Super Simple High Power LED Driver

About: Hi all, I'm a college student in the copenhagen technical college. I'm currently working with 3D printers and I'm building my own RepRap.

This Instructable will show you how to built a Constant Current for high power LEDs, using only two components.

High power LEDs are getting cheaper and cheaper, however the constant current drivers, to drive them are pretty expensive.

Here, I'll show you how to built a simple and cheap, yet very effective constant current source.

The image shows the constant current driver hooked up to a 1W white Luxeon LED.

EDIT: This LED driver supports PWM, which means that you can control the brightness of the LED(s). Those fancy and expensive drivers doesn't support that. I'll post some schematics and applications as soon as i have time.

Step 1: Get the Parts.

Here is a list of the the things you'll need.

a LM317 Regulator.
a Resistor (see next step).
a Heatsink for the LM317 (you don't need one as big as mine, I just took one i had laying around).
some Luxeon, or other brands of high power LEDs (see next step too).
some Wire to hook it up.
it will be a good idea to use a heatsink for the LED as well.

Step 2: How It Works

The LM317 regulator gives out a constant voltage of 1,25 volts between ADJ and Vout, so by adding a resistor between these two outputs, you'll get a constant current.

Ohm's law says that U/I=R, which means that Voltage divided by Ampere makes resistance.

so if you want to connect one or more luxeon 1W LEDs, which has a power consumption of 350mA, the calculation should look like this: 1,25 (the constant reference voltage of the LM317) divided by 0,350 (the LEDs power consumption) makes 3,57. So if the resistor is 3,57, constant current will be 350mA. The closest E12 value is 3,9 ohms, it will give you a constant current of 321mA. However you can't see any difference in the light output.

If you use 3W LEDs, which has a current consumption of 700mA, the calculation should be: 1,25 divided by 0,7 makes 1,78. The closest E12 value is 1,8 ohms, the output will be 694mA

the resistor must be at least 1W in both calculations.

Although the LM317 is rated for 1,5 Ampere, I wouldn't recommend it for applications that need more than 1 Amperes, because it gets very, VERY hot. the LM350 is equal to the LM317, but it's rated for 3 Amps

Step 3: Assemble It

I couldn't get my schematic drawing program to work, so here is a hand drawn.

The constant current source has a drop voltage of 3 V, so the supply voltage should always be 3 V higher than the LED voltage and can be up to 37V which is the maximum input voltage of the LM317.

Example: You are going to connect two white Luxeon LEDs with 3,42 forward voltage each (mostly mentioned as Vf in common datasheets). The input voltage can change from 9,84V (3,42 + 3,42 + 3) till 37V (3,42 + 3,42 + 30,6).

You can connect up to ten high power LEDs to this circuit.

The higher voltage you supply the LM317 with, the hotter it gets. so it wont be a good idea to supply it with unnecessary high voltage.

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    310 Discussions

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    assasinsareus

    3 years ago

    I like how simple this is. Probably not that efficient as a lot of energy will be wasted on heating up the LM317 device but still a good job.

    1 reply
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    odalcetassasinsareus

    Reply 8 months ago

    It won't waste heat if you use only the needed input voltage, not more. Excess voltage will be converted to heat. Too much heat in the LM317 means too much input voltage.

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    apatil23

    Question 1 year ago on Step 3

    If I want to use 10 Watt 12V LED chip, what all I have to change?
    Please help

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    VineethK7

    3 years ago

    why cant u put dot(.) instead of comma(,)
    it makes little confusing.... :/

    3 replies
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    JimTheSoundmanVineethK7

    Reply 3 years ago

    In Europe they use a dot where a comma would go, and a comma where a dot would go. Just accept that and change it accordingly when you make your own plans.

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    GilesB4JimTheSoundman

    Reply 1 year ago

    Only in continental Europe, in the UK we use the dot to delineate a fractional part. It is very confusing when looking at continental european data

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    RhealD1VineethK7

    Reply 2 years ago

    Why cant you look at the comma and see a dot?

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    DayFrame

    2 years ago

    Input voltage ?

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    JamesF153

    3 years ago

    Yup, low value resistor+lots of current = lots of power (heat) loss. Switching is the way to go. If you get creative, the power loss in the driver circuit can be cut down to almost nothing. A switched diode / capacitor combination would work quite well.

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    SabrinaV3

    3 years ago

    this circuit is a nonsense.

    using a high power resistor is a proof of very poor efficiency !. Most of energy is used to produce HOT in the R.

    switching regulator is the right way...

    NCP3065 just a little bit more complex

    http://www.onsemi.com/pub_link/Collateral/NCP3065.PDF

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    imaadak

    3 years ago

    can you suggest the other regulator ICs for the same job ()? i need to drive a typical 5W power LED at maximum efficiency as a strobe.


    Regards and Thanks.

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    AbdulA55

    3 years ago

    i made it, the wire melted, it's overheat

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    lex76

    3 years ago

    Where would I hook up the pot meter as in where do the pot meter wires go please help me out thanks

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    kjbkjbl

    3 years ago

    buen dia

    me podrian dar el cricuito para conectar un led de 50w para 12v usando lm317

    saludos

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    LostinAsia

    4 years ago on Introduction

    Good morning kind sir.

    I was skeptical that this circuit would work due to it's simplicity.

    However, it worked fine and I am quite satisfied.
    In the article you mentioned that the circuit supports PMW.

    Not sure how to do this, just can't get my head around it. lol

    I would like to control the brightness of the LED using PWW from a Arduino Mega.
    Not sure how to do this.
    Should I just connect the PWW output from Arduino to the ADJ pin?
    Thank you in advance.

    Charlie

    1 reply
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    assasinsareusLostinAsia

    Reply 3 years ago

    If I were doing I would probably use a P channel FET to switch the VIN signal or a N channel FET to switch the LED Cathode pin. Either should work. Make sure the FET is a logic level type FET so you can switch it on/off correctly with Arduino voltages. Also make sure the FET can handle the LED current.

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    qasimr1

    4 years ago on Introduction

    hy

    i have 3 volt source and 3 volt led with 20 ma current can u help me in designing driver ckt for this led thanx

    1 reply
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    assasinsareusqasimr1

    Reply 3 years ago

    You need a 1 Ohm 1/8W resistor in series to give you 20mA. Type LED calculator into google and the top link is useful.

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    martik777

    4 years ago on Step 3

    CC Led drivers are so cheap on ebay now, less than $1 each, for up to 3 leds, although I've been able to drive 4 from these. There are drivers for 4-5, 8-15 etc for not much more. I have been running these 24x7 for a couple years with no problems.