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Input impedance of GPIO vs ADC pin

Kenny Tay
Intellectual 970 points
Other Parts Discussed in Thread: CC430F5137

Hello folks,

I added a current sense resistor of 0.1 ohm between P2.4 on my CC430F5137 and VCC. Unfortunately, this resistor seems to be drawing a ton of current (~2mA). Looking at some similar threads, I saw that the input impedance of a GPIO pin is 2kOhm; is this correct? Where is this information on the datasheet?

Also, from some other threads, I saw that a pin configured as ADC has an input impedance of about 300kOhm. Is this true? Where is it in the datasheet? If so, does the following code set P2.4 as an ADC?

  ADC12CTL0 = ADC12ON;
  ADC12MCTL0 |= ADC12INCH_4;
  P2SEL |= BIT4;
  P2DIR &= ~BIT4;
  P2OUT &= ~BIT4;

  • 0
    Guru 23000 points

    Kenny Tay1 said:
    a ton of current (~2mA).

    Actually 2 ton (2,000uA).

    The User’s Guide and the Datasheet provides all of the information and parameters to calculate the input impedance in your particular situation.

    Kenny Tay1 said:
    does the following code set P2.4 as an ADC?

    No! Your device has Port Mapping availability, you need also to enable it for Analogue function.

  • 0 in reply to Leo Bosch
    Intellectual 970 points

    Thanks for the response. So, for the CC430F5137, is the input impedance for digital input equal to the pull down/pull up resistance on page 46 of http://www.ti.com/lit/ds/symlink/cc430f5137.pdf?

    Setting port mapping to PM_ANALOG worked for me. I was thinking about that but it was weird that the sample code didn't do that. I guess the sample code isn't written specifically for the CC430F5137.

    Anyway, thanks for the help!

  • 0
    Guru 227245 points

    An MSP digital I/O pin, when configured as high-impedance input (the default), has a maximum leakage current of 50pA. As long as the applied input voltage is between VCC and GND.

    When sampling analog values on a pin, there is an inrush current into the sampling capacitor during the sampling phase of the ADC. Due to the capacitor size (few pF), the multiplexer/switch series resistance (~1k) and the maximum sampling frequency (4 MODOSC cycles sampling, 13 cycles conversion), this results in an effective input impedance (average) of ~300-400k. However, this is a simplification of the real input behavior: a cyclic capacitor charge operation.

    However, if the ADC is off or not actually sampling the input channel, the switch is open and no additional load happens on the pin.

    Edit:

    If you enable the internal pull--up/down resistor, then of course, the input impedance is as low as these (~25-50k). But then, when pull-ups are used, input impedance is nothing to talk about anymore.

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