Different from the best answer of this one: Why does a resistor need to be on the anode of an LED? But well, I'm also an electronics noob ^.^

Recently I bought the official Arduino starter kit and played around with the common cathode RGB LED that came with this kit. In the project book, the instructions for the colour mixing lamp is to use separate 220ohm resistor for each of the RGB legs. But I thought, why can't I put the resistor on just the cathode of the LED?

I tried that by plugging the LED into the breadboard and connecting the cathode through a 220ohm resistor to the ground of Arduino UNO. When I touched any one of the R, G or B anode legs with a jumper wire connected to +5V, the respective colour lights up alright. However, when I bridge the G and B pins together, only the green LED lights up. When I bridge all three anodes together, only the red LED lights up. But if I connect the anodes to +5V through their separate 220ohm resistors(according to the instruction), the colours will combine.

Why is it so?

  • 4
    \$\begingroup\$ Essentially you are running multiple LEDs in parallel. Search for one of the many questions here that explain why parallel LEDs should have all their own resistor. \$\endgroup\$
    – PlasmaHH
    Commented Feb 13, 2015 at 10:07
  • \$\begingroup\$ @PlasmaHH well anyway through this question I have linked RGB LED and LEDs in parallel together for absolute noobs :p \$\endgroup\$ Commented Feb 13, 2015 at 12:00

4 Answers 4


The red led will hog all the current because it might only need 2 volts across it to begin conducting. The green and blue LEDs need a higher voltage but because they are all in parallel the red led dominates. Try measuring their respective volt drops when each operates.

It's like putting a 5 volt zener in parallel with a 10 volt zener. The 10 volt zener will never conduct.


For a single LED you are correct - which side you put it on does not matter. However, an RGB LED is a rather different animal since one side of all of the elements are tied together. This presents a bit of a problem. If you wire the common terminal directly to a power rail and put 3 resistors on the other side, it will work as expected. However, if you a stick one resistor pin on the common terminal, you will have some issues. If you only try to turn on one element at a time, it will work correctly. However, if you try to turn on more than one, then the LEDs will be in parallel and they will behave in an unexpected manner. If the forward voltages are the same, then the LEDs will split the current and light up at half brightness, approximately (not exactly because the current will never quite split exactly). If the forward voltages are different, then only the LED with the lowest forward voltage will light up as it will turn on and steal all of the current before the other LEDs will exit cutoff. Bottom line: don't put LEDs in parallel as they will not evenly share current.

  • \$\begingroup\$ Thank you for the detailed explanation, but are you sure when the forward voltages are the same the LED will light up at half brightness...? I thought they will just draw more current and become brighter than if they are wired in series (well I'm not sure about LEDs though)... Also what did you mean in the sentence 'If you you're the common terminal directly to a power rail'? \$\endgroup\$ Commented Feb 13, 2015 at 12:03
  • \$\begingroup\$ I'm assuming you posted this from a mobile device, because there are some awesome spelling mistakes! "forward villages?" "corny?" \$\endgroup\$ Commented Feb 13, 2015 at 15:35
  • \$\begingroup\$ If one doesn't need absolute maximum brightness but is trying to minimize cost, would it be reasonable to use one resistor and feed the individual-LED legs with non-overlapping PWM? \$\endgroup\$
    – supercat
    Commented Feb 13, 2015 at 16:43
  • \$\begingroup\$ Hah, I thought I got all of those. Blame it on autoincorrect. \$\endgroup\$ Commented Feb 13, 2015 at 18:38
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    \$\begingroup\$ @DoktorJ: The total duty cycle would be limited to 1, but if "white" requires e.g. 4:5:7 ratio there would be no problem with one of them being 7/16 brightness, provided that total duty cycle was no more than 1. \$\endgroup\$
    – supercat
    Commented Feb 13, 2015 at 20:33

The answer is simple: Its because if you have only one resitor and three leds, you have a "variable" load for a fixed value resistor.

If red and blue leds are turned on, this means a load of a value. If green and blue are lit, this means a load of another value, and so on. You cannot match all possible states with a single resistor, so you place three on the other side, allowing each resistor to deal with a single led.

If, for any porpuse, you would use the leds in a fixed load state, ie, keep the same leds lits forever, them a single resistor would be enough.


To put everything in the answer and comments together:

First, RGB LEDs are just three separate LEDs packaged into the same package, with their anode or cathode tied together. The other end is also tied together if only a single resistor is used. So that is three diodes in parallel.

Second, the voltage each LED need to begin conducting is called the forward voltage, usually noted as Vf.

In the ideal diode model, a diode does not conduct when the voltage applied to its terminals is lower than Vf or when it is reverse biased (negative to anode and positive to cathode), and conducts like a short circuit when the voltage is higher than that, while keeping the voltage between its anode and cathode constant.

The resistor is added to limit the current through the LED so it's no longer a short circuit. Else the LED will burn out. Or to think about it in another way, it doesn't add up for a 5V source and a 1.8V voltage source to be shorted together. Either the LED or the source will burn out so the voltage settles. The LED will usually be the one to burn out first as it usually tolerates a lower current.

In the example mentioned in the question, taking the following Vf for different LEDs:

Vfred = 1.8V Vfgreen = 2.1V Vfblue = 3.0V

When applying 5V to the anode of all three, the red LED will conduct first, so the cathode would be fixed at 1.8V below 5V.

The voltage difference between the other LEDs' anode and cathode will also be 1.8V since they don't get to conduct. As a result, they are as good as not in the circuit at all.


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