I must power a Wemos D1 mini at 5V (more on that below) (along with 4 LEDs and a DC motor) with a single cell AA battery. I have tried 2 boost converters (Vin 0.9V-5V) and none worked.

I need 5V for 2 of the LEDs, which are in series and 5V is perfect for them.

The motor will be driven with the 3.3V pin on Wemos D1 mini.

All the boost converters I tried deliver 5V unconnected. However, the moment I connect them to D1 mini, voltage drops around 1.5V. I tried with a power supply at 1.5V, same result.

I wonder, what I am missing. The specification says 600mA output current. I also tried to put a 100μF capacitor either to input or output side for sudden current draw. Supplying >3V worked fine. Dropping from 3V to 1.5V worked for a short while and then the output voltage dropped to ~1.5V. D1 mini draws ~70mA 5V.

I wonder if I am missing something?

Here are the converters I have tried:



EDIT: as a test I tried with MT3608, which requires >2V. I supplied it with 3V and got 5V out. It worked just fine. I guess all these cheap undocumented boost converters are bad.

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    \$\begingroup\$ What does the datasheet say? \$\endgroup\$
    – marcelm
    Commented Feb 20, 2021 at 16:59
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    \$\begingroup\$ All those loads boosted from a single AA cell? Seems like you're asking for too much current from a feeble battery. \$\endgroup\$
    – glen_geek
    Commented Feb 20, 2021 at 17:11
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    \$\begingroup\$ No data sheet = totally untrustworthy. Even if you could pull enough power from an AA battery to do all that. \$\endgroup\$
    – TimWescott
    Commented Feb 20, 2021 at 17:59
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    \$\begingroup\$ I actually have some experience with this. When you boost from 1.5V, well, it is never 1.5V. You are boosting up from 1.4 or so at the battery. Then any little loss in wiring or contact resistance drops that voltage even more so it is suprisingly easy to end up below 0.9V at the input to the DC-DC converter. That is the other thing that will get you. One product I worked on wouldn't run from the battery if you put a low quality ammeter in between the battery and device. Still, 70mA isn't that much. What you want may be possible, but you have to pay attention to many details all along the way. \$\endgroup\$
    – user57037
    Commented Feb 21, 2021 at 0:31
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    \$\begingroup\$ The other thing is that 6:1 is often considered the maximum boost ratio. If you want your device to function reliably down to 0.9 or 0.8V, you will be pushing the boost ratio limit. Which means you will be on the low side of the efficiency curve. It is very challenging but it can be done. You may need to design your own regulator or order an evaluation board (if you can find one) from some name brand IC vendor like linear technologies, national semiconductor, etc. \$\endgroup\$
    – user57037
    Commented Feb 21, 2021 at 0:35

2 Answers 2


What voltage is the battery when operating?

It will also be good to look at the battery voltage with a scope because the voltage may drop significantly when the switch in the converter is conducting but rise during the rest of the cycle. I find that it is essential to use a very low ESR capacitor at the input of the converter - a conventional small electrolytic is probably not adequate, try a tantalum or polymer electrolytic.

Make sure that the leads from the battery are heavy enough gauge, you can easily drop a 100mV or more across the leads.

Another source of voltage drop is the battery holder - the common ones for AA cells use steel springs on the negative end with a fairly high resistance - in one case I had a significant reduction in voltage drop by putting a short length of copper wire across the spring.

Also it can be very difficult to measure the input current at low voltages because the meter itself may drop 200-300mV which together with the voltage drop the cell will experience under load may drop the voltage below the 900mV limit and stop the device working.

  • \$\begingroup\$ For me this did the trick at least for powering up the microcontroller. Making sure that the voltage doesn't drop much until it arrives the booster. I guess, my initial test environment wasn't optimal. \$\endgroup\$ Commented Feb 21, 2021 at 23:56

I wonder, what I am missing. The specification says 600mA output current.

Well well.

These cheapo DC-DC converters always spec the biggest number they can find in the datasheet with the unit of "Amps" dangling at the end. Usually it's the big number in the title of the datasheet, which means...

Datasheets usually put in the title the biggest number they can find with the unit of "Amps" dangling at the end, which means...

It's the internal switch maximum current. Example.

So yeah, this is a "2A" boost converter which means the internal switch has a maximum current of 2A. This corresponds to the peak input current, at the peak of the inductor ripple current.

In your case, the peak switch current is 600mA, then the average inductor current will be about 450-500mA. Let's go with 450mA. So, with a 1V input from an AA cell, at this current it can take in 0.45W. Assuming 85% efficiency, there will be 0.38W at the output. With a 5V output, that's 76 mA. So at 5V output, it's a 75mA boost converter.

This is absolutely normal for a 600mA switch current limit.

The "output current 600mA" in the amazon product description is bogus, of course.


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