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

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.

• You haven't told us the LED current you want. That IC is rated at 200mA according to the link. – Brian Drummond Nov 16 '17 at 9:42
• @BrianDrummond I have edited the post to include the desired current. – Jann Nov 16 '17 at 10:09

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.

• I had mistakenly presumed that I had followed the answer that I had linked. I have updated the link and attached the proper circuit diagram also. – Jann Nov 16 '17 at 10:04
• Mosfet might also cause it to oscillate... – Trevor_G Nov 16 '17 at 10:05
• @Jann read my updated answer – Trevor_G Nov 16 '17 at 10:33
• Oh, I hadn't noticed that the MOSFET was way undersized. Should change it to a larger part as you have pointed out. Still, I cannot understand why a NPN will be better than a MOSFET in this case. As you have said, wattage is wattage. Or is it just better in terms of stability of the circuit? – Jann Nov 16 '17 at 11:11
• @Jann remember this is a linear circuit, the MOSFET is not saturated so does not add any benefits. Instead you have added significant capacitances to the equation which, with the addition of any significant impedance from wiring, and the sharper threshold, may make the thing oscillate. Horses for courses. – Trevor_G Nov 16 '17 at 11:18

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

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.

This is a simple low cost very efficient driver circuit.

Inductor

small low cost inductor with low DCR Colicraft MSS1038

Driver

This is an efficient Low Dropout Linear Driver

Adjustable Constant Current Regulator & LED Driver

150mA - 350mA CC Regulator On Semi NSI50150ADT4G