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I have STM32(F103VBT6) and trying to use RTC with battery to keep correct time, but after removing power from DC supply supercap ( Seiko ) loses power almost instantly.

How should going to battery be handled so I can keep correct time after getting power supply.

Diode: https://www.onsemi.com/pub/Collateral/MBR120VLSFT1-D.PDF

STM32: https://www.st.com/resource/en/datasheet/stm32f103vb.pdf

Tnx for helping!

enter image description here

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    \$\begingroup\$ Can you please provide links to the datasheets for the exact processor you are using and for the diodes? You do know that Schottky diodes are known for high leakage current, don't you? \$\endgroup\$
    – Elliot Alderson
    Commented Apr 9, 2020 at 13:53
  • \$\begingroup\$ Added datasheets. Thank you for the hint, but didn't took that into account if that is possibility ? \$\endgroup\$
    – user505160
    Commented Apr 9, 2020 at 14:00
  • \$\begingroup\$ I hope you are not just resetting the RTC when the MCU starts again. You haven't posted any code. What is the voltage on Vbat before you shut the main supply? \$\endgroup\$
    – Justme
    Commented Apr 9, 2020 at 14:55
  • \$\begingroup\$ Nope, but thank you for the hint. It seems that Schottky was the trouble. \$\endgroup\$
    – user505160
    Commented Apr 10, 2020 at 8:06

1 Answer 1

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The reverse leakage current of your Schottky diodes is much, much greater than the current required by the RTC in the STM32. The worst case current required by the STM32 is \$2.2\,\mu\text{A}\$, but the worst case reverse leakage current of the diodes is specified to be as much as \$15000\,\mu\text{A}\$ at full reverse voltage and high temperature. Even if we look at Figure 4 for the behavior with a reverse voltage of \$3.3\,\text{V}\$ the leakage current looks like about \$50\,\mu\text{A}\$.

There is a reason why that pin on the processor is called \$V_{BAT}\$ and not \$V_{CAP}\$.

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