I want to power up about 2 meters of led strip from a 16.5v dc battery (4s lipo).

I know there are dc-dc converters, but I rather avoid using them due to weight (rc plane)

The lights will not be used very often, probably they will stay on for about 30 minutes.

I have read somewhere that I could plug few diodes in series with the strip, could this be a good idea? Are there any precautions to put in place (do diodes overheat)?

I am basically looking for a solution that works, not for the best practice, performance or top reliability...

Thank you!

  • \$\begingroup\$ Thank you guys! You really gave me good advice. I think I will probably do some testing with the diodes as advised, even though at 11 grams/1 dollar a regulator doesn't look that bad anymore! \$\endgroup\$ Jul 27, 2014 at 18:44
  • \$\begingroup\$ I have found at home 2 diodes. They are 1n5404. Can I give these a shot? \$\endgroup\$ Jul 27, 2014 at 18:50

3 Answers 3


Common led strips take about ~18mA at 12V per segment (typically 3 LED). They are used in Automotive 12V rails, which is 12V nominal, but can go to 14V regularly, with larger spikes.

If you look at a segment, you will notice it has three LEDs and one resistor. Based on the color of the led and ohms law, you can figure out it's current at any given voltage.

$$I = \frac{V_s - V_f}{R}$$

Assuming a white or blue led with ~3.2V forward voltage drop, which is typically paired with a 120Ω resistor:

$$\approx 18mA = \frac{12V - (\approx 3.2V \times 3)}{120Ω}$$

At 14V, you get:

$$\approx 36mA = \frac{14V - (\approx 3.2V \times 3)}{120Ω}$$

This isn't exact. The higher the source voltage and current goes, and the LEDs forward voltage drop does change a bit. At ~36mA the LEDs are being over driven, shortening their life some, so you won't get 1000~5000 hours that they should give at 20mA. But its normal to overdrive them. And they will be brighter to boot, so you might even need less LED segments.

But now lets change the source voltage to 16V:

$$\approx 53mA = \frac{16V - (\approx 3.2V \times 3)}{120Ω}$$

53mA continuous is almost 300% of the typical recommended current of 20mA. But at 53mA, the forward voltage drop goes up as well, making the math tricky. The life of the leds will be much shorter, and they will warm up considerably.

Solution: As others have mentioned, you can use Silicon Diodes to drop the source voltage. You can use from 3 to 9 (Dropping 2.1V to 6.3V), depending on how bright you want the leds, how much current the strip should pull, your batteries voltage range (A 4S Lipo is 14V Nominal, 16.5V at max safe charge, 11.5V at lowest safe discharged voltage).

Best thing to do, is take one or two segments of the led strip and TEST THEM. Connect one directly at the charged 16V and see how long it lasts or how hot they run. Look at the resistor and do the math. Then start adding Silicon Diodes and see the difference.


The easiest way, then, is to find out how much current the LED strip draws with 12 volts across it, and then to connect 7 silicon (not Schottky) diodes rated for that current or greater (greater is better) in series with the LED strip and the LIPO.

Depending on how much current they have to handle, they could get pretty warm, so you might want to place them somewhere where air can blow over them to cool them.

If you use 1N4001 or 1N4002 diodes, figure about a 0.8 volt drop each with 750mA through them, so you'll need about 5 or 6 in series between the LIPO and the LED strip, like this:

      +|                                                              |+
    [LIPO]                                                       [LED STRIP]  
       |                                                              |
  • \$\begingroup\$ Thankyou, I measured the strip, it is not cut yet so measure 5 meters. At 12 volts it draws about 1.5 amps. I will probably need only half of it so I presume it will be 0.7 amps. So what I have to do is connect ground to - of the strip, then from the battery positive start with one diode after the other and go in led strip + . Is this correct? \$\endgroup\$ Jul 27, 2014 at 12:43
  • \$\begingroup\$ Yes. I'll edit my answer with a hookup sketch in a few minutes. \$\endgroup\$
    – EM Fields
    Jul 27, 2014 at 13:05
  • \$\begingroup\$ Greate! Also, do you think 1n5004 diode could be a good choice? \$\endgroup\$ Jul 27, 2014 at 13:30
  • \$\begingroup\$ Dunno on the 5004, they seem hard to get. A good/better choice might be cheap and easy to get 1N4001s or 4002s. \$\endgroup\$
    – EM Fields
    Jul 27, 2014 at 13:51

I know there are dc-dc converters, but I rather avoid using them due to weight (rc plane)

Although you say that you don't care for such a solution I would like to add it as an option.

A switching step down regulator module such as LM2596 weights just 11.5 grams (costs about a dollar from ebay).

The benefits of such a regulator are :

  • better output regulation (constant voltage at the output irrelevant of load changes) compared to the passive method (diodes)
  • high efficiency which will generate less heat and waste less energy from the battery
  • and you get the short circuit protection as a bonus

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  • \$\begingroup\$ Hi! Thanks! Do you need to keep it with airflow or it can stay buried in foam? \$\endgroup\$ Jul 27, 2014 at 18:48
  • \$\begingroup\$ @user1937747 I use it in free air and the coil and regulator may get warm depending on the input voltage and output voltage/current. Your input/output difference is not high but I can't give l a definite answer to you question. What I can tell you is that you will not need a heatsink. \$\endgroup\$
    – alexan_e
    Jul 27, 2014 at 19:19

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