I have 10 Bridgelux leds (forward voltage 3.5 to 4.5v, drawing 750ma max). I also have a 36v Mean Well HLG Driver, offering 1.12a. It seems to me that trying to run all 10 leds (in series) is cutting it a pit fine re: Voltage, and therefore I should look to run just 8 leds. I cannot find one of those graph things for these lights, so cannot guess as to which would be brightest (10 lower powered or 8 higher powered). 2 paralell circuits would not work as there aren't enough amps available.

Maybe, though, I am talking complete crap.

Which of the options would you take? Is there a secret option C (modifying the driver?) Also, do you know of any good reading materials, so as I can actually understand your answer?

At the end of the day, this is just for a fish tank. I would like to learn, though...

  • \$\begingroup\$ I would look at finding an actual LED driver instead of trying to just wire in the power supply. There's a decent chance that you'll blow the LEDs if you don't restrict the current. \$\endgroup\$ Mar 9, 2015 at 22:59

1 Answer 1


You are, presumably, talking about an HLG 40-36 http://www.meanwell.com/search/hlg-40h/default.htm right?

Look at the data sheet. It can regulate current over a range of 21 to 36 volts. Since you don't know the exact model of your LEDs, the best bet is exactly as you suggest - use 8 in series. If you do that, you must use the dimming function on the MeanWell, and set the current to less than .75 amps. If you don't do this, you seriously risk destroying at least one LED. (Of course, if you do, you'll know you need to crank down the current, and you have 2 spares. Although I doubt that's much consolation.) As it happens, you can drop the current as low as 0.67 amps, so you're good there.

In principle, you could have a go at using two strings of 5 in parallel. In the worst case (3.5 volts per LED) you would only pull 17.5 volts, which is less than the minimum output voltage of your driver, so you would have to add something like a 10 ohm, 10 watt resistor in series with each 5-LED string to drop the excess voltage, but you would want to add such a resistor anyways to keep one of the LED strings from hogging all the current due to some of them needing a lower voltage than others, so that's two birds with one stone. Doing this would provide about 0.56 amps to each string (with the MeanWell set to a full 1.12 amps). Whether this would be a net win or loss as far as total light goes is probably a wash. Assuming equal efficiency at either current level, running 10 LEDs at the lower current would produce 90% of the light of 8 LEDs at max current, and this is essentially unnoticeable. Running the LEDs at lower current will make them more reliable, and will be much easier for you to provide adequate heat sinking for them.

  • \$\begingroup\$ Thanks guys, that is helpful... I need to check my understanding a little bit more, eg, Vasquez... I thought my Mean Well WAS a driver and WhatRough... I thought amps were pulled, that is, a device will take only what it needs, hence, I did not think amp output required controlling? \$\endgroup\$ Mar 11, 2015 at 6:49
  • \$\begingroup\$ No, what happens is that the driver varies its output voltage to try to keep the current at the right level. Voltage and current are not independent. If you increase voltage you will (except for a few very specific exceptions you don't have to worry about) increase current, and vice versa. But keep in mind that there are limits, such as the 21 volt minimum for your driver. \$\endgroup\$ Mar 11, 2015 at 13:58

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