In the datasheet they mention the MT3608 can handle up to 2A however in some stress tests with a load tester it gets already very hot at 1A. It is so hot, it is impossible to touch the board, all components around it get very hot because it spread all of the heat via paths. The chip itself is that tiny it is almost impossible to attach a heatsink. Because I don't have a second one I have to ask this, is this normal or abnormal?

Board (arrow point at chip and I have the one with protection diode): MT3608

Test config:

  • Battery 18650 2600mah

  • boost 4.10V to 5.6V (drops down to 4.87 when heated up)

  • Load 1A



  • 4
    \$\begingroup\$ Note that cheap electronics in general tend to list numbers from "Absolute maximum ratings" as normal working conditions, while they are clearly not meant to be interpreted in this way. \$\endgroup\$ Jun 29, 2019 at 13:26

3 Answers 3


I have the identical board. I connected it to a power supply set to 4.1V, adjusted the booster output to 5.60V and connected a 5.6Ω 10W resistor. The output stayed solid at 5.60V while the input current rose to 1.56A. After 6 minutes the chip temperature was stable at 39°C according to my infrared thermometer, 23°C above ambient. The inductor was 41°C, the diode was 42°C, and the load resistor was 112°C!

4.1V * 1.56A = 6.4W input power, so the conversion efficiency was 5.6W / 6.4W = 87.5%.

So this board should do what you are asking of it no problem. However if the input voltage drops the current will have to increase to get the required power, and efficiency may suffer. At 3.7V my unit drew 1.8A and its efficiency dropped to 84%.

At these relatively low voltages and high currents, any losses in wiring or connectors can make things worse. I used thick wires and heavy duty connectors, and measured voltages directly at the board terminals. I also put 22uF capacitors across the input and output to help reduce voltage ripple.

Before putting the blame on your board you should consider two things:-

  1. Boosting to higher voltage requires more power for the same output current, and therefore higher input current. Higher current causes greater voltage drop in wiring and connectors (as well as inside the booster itself) making its job harder.

  2. When input voltage drops the chip draws more current to compensate, but this could cause the input voltage to drop even more. If the power source is too weak to deliver the required power then the booster will continue trying to draw more current until maxed out.

18650 Li-ion cells have typical 'knee' voltage at end of discharge of ~3.4V (it might start at 4.1V, but it won't stay there for long). To get maximum usable capacity out of your battery the booster must be able to work down to this voltage, and you need to keep input losses down with short wiring and low-loss connections.

The MT3608 datasheet only shows output current up to 1A at 5V, and you may need a similar boost ratio as the 3V to 5V curve shown (where efficiency is heading downhill fast at 1A). Therefore, despite being rated for 2A, under these conditions I think 1A is a more realistic maximum.

  • \$\begingroup\$ Hi! Overwelmed by all answers, thank you for your nice detailed answer. Thanks for trying, simulate the same situation. Your results are much better, do I have a faulty one? The 22uF caps sounds like a good idea. I will try the test again. Also test the voltage drop on the input. \$\endgroup\$
    – Codebeat
    Jun 29, 2019 at 12:01

Yes, this is a really tiny power supply, yet very efficient if used as intended.

It's the experience you gain from these failures about thermal resistance, power calculations, test methods with accurate measurements and thermal sensing, that's far worth more than the cost of this inexpensive board.

Next time consider a thermometer and record hot spots and use a voltage controlled power transistor on a heatsink to control load current.

The datasheet for this IC design it gives examples for only step-up to 12V and 18V for a reason.

  • The lower the input voltage, the lower the efficiency
    • It drops from 90% to 80% as Vin drops to 3V.
  • the higher the step-up voltage before limits, the higher the efficiency.
  • expect a 200'C rise per 1W loss on this tiny chip unless there is a huge 4x4cm copper heat spreader
  • \$\begingroup\$ Hi! Overwelmed by all answers, thank you for your nice detailed answer. 200'C, wow. So this is not faulty one. Why do they even design this that small if it is producing so many heat and not capable to handle all of the heat. What is the lifespan of such product? Days, weeks? Any suggestion for a better solution? \$\endgroup\$
    – Codebeat
    Jun 29, 2019 at 11:49
  • \$\begingroup\$ If it gets that hot , it won’t last long. See specs for max operating temp typ 85C The other Answer is also correct probably it only can handle 500mA output with 3 to 3.7V input \$\endgroup\$ Jun 29, 2019 at 16:32

Your load is 5.6V @ 1 A = 5.6W.
If overall efficiency is 80%, that means an additional
Power lost = (20%/80%) x 5.6W = 1.4 Watts is dissipated.
That's more than an IC of that size can handle at sensible temperatures with PCB copper (if any) heatsinking plus a small amount of direct radiation and convection.

Heatsinking can be obtained by adding a blob of thermal transfer compound and positioning a metal heatsink against the IC and compound.
For better insulation (and somewhat worse heat transfer) use an insulating washer as sold for use with larger ICs. This could be thermally conductive rubber or mica or ... .
The thermal compound needs to be in good contact with the IC **and* spread out around it to allow good contact over a reasonable area with the heatsink materials.

For retention you could make larger holes in or near two of the Vin and Vout pads and fasten with nonconductng screws. Or sandwich a suitably cunning metal piece and the PCB and wrap a cable tie around the whole "assembly". Prettiness may suffer but it should be doable.

  • \$\begingroup\$ Hi! Overwelmed by all answers, thank you for your nice detailed answer, tips and tricks. Any suggestions for a better config, other booster (larger IC), other board? There is not much room to change this board. \$\endgroup\$
    – Codebeat
    Jun 29, 2019 at 12:06

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