I created an LED driver board based on PAM2863ECR http://www.diodes.com/_files/datasheets/PAM2863.pdf

The board includes three identical driver circuits to drive three LEDs. I'm not able to wire the three LEDs in series in this application. There is also a regulator and switch to adjust intensity. The board is typically powered from 6 NiMH cells, so the voltage is normally 6-8V.

The board works fine and intensity changes as expected, however I'm not getting the efficiency I expected. I never expect to get the max value claimed on the front page of the datasheet (97% for this part), but I expected at least 80% which is what I based the battery life estimates on. Instead I'm only getting about 64%.

Running a single LED at full output draws 930mA at 1.54V (1.43W), while drawing 340mA @ 6.6V from the input (2.24W), giving just under 64% efficiency. This is about the same efficiency throughout the input voltage range. I also changed the inductor to different values of the same series part, from 22uH to 68uH, and the result didn't change. This is a 2-layer board, and I thought maybe 4 layers with ground and power planes might help, but I can't see it making that much difference.

I'm attaching a PDF of the schematic (in some readers you can click on the parts to see part numbers), and images of the PCB layout.

Any suggestions on what I could try to improve the efficiency? If not, any suggestions on replacement driver chips that you've seen good efficiency from?

PCB Top with silkcscreen

PCB Top copper layer

PCB Bottom copper layer



Diodes: https://www.digikey.ca/products/en?keywords=MBR140SFT1GOSCT-ND Inductors: https://www.digikey.ca/products/en?keywords=SRN6045TA-470MCT-ND

Image of the schematic if you prefer it to the PDF: enter image description here

EDIT 2: I have done some additional testing. First I tried adding caps across the LED. It gave no measurable improvement in efficiency.

I then ran a sweep of the input voltage and saw efficiency drop by about 1% with Vin at 9V compared to 6V.

I rewired the LEDs in series and saw a huge increase in efficiency, up to about 86%.

Finally I changed the 47uF inductor with a 33uF type having 50mOhm DC resistance. This brought the efficiency up a little more, close to 90%.

  • \$\begingroup\$ What are the diodes D1-D3? And what's the DC resistance of these inductors? \$\endgroup\$
    – user16324
    Feb 18, 2017 at 21:10
  • \$\begingroup\$ Diodes: digikey.ca/products/en?keywords=MBR140SFT1GOSCT-ND Inductors: digikey.ca/products/en?keywords=SRN6045TA-470MCT-ND DC resistance is 200mOhms \$\endgroup\$
    – AngeloQ
    Feb 18, 2017 at 21:13
  • 3
    \$\begingroup\$ So, 0.18V across the inductor's DC resistance at 0.9A vs 1.54V across the LED. That's about 12% right there. \$\endgroup\$
    – user16324
    Feb 18, 2017 at 21:18
  • \$\begingroup\$ Yeah, good catch - I wasn't paying enough attention to the DC resistance when selecting. Most of the lower resistance parts are larger than I could use, but there are some in the same size range that are better which I could try, thanks. \$\endgroup\$
    – AngeloQ
    Feb 18, 2017 at 21:26
  • 1
    \$\begingroup\$ With low output voltage, 1.5V in this case, the power wasted through the external diode becomes very significant proportionally. It is common to go synchronous (i.e. two active MOSFET switches instead of a diode and a MOSFET) for low output voltage. Don't expect 90%+. For example, even a 0.1V sense resistor is a 7% loss. \$\endgroup\$
    – rioraxe
    Feb 18, 2017 at 23:14

2 Answers 2


I'm thinking 4 Batteries may be better than 6. Efficiency improves when Vin is close to Vout. The minimum for the driver is 4.5 V. A fully charged cell will put out between 1.25 and 1.4V. You do not want to discharge them down to under 1.2V to avoid deep discharge which will significantly reduce lifespan. By reducing the number of cells you may also avoid deep discharge when the voltage drops below 4.5 and the chip shuts down. Better efficiency, better battery life, and less batterie$. Not a bad deal

Running the Emitters in series you will raise the forward voltage closer to Vin improving efficiency, you will eliminate two inefficient sources of waste. Greatly improve efficiency, reduces watts by 1 Watt, improve battery life significantly, lower the cost and reduce the PCB real estate.

You efficiency calculations do not seem correct. 930mA @ 1.54 volts = 1.43 times 3 circuits = 4.29 Watts. And that is just from the IR emitters. So add the expected 20% inefficiency absorbed by the driver and you are at 5.1 Watts. That is twice your input Wattage.

For more than you ever wanted to know about batteries check out Battery University BU-215 should be useful to you.

In light of the IR revelation, A lower input voltage would help if you can't run them in series. You are at 6:1 in/out ratio. You do need a good output cap. From pg 8 of datasheet: as the ESR of this capacitor appears in series with the supply source impedance and lowers overall efficiency... A minimum value of 10μF is acceptable if the input source is close to the device. Your input is 6x output.

Why do you use 3 separate drivers and drive them all with the same intensity? It would be much more practical all around to put them in series.

The datasheet recommends a minimum of 33uH. You may be better off with 47uH.

Are you sure about the 1.54V forward voltage? That sounds very low even for an amber or red LED unless being driven low by excessive heat. What is the temperature? Any other color than red, red-orange, or amber I would not believe that voltage.
If real what's the part number?

Their reference design uses a Wurth 744770133 33μH,3.2A, 64 mOhm Max, Qty 1 Digikey: $4.53.

enter image description here

Your max efficiency @980mA is specified at about 85%, 97% is for 2 LEDs @ 1Amp, using a 47uH inductor.

efficiency chart

This looks to be the least expensive 33uH(with a better resistance than Wurth) that Digikey has for a 33uH:Taiyo Yuden NS12555T330MN 33µH, 3.16A, 49.8 mOhm Max, Qty 1 Digikey: $1.96


I was looking at the specs on the "LED" you are using. It makes zero sense. Specifically: MCD=70@750mA, MCD=75@1000mA

There is something wrong with this supplier. They know not what they sell.
spec IR emitter

MilliCandela is a luminous measurement. Luminous only applies to human visible light. Plus it should state the steradians (sr) at that Intensity or Irradiance (their spec does not differentiate between the two). The proper measurement is is mWatt/sr (Radiant Intensity) or µMoles/sr (photon, quantum).

A page from my paper "Understanding LEDs"
enter image description here

They do not mention the manufacturer, probably someone on or buying off Alibaba. ColdfusionX is an eBay Store.

I would take a serious look at OSRAM IR Emitters,aka LED, (even as much as I hate Siemens). Check out SFH 4714 and SFH 4715

RE: 3 LED String vs. 3 Drivers:

"They" need extraordinary justification. I cannot image ANY reason for doing so if the current through each LED will always be the same. Failure rate is very low if the temperature is kept reasonable. If reliability is an issue, then lower the current and increase the number of Emitters.

  • \$\begingroup\$ Thanks @Misunderstood. The LED is actually IR, here's the link. Thank you for pointing me to the reference design also - not sure why I didn't find it before. Clearly having only one LED per driver hurts me quite a bit, and the Vin range makes it even worse. My space constraints (8x8mm) limit me somewhat, but I'll get a few inductors to try out. I'm wondering if a different driver IC might perform better. \$\endgroup\$
    – AngeloQ
    Feb 18, 2017 at 23:29
  • 1
    \$\begingroup\$ I was not thinking IR, that makes sense. I do horticulture grow lights, so I only go as as high as Far Red. The voltage being that low will hurt even more. Can't you run all 3 in series? 3 would look to the circuit like 1 or 2 blue LEDs. Notice the efficiency chart the voltage are all 3.55+. \$\endgroup\$ Feb 18, 2017 at 23:35
  • \$\begingroup\$ Yes, originally the plan was to wire them in series, but 'they' decided they wanted parallel instead. Maybe I'll try again to convince them to go series. \$\endgroup\$
    – AngeloQ
    Feb 19, 2017 at 2:57
  • \$\begingroup\$ Thanks for the extra information. I fought for series, but I will ask again. I agree 100% with you that it makes so much more sense, especially now that I can show it would increase battery life. Interesting information about the emitters also. Those were already spec'd when I started on the project, but maybe we could look for alternatives for that also. \$\endgroup\$
    – AngeloQ
    Feb 19, 2017 at 15:24
  • \$\begingroup\$ Just added 3 paragraphs to the top of my answer. \$\endgroup\$ Feb 19, 2017 at 16:24

This led driver is capable of low losses .The internal switching scheme is good in this respect .The internal N chan fet should be cool.The buck coil does have significant ripple current .This means that the magnetic field strength is moving up and down with the switching frequency .Converters of this family benefit greatly from a high Q inducter .Remember the relationship between Q and the ratio of energy stored to energy dissipated .Get the best coil that money can buy .There are still some switching losses so using more inductance will reduce the switching frequency and reduce losses .This means a bigger coil .The coil current has lots of ripple so on your diagram so does the led .The led does have dynamic resistance .You can ballpark the dynamic resistance of your led from the led data .Place a low esr cap like say a ceramic across the led .This cap will mean that the led gets smooth DC so its resistance will be less lossey.The cap should have a impedance that is much lower than the led dynamic resistance at the switching frequency .When I did this on a S trap buck I nailed things with a 10 microfarad 50V ceramic smd cap .I was running a 6 led white string .Your cap will be different .Also look at the input cap .All buck converters take current in lumpy pulses .Low Esr will help you here .The freewheel diode is a significant loss contributer here because the output volts are so low .Shottkey diodes with higher current ratings and lower voltage ratings will waste less volts .

  • \$\begingroup\$ Thanks for your comments and suggestions. I'll try a cap across the LED, and also see if I can find a better diode. \$\endgroup\$
    – AngeloQ
    Feb 18, 2017 at 23:36
  • \$\begingroup\$ Good catch. I did not notice there was no Output Cap. \$\endgroup\$ Feb 19, 2017 at 0:12

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