We have an LED that when connected to an Arduino board provides an optical power of 22.58 mW. The circuit is just the LED in the digital Pin with the Arduino powered by the computer using a USB cable. Our optical power needs to be a minimum of 226 mW, so it should be one order of magnitude greater.

Since P=IV, and the Arduino has a fixed voltage of 5V, I think it would make sense to try and maximize the current through the LED. I've read that the maximum output current from one pin is 20 mA while the maximum current the Arduino can provide in total is 50 mA. Would connecting three outputs in parallel to lead into the LED in series provide 50 mA instead of 20?

I've also looked into using a transistor to drive a higher current. With the example in this video, what would be the voltage drop and current across the resistor?

Wouldn't the current be ~10 mA(5/470) to account for the 470 Ohm resistor and 5V EMF source? How would adding the transistor increase the current driven through the LED?

I may have a misunderstanding of how the digitalPin relates to the circuit, but it seems to be an output of the transistor in parallel with the output to ground. Or based on what is shown at 0:52 does the digitalPin serve as the input for the current while the part of the circuit connected to ground and the part with the LED are in parallel? As I understand it, the 10k resistor is there to minimize the amount of current that travels through that part of the circuit, so my first assumption should be correct.

I found another source online with an example circuit using an LED and a transistor https://circuits.io/circuits/742344-arduino-npn-transistor-led-driver#schematic, but this seems to invert the principles of the video above. If the digitalPin is supplying PWM, then the current should be about 1.4 mA (V=IR, V=5, R= 3600) which makes it hard for me to understand why the LED would drop 1.6V.

Sorry for the number of questions and confusion, but if anyone could help me design a circuit to maximize optical power, that's the main goal here. I would like a better understanding of the examples I provided to figure out what modifications I could make to those circuits and my own to get the target optical power.

  • 2
    \$\begingroup\$ Daniel, how do you know the 226 mW figure? In all optics, it isn't only about the transmitter, but also about the receiver. And it is almost never just about power, but about power per steradian and how that varies vs angle. So that figure must come from somewhere. The first thing I'd like is to be convinced about is that this figure is absolutely necessary (that there is not another means of achieving the intensity you need by physical modifications.) This means that you know the emission distribution (Lambertian or otherwise) and so on. So where did it come from? Start there. \$\endgroup\$
    – jonk
    Nov 14, 2017 at 8:41
  • 3
    \$\begingroup\$ Can't be bothered with the videos, but what you want is (a) a high-power LED and (b) a transistor (probably MOSFET) to drive it, rather than the limited current output of an Arduino. \$\endgroup\$
    – pjc50
    Nov 14, 2017 at 9:54
  • 2
    \$\begingroup\$ The first question should be - is the LED capable of handling ten times the power? You need to design the circuit around the LED you're using. For example, in the second circuit, the assumption is that the LED requires 20mA at 1.6V. The LED drops 1.6V just because it does - it's probably a red or infra-red one. LEDs are not resistors; they do not obey V=IR. \$\endgroup\$
    – Simon B
    Nov 14, 2017 at 10:08
  • \$\begingroup\$ jonk, the we have a target output optical irradiance that requires ten times the optical power we currently have, so the 226 mW figure was taken from. I'm not familiar with optics at all, but we have a set area for optical irradiance and a maximum time frame of two minutes to meet our criteria for average energy. So, the optical power divided by the area multiplied by the time needs to meet our average energy criteria. \$\endgroup\$ Nov 14, 2017 at 15:44

2 Answers 2


Since P=IV, and the Arduino has a fixed voltage of 5V

Optical or Radiometric power measured in mW is not the same as the electrical Forward Voltage x Forward Current.

226mW of white light is about 50-60 lumens. You need about 350 mA to get 50 lumens out of a 50 lumen white LED.

To get 223mW out of colored LEDs you need a fair amount of lumens.
You must convert radiant mW to lumens by the wavelength of the color.

Lumens required for 223mW radiometric optical power

  • red 58 lumens
  • green 152 lumens
  • blue 6 lumens

Arduino has a fixed voltage of 5V

Supply voltage does not matter. An LED is a current regulated device and its voltage is a function of its forward voltage, typically 2V (red) to 3V (white. green, blue) . The other 2-3V (of the Arduino 5V) will be dropped across the current limiting resistor.

Wouldn't the current be ~10 mA(5/470) to account for the 470 Ohm resistor and 5V EMF source?

A 300Ω resistor will give you 10mA with a 2V LED.
A 200Ω resistor will give you 10mA with a 3V LED.

You need a 1 Watt high brightness high efficacy LED which will still take over 300mA with a 5Ω-8Ω current limiting resistor.


SWAG: Keep in mind A power supply the Arduino is NOT. Use an external power supply. Are you using a 12V power source (example) through the barrel connector? If so you will have a lot of heat to dissipate for 350mA LED (there are other leds available with higher output). If you are supplying 12V (example) you can add a buck converter to drop the voltage down to what you need. I would suggest using an N-Channel MOSFET with the source grounded, the gate connected to the Arduino pin and the drain connected to the anode of the led. With this configuration you can use Ohm's law (lots of calculators online) to calculate the resistance needed (they also tell you the resistor wattage). Subtract the LEDs voltage from the source voltage and use that value to calculate your needed resistance. There are a lot of possibilities, you need to give us more information such as power supply etc and a schematic if possible.


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