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MSP430FG4618 Port1 hex values different every time I plug the board into my computer

Seth Rhodes
Prodigy 50 points

Greetings,

  I am a student in Electrical Engineering at University of Central Florida and I am currently studying Embedded Systems.  For my lab, we are required to run through a series of prewritten example codes in both C and assembly.  I am understanding (at least I hope I am) the basics of the assembly instruction set.  My problem is this: every time I come back to my code after a day or so hiatus, the hex values that Port1 reads out are different. Sometimes, for the switches both being open (not pressed), the P1IN value in the register window reads 0x03, yet other times it will be 0x0B or even 0x8something.   One day my code will work, but the next day, the exact same code will no longer work until I check the register window for the new hex values and change my code accordingly. I'm not sure what is causing these values to change every time I come back to my code. I have asked my professor and he has no idea what I am talking about. Neither does anybody in my class. Any ideas why it is doing this?

  • 0
    Guru 74080 points

    Seth Rhodes said:
    I have asked my professor and he has no idea what I am talking about. Neither does anybody in my class.

    And neither do I :)

    Could you upload the program you are talking about and explain at which location in the code the problem occurs?

    Dennis

  • 0
    Guru 297210 points

    Read SCBA004: Implications of Slow or Floating CMOS Inputs.

    You probably forgot a pull-up or -down resistor on your inputs, or have configured unused pins on the port as inputs.

  • 0 in reply to Clemens Ladisch
    Guru 74080 points
    Ah, OK, now after Clemens' post I can imagine the problem. But if the pull up / down resistors were missing, how can your program react on the inputs at all? If you have them enabled on the respective pins and read the complete IN port with any unused inputs, then the program simply needs to be changed to mask only the interesting input pins.
  • 0
    Intellectual 785 points

    I am going to guess that your two switches are connected to bits 0 and 1, and that the remaining bits of the port are unconnected. I am also going to guess that you have configured all bits of the port as inputs and that the unused ones are thus floating.

    When you read the port, bits 0 and 1 reflect the value of you switches (always 1 when the switch is open and 0 when it is pressed if the switch closes to ground), and the remaining bits are random because the inputs are floating.

    To determine the state of the switch, you need to just look at the bits in question and ignore the other bits, rather than do a comparison on the entire value read from the port.

    if (~P1IN & 0x01) {
      // the first switch is pressed
    }
    if (~P1IN & 0x02) {
      // the second switch is pressed
    }

  • 0 in reply to Matthew Kendall
    Intellectual 785 points

    I hate this craptastic editor.

    if (~P1IN & 0x01) {
  // the first switch is pressed
}
if (~P1IN & 0x02) {
  // the second switch is pressed
}

  • 0 in reply to Dennis Eichmann
    Prodigy 50 points
    I'm not really sure where the problem would be in my code. As far as I know, the only interactions with port 1 that I have are P1DIR and P1IN. If I set P1DIR to 0x00, that is setting bits 0 and 1 to inputs, right?
  • 0
    Intellectual 605 points
    Do you happen to have an example screen-shot?

    This may sound stupid, but you are using a consistent baud-rate, yeah?
  • 0 in reply to Seth Rhodes
    Intellectual 785 points

    It sets all the bits to inputs; 0, 1 and the other six that are not connected to your switches.

    Your original complaint was that sometimes you read 0x03 and other times, with the switches in the same state, you read 0x0B. Those two values have exactly the same values for bits 0 and 1, it is the other bits that differ: 0x03 = 00000011, 0x0B = 00001011. Bits 0 and 1 (the least-significant bits, on the right-hand end of the binary representation) are 1 in both cases. It is other bits that are different.

    My guess is that you have not wired up the other six bits, and that they are thus floating, and that you therefore read random values from them

    The correct approach is that, when you wish to test the value of a particular bit, use a bitwise operator. For example (P1IN & 0x01) gives the value of just bit 0, ignoring the others, and (P1IN & 0x02) does the same for bit 1.

  • 0 in reply to Matthew Kendall
    Prodigy 50 points
    Okay, that makes sense. Thank you for your help! I also wasn't quite sure if the pull up/down resistors are configured in software or if they are a hardware thing.

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