led 3w power supply ?

[mark krawczuk]

i have some 3w star type leds, 700 ma, f.v 1.4 to 1.8v.
i need to run 8 of these from a 12.8v ( auto battery) power supply.
what would be the best way to connect these up ? in series or parallel ?
how much total current will it need ?

 

[Harald Kapp]

LEDs are not operated directly in parallel. If in parallel, a means of distributing the current equally is needed.
To operate them in series, you'll require 8×1.4 V to 9×1.8 V (11.2 V to 14.4 V). A 12 V power supply can barely supply the LEDs at the lower end but not if the pass voltage is at the higher end.
I suggest a mixture of parallel and series mounting: 2 parallel strings of 4 LEDs in series each. Plus you'll need a means to limit the current - LEDs cannot be connected directly to a vooltage source.
I recommend you read our resource on the topic.

 

[bertus]

Hello,

Have a look at a dedicated led driver like the LM3421.

Bertus

 

[jaksonlee]

These power LED stars offer an extremely high light output in an energy efficient way. The power LED comes mounted on to special PCB star that acts to draw heat away from the LED, therefore extending its operational life. This makes them perfect for many lighting applications and this sheet explains how.
The LED is a 3W LED and the forward voltage is rated at 3.5V to 4.0V so for the purpose of the calculations a forward voltage of 3.7V is used. The typical current of the LED will therefore be 810mA (from Power = Current x Voltage). The power source used with the LED must therefore be able to deliver at least 810mA. Since most power supplies will be higher than 3.7V a current limit resistor will be required. The next page shows an example of using this LED with a 5V power supply. The resistor also needs to handle 810mA flowing through it and as a result will need to be a power resistor. Please note that during use both the star LED and the resistor will get hot and shouldn't be touched.

 

[Harald Kapp]

The power LED comes mounted on to special PCB star that acts to draw heat away from the LED, therefore extending its operational life.

You still need a heatsink to get rid of the thermal energy "drawn away" from the LED.

 

[jaksonlee]

Unfortunately LEDs can't radiate this heat away in the same way as an incandescent bulb – instead, the heat needs to be drawn away from the LEDs by a heat sink normally located in the bottom of the bulb. The heat sink absorbs the heat produced and then dissipates into the surrounding environment

 

[Bluejets]

The heatsink the star LED is mounted on would be insufficient in it's own as Harald has pointed out.

 

[jaksonlee]

A frequently listed advantage of LEDs is that they do not produce heat. In a way, this is true, LEDs are cool to the touch because they usually do not produce heat in the form of infrared (IR) radiation. This obviously doesn’t go for IR specific LEDs.
IR radiation is what actually heats incandescent bulbs and other light sources, making them hot to the touch. Without IR radiation, LEDs are able to be placed in spots where the heat from other sources would cause a problem.
The ambient temperature and the drive current both affect the junction temperature of LEDs. Other influences are the nature of the light, whether it is steady state or pulsed, and then the one we are really interested in, LED wattage per unit area of heat sink.
Heat sinks are an important part of LED lighting because they provide the path for heat to travel from the LED light source to outside elements. Heat sinks are able to dissipate power in three ways: conduction (heat transfer from a solid to a solid), convection (heat transfer from a solid to a moving fluid, air in most cases), or radiation.

 

[bertus]

Hello,

Here is an article on heatsinks, based on the heat transfer on transistors:
https://sound-au.com/heatsinks.htm
You can even make heatsinks your self:
https://sound-au.com/articles/diy-heatsink.htm

Bertus