# Arduino High Power LED

#### Plecc

Joined Aug 26, 2010
22
HI all,

I would like dim a blue high power LED using an arduino uno.
The LED is 3W blue 4.8v 700mA.
Does anyone know of a simple constant current circuit that can be dimmed using the PWM from the arduino?

#### Plecc

Joined Aug 26, 2010
22
I was hoping to knock up something simple with bits around the house.
I found this site -
http://bildr.org/2011/03/high-power-control-with-arduino-and-tip120/
I have a TIP122 instead of the TIP120.
The only bit I'm unsure about is how to work out what resistor is required to run this LED on a 12V 1200mA power supply?

#### MrChips

Joined Oct 2, 2009
20,202
You can use the circuit diagram shown in the DLD101 data sheet to roll your own using discrete components.

• Plecc

#### Plecc

Joined Aug 26, 2010
22
Sorry - Forward Voltage is 3.8v not 4.8.

#### Plecc

Joined Aug 26, 2010
22
I'm a bit too much of a novice to be able to choose the correct components to recreate the IC circuit the resistors and the NPN transistor I have but Q1 I have no idea what it is.
It looks a bit like a variable resistor with a diode strapped across the back of it.

#### MrChips

Joined Oct 2, 2009
20,202
Q1 is a HEXFET
R1 = 4.7kΩ
R2 = 47kΩ
Rc = try 2.2kΩ
Rs = Vbe/ILED = about 1Ω for 700mA LED

#### wayneh

Joined Sep 9, 2010
16,237
It looks a bit like a variable resistor with a diode strapped across the back of it.
Huh? What looks like that?

That 3rd (lightbulb) circuit will work for you. Your TIP122 is rated to a higher voltage than the TIP120 but is otherwise identical, so it'll be fine too.

For the LED, I'd aim to start with about half the rated current and only raise it once I was convinced everything is hunky dory. So you need 300mA from an 8V drop in voltage (supply - Vf). Ohms law ∆V = I•R 8v = 0.3A•x x = 27Ω

You need to consider heat. Your LED needs a heat sink and your resistor will dissipate I^2•R watts, so 0.3^2•27 = 2.43 watts. I'd use a resistor rated to at least 5W. If you eventually get up to, say, 600mA, you'll need 0.6^2•15=5.4W (rough numbers). So a 10W rated resistor, to be safe.

This all assumes continuous operation. That's the safe way to design it. If you KNOW the duty cycle will never exceed, say, 50%, you could think about skimping on heat management. Not the way I'd do it, I'm just sayin'.

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#### ErnieM

Joined Apr 24, 2011
8,019
Here's two basic methods. One of these should work nice for you: Both are for 3.3V or 5V logic inputs. Most any NPN transistor will work, such as a 2N2222A.

The first is pretty simple and best when you're running your LED off a voltage larger then 5 volts. The transistor will be in the active mode to help dissipate off power. I used this method on a job where my power varied between 10 and 40 volts. You select R1 by the following:

R1 = ( Vin - Vbe) / Iled = ( 5 - .7) / .7 = 6.14 ohms

Power(R1) = I^2/R = .7 x .7 / 6.14 = .0798

The closest standard value I found was 6.19 ohms in 1/8 W or 1/4 W (better) versions.

The next version is a bit more typical and uses the transistor as a saturated switch. R1 is selected by:

R1 = (Vdd - Vled) / Iled = (5 - 3.8 ) / .7 = 1.74 ohms

Power(R1) = I^2/R = .7 x .7 / 1.74 = .28

Unbelievably 1.74 ohms is a standard value available in 1/2W (not recommended) to 1W (Mikey likes) versions.

(Parts found off Digikey)