The P-N junction of LEDs is always placed on the cathode, which is larger than anode. Is this design of performance concern or just a historical choice?

  • \$\begingroup\$ P is the anode and N is the cathode so you can't haveboth on the cathode. \$\endgroup\$
    – Andy aka
    Apr 30, 2014 at 8:45
  • \$\begingroup\$ I know. But why the industry didn't choose the reversed way ( i.e. placing P-N junction on a larger anode and use a smaller cathode)? \$\endgroup\$ Apr 30, 2014 at 8:58
  • \$\begingroup\$ @Andyaka: I believe they're talking about the physical construction of the LED package, i.e. placement on the anode or cathode pins rather than the junction itself. \$\endgroup\$ Apr 30, 2014 at 9:01
  • \$\begingroup\$ Your question makes no sense. The P-N junction can't be "placed" on the cathode, since the cathode is itself half of this junction (the N part). \$\endgroup\$ Apr 30, 2014 at 13:15
  • \$\begingroup\$ See also this EESE question. \$\endgroup\$ Jan 25, 2015 at 0:50

1 Answer 1


While it may boil down to "that's always the way its been done", a convergence of market forces, there is one logical reason to mount the diode part on the cathode pin.

Traditionally the cathode or ground pin of a chip carries the most current, and is thermally connected to the die. Heatsinking is done through the ground pin unless there is a dedicated pin(s) for heatsinking. 99% of the time, those heatsink pins are also electrically connected to the ground. This is the same for small leds as it is for larger multi watt leds or voltage regulators or motor drivers.

Since the led junction is where heat is generated, its attached to the cathode pin, and made larger to compensate. Most 5mm leds are only rated for 20~30mA continuous, but they can be pulsed at higher rates, and without heatsinking will fry.


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