I've wired up a 3x3 matrix as in the attached image. My understanding is that if I apply 3.3 volts to row C and connect column 1 to ground, the electricity should flow through the bottom-left LED (as in the top-right of the image), turning it on. What seems to be happening, though, is that the electricity bypasses the LED and flows directly to the ground, meaning no LEDs light up (as in the bottom part of the image). This makes sense to me since electricity usually follows the path of least resistance. But everything I am reading online seems to indicate that what I am trying should light up the LED. What am I missing here?

I should also note that I have a 100 ohm resistor at the beginning of each row.


  • \$\begingroup\$ How do you know electricity flows? Is there voltage across your series resistor, or did you connect an ammeter in series? Check your work carefully: continuity of connections, directions of diodes, etc. \$\endgroup\$
    – Kaz
    Nov 15, 2012 at 4:14
  • 3
    \$\begingroup\$ I don't think the grid is wired the way you think it is. The lower image... where you have the red "L"-shape... those wires don't connect - they simply cross. For fun, google "diode rom". \$\endgroup\$
    – Tony Ennis
    Nov 15, 2012 at 4:18
  • \$\begingroup\$ I just know that my LED won't turn on and if I disconnect column 1 just before the cathode it does turn on. Putting my own specific circuit aside though, I guess I'm just struggling to understand how a matrix works in the first place. In a working LED matrix, why is it that the current flows through the LED and doesn't bypass it and go straight to ground? \$\endgroup\$
    – Dustin
    Nov 15, 2012 at 4:26
  • \$\begingroup\$ It can't bypass it. The LED is the only path. The lines in the grid overlap, but don't connect. As far as your problem is concerned... if you've connected the intersections, you're doomed. If you have not, then I'd suspect your LED is installed backwards. \$\endgroup\$
    – Tony Ennis
    Nov 15, 2012 at 4:39
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    \$\begingroup\$ And @TonyEnnis is absolutely right. I tried it again with the wires crossed but not connected and it works (and it makes a lot more sense to me too!). I feel a little silly for not realizing this. I guess I just saw lots of pictures of LED Matrices where it looked like they were connected, although they must not have been. \$\endgroup\$
    – Dustin
    Nov 15, 2012 at 4:40

1 Answer 1


In your actual circuit, have you connected the wires and the LED leads at the junctions indicated by the black circles, or have you connected them wherever the lines overlap in your schematic diagram?

The correct reading of the schematic is to only have connections where indicated by the junction indicators (black filled circles). If you do so, then the third scenario in your diagram does not exist - the wires going horizontally and vertically do not actually touch at all.

One way to visualize this: The horizontal wires are all on the plane of your work table, the vertical wires are all in a plane floating an inch above the table, and the LEDs are the only connections between these two sets, and the only connections are the black circles.

You are correct, electricity will follow a path of least resistance to the greatest extent. In the schematic, there is no path of least resistance, i.e. short-circuit, between row and column wires, since no junctions are indicated. If you have interconnected the row and column wires, disconnect them now. Then, as you will see, the only path that allows electricity flow is through the respective diode / LED.

  • \$\begingroup\$ This was exactly the issue. Thanks for the clarification with the black circles as well, this will help for reading future schematics. Thanks again to Tonny Ennis as well for pointing out this same thing. Unfortunately I don't have enough rep to upvote yet, or I would upvote his comments! \$\endgroup\$
    – Dustin
    Nov 15, 2012 at 21:02

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