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Author Topic: Will this work? Run 2x as many LEDs with same current, using tri-state logic.  (Read 1383 times)
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It would help if you lab led the components, but. The middle two LEDs are the only ones with both resistors in their current path to have two lots of LED current through them. That explains why they are dimmer.

I have drawn it like this to help you see:-


« Last Edit: March 08, 2011, 05:52:09 pm by Grumpy_Mike » Logged

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It would help if you lab led the components, but. The middle two LEDs are the only ones with both resistors in their current path to have two lots of LED current through them. That explains why they are dimmer.



I understand LEDs 3 and 4 are the only ones where both resistors in their current path are shared with other circuits, and that's the cause of the problem.  But what I don't grasp is why sharing them causes a problem.

You mentioned a voltage drop across R5 being greater because it shares two pairs of LEDs.  Does R3 also have a greater voltage drop?  Why does sharing two pairs of LEDs cause a resistor to drop more voltage?  And how do I calculate the voltage drop?
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Quote
But what I don't grasp is why sharing them causes a problem.

Because it pushes more current through the resistors and so the resistors drop more voltage giving less voltage to go across the LEDs.

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Does R3 also have a greater voltage drop?
Yes but you only get this once. The middle two get it twice.
The voltage at the hot end of R5 is going to be higher than the voltage at the hot end of R1, so this backs off the voltage across L2.

Anyway I thing you have answers the original question, will this work, it's a no.
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Well crap...I thought I might have solved the problem there for a moment. 

I replaced the 47 ohm resistors with pairs of resistors, so there was one going to each pair of leds instead of a single resistor being split between them.  And that at last allowed me to light my array evenly, like I wanted.

But then I started to determine what states I should set the pins in to get different patterns, and I immediately found a problem I'd overlooked earlier because I didn't understand how the voltage would be pulled down at certain junctions.

If I want to light LED1 and LED3 for example, but not LED2, there is no way to do that.  I can set pin 1 high, and pin 2 low to power LED 1, but LED2 will still be pulling 5v from pin 1 because it is connected through LED1.  And if I try to light LED3 by setting pin 4 high, I must also set pin 3 low, which will provide a path to ground for LED2.
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You have 15 LEDs total, right?
You could use a different approach ... wire them up as a 5x3 matrix, and then multiplex them in software.
Just get rid of 2 of the rows in my 5x5 matrix detailed here:  http://www.instructables.com/id/Yet-Another-Daft-Punk-Coffee-TableDisplay/step3/Wire-it-up/
All you need is 5 resistors.
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No, you misunderstand.  I was trying to light a 10 led bargraph with 50mA as brightly as it would be if fed with 100mA, by wiring leds in series so as to waste less power.

See my new thread here, where I explain a method by which this can be accomplished.  The new method requires multiplexing, but achieves the goal of getting the same brightness for less current, and uses 75% fewer pins and resistors than you would with normal (non-multiplexed) methods to boot.
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