Charlie-plexing

[Demon]

Only one LED will be bright enough to be visible even though multiple paths are active at the same time, because the shortest path is the only one with enough current. All the other undesired paths share the same current; not enough to make an LED glow.

Cycling through the 56 LEDs on the 8 pins in the post above, a LED does not have enough current 7 out of 8 times:


From Step 4:

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OK, so a current flows through LED5, current isn't flowing through LED6 because it is reverse biased (and so are LED2 and LED4)....but there is also a path for the current to take from pin A, through LED1 and LED3 isn't there? Why are these LEDs not glowing as well.

Here is the heart of the charlieplexing scheme. Indeed there is a current flowing both LED1 and LED3, however the voltage across the both of these combined is only going to be equal to the voltage across LED5. Typically they would have half the voltage across them that LED5 has. So if we have 1.9V across LED5, then only 0.95V will be across LED1 and 0.95V across LED3.

From the If/Vf curve mentioned at the beginning of this article we can see that the current at this half voltage is much much lower than 20mA.....and those LEDs will not glow visibly. This is known as current stealing.

Thus most of the current will flow though the LED we want, the most direct path through the least number of LEDs (ie one LED), rather than any series combination of LEDs.
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http://www.instructables.com/id/Char...eory/?ALLSTEPS

Robert


(That Habs logo just "happened" to line up with the monitor.)