I'm connecting a DC PSU to a high power LED. I'm setting the maximal current, and a voltage. For the first few seconds, I'm in the c.c mode (voltage drops from its initial value to a lower one). But after a few seconds, probably since the LED heats up, something strange happens: both the current, and the voltage drop to very low values.

If the heating causes the resistance to increase, I would expect the voltage to increase.

A few more details:
It's an unbranded 100W LED I got from ebay (link)
Unfortunately, there is no spec.

I'm setting the voltage to 30V, and maximum current to 2.7A.
The LED heats up, there's a fan attached to it, that does its best to keep it cool.

Thank you.

  • 1
    \$\begingroup\$ More information will help us help you. Link to LED datasheet or at least specs part number ... / What current and voltage are you setting the psu to? Why? | What are the initial values OC / on 1st connection / after a few seconds / is the LED hot? | ???????????????? \$\endgroup\$
    – Russell McMahon
    Commented Sep 12, 2019 at 12:54
  • \$\begingroup\$ Is your LED Vf and current within spec of your power supply? \$\endgroup\$
    – winny
    Commented Sep 12, 2019 at 13:01
  • \$\begingroup\$ I've attached the additional information. Regarding the Vf: The power supply of the led supplies 40V. I'm intentionally using the PSU, and not the power supply since I found out you can use less the 40V to turn on the led (to a reasonable power for the cause), and hopefully, it'll heat less that way. Surprisingly, when I'm using the "official" power supply, there's no drop in lighting as appears when I'm connecting the LED to the PSU (when there's a drop in Voltage and Current, there's also drop in light intensity). \$\endgroup\$ Commented Sep 12, 2019 at 13:08
  • \$\begingroup\$ On t=0, the psu shows v=~21V, and A=~2.7A. But after few seconds, V drops to less then 5V, and C to less then 0.5A. And it flickers up and down in that region. \$\endgroup\$ Commented Sep 12, 2019 at 13:14
  • 1
    \$\begingroup\$ Re, "both current and voltage drop to very low values." Sounds like what I would expect if your power supply had a foldback mode, and your circuit tripped it. \$\endgroup\$ Commented Sep 12, 2019 at 15:07

2 Answers 2


Specs for linked device say 30-34V, that should be your Vfwd. LEDs usually have higher initial Vfwd until they warm up, then it drops slightly. Could be that 30V is not enough to start it up properly. Once substrate has warmed up, internal resistance drops, Vfwd with it, assuming CC. What PSU are you using? Could it be, that it has internal overcurrent protection and just flickers on and off? E.g. it turns on, sees too steep increase in current and just turns off as a self-protection measure?

  • \$\begingroup\$ I'm working with Korad 60V, 3A psu: sra-solder.com/… It says it has an overload protection. So if that's the case, should I increase the maximum voltage I allow (HIgher than 2.7A)? \$\endgroup\$ Commented Sep 12, 2019 at 14:30
  • \$\begingroup\$ You can safely exceed Vfwd voltage, as long as you limit the current. Vfwd means that it has voltage drop of said value, not that it is LED max operational voltage. Like a regular diode it can be used with larger voltages, you just "lose" those 30V across it. Say, if you have 60V source and put a resistor in series with the LED, you have to ensure that the resistor can dissipate (Vin - Vfwd) * I_LED power, because resistor will have to drop remaining 26-30 V @ given current. Just crank your voltage up to 40V, CC to 1A and slowly increase current. \$\endgroup\$
    – stiebrs
    Commented Sep 13, 2019 at 6:40

You may be getting into a "chase its tail" fold back mode with the psu.

Try setting say voltage at the top end of allowed range (34V in spec sheet), current limit low and then wind up the current limit. What happens?

I'd be suspicious of devices with identical V and I specs for such a wide range of colours. That doesn't mean the LEDs are not usable - but that the specs probably aren't.

  • \$\begingroup\$ Hi Russel, thanks a lot for your support. \$\endgroup\$ Commented Sep 12, 2019 at 14:57

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