Simple 12V circuit help

[jmoney6]

I've been playing around with an idea to create LED tail lights for a vehicle. From what i've been reading i'm going to run strips in parallel. I'm still very new to electrical engineering and I have a few questions.

Is there a calculation for resistors?
How can I make a series set dim, and become brighter (lights on/brighter when the brakes applied).

If some one has a great deal of experience with 12v led circuits that could help me out would be awesome!

 

[patelvimal20]

Definately, Resistor calculation is there.
Before that you have to take care of some important things:
Number of LEDs
Your source rating
the most important thing is that you told that you are planning for vehicale light so that you have to see the area also.

Please let me know the number of LED that you want to required in tail.

calculation is very much simple
based on Ohms Low V = (I) * (R)

Thank You,

Vimal Patel

 

[Li_ereunpisan]

You can use a 470 ohm resistor - 1k ohm, depending on the LED you are using

You must use a minimum resistor 2watt .. because the power supply vehicle out of the big current

 

[enjunear]

Rather than running one LED and resistor, you can connect several LEDs in series and decrease the amount of power that your current-limiting resistor has to handle.

Example:
You are using LEDs with a forward voltage of 2.0V and a nominal bias current of 15mA (0.015 amp).
Since this is for a car, assume your supply voltage is 13.8V (potential voltage from alternator while charging the battery).

If you connect LEDs in series, their voltage drop is summed (3 LEDs in series = 2V*3 = 6 Volts). The most LEDs you could connect in series would be 13.8V / 2V-per-LED = 6.9 LEDs... since you can't have a 0.9 LED, you have to round down to an even 6 LEDs (maximum... you can do less, your bias resistor will just get larger in value).

So, 6 LEDs @ 2.0V each = 12 volts dropped. If the supply voltage is 13.8V (max), then the limiting resistor will have to drop 13.8-12 = 1.8V. Now you can use Ohms Law (V=I*R) to determine the resistor value. You know the voltage across the resistor, and the current through it (will be the same as the current through the LEDs in series with it... 15 mA).

V = I*R.... 1.8V = 0.015*R, and R = 120 ohms (rule of thumb, if you can't find this exact value, round up to the next standard resistor value). Last, calculate the power dissipation of the limiting resistor. P=V*I, so P=1.8V*0.015A = 0.027 Watts, or 27 milliwatts (mW). You could easily use a 1/4W leaded resistor (or even some SMT resistors, just check their power handling and derate by 50% for safety).

Use the above process to calculate the resistance for other combinations using fewer than 6 LEDs... just substitute in your desired numbers and do the algebra. Once you have a string of LEDs that will run off the car's 12V supply, you can run multiple strings of LEDs in parallel (as you mentioned in your post). This is the most efficient way to wire lots of LEDs up, and minimize the number of resistors needed.

 

[Syncopator]

How can I make a series set dim, and become brighter (lights on/brighter when the brakes applied).

Yes, quite easily. I answered a similar question to this on another forum. The basic solution is here https://www.davidbridgen.com/brakes.htm and there's a link to the original question on the other forum.