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I am seeking some advise on a making an infrared light source. I am using IR LED SE-5455-2. I have found some calculators on line but the results vary. I have also read many threads here. I am still somewhat confused.

LED specs are Forward voltage drop = 1.7, reverse breakdown = 2, Peak wavelength = 930, forward continuous current = 100 mA.

I want to run 7 LEDs in series and power them with a 12VDC power supply that can deliver 1 amp.

How can I determine what I need for a current limiting resistor?

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  • \$\begingroup\$ Is your plan to use the IR LEDs as a continuous light source, or pulsed, e.g. for communication? Looking at the datasheet, it can be driven even harder than 100mA in pulsed applications. Also, based on graphs in the datasheet, if you run them at slightly less than 100mA, the forward voltage drop is reduced, so seven should be easier to control. Six in series would provide enough voltage headroom to switch them with a transistor, or use a constant current source. \$\endgroup\$
    – gbulmer
    Commented Oct 31, 2015 at 8:36

3 Answers 3

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It's entirely possible... and apparently math just wan't my thing yesterday. As a few others have pointed out since, the LEDs would take 11.9V, and a resistor to drop the remaining 100mV at 100mA is 1 ohm.

As far as the resistor itself, assuming one diode, we can do an example. The diode will drop 1.7V from the 12V source, leaving 10.3V to go away somehow (without a resistor this burns up the diode). The 10.3V happens to also be at 100mA, which will be very bright, but that's ok. Now, we just need a resistor that will drop 10.3V at 100mA, or 103 ohms. With that resistor, that diode, and the 12V source, voltages add to 0V, so everything's happy.

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  • \$\begingroup\$ Thank you for the reply. I was thinking the voltage of 7 LEDs in series would be 1.7 X 7 = 11.9. Did I understand that incorrectly. That was why I was running 7 in series, so it would be just below the voltage of the power supply. \$\endgroup\$
    – laserman2431
    Commented Oct 31, 2015 at 2:07
  • \$\begingroup\$ So there will be a 100mV drop in your resistor, with 100mA current you get a value of 1 ohm resistor. \$\endgroup\$
    – Pedro Nadolny
    Commented Oct 31, 2015 at 2:13
  • \$\begingroup\$ Thank you. That makes sense. What about the watts for the resistor? I assume I would need a resistor rated at least .017 watts. Is there a problem with using a resistor that is rate at much higher. Say maybe 1/4 watt? \$\endgroup\$
    – laserman2431
    Commented Oct 31, 2015 at 2:17
  • \$\begingroup\$ @laserman2431 - At .250 watts, a 1/4 watt resistor is not "much higher" than you need. In fact, that's what I'd use. .25 watts is simply the maximum power that the resistor can gracefully handle, and any lower power dissipation simply means that the resistor will run cooler. And just as a matter of good practice, I'd recommend running the LEDs at somewhat less than the rated maximum current. \$\endgroup\$
    – WhatRoughBeast
    Commented Oct 31, 2015 at 3:11
  • \$\begingroup\$ To avoid confusion for the OP and future readers, please correct the calculation. 7 x 1.7V = 11.9V and not 13.9V. As a result, the first paragraph will likely be confusing too. \$\endgroup\$
    – gbulmer
    Commented Oct 31, 2015 at 8:25
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Great success. Thanks for all the help. I ran 7 in series with a 33 Ohm resistor rated for .5 watts. The light source worked perfectly and the LEDS are nice an cool to the touch.

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So these LEDs basically take 1.7V a piece to light up. 1.7*7 is 11.9, which is pretty close to what 12V supply will output. Your best bet is to run them in parallel here, so that you can actually see them.

So if you were to run them all in parallel, you've got these specs:

1.7V drop
700mA

so 12 - 1.7 = 10.3V, at 700mA. This needs to be dissipated in the resistor, so you'll need a (10.3 / 0.7)Ω = 14.7Ω resistor to light them up. This beast will dissipate about 10W at 700mA, so you might just want to use a resistor for each LED, or maybe a voltage regulator.

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  • \$\begingroup\$ Thank you for the reply. I'm confused about the math on the voltage. Isn't 1.7 X 7 = 11.9? \$\endgroup\$
    – laserman2431
    Commented Oct 31, 2015 at 2:14
  • \$\begingroup\$ *facepalm* edit made. \$\endgroup\$
    – Daniel
    Commented Oct 31, 2015 at 2:23
  • \$\begingroup\$ I should have explained the whole plan. I want to power several strings in parallel. That's why I was thinking I would make series strings with the maximum I can have and then run the series strings in parallel. If 11.9V is too close to 12V, would you recommend that I use 6 LEDs in series and make up the difference (1.8 V)with the resistor? \$\endgroup\$
    – laserman2431
    Commented Oct 31, 2015 at 2:32
  • \$\begingroup\$ Yes. That would be better. \$\endgroup\$
    – Daniel
    Commented Oct 31, 2015 at 2:57
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    \$\begingroup\$ Measure the actual voltage drop on the LEDs you will be using, at the current you wish to use, before committing to any particular arrangelment of LEDs and resistors. The forward voltage of the LEDs will vary slightly with current. \$\endgroup\$
    – Peter Bennett
    Commented Nov 1, 2015 at 1:59

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