I have a PCB with the schematic below. I have a coin cell battery for backup when the PCB main power is removed. From my understanding, the VBAT_PMU (Pin 51) on the MCU is dedicated for the Always‑On (PD_AON) domain.
When I measure the current draw of the battery, with no main power (only the battery), I get around 102uA. Which is far more than expected for the RTC to consume. Event if D12 is removed. Vbat only is connected to the VBAT_PMU pin and nothing else.
What is the expected behavior here? I would expect the VBAT_PMU to only draw enough current for the RTC to keep time while power is down. 102uA would drain a CR1632 coin cell battery pretty quickly.
How do I power the MCU so the battery only powers the RTC and draws less than 1uA when main power is removed?
On LPC55S28, VBAT_PMU is the supply for the always-on domain , which includes the PMC, RTC, and OS Event Timer, and that domain remains powered as long as a valid VBAT supply is present. The RTC can continue running in deep power-down from that domain.
VBAT_PMU by itself does not guarantee “RTC-only current.” To get sub-µA backup current, the device must be in deep power-down mode.
So your measured ~102 µA with main power removed is not consistent with RTC-only deep-power-down operation .
For LPC55S2x/LPC552x, the datasheet shows typical total supply current of 110 µA in deep-sleep mode at 25 °C and 3.0 V, while deep power-down with RTC oscillator disabled is 590 nA , and with the RTC running from an external crystal is 790 nA .
To get less than 1 µA from the coin cell on LPC55S28, power only VBAT_PMU from the battery and make firmware enter Deep Power-Down before main power is removed.
BR
Harry
