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Power LED's - simplest light with constant-current circuit

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Step 2: Specs & Function

Here i'll explain how the circuit works, and what the maximum limits are, you can skip this if you want.

Specifications:

input voltage: 2V to 18V
output voltage: up to 0.5V less than the input voltage (0.5V dropout)
current: 20 amps + with a large heatsink


Maximum limits:

the only real limit to the current source is Q2, and the power source used. Q2 acts as a variable resistor, stepping down the voltage from the power supply to match the need of the LED's. so Q2 will need a heatsink if there is a high LED current or if the power source voltage is a lot higher than the LED string voltage. with a large heatsink, this circuit can handle a LOT of power.

The Q2 transistor specified will work up to about 18V power supply. If you want more, look at my Instructable on LED circuits to see how the circuit needs to change.

With no heat sinks at all, Q2 can only dissipate about 1/2 watt before getting really hot - that's enough for a 200mA current with up to 3-volt difference between power supply and LED.


Circuit function:

- Q2 is used as a variable resistor. Q2 starts out turned on by R1.

- Q1 is used as an over-current sensing switch, and R3 is the "sense resistor" or "set resistor" that triggers Q1 when too much current is flowing.

- The main current flow is through the LED's, through Q2, and through R3. When too much current flows through R3, Q1 will start to turn on, which starts turning off Q2. Turning off Q2 reduces the current through the LED's and R3. So we've created a "feedback loop", which continuously tracks the current and keeps it exactly at the set point at all times.

 
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arij2 years ago
Hello. I am new user.
Is this circuit with different resistor, suitable for 5W ir power led?

Forward voltage: 1.5-1.7V Forward current: 2400mA Wave length: 940nm
Power source 12-14V dc

Thank you
Ari
Does this driver work with PWM?

Thanks for taking the time to put this together!
mechatr0nix2 years ago
Somebody on this page asked for the components for the 3 watt led. I am also interested in that. Could you please post the values of those components in the comments section in this step.
Can I use my IRF510A MOSFET for Q2, and an MJE3055T NPN transistor for Q1? Also, what value resistor should be used to get approx. 1.2 amps on the LEDs?
The FET and Transistor should work and figuring out the resistor is easy.

The transistor will start to turn the FET off once the base voltage starts to get above 0.7V so the resistor you would need would = .7V/1.2A = ~0.58 ohms, the resistor need to be above 1 Watt though to handle the current.

Rufus
yohanc7773 years ago
Hello Dan.I like your circuit its quite simple and neat.However im using a 3 watt led from luxeon.which needs 700 mA. What do i do to Q2 to power my led to 700 mA instaed of 200 mA. Kindly let me know yohanc777@gmail.com

richie14794 years ago
Is a good design
it is best...
thanks
i am so happy there are people like you on instructables that use the normal symbols for building a circuit draw up like this.
Normal?
 agreed
If I was wanting to just draw a constant current from the power supply (batteries in this case) could I replace the LEDs with a dummy load, say a power resistor, and have it still work fine?

Is there a better, but still simple circuit for doing this?

Thanks for any help.
1loloo4 years ago
i have eight 10W, 7-8V leds. What power supply i need and what change in this circuit i have to make in order to leds give light safely?
cryptopsii4 years ago
Won't work because a darlington is a sort of dual
bipolar transistor and you need a FET (Field effect transistor)
A bipolar transistor drive current if current flow throught
it's base. A FET drive current if Voltage is applied to it's
gate  (it remplace base). And  a Mosfet is different that a fet
and won't work either.
Umm, a MOSFET is a FET, FET stands for Field-effect transitor, MOSFET stands for metal–oxide–semiconductor field-effect transistor, its formed with a thin layer of a glass-like substance between the gate and the drain/source, as apposed to a JFET, which stands for Junction field-effect transistor, which is formed with a revesed biased p-n junction between the gate and drain/source.  JFETs tend to have higher leakage currents, around 10^ -9 amps, while MOSFETs leak only at around 10 ^ -14 amps, MOSFETs tend to stand much higher currents and be cheaper and smaller to make, while JFETS can work at much higher voltages and switch a bit faster.  Your processor is made out of millions of MOSFETs.  Both can work the same if there specks cover it, but JFETs can rarely handle much current
starwarts5 years ago
Hi Dan. Do you think I can use TIP122 NPN darlington pair instead of the MOSFET? I dont have them available locally while TIPs are lying handy. What should I be concerned about if I use TIP122?