I understand the principle behind charlieplexing, however I don't understand how the circuit below works. Say I wanted to illuminate LED5, wouldn't I set Pin1 to OUTPUT-LOW, Pin2 to INPUT, and Pin3 to OUTPUT-HIGH? Current will then flow through LED5. However, won't current also flow through LED4 and LED2?

Charlieplex circuit

  • 1
    \$\begingroup\$ Microchip app note on Charlieplexing. \$\endgroup\$ – Nick Alexeev Jul 11 '13 at 18:30
  • 1
    \$\begingroup\$ Important to realize that you have a third output state: HIGH IMPEDANCE. It is neither logic HIGH nor logic LOW, it does not conduct / source / sink current, it appears as 'not connected' \$\endgroup\$ – jippie Jul 11 '13 at 19:02
  • \$\begingroup\$ @jippie: Could you explain? As far as microcontrollers go, I am only familiar with the Arduino and Digistump(Attiny85). I am aware that I can set a pin to OUTPUT (High or Low) or INPUT (said to be in a high-impedance state). \$\endgroup\$ – Stephen Melvin Jul 11 '13 at 22:47

Turns out the answer was on the wikipedia page.

Current will still flow across alternate paths (an alternate 2-LED path exists for every pair of pins in the 3-pin diagram, for example), but the reduced voltage drop across those LEDs in multi-LED paths will not be enough that they actually significantly illuminate.

So, the other two LEDs are lit, just not in a significant way.

| improve this answer | |
  • \$\begingroup\$ Keep in mind that LEDs are very nonlinear devices, and there is a minimum voltage required (the band gap) in order to produce visible light. The total drop across LED2 and LED4 is limited to the forward voltage drop of LED5. With only half of their nominal forward voltage across them, LED2 and LED4 will be passing only leakage current, and will not be able to generate visible photons. \$\endgroup\$ – Dave Tweed Jun 6 '18 at 20:26
  • \$\begingroup\$ @DaveTweed Why is the drop across LED2 and LED4 limited by the drop across LED5? Is this a special property of LEDs? If I replace the LEDs by resistors, current would flow through both paths, and the voltage drop would be the same. \$\endgroup\$ – Michael Knudsen Jun 7 '18 at 18:14
  • \$\begingroup\$ @MichaelKnudsen: "Why is the drop across LED2 and LED4 limited by the drop across LED5?" Because LED2 and LED4 in series are in parallel with LED5. Parallel circuits always have the same voltage. If LEDs were identical resistors, then LED2 and LED4 would have half the current of LED5 -- but they're not. \$\endgroup\$ – Dave Tweed Jun 7 '18 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.