Need Help: Wiring LEDs with 2 source voltages
Note: The Voltage sources are indicated the wrong way around in the question's schematic, going by the LED direction shown (negative ground circuit).



First the always-on running light case. For the LED resistance calculator, use:


supply voltage = 11.5 Volts
Vf = 2.2 Volts ( x number of LEDs)
current = 75 mA


The entire forward voltage of all the LEDs would appear at the supply leads (not half), but you would want only about half the rated current to flow. 



Now, the brake lights:

In order to prevent current from flowing from the running light lead to the 0.5 Volt brake light line when the brakes are off, you would need a diode on the brake light line, connected so as to be forward biased, same direction as the LEDs. A 1n4001 diode should do fine.







simulate this circuit &ndash; Schematic created using CircuitLab

For the LED calculator, use:


Supply voltage = 11.7 - 0.7 = 11 Volts (the diode drops around 0.7 Volts)
Vf = 2.2 Volts ( x number of LEDs)
current = 75 mA


The reason for doing this is, the currents from the two sources add up in going through the LEDs. Hence, when both supplies are high, 150 mA will flow. When just the running lights are on, 75 mA will flow. The voltages do not add up between leads.



As an added twist, if the LEDs need to light up only at one-third intensity for running lights, and full intensity for braking, this is easy: Just take 50 mA for the running lights calculation, and the remaining 100 mA calculation for the brake lights calculation, in the bullet points above.