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I am trying to run four small, 1 watt LEDs with the help of a charger that supplies 5 volts at 2 amperes current. The LEDs require 3.4 volts to run. How do I run them without the LEDs overheating? I would provide other information if necessary for calculating.

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UjanRoy
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  • Looks like about 300 mA each so 4 of them would be 1.2 A. Your supply should be fine. And you have enough overhead that you can use a two-BJT constant current limiter for each. Would take 8 BJTs and 8 resistors, totaled. (I think). This link tells you how to do a proper design for them. Or you can just use resistor limiters and use four. But they don't regulate nearly so well. Not at that low overhead. You'll have to decide what's important. – periblepsis Nov 30 '23 at 10:24
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    @periblepsis I suggest that you consider making that an answer. As a comment it is liable to be lost with time, and it's useful, even without any expansion. – Russell McMahon Nov 30 '23 at 12:17
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    Flagger: The SE system is deficient in not allowing some flags to be accepted nor commented on. If a flag is rejected this means that it has been noted. In this case the only other choices are not suitable. – Russell McMahon Dec 01 '23 at 10:07
  • @RussellMcMahon Just noticed. The OP isn't responding to anything/anyone. So I'm not yet motivated to write, I guess. – periblepsis Dec 02 '23 at 05:01
  • Cheers @RussellMcMahonRussell I'm responding so that maybe my interjection might persuade a deletion of the comment. – Andy aka Dec 02 '23 at 15:09

2 Answers2

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Do some math and use ohm's law.

1 watt at 3.4 volts implies a current of 294 mA through each LED. If you use a dropper resistor, that resistor would need to drop 1.6 volts whilst carrying 294 mA. That means it has a value of 5.44 Ω.

Each resistor would dissipate 1.6 volts × 294 mA = 0.47 watts.

Total wattage of 4 resistors is therefore 1.8816 watts. Add that to the 4 watts from the LEDs and you get a total power used of 5.8816 watts. That's within the capabilities of your 5 volt supply.

And, the current from a 5 volt supply is 1.176 amps so, your power supply is fully capable of meeting the requirements.

How do I run them without the LEDs overheating?

You would use heat-sinking techniques.

Andy aka
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  • Will using a single 6 ohm resistor and connecting the LEDs in parallel work? And I don't have a heatsink is there anyway to build something close to a heatsink using things lying around at home? And also could you clarify what do you mean by my 5 volt supply being 1.176 amps, it's 2 amps. – UjanRoy Dec 02 '23 at 14:37
  • No, not at all advisable unless the LEDs are batch selected for identical characteristics. The current drawn by the circuit would be 1.176 amps. You should be able to make a heatsink from aluminium sheet but, it needs careful thought so, consult the data sheet for the LEDs – Andy aka Dec 02 '23 at 15:06
  • https://i.imgur.com/7HS0U8n.png Will this circuit with heatsink work? – UjanRoy Dec 03 '23 at 03:50
  • My previous comment addressed this question @UjanRoy – Andy aka Dec 03 '23 at 09:24
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Put all 4 LED's in series and use a high voltage Joule Thief circuit, like this one:

enter image description here.

The key part of this puzzle is that you don't have to run them at one Watt each.

The resistor R2 takes the place of the usual 1K resistor of the canonical Joule-Thief, and can be replaced with a current-limiting resistor plus potentiometer. Start with a 100K (or 1 Meg) pot at max resistance, and slowly bring the resistance down until you get some heat at the LED's, then dial the brightness back down until they are cool again, then go a bit further for some margin.

If you underpower the LED's, they operate with higher efficiency and you may not even need to heatsink them. A common underpower figure is one quarter the rated power. Underpowering them will also make them last longer.

If they are quality LED's, they will be plenty bright even if underpowered. That's how to do what you asked for.

Because this is a switch mode power supply, a basic boost converter under the hood, it avoids the heat creation that current limiting resistors would usually give you.

For higher power outputs, using this SMPS, or another one, is really the only way to go.

If you're firmly decided on full power, you'll probably have to heatsink them.

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