Luxeon 1 watt lighting system

[fever]

Hello all,
last week i bough some 1 watt LED's of luxeon.i made differ4ent circuits to drive them.imean simple resistor (as 1 watt need 300ma-i used 8 ohm/2 watt resistor) and LM317 based linear current regulator.they worked great.
but now i want to use lot of them on mains(230v) to power my work shop completely.
so for holding purpose.iam making a hexagon shaped aluminum pulg.on which LED's can be mounted(also work as heat sink) and which can be easily plugable in to current electrical system.
Each AL holder is capable of holding 7 LED's(6 on face one on button)
so i need some help to build the circuit.as 1 led need 300ma-7 of them need -2.1A.
how can i do this on 230vAc supply.

i desperately need some help here.
Thanks and regards

 

[Diver300]

You should connect the LED in series.

Luxeon LEDs seems to have typical voltage of 3.7V but there is a big range of voltages. If you put them in parallel, some may have less voltage and take more current.

If you put them in series the currents will all be the same. Also the current stays at 300 mA. You should allow for up to 30 V.

The easiest way to provide the power is to have a 230 - 24 V transformer. After rectification and smoothing you will get about 32 V DC. This will run all 7 in series with one current regulator.

To stay safe, you have to use a transformer.

 

[fever]

thanks for the response diver300.
according to the calculation,for 7 LED's with forward voltae of 3.7V each,the total voltage will be 25.9(26) volts.
but iam feeding 32vDC isn't dangerous?

ohh i shld use the current limiting resistor accordingly..am i right?
i just used this wizard
**broken link removed**
current is 300ma,source voltage-32v and No of led's-7
it gave me 22ohms resistor.
can i go with these values?

 

[Diver300]

Yes, you should never feed LEDs with a fixed voltage. You should always control the current.

The problem with a resistor is that the current can change a lot if the voltage isn't exactly what you expect.

I would suggest some type of constant current circuit. Then the current will be the same even if the voltage varies. You said that you used a circuit based on an LM317 and that should work with the larger voltage. You need at least 3 V drop with the LM317, and you need a resistor of about 4 ohms between Vout and Vadj.

 

[MrAl]

Hello,

I have run Luxeon 1 watters myself off of wall warts in the past.

One setup used a 9vdc wall wart and two Luxeon 1 watt whites in series.
I think i used a 10 ohm resistor, which means about 300 to 350ma.

Another setup uses a 5vdc wall wart, one Luxeon, with a 5.6 ohm resistor. I get
about 300ma though that one.


The idea when using a simple resistor is to make sure you have enough
voltage headroom, and to test the design on paper by letting the power
line go to high line (15 percent higher) and check the current flow. You
want to make sure the current doesnt go way up when the line goes up.

For example, with my 5v and 5.6 ohm design when it is operating
normally the voltage would be 5vdc and so that would cause
268ma on paper, but at high line it would cause 402ma to flow
through the LED, which is quite a bit higher. Luckily, this wall wart
is regulated internally to 5v which keeps it constant.

The other one with two LEDs is not regulated, and the output is normally
closer to 10vdc, so that allows 300ma through the LED normally, but then
at high line this would change to 450ma, which is quite a bit more but it
can tolerate that for a short time.

BTW, one of the other important things is heat sinking. The Luxeon Star
1 watters require a heat sink to operate at full power or else they get
extremely hot. Even a piece of angle aluminum works ok, attached with
some thermal epoxy. I even used PC11 one time keeping it very thin.

 

[Hero999]

Is size a problem?

You can buy special ICs for this kind of thing but you'll need to connect the aluminium plag to earth for safety reasons.

 

[Ubergeek63]

Is size a problem?

You can buy special ICs for this kind of thing but you'll need to connect the aluminium plag to earth for safety reasons.

Which you can not do unless the system is running on a transformer anyhow since the Luxeon stars base is not electrically neutral.

 

[Hero999]

I didn't know that. The high powered LEDs I use have isolated heat pads.

If this is the case then you'll need insulators, provide sufficient insulation between the user and the heatsink.

 

[Ubergeek63]

I didn't know that. The high powered LEDs I use have isolated heat pads.

If this is the case then you'll need insulators, provide sufficient insulation between the user and the heatsink.

It is a pain... if you look at the Luxeon star data sheets they warn about it.

 

[steev]

i have a question here.
yesterday i bought some 1 watt Led'S and powered them using my 12v wall wart.3 stars in series connected to 12v wall wart.and i didn't use any resistance as 3*3.7=11.1v.and this setup fried my first LED.i don't understand where i did wrong.
the LED cal posted by fever gave me the resistance value of 3.3Ohms.so this is my mistake?
and if i want to use LM317 with supply of 12v and for 3 stars.what shld be my component configuration.

 

[audioguru]

i have a question here.
yesterday i bought some 1 watt Led'S and powered them using my 12v wall wart.3 stars in series connected to 12v wall wart.and i didn't use any resistance as 3*3.7=11.1v.and this setup fried my first LED.i don't understand where i did wrong.
the LED cal posted by fever gave me the resistance value of 3.3Ohms.so this is my mistake?
and if i want to use LM317 with supply of 12v and for 3 stars.what shld be my component configuration.

The LEDs are not 3.7V each. They are somewhere from 3.3V to 4.2V each. If the LEDs are 3.3V then their total is 9.9V and if the 12V supply was a little high at 13V then the current-limiting resistor had 3.1V across it. If the 3.3 ohm resistor was a little low at 3.1 ohms then the current was 1A which is way too high.

That is why LED manufacturers recommend using a regulated current source. An LM317 plus one resistor makes a regulated current source but it needs 3V extra to work. If the LEDs are 4.2V each then their total is 12.6V plus 3V for the LM317 current regulator is a total of 15.6V minimum.

 

[steev]

Thanks gura,
so lm317 seems best idea.i made a sample diagram to connect 3 Luxeons using LM317.pls have a look.
(considering the LED forward voltage is 3.8v and 3v for LM317)
thanks

 

[audioguru]

Hi Steev,
I don't know how you determined the forward voltage of your LEDs as 3.8V unless you measured them. I don't know which colour you have but the datasheet spec's a forward voltage range for white as from 2.79V to 3.99V at 350mA when cool. the voltage at 320mA and when warm will be a little less.

If your LEDs are actually 3.8V or less then your circuit looks fine but the input of the LM317 should have a 0.1uf ceramic disc capacitor to ground.

 

[steev]

thanks guru,
iam using white LED.and my seller told me to check particular data sheet where it was written 3.8v as forward voltage.
so i hope 15v at the input of LM317 along with a 3.9Ohms resistor will drive my 3 LEd's in series with out any damage.
is there any for formula to calculate the power dissipation of resistor(i found only 1/4watt resistor.so is this gona work? )

 

[audioguru]

The current from the LM317 is 1.25V/3.9 ohms= 320.5mA so the 350ma rated LEDs will be fine if their heatsink is big enough.

The power in the 3.9 ohm resistor is 321mA x 321mA x 3.9 ohms= 401mW so a 1/4W resistor will fry and a 1/2W resistor will get pretty hot.
The LM317 will also get pretty hot and will need a heatsink.

Here is the 1W Luxeon datasheet:

 

[Leftyretro]

Here is the hot set-up for driving your 1 watt LEDS. This can drive from 1 to 5 series connected 1 watt LEDs if you provide the stated source voltage.

**broken link removed**

Lefty

 

[Ray@]

Very new to this...I read you responses and you seem to know what your talking about...I have 1 watt LED white lights...I wish to put them in a series...They are listed as Forward Voltage : 3.2V~3.4V..Forward Current 300Ma..Luminous Intensity 60 Lumen...Color Temp 5500K I have been looking for a resistor to put on them for a 12 Volt power supple as I plan to put them in a lightbar on a vehicle...any suggestions

 

[Diver300]

Very new to this...I read you responses and you seem to know what your talking about...I have 1 watt LED white lights...I wish to put them in a series...They are listed as Forward Voltage : 3.2V~3.4V..Forward Current 300Ma..Luminous Intensity 60 Lumen...Color Temp 5500K I have been looking for a resistor to put on them for a 12 Volt power supple as I plan to put them in a lightbar on a vehicle...any suggestions

How many LEDs do you want to light?

 

[Ray@]

I have 50 to put together..
I was planning to put 25 on one side and 25 on the other..

 

[Diver300]

You could put 2 or 3 in series. 2 would need about 6.6 V, and 3 would need about 9.9 V.

You should allow for 14 V with the engine running, so the voltage remaining is either 7.4 V or 4.1 V so the resistor you should use is 24 Ω for 2 LEDs or 14 Ω for 3 LEDs. The power rating is 2.5 W or more for the 2 LEDs and 1.5 W or more for 3 LEDs.

Now when the voltage falls to 12 V when the engine stops, the current will fall more if there are 3 LEDs in series. For 2 LEDs, the current will be (12 - 6.6)/24 = 225 mA. For 3 LEDs, the current will be (12 - 9.9)/14 = 150 mA

So if you have 2 LEDs in series, you need 25 sets of 2 in series so you need 25 resistors of 24 Ω, and the total current will be 7.5 amps. The current will drop to 3/4 when the engine stops.

If you have 3 LEDs in series, you will have 16 sets of 3 in series, and one set of 2 in series.

You will need 16 resistors of 14 Ω and one of 24 Ω and the total current will be 5.1 amps. The current will drop to 1/2 when the engine stops.

The alternative is a dedicated LED driver IC, which needs an inductor. Something like this https://www.diodes.com/datasheets/AP8801.pdf would let you run 3 LEDs in series with no dimming at 12 V.

There are also boost converters that let you put more LEDs in series.