I just know the basic electronics, not an expert in the area :)

When I dismantled a broken LED lamp, I could see the circuit below (section after the rectifier, the DC part). 15 LEDs in series. There are two such lines. If they were just parallel lines I know how the circuit works, but they are inter connected here.

enter image description here

Two of the LEDs in a row were broken, So I thought of adding an equivalent resistor value as I did not have the right LED with me. Then the bulb started blinking when powered. Then instead of resistor I just shorted the position of broken LEDs and now it works.

Can anybody explain me the working of this circuit, especially what happens when the parallel lines are interconnected?

  • \$\begingroup\$ Welcome to EE.SE! What resistor value did you insert and where? \$\endgroup\$
    – winny
    Aug 29, 2018 at 9:03
  • \$\begingroup\$ I removed the 2 LEDs in 2nd row(so the circuit is open then). Then I added a 470ohms in one LEDs position, thinking it will match the excess voltage to be dropped when I remove the LED. \$\endgroup\$ Aug 29, 2018 at 12:36
  • \$\begingroup\$ This app note explains the wiring of LEDs very well: dammedia.osram.info/media/resource/hires/osram-dam-2496697/… \$\endgroup\$ Aug 30, 2018 at 20:41
  • \$\begingroup\$ Thanks, well explains and matches with the circuit I have. Thanks \$\endgroup\$ Sep 3, 2018 at 7:12

1 Answer 1


LED bulbs usually use a current source. That is a driver that provide a constant current through the LEDs. Putting the LEDs in series is then the logical way to connect multiple LEDs to the same driver.

The reason the leds are put in in a series string of parallel pairs is to add some redundancy if one LED fails and to avoid mismatched LEDs making one string brighter than the other.

If you short out a single LED the voltage across them all will be dropped by the driver so the current matches what it was programmed to emit. The matching led will not light up anymore.

The reason it started blinking when you replace an LED with a resistor is because the driver will have open circuit detection and reset itself if it cannot supply the voltage needed for the current it wants to drive.

  • 1
    \$\begingroup\$ >> The reason it started blinking when you replace an LED with a resistor is because the driver will have open circuit detection<< But, when I add a resistor the circuit is closed, not open, right ? \$\endgroup\$ Aug 29, 2018 at 12:37
  • \$\begingroup\$ Yes but the voltage needed to push the required current through is too high for the driver to handle. \$\endgroup\$ Aug 29, 2018 at 12:45
  • 1
    \$\begingroup\$ So is it like, normally the the LED driver produces 75V at 15mA(e.g.) and when I add a resistor it can detect that something is wrong that it is not able to drive this 15mA to turn the LEDs ON ? \$\endgroup\$ Aug 29, 2018 at 13:19

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