LED Matrix Design

Question

I have a LED arrays designed on Aluminium heatsink PCB. I have SMD White LEDs whose specs are as follows and differ from already existing LEDs mounted on PCB :-

• Forward Current : 60mA and Max. 90mA
• Wattage : 0.2W

Link to Datasheet of LED :- http://www.edison-opto.com/en/product/plcc_2835_series

Part Number : 2T03X2WW11000002

The Forward voltage of LEDs are binned by structure

As per the distributor of these LEDs I will be getting either VC1 VA2 VB2 or VC2 i.e the Forward voltage of LEDs will differ by every lot from 3.0V to 3.4V.

I have regulated power supply rating of 17V@800mA, which has tolerance of +/-5%.

I will be using 20 White SMD LEDs of 0.2W each. Here is schematic for that

XORB is N channel-FET which VDS of 30V and IDS of 5.8A.

Do I need resistorS to drop the voltage in LEDs? Will circuit have any effect if Forwards voltage will vary? Because Voltage bin group will vary from manufacturer itself.

Also any thermal issues in circuit which I need to take care of?

Answer 1

The constant current source is of little value in a matrix of LEDs. The normal technique to ensure appropriate current is to have each series string controlled by a current regulator.

But recognizing you are working with an existing design, I would recommend simply removing T1 in your schematic and change the parallel resistors to 10 ohms, 1/2 watt each. Ideally these should be located on the drain side of the FET. These changes will allow your FET to run quite cool since now it is only switching while the two resistors handle the current limiting.

If your supplier can ship you LEDs that are binned, you will find that if you build a matrix from the same bin, the brightness will be more uniform. You can then also optimize your current limiting resistors for each bin. I used to do this for a product line that used millions of LEDs per year.

Answer 2

Also any thermal issues in circuit which I need to take care of?

Almost surely. Some simple math will be able to tell you how much power is being dissipated on the mosfet.

I would also put a serial resistor on the line to the controller: or it will be shorted by the bjt.

edit: for power dissipation estimation

the worst case (from the mosfet's perspective) is for the leds to drop 3.0vx5 = 15v. That means the mosfet drops 17v-3.0vx5 - 0.7v = 1.3v.

The max current is 0.7v/(3R/2) = 0.5amp.

mosfet dissipation is 1.3v * 0.5a = 0.6w.

SOT23 will have a thermal resistance of 250 - 300c/w. so the temperature rise on the mosfet is 150c.

it is toast. and with no room to spare.

Answer 3

With a corrected circuit available, the issue makes a great deal more sense. The circuit is indeed a constant-current driver, and with the components shown will draw about 1/2 amp. To use this circuit with the LEDs specified, replace R4 and R16 with a single 3 ohm, 1/4 watt resistor, although a 1/2 watt resistor will give more margin for heating.

With the LEDs specified, the worst-case Vf will total 17 volts. Add to that the 0.7 volts across the sense resistor, and this suggests that the circuit will not work. Either use strings of 4 LEDs in series, rather than 5, or increase your power supply to 20 volts or better.

Note that the sense resistor (R4/R16) will always have about 0.7 volts, and the remainder of the voltage will appear across the MOSFET.

The power dissipated in any component is simply the voltage across it times the current through it. You can calculate this for all of your components. Whether this is a problem depends entirely on the component and how it is mounted. You have provided insufficient information to address the question. In general, though, heat-sinking LEDs is a good idea.

EDIT - Upon viewing the other answer, it is clear that I should have looked up the FET you are using. Shame on me. The part you specify is completely unsuitable for your purpose, since it is not capable of handling the power which it will dissipate, and I do believe that counts as a "thermal issue". You'll need to go to a FET in a bigger package, such as a TO220. You'll need to calculate the power dissipated, then refer to heat sink specs when choosing a heat sink for it. Depending on how you package your circuit, you'll also need to beware putting a board with a local heat sink on it inside a box, since this will prevent cool air from getting to the heat sink, thereby reducing its efficiency. Whether or not this will be a killer problem depends on the power dissipated, the heat sink chosen, and details of the enclosure.