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I'm trying to use a boost converter from Pololu that steps up my 1.25 input voltage from a AAA battery to 5V in order to power an Arduino. I have found that my Arduino needs .03A to be powered.

I understand that if the boost converter were 100% efficient and when using (V*I)in = (V*I)out, I would need .12A to be drawn from my AAA battery into the boost converter. However, when measuring the current drawn directly from the battery using a multimeter, I find that 1.05A is leaving my battery and entering the boost converter, while .03A is leaving the boost converter and entering the Arduino. This implies that my boost converter is about 11% efficient. The voltage seems to be stepped up fine from 1.25V to 5V.

This efficiency is concerning since the Polulu component has an input current limit of 1.2A, and I also need to drive a small motor. I am wiring the positive and negative terminals of my AAA battery to "Vin" and "GRD" on the boost converter respectively, and then wiring the "Vout" and "GRD" pins to the "5V" and "GRD" pins respectively on my Arduino. I am not using the "SHDN" pin.

So, am I wiring the boost converter incorrectly, or is this efficiency typical?

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    \$\begingroup\$ That is concerning. Besides the inefficiency, that 1 ampere drain will slurp a AAA cell empty pretty darn fast. \$\endgroup\$
    – JRE
    Commented Jan 14, 2021 at 20:13
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    \$\begingroup\$ Measure the voltage going into module. If you have it wired through a breadboard then the contact resistance might be causing the voltage to drop. Better yet, remove the breadboard from the equation. Connect the bettery directly to the module and see if it gets better. \$\endgroup\$
    – JRE
    Commented Jan 14, 2021 at 20:31
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    \$\begingroup\$ I suggest you test the upconverter separately, connect battery to upconverter, leave the output of the upconverter unconnected (open) do you get 5 V at the output of the upconverter? Measure the current taken from the battery. If that's all OK, connect a load resistor at the output of the upconverter use 5 V / 30 mA = 170 ohm, 180 ohm or 2 x 100 ohm in series is close enough. Now almost 30 mA should be taken from the 5V. Is there still 5 V at the output of the upconverter? \$\endgroup\$ Commented Jan 14, 2021 at 20:47
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    \$\begingroup\$ Do realize that most Arduinos use a ATMega328 mcu which can run on a supply voltage of only 1.8 V but you might have to re-program the "brown out detection" setting for that to work. It depends on what the Arduino is controlling if you can actually go as low as 1.8 V, you're not showing a schematic so I cannot judge. You appear to want high efficiency and for that a lower voltage at the output of the upconverter will help. \$\endgroup\$ Commented Jan 14, 2021 at 20:51
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    \$\begingroup\$ First you say there's 50 mA at the output with a 100 Ohm load then it's 3.2 V ??? Stop measuring currents for a moment and measure voltages. Are all voltages correct? Measure the battery voltage directly at the battery's contacts but also directly at the upconverter's connectors. Are you using only soldered connections? I suspect that all your current measuring introduces a voltage drop (burden voltage) so that even if the battery is 1.5 V there could still be a much lower voltage at the input of the upconverter. \$\endgroup\$ Commented Jan 14, 2021 at 21:55

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Take a look at the curves they published, What you are getting sounds somewhat close. I know you are losing voltage in the current measurement, that is the way they work. The max with one volt in is 150mA at about 60% efficiency. I would assume you are below 1 volt which takes you completely off the curve. Consider putting two batteries in series, your efficiency will go up a lot and it might even work.

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