One of the first things students are taught about LEDs is that each LED in a parallel circuit must have its own resistor. This question describes this perfectly and the accepted answer is great. However, often those cheap LED fairy lights use a single resistor at the battery box for the entire circuit, and that's it.

Why don't the theoretical properties described in this answer apply to LED string lights? How are they manufactured without a resistor for each LED in the circuit?

  • \$\begingroup\$ They use dedicated ICs and circuitry to limit the current. this Q/A explain the function of one of many ICs. \$\endgroup\$ Commented Nov 18, 2020 at 7:04
  • \$\begingroup\$ It's almost like the theoretical properties don't apply to reality as well as the theorists would like them to. \$\endgroup\$ Commented Nov 18, 2020 at 13:58
  • 1
    \$\begingroup\$ the tiny shite wiring used in cheap fairy lights is a resistor. \$\endgroup\$
    – dandavis
    Commented Nov 18, 2020 at 19:27

1 Answer 1


From a different answer to the question you referenced:

The above is written with assumption that you will chose the resistor in such way that is sets the current so that the current is n times the current you want in each diode where n is the number of diodes and that the current is actually larger than the current which a single diode can safely conduct. What then happens is that the diode with lowest forward voltage will conduct most of the current and it will wear out the fastest. After it dies (if it dies as open circuit) the diode with next lowest forward voltage will conduct most of the current and will die even faster than first diode and so on until you run out of diodes.

The bad things all happen if you are driving multiple LEDs in parallel at a current at or above the total rated current for all the LEDs together.

Say you have 10 LEDs rated for 20 mA, and you connect them all in parallel with one resistor to limit the current to 200 mA (maximum current to all LEDs.)

One LED might conduct a little more than the others, so it will warm up and conduct more. As it gets hot it takes up more of the current, until it is trying to conduct too much. At that point, it burns out and leaves 9 LEDs trying to share the current intended for 10 LEDs. One conducts a bit more, gets hot, and so on until all the LEDs are gone.

Now lets try it a little differently. Rather than try to run all the LEDs at maximum brightness, you allow only the rated current for one LED.

All 10 LEDs now have to share 20 mA. They aren't as bright, but that's OK. Each LED gets 2 mA.

The forward voltages for all the LEDs are closer together at low current, so it is less likely for one LED to "snarf up" significantly more current than the others. This also makes the (cheap, not brightness matched) LEDs more consistent in brightness.

Even if one LED does take more current than the others, it can't get hot and "run away." The 20mA is lower than the rated current, so if one LED "hogs all the current" it won't burn out.

Check the battery box on a cheap string of LEDs, and I expect you'll find a large resistor that reduces the current to something lower than the maximum allowed for one LED.


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