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Step 3Assemble 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.
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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I m making a 90W Led panel,Using 1W led(rating 350mA,3.2V)
In which design shud i make it??
And for Led lights panel,Constant current source is better or Constant Voltage source??????
n do the series or parallel combinations of LEDs affect the efficency or temperature of the panel....
Please Email me the design at
good_boy_9211@yahoo.com
|--CCDriver--LED-LED-LED-|
|--CCDriver--LED-LED-LED-|
12V--|--CCDriver--LED-LED-LED-|--Ground
|--CCDriver--LED-LED-LED-|
|--CCDriver--LED-LED-LED-|
I'm planning on using the LM1084it-adj vs the LM317.
So far I've figured based on datasheets @ 700ma, I should have about 2v total drop with the 1084 (1.25adj+ ~.8@700mA). Is this correct?
My source is going to be a computer PSU converted to a benchtop PSU, using the 12V out. Based on my LED's datasheets, I should be at about 2.2V per Red or Yellow @700ma, based on this, the 1084 will be dissipating 1.2V@700mA with 4 red or yellow LEDs wired in series (total of .84W). For the blue I've figured ~3.6V for 700mA, with 2 in series the 1084 would have to drop ~2.8V@700ma (1.96W).
Forgive me, I am a relative newb to electronics, does my math seem correct? Am I misinterpreting the datasheets when it comes to how much voltage the 1084 drops under varying loads?
There is no point to doing it this way. Simply using a series resistor will do just as well, just as efficiently. The only thing nice about this circuit is an easy way to calculate the current regardless of (an unknown mystery LED) what the forward voltage of the LED is.
This doesn't inherently "support PWM" either.
Above all, LM317 is designed to be a voltage regulator. The best way to drive an LED, whether it be by PWM or not, is by more directly regulating the current but ideally, not using a linear voltage drop stage which is what the LM317 is.
It is a good teaching circuit to learn about LM317, but for powering an LED, not so much.
Check out http://www.dealextreme.com as they have a selection of dirt cheap LED drivers for most 1W to 5W applications. And good prices on the most efficient 3W Cree LEDs too.
To put it another way, usign a $4 driver and $6 LED, you would have over triple the light output using less than 1/3rd the power of running this circuit with one LED off more than roughly (too lazy to do the math) 12V, and eliminate the need for a massive heatsink.
LED drivers are purpose designed to do the job best. DIY projects are great but also appreciate when professional EEs pick the right parts, when a factory can get it all made, and delivered to you at a good price ready to do exactly what you need.
Modern tech combined with the internet for finding and ordering products is very very great, even if you want a DIY project there isn't a need to do everything from scratch anymore and with LEDs, to regain the efficiency is very desirable for not only efficiency but to not have to deal with all that heat in a project casing.
Also, since i have 6 in serial, the total voltage drop is 6x3.2(LED)+3(LM) = 22.2V, but my source is 24V, what do i do with the extra 1.8V? same goes for the 6.5A output from the source, my parallel config is only 4.2A (6x700mA), what happen to the other 2.3A?
thanks
So, 3x2,5V(leds)+3v(lm317)=10,5V
What I have to do with the rest 1,5V voltage?
Forward resistor before the leds by the ohm-law?
Apologize me. My english is so weak.