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MSP430FR6989: Interface 5V input on interrupt pin

Punit Chauhan
Prodigy 230 points

Part Number: MSP430FR6989

I am working on a low power application using MSP430FR6989.The uC is working on a CR2032 battery[it is always in ON mode] and several other peripherals[few op-amps,LED etc]  are working from a 5V auxilliary supply that is generated from a onboard 9v Battery.To indicate to the uC that the 5V Power is available i am using a Resistor divider network as shown below[Please ignore my drawing skills!]

To uC

     |          

5V^^^^^^^--------^^^^^^^^Gnd

1.5M               1M

The problem is that when the 5V supply is not available, the current drawn from the battery is close to ~7mA.if the 5v Supply is available then the current reduces to about 500uA.The 5V is no where directly connected to uC apart from this.I am giving this to Pin P2.4 configured as Input without pull ups and in interrupt mode.

What could be the reason for this??

  • 0
    Guru 74080 points

    Hi Punit!

    These are very high values for the two resistors. If the 5V supply is not connected, the 1M resistor to GND might have the same effect as if the pin would float, picking up everything it gets from the surrounding. This could lead to permanently switching the input high and low which causes a current draw. The worst case happens when the input voltage is in between Vcc and GND, reaching the threshold levels of both input transistors. Then the upper one is open and the lower one is open as well, providing a low impedance path from Vcc to GND.

    Here is a document providing some information about this issue.

    You should lower resistor values to something in the 10k or 100k range.

    Dennis

  • 0
    Guru 320930 points
    To confirm what Dennis said, could you measure the divider output with the ADC?

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