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Old 11th March 2009, 01:19 AM   #1
Default Resistors and parallel LEDs

I'm wondering if someone is willing to send me in the right direction regarding my voltage/resistance calculations...

So using that rotary switch you guys suggested yesterday (thanks again for that btw) I've realized I can simplify my circuit down to just three different switch settings - one with 16 LEDs on, one with 12 on, and one with 8 on [Pictured below].

Each LED is rated for 20mA at 3.2V, so when the switch is set to light all 16 LEDs I need to put 320mA across the parallel LED array; likewise when 12 are lit I need 240mA, when 8 lit 160mA. What is the best approach to calculating the size of the parallel resistor for each LED branch, and the size of the main series resistor for each of the three switch settings? I've tried reducing each parallel branch into a single load (3.2V @ 320mA, 3.2V @ 240mA, and 3.2V @ 160mA) I don't think that's right...

Hope this all makes sense. I included the full circuit I redrew at the bottom for context.

Thanks.
Attached Thumbnails
Resistors and parallel LEDs-led-combos.jpg   Resistors and parallel LEDs-fullcircuit.jpg  
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Old 11th March 2009, 01:25 AM   #2
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First of you do not need the diode in series with each LED. I would eliminate the single common resistor and just put a 270Ω resistor in series with each LED. It will limit the current to about 10mA. You do not need to run the LED's at 20mA. I think you will hardly notice the difference in brilliance between the 10 and 20mA.
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Old 11th March 2009, 01:41 AM   #3
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I do need the diodes because each of these three images is just a simplified subset of the larger circuit I attached; there are only 16 total LEDs and the 14 positions of the rotary switch light up different ones.
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