Reducing heat dissipation in linear current source used for driving a LED

Question

I'm currently using a linear current source to drive a single LED as in here. In my case (following the notation in the link), VDD is directly connected to a single cell lithium polymer battery, i.e. Vcc = 3.0 ~ 4.2 V. For the transistor and MOSFET, I am using MMBT3904LT3G(Digikey link) and 2N7002P,235(Digikey link), respectively. Rs is set to 2.2 Ohms. My desired current is approximately 250 mA.

Here is the circuit diagram from the site:

The problem is that there is so much heat dissipated it is hard to touch the PCB board. There are 3 LEDs (and corresponding circuits) in total on a 50 mm x 50 mm PCB (FR-4, 1.6 mm thickness). I am measuring 565 mV (which actually keeps dropping as the PCB heats up) across R2 and a voltage drop of 2.14 V across the LED at Vcc = 4.2 V. I am guessing that most of the heat is generated from Q2, since it has to drop the rest of the voltage which is approximately 1.5 V.

I have considered using a step-down switching regulator to drop the battery voltage to around 3 V, which I certainly think will help the problem. I have also looked on Digikey to find dedicated ICs such as this one. However, both require an inductor that I have heard to be expensive. There also seems to be charge pumps that I am not sure will be effective in my case (LED forward voltage is always lower than the battery voltage).

What would be the best way to go? In the end, I am looking for a cost-effective solution, but now open to any suggestions. And could you clarify whether charge pump LED drivers (such as this) will be effective for my case.

Answer 1

If it is "hard to touch the pc board", then by a common rule of thumb (or finger) your board is running at about 60 C. That is generally considered the "must let go" temperature. While that is indeed pretty hot, it's within the tolerances of the parts you're using. If you put the pc board in a small enclosure it will get hotter still. Air flow (cool air) is your friend.

As for reducing the temperature, you are out of luck. Your LED drops 2.15 volts at .25 amps. A Vc of 4.2 volts means that Rs and your FET will collectively drop (4.2 - 2.15) volts, or about 2.05. 2.05 volts times .25 amps gives .5 watts. 3 times that is 1.5 watts. And yes, I'd expect 1.5 watts over a 4 square inch pcb to get pretty hot. Rs is dissipating 1/4 watt, and your FET is dissipating 3/8 of a watt, which is right up at its limit.

ANY linear regulator doing your job will dissipate the same amount of power. The only question is how you deal with it. If you want to stick with your circuit, you'll need to learn about heatsinks.

Answer 2

Assuming I am looking at the right circuit, which you should have reposted here so people do not have to go back and forward....

Also assuming you wanted 0.25A per LED and they can handle that.

Your choice for Q2 is way undersized at 350mW you need at least a 1W part if not a 2W and a heat-sink would not hurt. The 2N7002P is also a MOSFET. I do not see any MOSFET circuits in your link. So this whole answer may be bogus. A MOSFET may cause this circuit to oscillate, and is not buying you anything.

R2 also needs to be at least a quarter Watt, which will be at 20C above ambient, a half Watt would be cooler.

The LEDs themselves, at half a watt each, are your worst offenders and also need special treatment to get rid of heat.

As for reducing heat.

With a linear solution at that range of voltages in, you don't get a choice, you need to dissipate @3W, which is not that bad for a two inch square. You can better manage the heat, but the wattage is what it is. A switching driver would be better for this high a current, and the inductors are not as expensive as you might think, not compared to a ton of heat-sink stuff. There is nothing you can do reduce the LED heat generation though, other than run them dimmer.

Anyhow, fix Q2 first to a larger NPN part that wont get so hot temperature wise.

Answer 3

You could possibly reduce the current by using a lower wavelength red.
1 Watt of radiated flux at a wavelength of 625nm will produce 180 lumens. At a wavelength of 620nm would produce 260 lumens (45% increase).

Red LEDs range from 30 lumens to over 100 lumens per watt. Using a high efficiency red LED (100+ lumens per watt) will reduce the required current.

Cree 623nm XPEBRD-L1-0000-00902 = 109 lm/W
OSRAM 626nm GR CSHPM1.23-JUKQ-1 = 101 lm/W

Lumiled Luxeon Color C 629nm L1C1-RED1000000000 = 53 lm/W

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At 250mA the LEDs are going to heat the board beyond 50°C (touchable but uncomfortable) without significant thermal management.

I measured the temperature of a string of 16 white LEDs spaced 0.75" on a 0.7" wide PCB with no heatsink. The temperature was measured on the bottom side under an LED.

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This is a simple low cost very efficient driver circuit.

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Inductor

small low cost inductor with low DCR Colicraft MSS1038

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Driver

This is an efficient Low Dropout Linear Driver

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Adjustable Constant Current Regulator & LED Driver

150mA - 350mA CC Regulator On Semi NSI50150ADT4G