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I'm new to this forum and glad to be here! I'm building a 1 watt led circuit (planning to...) most of the circuits I found online, make a lot of heat and I wish I could find a better one to get bright and power friendly flashlight for my "power bank".

Recently, I've built a 555 based LED driver which is super efficient and bright enough for smd leds it is actually blinking at high frequency so you won't see by naked eye, it is so beautiful. Anyway the problem is that it cannot drive 1 watt leds so I'm wondering if I could do something about it, I bought lm317 for 5.1 to 3.3 volt conversion but I’m pretty sure lm317 is not the best idea (cause of voltage conversion and wasted power by heat) Can I mix ne555 and lm317 or even a transistor? So again it blink at high frequency I mean using pin 3 of 555ic as trigger for the transistor or lm317 to bright the 1 watt led Vcc=5.1volt @2amps

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    \$\begingroup\$ Show your schematic. This website has a function to develop and post a schematic. Just click "edit" on your post. \$\endgroup\$ – DKNguyen Jul 12 at 20:11
  • \$\begingroup\$ Just use something like this: ti.com/lit/ds/symlink/tps54201.pdf OR this: ti.com/lit/ds/symlink/tps92512.pdf \$\endgroup\$ – John D Jul 12 at 20:42
  • \$\begingroup\$ No linear design or switching on/off will be efficient from 5 to 3.3V or 300mA CC but given the surface power density of the LED, you need a good heatsink for the 1W LED at least 10 sq. cm. A simple resistor from 3.3V to drop V/R=0.3A is most efficient. Since LED resistance will be about 1 Ohm above 2.8V +/-50% you need to calibrate to choose the best series R from 3.3V but will be in the 0.1 to 1 Ohm range \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 13 at 2:08
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Edit: A homemade 555 switching regulator is not the most efficient (switching device is forced to drop voltage that the LED can't use) or safest (no current protection) way to light your LED. The least efficient way is to put an appropriately sized resistor in series for your particular power supply. The most efficient method which might also give you dimmer control would be a buck regulator (uses a magnetic field to "level shift" your voltage efficiently, if you need less voltage ie LED forward voltage, the buck converter will draw less current from the source so Pout/Pin is high).

You may also be using the 555 because you have heard that the pulses appear brighter to the human eye than continuous light, see this thread: Does pulsing an LED at higher current yield greater apparent brightness?

So to recap: efficient = buck converter; easy = resistor (but for what it's worth, the lower the voltage supply the more efficient it'll be because the power supply is likely acting like a pre-set buck regulator before you attach your big hot resistor to it); fun but not reliable = 555 with semiconductor output stage, see this thread for details on how to drive an LED without current protection: Do I really need resistors when controlling LEDs with Arduino?

Original reply: You can absolutely use a 555 to drive the base of a transistor! You probably want to set up an emitter follower (boost the current) output stage if you can afford the .6 volt drop.

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  • \$\begingroup\$ Yes but why is this better? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 13 at 4:48
  • \$\begingroup\$ I interpreted OPs question as asking how to control the power flow to the high power LED using the 555 as a signal generator. I don't think the 555 can source enough power at 5 volts to singlehandedly drive the LED so a transistor current amplifier or MOSFET would be great solutions as long as the power supply can source enough current at a voltage above the LEDs conducting voltage. \$\endgroup\$ – Kent Altobelli Jul 13 at 5:01
  • \$\begingroup\$ A BJT 555 can drive 200mA max and CMOS 555 only 100mA absolute max. But why bother when a simple R can work better from 3.3V \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 Jul 13 at 5:08
  • \$\begingroup\$ He was asking for an efficient circuit and in my mind whenever you add a resistor in series with the load, you're instantly losing efficiency. I like your analysis and I think your idea is much better for preventing the LED from burning out since there's no current control if you hang an LED off an amplifier with no current limiting. Perhaps a combination of both techniques would yield an efficient but protected circuit? \$\endgroup\$ – Kent Altobelli Jul 13 at 5:16
  • \$\begingroup\$ Oh I see the problem, I was also assuming the OP cared about being able to dim the LED using PWM since he was using a 555. For constant illumination, yes it would be hard to save the efficiency of the remaining 2 volt drop with a homemade buck converter. \$\endgroup\$ – Kent Altobelli Jul 13 at 5:23
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No linear design or switching on/off will be efficient from 5 to 3.3V or 300mA CC but given the surface power density of the LED, you need a good heatsink for the 1W LED at least 10 sq. cm. (Thermally well-attached.)

A simple resistor from 3.3V to drop deltaV/R=0.3A is most efficient.

Since LED resistance will be about 1 +/-50% Ohm (for 1W) above 2.8V threshold for dim, you need to calibrate to choose the best series R from 3.3V but will be in the 0.1 to 1 Ohm range. Do you know Ohm’s Law? This is mandatory to learn for choosing R.

I am assuming your LED is white.

P.S. if your LED is >50% tolerance on series bulk resistance , then it is poorer than nominal quality efficiency so let me know your results.

If you use the LM317 Rja * 0.3A * (5.1-3.3) compute heatsink for 85’C max.

You can also use a 3.6V Li Ion cell but if you can’t manage this or 3.3V then wish to make a custom PCB including copper heat sink for LED for the required 2x sided area, then follow @JohnD ‘s datasheet.

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