Graph one shows Fv at 15mA. So resistor required is (12 – 4 * 1.92) / 0.015 for red which is 288 <a href="#">ohm</a>. Resistor for green is (12 – 4 * 2.1) / 0.015 which is 240 <a href="#">ohm</a>.
Excellent! See, nearly that elusive Eureka moment
So in the circuit there would be a common resistor of 240 <a href="#">ohm</a> which would be used by all LEDs and each red would require an additional 12 <a href="#">ohm</a> resistor run in parallel to the red LED so that particular red LED is then using 15mA when lit.
As Audio wrote, this is wrong. You must have miscalculated somewhere. Oh boy. here I go again. Doesn't electricity like logic? if red needs 288 Ohm for 4, and green needs 240 for 4 to both have 15mA then that's 48 difference which, divided by the 4 red LEDs is 12 per LED.
This would then give the Luminous intensity of each LED colour as shown in Graph 2 which, at 15mA, are virtually the same.
As Audio wrote, this is not necessarily correct. You will have to experiment with the LEDs to see how bright they are for various currents. Ok, so Maplins tomorrow.
Then, as was said earlier, another resistor is placed in parallel to the red LEDs as the voltage is different therefore it has to be allowed for. This is the circuit diagram I did in visio with the extra resistors in parallel. I think this is where you meant them to go.
Yes, we just have to find the right resistance value. Ya-hey... Eur...