The computer power supply is very likely to have a 3.3 Volt output, and you should use that for the LED power. Even the smallest ATX type power supply should have no problem delivering 5A on the 3.3 Volt rail (But check the specs).
If you have to use a 5 Volt rail, the problem won't be your power supply, it will be the LED current sinks. The forward voltage drop of the LEDs will be between about 2.1 and 3.3 Volts depending on the color. If you apply 5 Volts, the 'extra' 1.7 to 2.9 volts are going to have to be lost as heat in the controller current sinks. The controller has 24 of these sinks and it is going to become very hot very soon.
If you can't dissipate enough heat from the controllers, a simple fix could be to use one or more linear regulators to drop the voltage from 5 to 3.3 Volt or so. It might be easier to dissipate heat from the voltage regulators than from the controller chips.
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 – 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.
Best Answer
8 parallel strands of 2 LEDs each and a 150ohm resistor. All loose anodes should be connected to V+. The loose cathodes should be connected to the collector of a TIP31 NPN transistor. The emitter should be connected to ground. The base should be connected to a 100ohm resistor, and the other side of the resistor should be connected to a digital output. 16 LEDs, 9 resistors, 1 transistor, 1 digital output.
(12V−(3.2V⋅2+1.2V))/30mA = 146.66667 Ω
30mAâ‹…8 = 240 mA
(5V−1.8V)/(30mA⋅8/10) = 133.33333 Ω
simulate this circuit – Schematic created using CircuitLab