I'm building an assembly that will drive some infrared LEDs. Although the final product will use three, I'm just playing with one for testing purposes at the moment.

The spec sheet for the LEDs I'm using can be found at http://www.osram-os.com/Graphics/XPic0/00083225_0.pdf

If I'm reading that sheet correctly, the maximum current for this LED is 100mA. The information I have found online suggests that you shouldn't run them at their maximum output except in short bursts so I was going to go with 50mA.

I used the calculator at http://www.digikey.com/en/resources/conversion-calculators/conversion-calculator-led-series-resistor?WT.srch=1&WT.medium=cpc&WT.mc_id=IQ63967658-VQ2-g-VQ6-36682054515-VQ15-1t1-VQ16-c

I'm using a battery case that holds 2AA's so I entered 3 as my supply voltage, forward voltage of 1.5 (from the spec sheet), and 50mA as the forward current. According to the calculator I should be using a 30Ohm, which is what I have.

I pointed the camera on my cell phone at our TV remote and verified that it does indeed "see" IR, but when I test my circuit I get nothing. I suspect I'm interpreting something on the spec sheet wrong and sized the resistor incorrectly. Can someone set me straight?

Here is some additional information per your requests. First, a pic (albeit fuzzy) of the circuit... Circuit

Since it is a little fuzzy, here's a quick tour. Battery to power rails. Resistor from negative power rail to A9. Negative leg of the LED (verified per spec sheet) in E9, positive leg in F9. Jumper from H9 to positive rail. I put some wires on the power rails on row 15 to make it easier to measure voltage across the battery.

When I measure across the battery I get 3.153v, across the resistor is 2.073, and across the LED is 1.077. Everything is around the values I expected, I'm beginning to wonder if my assumption that my cell phone should be able to see the light is flawed. It does see activity from multiple remote controls I have here so I would have thought it would see this as well. Going to pick up a red LED today for testing purposes but if this new information causes any light bulbs to go on for any of you (no pun intended) I'm always happy to hear anything you have to share. This is a learning experience for me, many thanks to the community.

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    \$\begingroup\$ Reversed connection perhaps? Is the 30 ohm 1.5 Watt resistor warm? Since Vf is 1.5V maybe try with just one AA battery first. \$\endgroup\$
    – MarkU
    Commented Jan 16, 2015 at 4:12
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    \$\begingroup\$ Including a photo of your setup would be very helpful as well. \$\endgroup\$
    – Dan Laks
    Commented Jan 16, 2015 at 5:14
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    \$\begingroup\$ Use a cheap red LED first to ensure all is OK. Reverse polarity of LED seems most likely. You can check correct polarity with an Ohm-meter but note that some LEDs are VERY sensitive to incorrect polarity and may die at 9V test-meter potential. Most survive 3 Volts. Better stilol - have a data sheet and use correct polarity initially. \$\endgroup\$
    – Russell McMahon
    Commented Jan 16, 2015 at 12:35
  • \$\begingroup\$ Do you have a multimeter? Confirm you have the voltages and currents you expect. \$\endgroup\$ Commented Jan 16, 2015 at 14:46
  • \$\begingroup\$ Original post edited with additional info, thanks. \$\endgroup\$ Commented Jan 16, 2015 at 15:00

1 Answer 1


The LED voltage, current, and resistor equation will be (3v - 1.5v) / 30 ohm = 0.05 A (or 50ma), so your calculation is correct.

To troubleshoot this measure the voltage across each component when the circuit is powered, across the battery should be 3v, then 1.5v across the resistor, and 1.5v across the LED.

If the LED shows 3v it may be connected in reverse, or is burned out open.

If the LED shows near 0v it may be burned out shorted.

(Note that the voltages might be off a slight bit as the LED forward voltage at 50ma seems to be nearer to 1.4v per the forward current chart, see data sheet page 6).

Also be sure to observe the warnings in the data sheet about eye safety. These LEDs are bright and emit concentrated light, even though they are infrared they can still be dangerous to look at directly.

  • \$\begingroup\$ Original post edited with a pic of circuit and the voltages you asked me to measure. Thanks. \$\endgroup\$ Commented Jan 16, 2015 at 15:00
  • \$\begingroup\$ The LED voltage seems a bit on the low side, though not by much. With 2.073v on the resistor the current should be 69.1ma. Per the forward current chart that should give about 1.4v on the LED. But the chart may be assumed to be typical & it actually represents a pulsed operation. But even at that voltage the LED should produce IR output. How does you phone camera see IR from the TV remote? (as a spot on the screen or as data?) A TV remote puts out IR pulses, your breadboard setup is a constantly-on IR source. You may need to build an IR detector just to prove your emitter operation. \$\endgroup\$
    – Nedd
    Commented Jan 17, 2015 at 12:07
  • \$\begingroup\$ Turns out the circuit was fine. Although my phone can see IR light from every remote in my home it didn't see this. But when I removed the IR filter from a PS Eye web cam (doing this to build a FreeTrack rig) that cam saw it fine. Your troubleshooting steps were invaluable and you were on the right track with you comments about needing a different method of detecting the output so I'm accepting your answer. Thanks very much for the help. \$\endgroup\$ Commented Jan 19, 2015 at 13:24

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