I got a new IR led, emitting light at 850nm. It says 5v reverse voltage and 1.5 forward voltage. How many volts should I give it? I have DESTROYED three other IR LEDs because of giving them too many volts. What should I do?!

Here is the full information on the LED, typed up:

Max. Reverse voltage:5v
Max. Continuous forward current: 100mA
Max. Forward Voltage: 1.5 typ., 1.8V max.@50mA

So, if I give it 3.3v@50mA raw, it will destroy it, correct? Therefore, I guess I need a 10k resistor?

  • \$\begingroup\$ What kind of driving resistor are you using? What voltages did you try? \$\endgroup\$
    – I. Wolfe
    Commented Apr 6, 2015 at 23:27
  • \$\begingroup\$ Please add the missing information to your question: what voltage do you want to run them from - and - how much current do they need? \$\endgroup\$ Commented Apr 6, 2015 at 23:28
  • \$\begingroup\$ Possible duplicate of Does LED brightness change with voltage? \$\endgroup\$
    – winny
    Commented Dec 20, 2017 at 6:54

2 Answers 2


I'm pretty sure that there are multiple answers on this site with the information that you require.

But: decide what voltage you want to run the LEDs from. Subtract 1.5V - you now know the voltage across the series resistor that you need.

Now divide that voltage by the desired current. If you divide by Amps, you get the resistor value in Ohms. If you divide by mA, you get the resistor value in Kilohms.

  • 1
    \$\begingroup\$ And that resistor value is obviously the minimum amount of resistance necessary for me to not destroy my LED, correct? \$\endgroup\$ Commented Apr 6, 2015 at 23:39
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    \$\begingroup\$ When you said "subtract 1.5v", is that because that's the typical voltage? \$\endgroup\$ Commented Apr 11, 2015 at 23:31

As with any LED, you should be concerned with the LED current, much more than its voltage.

The voltage across any forward-biased LED (any diode, really) will be governed by the diode, and will only vary slightly with current.

Look on the LED's data sheet for the recommended operating current, or, failing that, the Maximum Recommended current. If you can only find a Maximum Recommended current, operate the LED at 80% or less of the Maximum.

You normally control the LED current using a resistor in series. The resistor value is calculated by subtracting the LED forward voltage from the supply voltage, then using the resulting voltage and desired current in Ohm's Law.

-----------------------Added, after the OP added vital data---------

With 3.3 Volt Vcc, the Atmel ATMega328's recommended output current seems to be 10 mA or so, so you can't drive the IR LED directly from an I/O pin. Instead you will have to do something like:


simulate this circuit – Schematic created using CircuitLab

The 18 ohm resistor should limit the LED current to about 80 mA - safely under its maximum rating of 100 mA. 20 or 22 ohms would reduce the current a bit more, if you want to be safer.

The transistor's emitter must be connected to the Arduino ground.

  • \$\begingroup\$ OK, so if I run it from an Arduino Uno 3.3v output pin, what should I do? \$\endgroup\$ Commented Apr 6, 2015 at 23:30
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    \$\begingroup\$ You do what I said in my answer. Your first step will be to determine the desired operating current for your LED. \$\endgroup\$ Commented Apr 6, 2015 at 23:32
  • \$\begingroup\$ I safely powered it from my Uno using the 3.3v output pin, which, according to arduino.cc, is going at 50mA. I put one wire from the 3.3v output to my 330 ohm resistor(smallest one I have), and ran it from there. Worked fine, showed up on my phone's camera as a little purple-ish dot. \$\endgroup\$ Commented Apr 7, 2015 at 11:57
  • \$\begingroup\$ Can you explain what equation you used and where to see their values on the transistor? I have that transistor and also a 2N2222 but I have 5V Vcc and 40mA from the arduino. \$\endgroup\$
    – shinzou
    Commented Oct 3, 2017 at 19:43
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    \$\begingroup\$ Hm, since you want to drive the LED at 80 mA and you want to saturate Q1, R1=1kOhm is too high. I arrive at R1 = (2.6 V - 0.9 V) / 0.008 A = 212.5 Ohm. Note that an ATmega328p output pin drops voltage with higher current, thus I'm using 2.6 V instead of 3.3 V (cf. datasheet). And 0.9 V is the voltage drop at the Q1 base (cf. V_BE(sat) in a 2N3903 datasheet). Assuming that Q1 saturates at I_C / I_B = 10, as specified in the datasheet. For R2 I arrive at R2 = (3.3 V - 1.5 V - 0.3 V) / 0.08 = 18.7 which is what you already have in your schematic. (i.e. V_CE(sat)=0.3 V, V_F_LED=1.5 V) \$\endgroup\$ Commented Feb 20, 2022 at 16:15

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