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MSP430G2332: ESD diode current when MSP is not powered

Oba
Genius 16255 points
Part Number: MSP430G2332

Hello,

 

I have a question for internal ESD diode current of MSP430G2332.

 

I found some similar posts on E2E and read SLAA689 application note, but I don’t have confidence whether the below case is safe or not.

 

The below is the case what I want to confirm.

 

A sensor output is maximum 7V. This sensor output is connected to AD input of MSP430G2322 through 1MOhm(High)/820KOhm(low) resistor divider.

This sensor is always powered even if MSP is NOT powered. It means a certain voltage could be input to AD input of MSP although MSP is not powered.

But due to the resistor divider, the current is less than 7uA. It is less than internal diode current limit. Is this case safe?

 

Regards,

Oba

  • 0
    Genius 4350 points
    Your fellow TI guys will give you the official answer, but even if the diode will handle the current, it would be interesting to see if that current might charge up the capacitor on Vcc enough for the G2332 to turn on, or try to, or perhaps cycle on and off. A while back a friend of mine flashed new firmware to a G2231 in-circuit via a JTAG header. The flash was successful, and then he noticed that the G2231 had been unpowered during the flash operation. It turned out that current was flowing into /RESET, through the ESD diode to Vcc, and that was providing enough power there to power erasure and flashing, which is a bunch.
  • 0 in reply to George Hug
    Expert 1335 points

    Goerge, this happens to me so often too.

    As MSP430 are extremely low-power you should be extremely careful in your design, because any IO pin that might have some voltage above minimum Vcc can possibly drive the whole MCU through the clamping diodes. EEVblog has an excellent video about it.

  • 0
    Guru 274520 points

    The internal diode current limit tells you when the diodes will burn out. 7 µA is below this limit, so the chip will not be damaged.

    However, there is no guarantee that the chip will work correctly.
    If the MSP tries to draw more than 7 µA, the additional voltage drop over the 1 MΩ resistor will activate BOR, but if there is a capacitor on Vcc, it might have anough charge to prevent the voltage from dropping too much, and to allow the MSP to run with an average current of at most 7 µA.

  • 0 in reply to Clemens Ladisch
    TI__Genius 13215 points
    Hi Clemens,

    I would agree you opinion. And I will also check with the MSP430 design team to confirm this.

    Thanks
  • 0
    TI__Genius 13215 points
    Hi Oba,

    I would tell you that you shouldn't tie a voltage more than –0.3 V to VCC + 0.3 V on any of the Pin, which is described in the P17 of the device datasheet.
    When the MCU is not powered or just powered up before the Pin is configured, the Pin state is uncertain so that the voltage level on the AD pin may be over the max voltage applied for Pins.
    On other side, I would agree with Clemens that the MCU would active unexpected BOR with the 7V tied to the AD Pin even the MCU is not powered.

    I would suggest you using the resistance voltage divider to divide the voltage to meet the datasheet Pin max in.

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