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LED Project on my Car

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Broken Gadget

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Greetings.
I'm working on installing custom lighting in my car using LED's. The ones I have are 3v LED's.
The ovious power source is 12v from the car.
Is there an off the shelf converter that I can use or any specific wiring/circuitry that I can use to power the lights?
Note that I'm trying to refrain from using a lighter adapter, I would rather wire it directly from the fuse box using an independent switch.
Thanks:confused:
 
Leds need a fixed current, not voltage. The 3 volts is the forward voltage drop of the led.

You can and should use a current limiting resistor inline with the led. This will allow you to connect it directly to the 12v scource.

How to calculate resistance in ohms
This is a nifty site for calculating the value of the resistor.

Remember the voltage of a car battery will be around 13.8v when the car is running. Size your resistors accordingly.

How many leds are you using? You can connect several in series to save on resistors and current draw. Whats the current (mA) rating of your leds?
 
More information

I'm looking at some types in the Internet, but this may be the one I decide for:

Emitted Colour : Blue
Size : 5mm
Lens Color : Water Clear
Peak Wave Length (nm): 465-470
Forward Voltage (V): 3.4 - 3.6
Luminous Intensity (MCD) : 10,000
Viewing Angle : 20-25 Degree
Max Continuous Forward Current : 30MA


If I decide for something else, what variables should I look for and how do I calculate? BTW, I plan to use 4-6 LED's in the sequence.

Thanks for the help

BG
 
You could use 6 LEDs, split into 2 groups of 3 LEDs in series with a resistor.

The max current is 30mA. Its a good idea to use less then the maximum value. It will make the LEDs last longer. When you get them (if you have a assortment of resitors around) you can play with them and see how bright they are at 30mA compared to 20mA or even less.

With the car off, the battery will be around 12v.

3 leds in series = Vf 3.5*3 = 10.5

12v - 10.5v = 1.5v

(Ohms law here)

20 mA = .020 A
R = V/I (V = IR rearranged)
R = 1.5/.020 = 75 ohm, should be a standard value
Wattage = V*I = .020 * 1.5 = .003 Watts - a 1/4 watt resistor will be fine


at 13.8v your current will rise to 44mA and the leds could burn out.
There are a couple ways around this.

Size the resistors for 13.8v

12v - 13.8v= 3.3v
20 mA = .020 A
R = V/I (V = IR rearranged)
R = 3.3/.020 = 165 ohm, they don't make a 165 ohm, but they make a 150 ohm
Replacing 165 with 150, If = 3.3/150 = .022 A (22 mA)

This will keep them safe with the engine running, but might make them dim with it off.

You could also use a voltage regulator set at 9v. This will keep the voltage stable running or not. You can redo the calculations using 2 leds in series and a 9v supply.

You could also use a resistor on each LED. This will make the swing in the amount of current less, but isn't as effecient.

With a 510 ohm resistor

13.8V = 20mA
12v = 16.7 mA

Sorry, i got a little long winded there. I am bored:eek:
 
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The LEDs have a very narrow angle and will look like searchlights.
 
Hi BG,

if you want constant LED brightness independent of the car alternator you should think about using a low dropout fixed voltage regulator such as an LM2940CT12 (1A), LT1083CP12 (7.5A), LT1084CP12 (5.0A) or LT1085CP12 (3.0A).

All regulators are available with TO220 package.

Using a fixed voltage of 12V connecting 4 LEDs in series is not recommended, since they won't get enough forward voltage to lit.

A good side effect of using a low dropout voltage regulator is a passive battery check: If the car battery is depleted below 12.5V none of the LEDs will light up.

The calculation for three LEDs in series: 12V-(3X3.6V/0.03A)=40Ω. Use the next higher standard value (47Ω) as a current limiting resistor.

Connecting single LEDs: 12V-3.6V/0.03A=280Ω (use standard value of 300Ω (E96) or 330Ω (E48, E24 row)

Regards

Boncuk
 
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P = I²R or E²/R, not VI, right? Never heard of using VI to calculate power.


have you finish your EE degree? this is no offense. by ohm's law you can use Power=voltage x total current. anyways Broken gadget. have you tried connecting a resister to your series LED setup? i did one in my car a few years back and it works fine.

let me know if you need further assistance.
 
:confused: Where did you see "VI" used to calculate power?
"Wattage = V*I = .020 * 1.5 = .003 Watts - a 1/4 watt resistor will be fine"

Im not saying its wrong Im just wondering how you use Ohm's Law to find power, when Ohm's law deals with V I and R? Again Im not trying to say youre wrong just understand what youre putting. We were just always taught to use P = I^2*R to find power.
 
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"Wattage = V*I = .020 * 1.5 = .003 Watts - a 1/4 watt resistor will be fine"

Im not saying its wrong Im just wondering how you use Ohm's Law to find power, when Ohm's law deals with V I and R? Again Im not trying to say youre wrong just understand what youre putting. We were just always taught to use P = I^2*R to find power.

I would suggest you were taught VERY strangely?.

The standard formula is W=VxI

The more complicated formula you were taught is simply generated by the one above and ohms law (hence why ohms law was mentioned), transposing the two formula's to use the two you know, and find the third you're wanting to know. You could just as easily use ohms law to get V and I, and use the standard formula though, but it makes it a two stage process then.
 
Heh I guess so. Perhaps W = V*I was mentioned in passing, but we were taught to use the more complicated one. Strange. Thanks for clearing that up Nigel
 
Please do not use 12V or 13.8V as the car battery's voltage. Use 14.4V or 15V for your calculations to be safe. When the motor is running it's around 14.4V most of the time. It spikes higher if there's a load dump. Automotive components such as voltage regulators are rated for around 36V, but don't use this 36 for calculations or we won't see the LED light up:)
 
Please do not use 12V or 13.8V as the car battery's voltage. Use 14.4V or 15V for your calculations to be safe. When the motor is running it's around 14.4V most of the time. It spikes higher if there's a load dump. Automotive components such as voltage regulators are rated for around 36V, but don't use this 36 for calculations or we won't see the LED light up:)

This is where i have found frustration in LED/car circuits. I had a topic awhile back and someone posted a document showing load dump spikes could be in excess of 100v. Yet I have seen planty of LED in cars with simple series resitors, not fancy regulator/driver circuits.

13.8 seems to be the norm for automotive voltage. But I agree it can be more or less. I have also seen the voltage of a 12v battery at 11.8 or 12v with the engine off. That makes it hard to use a LDO 12v regulator.

What really is the best way to protect citcuits in a automotive enviroment? Are there any engineers here that work with automotive circuits? How are the sophisticated ECM and other delicate circuits in modern cars protected?

In the real world, alot of basics (like simple current limiting resistors) seem to work fine, atleast for a while, but a real solid solution to this type of problem would be a great sticky on the boards here or a great stand alone website that could be referred to.

I have asked questions like this here and it seem about once a week someone posts here about wanting to add a custom do-dad to their car.
While it seems so simple at first, the more you dig, the cars are nothing close to a good souce of power for LEDs or any other electronics.

Are there any buck/boost ICs out there that can withstand the harsh voltage spikes and provide a stable 12v at a low current?
 
This is where i have found frustration in LED/car circuits. I had a topic awhile back and someone posted a document showing load dump spikes could be in excess of 100v. Yet I have seen planty of LED in cars with simple series resitors, not fancy regulator/driver circuits.

People try to be far too clever - brief spikes make no difference to an LED.
 
have you ever heard of ohm's law?

Yes, I did. But that was not the point. Nowhere in the entire threat there was a power calculation.

To calculate for power I usually use V*I or U*I. VI roman six?
 
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