As the LPC1765 datasheet(chapter 8.30.2) says, the LPC17xx include 2-stage monitoring of the voltage on the VDD(REG)(3V3) pins. If this voltage falls bellow 2.2V or 1.85V. The BOD asserts an interrupt signal to the Vectored Interrupt Controller.
But the MCU supply voltage range is 2.4V to 3.6V. How the BOD works? What is the sense of 2.2V and 1.85V level?
The supply voltage as you mentioned is 2.4 to 3.6V, so the voltage should ideally not go lower than 2.4V. If this voltage falls enough as to reach 2.2V then an interrupt from the Brownout Detection can warn the system as this could mean for example that the battery voltage is too low, and you may want to take some action.
For reference, the RTC module will continue working down to 2.1V, so this would be the lower safe voltage (albeit not desirable). If the voltage would keep lowering to 1.85V the Brownout Detection would assert a reset to inactivate the LPC1765 as flash operations would not be reliable at this voltage level.
More than a low battery level, these would be thresholds to where the supply voltage is critically low, with the 1.85V interrupt shutting down the microcontroller
I hope that this information helps.
That’s correct. While operation under 2.4V is outside of the operating parameters, the microcontroller won’t shutdown as soon as you go lower than 2.4V. That’s why even at 1.85V the microcontroller can still get the interruption.
However, 1.85V does immediately shut down the microcontroller to protect it. On most implementations would want to use the 2.2V interrupt to finish any critical tasks and safely shutdown before the operation of the microcontroller is compromised.
Would you mean 2.2V or 1.85V, MCU system still can get the interrupt from the Brownout Detection? If so, what about the status of clock and peripherals? I thought when the supply voltage falls down to 2.4V(2.4 to 3.6V), mcu system had been shut down completely.