17-01-13, 02:28 PM
OK, so if we use 2.1V as the value, that gives 10.5V drop across a chain of 5 LEDs. If we stick with 13V as the supply for the moment (in order to make sure the LEDs don't get over-driven it's best to take the highest voltage you're likely to get) that gives 2.5V across the resistor and 25Ω would be the right value to give 100mA. Then, if the voltage drops to 12V the current will reduce to 60mA.
If you suit the resistor for a 12V supply you'd use 15Ω and should the voltage rise to 13V (with the headlights off and at high revs for example) then the current would rise to 167mA.
In fact, the current swings wouldn't be quite that severe because the forward voltage drop of the diode would also increase... it's what is limiting the current with your present circuit.
As for the tail light, I'm not sure what current you'll be getting at the moment. With such a narrow voltage margin for the resistor to use it's well into the knee region of the LED's curve. Perhaps 30mA per LED would be a good starting point, with 5 in series and assuming a 1.9V forward voltage at that lower current that gives 9.5V.
A 12V rail then would be reduced to 11.5 by the Schottky diode (think you've drawn that in the wrong place for the LED strip, it's there to stop the clocks and sidelight lighting up with the brake when the headlights are off). So the total resistance a chain needs would be 67Ω... however, there is already 25Ω present so the remaining 42Ω could be provided by a single resistor of 3.5Ω (twelve 42Ω resistors in parallel). The nearest preferred values are 27Ω and 3.3Ω but you might need to experiment with the current for the tail light.
If you suit the resistor for a 12V supply you'd use 15Ω and should the voltage rise to 13V (with the headlights off and at high revs for example) then the current would rise to 167mA.
In fact, the current swings wouldn't be quite that severe because the forward voltage drop of the diode would also increase... it's what is limiting the current with your present circuit.
As for the tail light, I'm not sure what current you'll be getting at the moment. With such a narrow voltage margin for the resistor to use it's well into the knee region of the LED's curve. Perhaps 30mA per LED would be a good starting point, with 5 in series and assuming a 1.9V forward voltage at that lower current that gives 9.5V.
A 12V rail then would be reduced to 11.5 by the Schottky diode (think you've drawn that in the wrong place for the LED strip, it's there to stop the clocks and sidelight lighting up with the brake when the headlights are off). So the total resistance a chain needs would be 67Ω... however, there is already 25Ω present so the remaining 42Ω could be provided by a single resistor of 3.5Ω (twelve 42Ω resistors in parallel). The nearest preferred values are 27Ω and 3.3Ω but you might need to experiment with the current for the tail light.