Achieving low power with Seeeduino Cortex-M0+ powered from 12V battery

Question

I have Seeeduino Cortex-M0+, which is similar to Arduino Zero, both using Atmel SAMD21 MCU. It is powered from 12V battery directly to 12V PWRIN jack. Current draw for LED blink test is 18mA. I tried two low power libraries:

• https://github.com/rocketscream/Low-Power
• https://github.com/adafruit/Adafruit_SleepyDog

with:

• LowPower.standby(); delay(5000);
• Watchdog.sleep(5000);

In both cases, the current draw dropped only to 12.7mA, which is still huge. I checked the datasheet of MP1496DJ-LF-2 step-down, switch-mode converter used by this board and its quiescent current is below 1mA, so it doesn't look like the source of high current consumption. I've read posts about Arduino Zero running on 5mA from 12V input in low power. What else can I do in order to get the current closer to 5mA?

I cross-posted at https://electronics.stackexchange.com/questions/607000/achieving-low-power-with-seeeduino-cortex-m0-powered-from-12v-battery

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I did a few experiments with external DC/DC converters feeding 5V pin on Seeeduino Cortex-M0+ from 12V battery, bypassing its MP1496DJ-LF-2 power stage. Here are the results:

DC/DC model	sleep current [mA]	LED blink current [mA]
MP1496DJ built-in baseline	12.7	18.0
LM2596S generic	8.6	14.8
Pololu D24V22F5	5.0	11.2
Pololu D24V5F5	3.6	9.3

Answer 1

As @devnull at Electronics Stack has answered, the quiescent current of 5~10 mA from the LDO could be enough to make up for the lost currents.

I don't have any specific suggestions on code or parts, but I can give you some tips regarding methodology. If you're working on the software side to reduce current consumption, you might be better off eliminating the possibility of hardware imperfections entirely. I'd power the board with 3.3V directly and preferably cut the traces to 5V or other unused circuits. Reduce the current consumption to the lowest possible on software side, and then work on the hardware side. You could be better off with an external LDO or a 12V to 3.3V converter with a low quiescent current to power your board.

From my experience, currents can leak from even the most unexpected circuits. There could be a faulty or crappy capacitor with higher than usual leakage current, resistors in voltage dividers with resistance not high enough drawing a couple of mA in total, or some random transistors we didn't think of drinking 1~ mA per chip. Or combinations of everything above being responsible for only a few hundred uA per part, but totals to a few mA in total. It's best to make a board with the most barebone parts or get one.