I've done some research but can't seem to find an answer to my specific problem.

I'm using an Arduino as a beam-break sensor. In order to do that it modulates the IR LED at 38kHz by code and listens for this signal on the sensor pin. Now, I want to increase the range of this system to 15m using a retroreflector on the other side. In order to do that I need more power on the LEDs than the 20mA the Arduino can output.

As I understand it I will need a transistor to do this.

I have ordered these LEDs

And these TIP120 transistors

Will I be able to accomplish having 4 LEDs running through this transistor at 100mAh each and then have the Arduino modulate the transistor at 38kHz so the combined output of all LEDs is modulated?

I already have a 12v power supply in place so that will be the supply voltage for the system.

  • \$\begingroup\$ IR LEDs usually only drop about 1.5V. So you could just wire 4 in series, and select an appropriate current limiting resostor. \$\endgroup\$
    – Simon B
    Dec 3, 2018 at 16:22
  • \$\begingroup\$ I was thinking that as well. My problem, however, is if I understand the use of the transistor correctly and if I need anything else for this to work \$\endgroup\$
    – BlueSails
    Dec 3, 2018 at 19:01
  • \$\begingroup\$ You will also need a couple of resistors. Driving a transistor from an Arduino is such a common thing, you can just Google it. Find an example of "low side switching". \$\endgroup\$
    – Simon B
    Dec 3, 2018 at 23:43

1 Answer 1


In addition to the transistor and LEDS, you will need a couple of resistors. It will be easiest to wire the diodes in series, with one resistor for current limiting. you should also add a resistor before the base of the transistor.

See this Stack Exchange answer for an example. Circuit A - "low side switching" - is almost certainly better.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.