Remember that three watt led module I built a while back? Well, I finally got around to building a driver/interface for it so I can use it with my rgb led controller.
The time has come!
Specs:
Three LM317T adjustable voltage regulators, configured in constant current variable voltage mode.
Six one-watt 10 ohm resistors, wired in parallel for 5 ohm loads on the adjust pin of the regulator.
Three 1uF smoothing capacitors on the regulator output, one 100nF filter capacitor on the power connection.
Each channel is configured for 250 mA, which is 100 ma less than my leds can handle, but, this is good enough for a start - I have proper sized resistors on order. I will probably make a PCB for the second iteration of this project, and I’ll be using a single resistor instead of two in parallel.
The six pin connector goes to the LED module, the five pin connector is the power and control lines from my rgb led controller.
I started out by laying out the three biggest components of the project, and getting them secured to the protoboard. I’m using a protoboard with two supply busses on it, they’ll come in handy for carrying nearly an amp of current to my regulators. You can see the Vin pins of the regulators all bent ‘up’, overlapping the central power bus.
Now for some resistors. These resistors set the current output based on voltage on the adjust pin. You can clearly see two 10 ohms resistors. I have wired these in parallel, giving me a result of 5 ohms, which sets the drive current at 0.250 A
The beginning of the soldering; tacking parts down one at a time.
These three red wires each carry the drive current for one color, red, green and blue to the led module connector.
Same wires, different angle!
I’ve installed some 1uf smoothing capacitors and a 100nF filter capacitor as per the LM317T datasheet… they help make the leds fade smoother under PWM control.
I’ve got a haloween task planned for this driver, stay tuned!
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Hello, Your site is very detailed and useful. It’s a nice collection of good articles that I enjoy very much!
I am planning to build some LED drivers like yours. Could you please help me?
My input voltage would be 9-16V (from a car). I am replacing my tail light bulbs with 3W Lexon LED (For each, typ vf=2.20v; typ current=700mA).
And I will use three 3W LEDs to replace one filament bulb.
Using LM317T, if the resistor is 2ohm, the regulated current is about 700mA. My question is: Is it possible to wire all three Luxeon LEDs in series with the driver, if the resistor is rated at 10W?
Or, is it better to have an invidual LM317T to power each LED? In which case, I can source 2ohm resistors rated at 3W. Will there be less TOTAL waste heat and/or energy using this method?
Which is the better option?
Many Thanks for your help in deed! I look forward to your reply!
John
May 13th, 2006
I Just realise I made a utterly stupid calculation on the Watt ratings…. (haven’t done electronics before, Just started learning! :P )
John
May 14th, 2006
John,
You have the right idea - wire as many luxeon in series as you have voltage available (remember the lm317 costs you around 3v in overhead). A separate regulator for each LED = lots more waste. Since the LM317 has a dropout voltage of approximately 3v under load and heated up, you can use this factor along with the input and output voltages to calculate the power dissipation in the regulator:
Iout(Vin - Vout) + Iout * Vdo = 0.7 * (14 - (2.95 * 3)) + 0.7 * 3 = 3.6 + 2.1 = 5.7 watts.
Next we do the resistor:
I^2*R = 0.7 * 0.7 * 2 = 0.98 watts
This means you have to dissipate six watts on the regulator and one watt on the resistor. Seven watts is the reason I hate linear regulators … the numbers just get much worse if you try to drive your luxeons at their full rated power … 700mA will be under-powering your luxeon to about 50% of their 1400mA rating. You can help reduce waste in the regulator by matching the output voltage closely to the input voltage; in that case, a low dropout regulator helps a great deal. The LM1085 is pin compatible with the LM317 but handles up to 3a and offers a 1.5v dropout voltage, half of the LM317. Let’s try this:
0.7a * (14v - (2.95v * 4)) + 0.7a * 1.5v = 1.54 + 1.05 = 2.59 watts
So now you only have 2.6 watts on the regulator … less than half the waste of the above example, just by adding one more LED and using a more modern regulator!
Another thing to remember, the LED (like all light bulbs) is a heater than happens to give off some light, make sure you keep those bad boys cool!
justDIY
May 14th, 2006
Thanks so much for the detailed reply JustDIY!
I am getting the cheaper ProLight Luxeons, so driving at 700mA is what they recommend…. 1400mA would probably burn those LEDs badly! :P
And I will certainly go for the LM1085 option. But is the adjusting resistor’s value calculated the same way with it? And is a huge heatsink required on the LM1085 itself?
And what happens if the voltage of the car falls to about 10V (which could happen sometimes in my old car, especially during start up), will the LEDs fail to light at all?
I will mount these 4 LEDs on an old Pentium II processor forced air heatsink, which measures 5cm x 11cm x1.5cm, do you think it will be enough to keep the LEDs cool?
Actually, how hot does these LEDs actually gets when running? I have never bought them before, so I am just putting a bet on making something that would work….
Thanks so much for helping!
John
May 14th, 2006
Oh… forgot to ask, how do you work out the capacitor values for the current regulator? And is it required for my application when I only need instant mono colour light from the LEDs when I step on the brake?
John
May 14th, 2006
And on the calculations with LM1085, you say that the regulator would use 2.59 watts and the resistor would take 0.98 watts.
So am I right to say: Total wattage including everything would be…
2.59watts + 0.98watts + (4* 3W) = 15.57W?
Where the (4* 3W) is the wattage of the four LEDs…
Have I done some stupid calculations like before again?
John
May 15th, 2006