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Topic: Tri-state Multiplexing (Read 1 time) previous topic - next topic

scswift

#5
Mar 10, 2011, 05:01 am Last Edit: Mar 10, 2011, 05:07 am by scswift Reason: 1

But if you're lighting the LED's w/ a 50% duty cycle, they're on only half the time, so aren't you getting just half the brightness?


No.

Let's say you have an array of 8 LEDs, lit the traditional way.  You put 10mA through each LED, and use a total of 80mA.
Now let's say you wired that same array with each pair of LEDs in series.  You'd still put 10mA through each LED, but now you're using only 40mA.

So far so good.

Now let's put 20mA through each pair of LEDs in the array wired in series.  You're now using a total of 80mA, just like the traditional array, but each LED sees 20mA, so they're twice as bright.

Makes sense right?

But we can't animate the array wired in series.  Not if it's wired the traditional way.  So we wire it using the method I outlined above.  And as a consequence of that, we have to multiplex it on a 50% duty cycle, with each pair of LEDs lit half the time.  That halves our brightness.  But it also halves our current usage.


So you're now using a total of 40mA, and each LED sees an average of 10mA.  Half the current as the traditional array.  But the same brightness.
Or, to put it another way... twice the brightness of a traditional array using that same 40mA, with each LED seeing only 5mA.


Quote

Or does the nonlinearity of the eye make it appear brighter than 50%?


No, that's not what's at play here.

defsdoor

Aren't you assuming that there is no drop across each LED ?

scswift

#7
Mar 10, 2011, 03:31 pm Last Edit: Mar 10, 2011, 03:42 pm by scswift Reason: 1

Aren't you assuming that there is no drop across each LED ?



The drop across each LED is why there's two different values of resistor used.



Imagine the above circuit without LEDs 3 & 4.  

There are then three possible paths (when the input state is used to disconnect certain pins) for current to take:  
1. Pin1 to Pin2 through LED1, where there is a total of 317 ohms of resistance.
2. Pin 2 to Pin3 through LED2, where there is a total of 317 ohms of resistance.
3. Pin 1 to pin 3 through LEDs 1 & 2, where there is a total of 94 ohms of resistance.

Now go to the LED array wizard:
http://led.linear1.org/led.wiz

Enter 5v source, 2.1fv, 10mA, and 1 LED.  It tells you you need a 330 ohm resistor.
Now enter the same, but for 2 LEDs.  It tells you you need an 82 ohm resistor.

Both these cases will light the LEDs with 10mA.  Both cases will be just as bright.  But in the case of 2 leds, more power is used to generate light, and less power is wasted in the form of heat, in the resistor.

To get the values in the above circuit, I did 82/2 = 41 and chose the closest value resistor I had on hand, 47 ohms.
I then subtracted 47 ohms from 330 ohms, and got 283 ohms.  The closest value to that being 270.

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