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I am constructing a circuit to drive a powerful LED that requires about 10 amps at 12V. I also have a battery that happily supplies that. For obvious reasons, that LED requires a cooling fan, and I've chosen one that takes 0.05 amps at 12V. How can I run this fan with the existing circuit?

EDIT:

I will probably be using a boost converter to control the current. It will be boosting from ~10V to the 12V required for the LED. I will also be using it to dim the LED.

Specifications for the boost converter:

  • Input: 8.5-50V, 15A max
  • Output: 10-60V, 12A max
Daniel Williams
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    Just hook it up, the fan will draw only what it requires (unless it shorts out). The important thing is that the voltage is the same. – Ron Beyer Dec 23 '19 at 21:00
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    We see similar questions several times a day. Is there a canonical duplicate? – Eugene Sh. Dec 23 '19 at 21:00
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    @Daniel, a thought experiment: I have a mains light bulb that takes 0.1 A. What will happen when I connect it to the mains supply in my house that's connected to the national grid and could supply hundreds of amps? – Transistor Dec 23 '19 at 21:13
  • @EugeneSh. - Hi, Good question - the answers to this previous question ("[Choosing power supply, how to get the voltage and current ratings?](https://electronics.stackexchange.com/q/34745/101852)") would answer the new question asked above. However from its title, it might not seem like an *obvious* duplicate. (The linked question includes the sub-question "*I don't want a 10 A supply to damage my 1 A device*" which is obviously the same as the topic here about the fan.) Is it a close-enough duplicate to be applied here? – SamGibson Dec 23 '19 at 21:38
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    What kind of LED is this? Is this an LED *emitter* on a tiny PCB (yes, they do make them that large) or is this a lamp assembly built as a product with LEDs, lenses, driver, case and cable already assembled? Can you link the thing it is? – Harper - Reinstate Monica Dec 23 '19 at 23:33

2 Answers2

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You just connect the fan in parallel with the LED, and it will draw 0.05A. You don't have to do anything to limit the current - the fan does that for itself.

Edit: If your supply is 10V, it would be easiest to just connect the fan to the 10V supply. It's close enough that the fan should run fine.

Simon B
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  • So, for example, if I reduce the current to 5A (assuming the same voltage) the LED will draw 4.95A and the fan will still draw 0.05A? – Daniel Williams Dec 23 '19 at 21:11
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    @DanielWilliams Not necessarily, what will probably happen is you will trip a breaker or some other kind of circuit protection. The LED will try to draw more than the supply can deliver. How are you limiting current? By a resistor at the LED, or an adjustable supply? – Ron Beyer Dec 23 '19 at 21:12
  • I was planning on limiting the current with a boost converter. – Daniel Williams Dec 23 '19 at 21:18
  • I would be careful there, "limiting the current with a boost converter" I assume means either you buy one that's says "5A supply" or "10A supply" on it. What happens after you pass that is the voltage starts dropping and/or the component can heat up and burn out. You need to make sure that what you draw from it doesn't exceed it's rating. The LED will try to pull everything it can (and more) unless you limit it with a resistor (or the LED has limiting built-in that you can adjust). – Ron Beyer Dec 23 '19 at 21:23
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    @DanielWilliams I think you need to go back to the original question and add in all the important details you haven't told us. – Simon B Dec 23 '19 at 21:23
  • Define Vbat and LED V vs I specs . Fan can be temp controlled or just enabled by 12V , LEDs can be PWM or buck boost but we need specs – Tony Stewart EE75 Dec 23 '19 at 22:07
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I'm going to assume you mean a constant current power supply delivering 10A at about 12v. What happens will depend on the voltage across the LED at the temperature you plan to run it at. Probably as the led heats up the forward voltage will drop and the fan will spin a little slower. Alternatively, you may end up running above 12v, so make sure the fan is ok with that.

user1850479
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