This circuit was sent to me by a friend for driving a Luxeon star, or any other power LED.
Simple to build and, electronicaly speaking, simple in its operation.

The circuit can be powered from 2 to 18 Volts and has a 0.5 Volt dropout. ie if you shove 10 Volts in you'll get 9.5 Volts out so allow for this in any calcs.
It will supply a constant current of up to 20 Amps with a suitably large heatsink of course. If you tailor your input Voltage as close as you can to the LED supply Voltage then there will be minimal heat generated.
The values shown here will run a Luxeon at 960 mA from an Oldham cell or anything else you want to connect it to, and the current in amazingly stable. So if you don't mind losing a bit of power in heat then you can run your LED on a variety of Voltages without having to alter any component values.
R3 sets the current and with R3 = 2.2 Ohms the current available to the LED will be 225mA.
With a 20 Amp ceiling, the circuit will easily handle multiple connected LED's. :thumbup:
And with the component count being so small, it can be built into most single LED lamp enclosures. All parts are easily available (no waiting for the post from Hong Kong) and the cost is less that £2.00 :thumbsup:
The FET (field effect transistor) Q3 constantly tracks the current and limits it through Q1 so if LED wants to suddenly draw scuds of power (for any reason) then Q3 shuts down to allow only the set current. A simple yet clever circuit.
I saw this circuit driving a very bright luxeon, flavour unknown, from 4 AA's and the heatsink used for the driver was smaller than an SD memory card.
As I see it, using the maximum voltage, 18 V at 20 Amps that gives a potential 360 Watts 😮 Though I think you would need to seriously 'beef up' R3 (suggestion - replace with electric fire)

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This is a linear current regulator circuit, and it actually works like this:
- The high current path is through Q2 and R2.
- R2 is sensing the current, the voltage across it increases linearly with current (as per Ohms law)
- Initally Q2 is turned on by R1 and Q1 is turned off
- As the current through the LED, Q2 and R1 increases, the voltage at the top of R2 (and thus the base of Q1) increases.
- As this voltage reaches the turn-on voltage of Q1, Q1 starts to conduct and reduces the voltage on the gate of Q2.
- Q2 will start to turn off, thus reducing the current.
- A happy equilibrium will be reached
- R2 sets the current.

It is very important with a linear circuit like this to optimise the input battery voltage. Any excess battery voltage will be dissipated as heat in Q2 and R2. As a result, efficiency is terrible with anything other than a very close battery voltage.
In fact if the battery voltage is that close, you might as well just use a series resistor if you are using NiCd / NiMH or LiIon as the terminal voltage holds up pretty well until almost flat.

Note you can do very similar with e.g. an LM117 voltage regulator, but it will suffer from the same inefficiency issues.

The efficiency / power dissipation issue is why you will never see a circuit like this in any real product.

However, it will work just fine!

Chris.