Electronic – Home solar system – COB LED Drive

For best overall LED efficiency (which is your stated aim) you want high efficiency LEDs plus high efficiency drivers. There are may high power LED modules available but they in many cases use LEDs that are less efficient than the best available. No amount of efficient driving of a very low efficiency LED will make up for its low efficiency.

You mention 0.8A x 40V (which will use multiple LED dies or separate LEDs) = 32 Watts. This input power could be obtained with about 5 to 8 of the top available LEDs and it will often be better to use the best available if top efficiency is wanted.

It will probably be most cost-effective to buy a commercially available LED driver rather than making your own, looking for good efficiency. ebay lists a vast quantity of possible candidates

You also need to specify your energy source. Is this 110 VAC mains, or a 12 Volt battery or?

The standard way of driving LEDs correctly and at maximum efficiency is straightforward (fortunately).

• Determine required LED operating current.

• From data sheets, determine maximum voltage drop that LED will have at desired current.

• Provide a switched mode constant current source that will produce the required current at at least the maximum possible voltage.

• Extra points:

  • Minimise LED temperature with cooling

  • Operating an LED at below its maximum current ratings will result in somewhat increased efficiency - perhaps 10% to 20% more light out per Watt in at 20% of full ratings compared to full power.

  • LED lifetimes increase with decreasing temperature and with decreasing current. While more current will usually result in more heat produced, These effects are independent of each other.

To get an LED driver of essentially any desired rated current. use of an external MOSFET of suitable rating, driven by a controller that sits, will allow whatever current the MOSFET can handle. This can be very large if desired.

For very best efficiencies use "synchronous rectification" - this replaces rectifier diodes with switched MOSFETs and gives lower losses.

Efficiencies around 85% from a low voltage DC source to an LED string will be achievable.Over 90% in selected cases and with care.

Comment on the above and we can provide better information.

The LEDs are designed for a constant current supply which produces whatever voltage is necessary so reducing the current would be effective in dimming the LEDS.

Rohat calculated that the supply output would be a constant 700mA and for 3W per light you need 86mA each. Putting 8 lights in parallel would give 700/8=87.5mA per light which is about right, but that assumes they will share the current equally. The problem is that LEDs have a fairly sharp turn-on so whichever bulb has the lowest forward voltage will take almost all the current and die, then the next lowest would blow and so on. Unfortunately, whichever takes the highest current will heat up most and that will probably drop the voltage futher giving thermal runaway.

What you would need to do is add something to equalise the currents, a series resistor as shown would do but I suspect the value needs to be higher than shown in the circuit, at least the maximum forward voltage variation divided by the current variation you can tolerate. As the LEDs can take up to 350mA, that is fairly easy but you probably want to keep the brightness of the bulbs fairly similar so for example 10% brightness variation implies no more than 8.6mA difference between the bulbs. Higher values of resistor will maintain more even brightness but will increase heat dissipated in the resistors. The advantage is that including 8 resistors is a lot cheaper than separate drivers.

P.S. You probably realise that the supply voltage need only be 35V plus the resistor drop so a 144V-capable supply is much more than you need, it is really intended for driving multiple bulbs in series so that all get exactly the same current and hence brightness.