I read a few posts here about resistor in led applications. The thing is, the resistor was always use with a power source delivering a higher forward voltage than the led could tolerate(ex: a 9volt battery hook up to a 3w 3v led).But what about a power source that the voltage can be adjusted to fit the led? My science courses are long gone behind me so I want to make sure my project doesn't explode on my face ;).

Right now I'm building a led lamp for a photography project using a 400w 36V 11A Single Output Switching power supply AC to DC SMPS which has a voltage adjuster so I can go from 32.4 to 39.6V. I'm using 168 3w Leds for my lamp, with a voltage rating of 3.2 to 3.4V and 700mA of current. I hooked up 12 leds in series so I can put 36v to them and not pushing them too hard. I have 14 series of 12 leds in parallels. The leds are soldered to a copper board that I cleaned with acid with the proper circuit to have 14 series of 12 leds in parallels.

My question is do I need to include resistors for safety let say a led in a series dies so it doesn't blow up the rest of the led in that series? Is it possible to put a resistor at the end of each series or I need to put a resistor after every led?

I took some measurements:

-The lamp is drawing approximately 4A at the power supply which I find odd since it should draw more power if the led are supposed to draw 700mA

-Power supply is pushing 36V precisely

-Measurements at the led give me 3v as it should

Maybe I'm wrong here but normally to calculate a resistor resistance you have to do R= V/I so 36V-(3V*12)= 0/.7A =0R

But the thing that is bugging me is the power supply is rated for 11A which is more of what the leds should take. Why my concern for a resistor somewhere in my lamp design.

Thanks for the help! I hope my explanations are clear enough!

  • \$\begingroup\$ LEDs almost always die open. \$\endgroup\$ Commented Aug 23, 2016 at 17:33
  • \$\begingroup\$ Take a look at the I-V curve in the datasheet for your LEDs and you will see why it works, and also why you have 4A. \$\endgroup\$
    – Majenko
    Commented Aug 23, 2016 at 17:33
  • \$\begingroup\$ Would love to but those leds were bought on ebay. The only information I have is 3.2 to 3.4V and 700mA. There's no datasheet on the listing. Since I'm driving them at only 3V and the current draw is exponential to the voltage, maybe I'll get more Amp draw at the power supply if I run the leds at 3.3-3.4V. But my question mainly is do I need a resistor in this lamp design since I can control the voltage and if so which one? Thanks! \$\endgroup\$
    – John Smith
    Commented Aug 23, 2016 at 17:46
  • \$\begingroup\$ See my design walk-through just posted at the following link. It's not an exact answer, but it covers some thinking you might apply to your case. See: electronics.stackexchange.com/questions/253766/… \$\endgroup\$
    – jonk
    Commented Aug 23, 2016 at 18:07
  • \$\begingroup\$ Thanks Jonk! So if I understand correctly I need to add a current limitation to my lamp? Do you think a booster like this one would do the job to set the current and the voltage for the lamp? Would one do the job or it's better to fit one for every series? ebay.ca/itm/… \$\endgroup\$
    – John Smith
    Commented Aug 23, 2016 at 20:11

2 Answers 2


If you ensure that the supply has no large ripple, or noise, or accidentally gets adjusted up, or starts to drift due to age, bad parts, or temperature then yes. It's not the most secure way.

Any number of things could start to go wrong, but it is in theory and practice okay.

*Right off the bat, your max normal load would be 168 * 3W or 504 Watts, which is 100 Watts more than your supply is rated for...*

And at less than 700mA, the VF will be lower as well. You may see 3V at 600mA, or 2.5V or something. So 12 in series may have a VF of less than 36V...

Proper heat sinking of the leds would be needed to ensure that a high temperature doesn't drop their VF causing thermal runaway.

As to your supply being precisely that voltage, is that true under load and taking into account any ripple?

The preferred way, especially for this high power, would be a constant current supply, not a simple resistor btw.

  • \$\begingroup\$ Thanks for the answer! I never intended to run those leds to their max output since I read that a 60-70% efficiency is better on the long run with high power leds. I was planning to run them at 60-70% capacity. So If I understand right it will be better to have a constant current supply hooked up to my lamp? Could this work? (ex: my power supply I have now plugged into the constant current power supply to the leds? ebay.ca/itm/… \$\endgroup\$
    – John Smith
    Commented Aug 23, 2016 at 18:00
  • \$\begingroup\$ And for heat sinking, I bought enough heat sink for each leds (those 8.8mmX8.8mm aluminum heatsink) with 4X 140mm fans blowing on the leds and heatsink \$\endgroup\$
    – John Smith
    Commented Aug 23, 2016 at 18:07

The current is a highly non-linear function above the forward voltage. You need a constant-current source (linear or switched) to drive this safely and efficiently at the rated current.


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