CAR DC with Alternator

Atom, I hope you used a resistor. Did you measure the current?

Every couple of days we see another person that puts 3 volts on a 3 volt LED. Often the current is not close to what they think. I hear said "my LED works with 3.4 volts so you are wrong about the voltage range." As a hobby one or ten LEDs will work at 3.4 volts. We used about 8,000,000 LEDs last year and the voltage varies from 2.5 to 4 volts. Put 3.4 volts on the LED and the current might be 1/8 or 3x that you want. My LEDs need to last years so I won't run them at 1.1x the rating and we rate the light output so 0.9x will not do.

I know your LED will work at 3.0 at the rated current and there fore I am wrong.......gain.

I didnt measure the current but the drop is in fact.. 3.9v - 4v . The issue is if the car outputs about 14v and i use a zenor to drop the excess the resistor will be 200 ohms for the zenor. And no resistor for leds. Am i correct

14 / 70mA = 200 ohms... To allow maximum brightness.

How about these: https://www.mouser.com/Search/Produ...LPvirtualkey59500000virtualkey595-TL750L12CLP
or better yet: https://www.mouser.com/ProductDetai...GAEpiMZZMug9GoBKXZ752Mey0D%2b9EP3iyb9w6SVWLw=

If the car battery is 14V and the three LEDs are 4V each (12V total) and the resistor feeding the 12V zener diode and LEDs is 200 ohms then there is 2V/200= 10mA in the zener diode and almost no current flowing in the LEDs so they will be very dim. They will not light if the battery is less than 14V and the battery will probably be less than 14V.

LEDs always need a current-limiting resistor.

Ok so answer this... if i use a 12v regulator with a low dropout voltage...

1. Would it work? (drop out is 0.6v @150mA which is the max 0.2v @ 10mA)
2. would i need a resistor since the leds add up to 12v ? ("LEDs always need a current-limiting resistor. ")
3. if i do need a resistor ... how would i calculate it? I applied 12v here to 3 leds in series and they each show 4v dropout... 1 was 3.9v and 2 was 4v

AtomSoft said:

Ok so answer this... if i use a 12v regulator with a low dropout voltage...

1. Would it work? (drop out is 0.6v @150mA which is the max 0.2v @ 10mA)

Click to expand...

The output is 12V plus and minus 4%. The battery must be at least about 0.5V higher than the output.

2. would i need a resistor since the leds add up to 12V?

Click to expand...

You didn't measure the current so with 12.0V the LEDs might be ready to burn out. The voltage of LEDs changes when the temperature changes then their current will also change and maybe cause them to burn out or not light.
LEDs ALWAYS need a current-limiting resistor.

3. if i do need a resistor ... how would i calculate it? I applied 12v here to 3 leds in series and they each show 4v dropout... 1 was 3.9v and 2 was 4v

Click to expand...

The regulator voltage is too low or the voltage of your LEDs is too high.
You measured only 3 LEDs and the rest of them might have lower voltages.
Use 2 LEDs in series then their total voltage is 7.9V or 8V. The current-limiting resistor is (12V - 8V)/20mA= 200 ohms and the current will barely change if the regulator voltage is plus and minus 4% and will also barely change when the temperature changes.
If the LEDs are actually only 3.4V then their current will be (12V - 6.8V)/200 ohms= 26mA which is fine.
If the LEDs are actually 3.0V then their current will be 30mA which is high but is allowed on the datasheet. The 200 ohm resistor will barely get warm.

Regulators have a dropout voltage.
LEDs have a forward voltage drop.

Awesome! thanks. Ill take the 2 in series approach. Seems way simpler and more safe. Thanks a ton and sorry if i seemed like a pain
EDIT: Also if i use 2 in series the current needed goes up... just noticed... but still below the 150mA the reg will supply so im ok i guess

AtomSoft said:

if i use 2 in series the current needed goes up... just noticed... but still below the 150mA the reg will supply so im ok i guess

Click to expand...

The max allowed current in the LEDs is 30mA, not 150mA. The current-limiting resistor must be calculated for 20mA when the LEDs are 4.0V each then if their forward voltage drop is only 3.0V each the current will not exceed 30mA.
Each string of 2 LEDs needs its own current-limiting resistor.
Use more strings of 2 LEDs with resistor if you need more brightness.

Im referring to the regulator not the led

Here is how ill do it...

ALL low dropout regulators need an output capacitor. Most regulators also need an input capacitor. They are not shown on the datasheet but it says to use them. The input capacitor is 0.1uf and the output capacitor is 10uF.

Oops almost forgot those heh... Thanks again.

How does this look:

G'day Atom,
Just to throw a spanner in the works mate, a few years ago there was a series of threads on dealing with led's on off grid arrays where the battery voltage will swing and led's with just a simple resistor burn out.



The above circuits was designed by a member called 'Commanda' and personally I've built several of the circuit here for use both on 12 and 24 volt battery systems.

Rg is generally 1M and Rs is the resistor that sets the current in the led's. For 20 -30mA led's usually a 2n7000 fet is all thats needed and a bog standard npn. If the voltage drop of the led's match's the input voltage then the resistor above the led's can be omitted. As a trail on the 12 volts ones I made I went from turning the led's on @ 10 volts and went all the way upto 20 volts for an extended period and NOTHING got warm.

The main point is the forward voltage drop of each led should be matched and I find my fluke 865 shows the voltage drop on each led to 0.xxx accuracy.

All I can say is make up a circuit and try it out, I'm sure you'll find it purfect for the job your doing.

Cheers Bryan