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You have a total power requirement for all LEDs (including the RGB LEDs) of less than 4W. Even if you double that to accomodate resistor losses, you don't need anything more than a 10W wall wart. A 12V 1A or a 24V 500mA supply would be sufficient.
As to method:
https://www.electronicspoint.com/got-question-driving-leds-t256849.html
Can someone help me with rhe circuit.what resistors capacitors to use and such?
all 10 white leds turn on together but I can cut that to 9 and just add the last one with its own resistor.the RGB leds (9) will all be on at the same time but different colors (no mixes,just basic RGB).The reds (16) will be on only when switched but all wil be on..thank youI gave it a try and put in your numbers:
Source Voltage: 12
Diode forward voltage: 3
Diode forward current (mA): 20
Number of LEDs in your array: 10
What I got was a few options for arrays. I might not be understanding the whole problem here. Do you have to have each diode capable of turning on and off independently of each other?
~Bill
With 3V/led, three leds = 9V To get 20mA and a voltage drop of the extra 3V, a series resistor R = V/I = 3/0.02 = 150 ohm. The three leds should be connected in series with the resistor.
Three strings will allow nine leds and the current drain will be 3 * 20mA = 60mA
You can work out the resistor if you want to add the odd led.