now. \$\endgroup\$ – rdtsc Sep 14 '15 at 12:37
0
\$\begingroup\$

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

The 555 won't be able to output enough current to drive the LEDs; you'll need a beefier solution.

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.

Similarly, a MCU won't be able to output enough current to drive the LEDs.

What you'll need to do is use a MOSFET; a simple and cheap FET like a IRLB8721 will do. You can use the output of a 555 or a MCU to drive the gate of the FET; if you find that you can't switch fast enough, a FET Gate Driver like a TC4426 can take the signal from a 555 or MCU in and then output a strong enough signal to drive the MOSFET.

If you're finding that

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.