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I am using a PCA85073A RTC in my design. On some boards, the supercapacitor discharges very fast (5V to 1.1V in half an hour.)
This happens when we turn off 5V_uC.

The circuit is given below.

enter image description here

All the signals on the right hand side are going to a microcontroller. On the I2C line one more device is connected. A 100K pullup is connected for RTC_INTP, RTC_SCl, RTC_SDA. Below is the block diagram of right hand side signals. All the signals are going to the microcontroller.
The FET RESET signal comes from the microcontroller.

enter image description here

When 5V_uC is off, the RTC runs the from supercapacitor.
The microcontroller is also off at that time. The clock output of the RTC will be in enabled state.

I am seeing only two paths for the super capacitor to discharge. One is via the MOSFET and the other via C260 (if C260 is shorted.)

Will the discharge happen via any other paths?

---- UPDATE There is a mistake in this question:

I discussed it with the system team and got a complete picture; here's the improved new question.

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    \$\begingroup\$ Maybe reverse leakage through D64? Perhaps try removing that component on one of the faulty boards and see if the cap holds up any better \$\endgroup\$
    – Frog
    Commented Apr 17 at 5:21
  • \$\begingroup\$ There clearly is a path. What you have measured about the circuit? Also there is no info how exactly those signals on the right connect to other systems, and if these systems are powered or unpowered. And there is no explanation for the reset signal that controls the discharge FET. What is the purpose of the clock signal, how is it used and is it ever turned on or off? \$\endgroup\$
    – Justme
    Commented Apr 17 at 5:30
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    \$\begingroup\$ Is the CLKOUT enabled? Are there signals being output on the I2C lines? These could result in significantly higher current draw from the IC, depending on what is on the other end of the lines (per p.38 of the datasheet). \$\endgroup\$
    – Tony
    Commented Apr 17 at 5:38
  • \$\begingroup\$ @Justme,I updated the question \$\endgroup\$
    – Confused
    Commented Apr 17 at 6:18
  • \$\begingroup\$ @Tony,When 5V_uC is off the RTC will be working from supercapacitor and the clock out will be enabled \$\endgroup\$
    – Confused
    Commented Apr 17 at 6:20

1 Answer 1

Reset to default
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This explains the issue.

When 5V_uC is off the RTC will be working from supercapacitor and the microcontroller is also off at that time.The clock output of the RTC will be in enabled state.

You can't feed logic signals or voltages into a MCU that is unpowered.

The MCU input protection diodes try to clamp the voltage that the RTC clock pin outputs to MCU, and current flows through the MCU protection diode to output 5VMCU node.

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    \$\begingroup\$ leaping ahead to a solution, or at least a test of the diagnosis, a diode in the RTC_CLK line to the MCU, and a new pullup to MCU_5V, would allow the RTC_CLK_IN to be pulled low through the diode, and be pulled high when the MCU is powered, and would isolate the MCU load when it's unpowered. \$\endgroup\$
    – Neil_UK
    Commented Apr 17 at 9:09
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    \$\begingroup\$ @Confused Which signals you mean? Why would you break all instead of the one that causes the problem? \$\endgroup\$
    – Justme
    Commented May 29 at 4:59
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    \$\begingroup\$ @Confused I quoted the cause of the problem in my answer and it's being discussed in these comments. The only signal that's active when MCU is powered off is the square wave clock. \$\endgroup\$
    – Justme
    Commented May 29 at 5:11
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    \$\begingroup\$ @Confused It depends. It simply moves the problem of current leaking through IO protection of unpowered MCU to a problem of current leaking through IO protection of unpowered analog mux. Unless your mux can handle signals on inputs while unpowered. \$\endgroup\$
    – Justme
    Commented Jun 7 at 8:52
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    \$\begingroup\$ @Confused why would you even consider a relay? Simply used the mux correctly or use a digital chip that can tolerate signals on input while unpowered. \$\endgroup\$
    – Justme
    Commented Jun 7 at 9:08

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