I purchased a number of 50W 5B10C high power leds. Link at end. These need at least 25V to turn on and I am using a setup with a 21V LIPO battery and a 150W boost converter stepping this up to 30V. I fully expect the battery and boost converter to be able to deliver 150W if short circuited.

Having no constant current driver, I took a chance on blinking one of the leds by connecting it directly to the 30V, and it worked fine, as expected for very short blinks.

Eventually I extended the duration, and in the end I just kept power on, and the led did not burn out, or even get overly hot.

The question here, since I am apparently missing something, is how I am getting away with this? Assuming the boost converter is actually able to deliver 150W, as per data sheet, would it not be expected that the led burned out in short order, driven as directly as this?

edit: In the end, it probably comes down to a balance of internal battery resistance and resistance in the led, so with the right voltage, it can reach a balance.. by luck.


  • \$\begingroup\$ Try to add capacitor or RC-filter like 1.2.4 learnabout-electronics.org/PSU/psu12.php \$\endgroup\$
    – nick_n_a
    Sep 4, 2020 at 13:34
  • 1
    \$\begingroup\$ What is the voltage and current going to the LED, while it is running? Actual measured values. \$\endgroup\$
    – rdtsc
    Sep 4, 2020 at 14:29
  • \$\begingroup\$ The voltage is 30V and the current is initially 0.8A and rising, as I keep the LED running. I let it run to the point where the current was 2A, which would be 60W. \$\endgroup\$
    – JoeTaicoon
    Sep 4, 2020 at 16:14
  • 1
    \$\begingroup\$ Since \$ P =V I \$ and V = 0 in a short-circuit the power delivered in a short-circuit is zero, not 150 W. \$\endgroup\$
    – Transistor
    Sep 4, 2020 at 20:43
  • \$\begingroup\$ Looks like parasitic resistance is saving your bacon. But why would you risk damaging the LEDs in the first place? Get a constant current driver! \$\endgroup\$
    – winny
    Sep 4, 2020 at 21:44

1 Answer 1


I think what you are seeing is that because your 30V power supply is close to the 25V typical Vf for your module, your LED module is driven to a point on its Vf vs Current curve where it doesn't burn itself out.

Below is a typical LED Vf curve from a good article from Meanwell about LED drive voltage.

enter image description here

You can see that as the current increases, the drop across the LED increases. If your 30V does not put the operating point above the maximum current, then your LED runs just fine.

One problem is that Vf varies on a part-to-part basis. Parts from a different batch could run at much different brightness or even burn out at 30V.

Another problem is that as shown below, Vf goes down with temperature. So as your LED warms up, with a fixed voltage drive the operating point will move to higher current where the Vf is again 30V. This will cause it to heat up more, reducing Vf more, and potentially burning out the module. This is called thermal runaway.

enter image description here

  • \$\begingroup\$ You answered the main question clearly and in good detail, so I will mark it as the answer. Turns out I was a bit naive initially, and thought a LED, as it exceeded Vf, would effectively short circuit on its own with a vertical I/U curve. I should have researched it better before asking, and only realized this after posting here, but now your answer stands as another good one, explaining this issue, so thanks. Did manage to burn out my led after increasing the voltage to some 35V, exiting the safe zone, you mentioned in your answer. \$\endgroup\$
    – JoeTaicoon
    Sep 5, 2020 at 9:05

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