In the Arduino Start Kit, the first circuit is very basic: 5V power, 220 ohm resistor, and an LED with a forward voltage of 2.2V. However, here's two things I can't reconcile:

  1. The booklet says that current can be calculated via Ohm's law: I = V/R, I = 5V / 220 ohms, I = .023 amps.

  2. However, I know that, when starting from just the power supply and LED, calculating the resistance required uses Ohm's law as well: R = V/I. This time, though, you need to subtract the voltage drop across the LED (forward voltage) from the power voltage, so we subtract 2.2V from 5V giving us 2.8V. We then use 2.8V as the voltage in the R = V/I calculation (and then provide the LED's desired ideal current as I, etc.)

How do I reconcile these two things? Based on what I know in #2, shouldn't the circuit's current be calculated using the voltage across the resistor, 2.8V, instead of the 5V power supply as is used in #1 above?

  • \$\begingroup\$ The formula in #1 applies to resistive elements such as resistors. Diodes, including LEDs, have a more complex relationship between voltage and current, so you use #2 which is a rough approximation of the true diode behavior. \$\endgroup\$ Commented Sep 18, 2021 at 14:53

3 Answers 3


Your method 2 is correct.

You might find it useful to learn about load lines. These are useful when the device (the LED) doesn't behave in a linear fashion like a resistor.

enter image description here

Figure 1. Various loadlines for a 5 V supply and various LED types. Image source: Loadline resistance graphic tool.

  • Each loadline is drawn from 5 V, 0 mA to 0 V and the current calculated for the resistor on its own (method 1 in your question).
  • Now, to decide the resistor value choose the current that you require and go horizontally across to the colour of your LED. Pick the nearest resistor load line.

e.g. I want to run 10 mA through a blue LED.

  • Run your eye across the 10 mA line until it intercepts the blue LED curve.
  • We can see that this is on the 270 Ω load line so that's the one to choose.
  • Meanwhile you can also see that the 270 Ω would give about 12.5 mA to a red LED.
  • \$\begingroup\$ I am a bit on doing it the easy way. Here is an online calculator that does a nice job. \$\endgroup\$
    – Gil
    Commented Sep 18, 2021 at 16:32
  • \$\begingroup\$ Ah, but without understanding - oh, and without a link! \$\endgroup\$
    – Transistor
    Commented Sep 18, 2021 at 16:36
  • \$\begingroup\$ This is a terrific resource. Thanks! \$\endgroup\$ Commented Sep 18, 2021 at 17:39

When calculating the current through an LED and a resistor in series, you need to take into account the voltage drop across the LED. So method 2 is the correct one.


You need to allow for the LED voltage drop, whether you have a simple LED, resistor and battery circuit, or the LED is controlled by an Arduino or other device.

There is no "ideal" current for a LED. There will be an Absolute Maximum current, beyond which the LED may be damaged.

LEDs will work happily at currents much below their rated maximum current, but will be dimmer at lower currents. I once had to run a green LED at somewhat under 1 mA to get it dim enough for my needs.


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