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Recently I obtained this setup-regulator module (archive link) from Ali Express for a 3.3V project to be powered by two 1.5V AA alkaline batteries.

The seller had this to say about the module:

Input voltage 0.8 ~ 3.3V,output 3.3V Maximum output current: 500 MA,

Start Voltage 0.8V, Output Current 10MA

INPUT 1-1.5V, OUTPUT 3.3V 50-110MA;

INPUT 1.5-2V, OUTPUT 3.3V 110-160MA;

INPUT 2-3V, OUTPUT 3.3V 160-400MA;

INPUT above 3V, OUTPUT 3.3V 400-500MA;

DC-DC Boost module working frequency 150KHZ. efficiency is normal 85% . 2.54mm pin pitch, Arduiuo Breadboard friendly. Excluding Pin Size 11mm x 10.5mm x 7.5mm(Very small) Weight : about 1.2g( Very light)

The seller does not specify the regulator used. I can read the numbers 2108A 1515/33 off the chip.

Photo of regulator IC

Based on the wave symbol I expect it to be the ME2108A of the DC/DC Step up Converter ME2108 Series (archive). The characteristics seem to match, as do the example circuits.

My application draws about 75mA when on, and about 100uA in sleep mode, which would seem to be in-spec. However, upon connecting everything I found out it was drawing significantly more current than I was expecting, even with the device in sleep mode. To troubleshoot the issue, I made a minimal linear test circuit.

schematic

simulate this circuit – Schematic created using CircuitLab

Here M1 is the regulator module mentioned in the listing above. The diode D1 drops the voltage just a bit, so it's clearly under the desired 3.3V. The regulator should boost the voltage to 3.3V, which when passed through a 47R resistor should result in about 70 mA of current, which is similar to the draw of my project. Between the lower input voltage (I measure about 2.65V) and the regulator effciency (which is advertised as 85%), I would expect to see about 100mA going into the regulator. Instead I measure over 230mA going through R_sense.

The second graph on page 8 of the datasheet suggests that for an output current of 70mA and an input voltage of 2.65V, I should be seeing an efficiency of between 85% (@3V) and 77% (@1.5V). Instead I'm seeing about half that.

What's going on here?

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    \$\begingroup\$ Is the inductor getting warm? (200-300mW may be enough to sense with a finger, but a small thermocouple would be better). Check that the diode is actually a Schottky as well. If they used a 1N5819 rather than 17 and a crappy cheap inductor that would about explain it. As the film noir approximately said "Forget it, Jake. It's China" \$\endgroup\$
    – Spehro Pefhany
    Apr 13, 2017 at 2:44
  • \$\begingroup\$ Not really an answer, but it's very common for very cheap products ordered over the internet to have exaggerated specs. Components can be underdimensioned (popular for inductors on switching converters), counterfeit, poor quality, or omitted entirely. \$\endgroup\$
    – marcelm
    May 15, 2017 at 10:31
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    \$\begingroup\$ Another remark: I find the efficiency graph on page 8 of the datasheet suspicious in that the graph continues down all the way down to Iout=0, but its efficiency doesn't drop below 85%. As Iout tends to 0, so does efficiency for all converters I've ever seen. So either this one is magic, or its graph is inaccurate. \$\endgroup\$
    – marcelm
    May 15, 2017 at 10:33
  • \$\begingroup\$ @Marcel pwm drive gives higher efficiency at higher duty cycles, but pfm drive maintains the same efficiency across the load range with a small upturn at low load as conduction losses dominate. This is because the ratio of switching time to on time stays constant. \$\endgroup\$
    – K H
    Sep 15, 2018 at 18:12
  • \$\begingroup\$ That said it's definitely buyer beware or buy and test on Ali. \$\endgroup\$
    – K H
    Sep 15, 2018 at 18:13

4 Answers 4

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One possibility is that you don't have an input decoupling cap, so VIN to the IC may be dropping quite a bit (in an AC sense). The resulting high VIN ripple may also be causing you to get an erroneous input voltage reading.

However, this doesn't necessarily explain why you're also seeing low efficiency in your actual application circuit - unless you are also lacking input decoupling there as well, and your input source isn't close and low impedance.

Dave

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    \$\begingroup\$ In a boost converter the input current should be continuous so minimal filtering is required. \$\endgroup\$
    – Bruce Abbott
    Feb 6, 2017 at 2:04
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Actually, I missed the obvious - you NEED to have a good filter cap on the output. There isn't one shown in your test circuit. The output current of a Boost converter is discontinuous. It won't work without a proper output filter cap. See the examples in the IC's datasheet.

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If it shares the same design as the ZXLD381, it could be a design issue.

Looks like your input voltage is close to the output voltage. Here's what the ZXLD381 datasheet has to say about output voltage related to input voltage:

[snip] a feature of the pulse control circuit that requires the load voltage to be at least 0.8V greater than VCC. (The device will function with load voltages smaller than this but output current regulation will be impaired.)

BTW I am also unhappy about this same circuit, trying to build one around ZXLD381 with a larger inductance but no luck so far.

As an idea, what happens if you supply only 1.5V to the boost? If the issue is a design flaw, then the current should drop dramatically @ 1.5V VCC.

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You need 22uF capacitor on input & output. It decreases power consumption by almost 90%!

No load current test: without caps = 2.20mA, with caps = 54uA caps connected & current draw

There is another board design using the same IC (ME2108), that shows having a cap on input. enter image description here

Input actually already has caps on the board you linked, but some* said that it made a difference to add another.

**Link to the discussion on this board's power consumption.

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