Wiring LEDs

I've just bought a bunch of LEDs that I want to use to replace some 12V lamps. The specification is:

Emitted Colour : White
Size (mm) : 5mm
Forward Voltage (V) : 3.2 ~ 3.8
Reverse Current (uA) : <=30
Luminous Intensity Typ Iv (mcd) : Average in 13000
Viewing Angle : 20 ~ 25 Degree
Max Power Dissipation : 80mw
Max Continuous Forward Current : 30mA
Max Peak Forward Current : 75mA
Reverse Voltage : 5~6V

My question is this, can I wire 3 or 4 of them in series to work with a 12V supply that may vary between 11V and 14.5V, or should I wire (perhaps more of) them in parallel with a resistor. If so, what value?

In what sense does the power supply vary? What kind of supply is it?

Putting three of the LEDs in series would eat a maximum of 3*3.8V = 11.4V to make them conduct. Despite not all of them having the maximum forward voltage, they would be very dim unless off when the supply voltage drops close to 11 volts.

With this setup, you would probably need to wire the circuit so that you have strings with two LEDs and a series resistor in each string, and the strings are parallel connected to the power supply. Still this depends on what kind of power supply you have, and how much power it can output. What wattage were the lamps you're looking to replace? Is the power supply the same that was used for them?


Mankku

It's a battery bank on a boat, about 500AH!

So the best way is with a series resistor on each LED?

You must have a series resistor in each string of LEDs, be they one LED or many. It must be properly dimensioned to allow enough current to make it shine, but not too much and not too little. If the resistor is left out, there will be a quick current surge which will burn the LEDs out in an instant.

Mankku

Ok so if I got this right then the battery starts out at 14.5V when fully charged and then gradually decreases...

In that case you can connect LEDs in strings of three with a series resistor, and make a number of such strings and connect them in parallel. They will shine brightly when the battery is full and will become dimmer as it discharges. When the battery is about 11-12V they will turn off (or will be so dim it seems they've turned off).


Mankku

2 LED's per string will give a voltage drop of 7.6 total.

The led's will easily have enough voltage even at 11v (minimum).

We can then calculate the resistor for each string using the maximum voltage (the led's will be a bit dimmer at the 11v, but we don't want them to burn out at 14).

That makes it 14.5-7.6 = 6.9V/0.030 or 230ohms.

So, series strings of 2 leds each, with a 230ohm resistor also in series. Any amount of parallel strings can be connected.

Cheers

Actually, I guess the range is closer to 11.3V to 14.3V. The maximum voltage is only there when the battery is charging, it normally drops to 12.8V, then discharges.

Would it be better to have the LEDs all in parallel, with a single dropper? Or each with their own?

What do they do in those LEDs that you can use to replace bulbs in cars?

This post has been edited by Big_Nige on July 13, 2007 03:13 pm

Leds aren't all made equal. Some leds will conduct more current than other resistors (though the total current is still limited by the resistor), these leds will burn out, leaving fewer remaining leds to carry the extra current, which will cause them to burn out faster and so on.

The solution that danga1993 provided should solve this anyway. Strings of two LEDs with series resistor in each will do.

If you mean connecting LEDs in parallel and put a single resistor in series, that wouldn't work. First of all the resistor would need to handle a hellish amount of power (current x voltage drop) meaning it'd be as big as your battery pack (ok not really but you get my point )... Second I believe (never tried it myself) that LEDs connected in parallel will pop each other little by little, as they all have individual forward voltages... That's why you need to put a series resistor for each string of LEDs.

The fact that we recommend putting two LEDs in each of your strings is that it is more energy efficient and reduces the voltage drop over and thus power dissipation in the resistors, enabling you to use smaller-size resistors.

This I cannot answer because I have no knowledge in that... Perhaps one of the forum experts can enlighten us?


Mankku

The 'luxeon' range of high power LEDs are commonly used in car lights.

Thanks for all the replies. I've decided to buy a ready-made LED bulb, but will use the LEDs for something else.

Thanks, good to know... You use them basically like any standard LED, apart from allowing larger current flow?


Mankku

indeed - and they require a decent heatsink too as the power disippation is considerable.

LED technology is getting better all the time and brightness (efficiency) is always on the increase. Check the Luxeon website for full details.

Sure will!

Back again, similar project ...

Bearing in mind that the object is to replace normal bulbs with LEDs to save current drain, would a voltage regulator at around 7.6V be any use?

You can run LEDs from a regulated voltage source, without resistors. But, the LEDs would have to be carefully matched as to their forward voltage drops, as they all vary a bit. If you have a mix of voltage drops, the lower one will hog all of the current & burn out. Then the next lower one will hog the current & burn out. And so on. Matching LEDs is done by measuring their forward voltage drops & only using those that have identical drops. It's easier to use a dropping resistor in each string.

Thinking about it, what I really meant was use a regulator to fix the voltage below my minimum, then use resistors. Assuming my minimum is 11.3V, and knowing there is a minimum drop, what would be the highest value I could regulate to?

Would the regulator dissipate more heat (=lose energy) more than reisitors?

Thinking about it, what I really meant was use a regulator to fix the voltage below my minimum, then use resistors. Assuming my minimum is 11.3V, and knowing there is a minimum drop, what would be the highest value I could regulate to?

Would the regulator dissipate more heat (=lose energy) more than reisitors?

Minimum would be 8 or 9 volts. A linear regulator will have the same loss as a resistor, because all a linear regulator really is is an active resistor that changes value with changing input voltage.

Way back in my distant past, ISTR that you could wire a regulator to be a (limited) constant current source, with a large value resistor in the reference (I think?)

With no need for a dropper/current limiter in series, wouldn't this be ideal to run LEDs?

The difference must still be "absorbed" by the regulator.

For any linear device (the regulator in this case), multiply the voltage drop across the device by the current passing through the device to find approx how much power is lost/dissipated through heat (in Watts).

Probably the most economical route (power-wise) is to either use LED strings with a resistor, or go with PWM and single LED/resistor units. Neither is perfect because LEDs aren't perfectly matched.

PWM? Pulse-width modulation? How would this work? [Edit: got it now]

My main concern is that the voltage fluctuates between 11.3V and 14.3V so using a resistor means sometimes they are dim, sometimes very hot!

Hmmm... wonder how hard it would be to set up a PWM and have it auto-compensate for when the voltage drops?

Otherwise, you're probably going to be stuck regulating it around 9V if you want to keep the LEDs at a steady level of brightness. Or perhaps near 11V with an LDO regulator.

I didn't even know such a thing existed, thank you.

So my options are:

[1] use a 12V LDO with 3x LEDs and a series 20R
[2] use a 9V LDO with 3x LEDs, and a 2.4V ZD in the ground feed to the LDO-R to boost the output to 11.4V. This method would have no series R with the LEDs.

Will [2] work to give 11.4V? Will it blow the diodes?

Someone else will have to chime in on this, I have no LDO experience.

Use a CCS (constant current source).

Rouslan