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2005/09/15 08:40:30 (permalink)
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Unused Pins

What is the recommended practice for unused pins in a design? I would suspect that the most robust design would be to ground pins which are configured as, or default to, inputs, through say a 1k resistor. However, I would also like to bring all the unused pins to a ribbon cable connector, serving as an I/O expansion port which could be used on future projects.

Any recommendations?

Thanks. I'm currently laying out the board for a 18F4580.
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    RE: Unused Pins 2005/09/15 08:58:15 (permalink)
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    Unused pins can be declared as inputs if they see a low-impedance valid digital voltage. This can be accomplished by use of a pull-down or pull-up resistor. Direct connection to VDD/VSS in a development or not still mature code is not recommended, since glitches in the code can program the pins as outputs and high current draw could result.

    Unused pins should not be left in the analog input default configuration. Unused analog inputs always couple noise to the analog mux front-end.

    Unconnected digital inputs can saturate the input inverters and cause high-current flows that cause chip overheat, burnout, or i/o port function disruption.

    Unconnected analog inputs can trigger input MUX circuits ON and badly distort signals from other analog inputs, even though the unused input is not selected.

    If your code is stable, and you don't want to use pulling resistors, the best practice is to configure unused pins as outputs and leave them unconnected.
    < Message edited by j_doin -- Sep. 15, 2005 12:59:18 PM >
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    RE: Unused Pins 2005/09/15 10:07:14 (permalink)
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    An excellent summary - thanks, J_DOIN!
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    RE: Unused Pins 2005/09/15 10:14:42 (permalink)
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    In addition, since you are designing your circuit board, you may like to check some very good Analog Devices articles about PCB layout at this other thread.
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    RE: Unused Pins 2005/09/15 10:54:18 (permalink)
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    ORIGINAL: j_doin
    If your code is stable, and you don't want to use pulling resistors, the best practice is to configure unused pins as outputs and leave them unconnected.


    One question, would that not cause power consumption to increase. Even though there is no load on the pin, turning on the drivers for the output gates wouldstill consume some power. IIRC correct I believe the reference manual suggests leaving pins as inputs to reduce power consumption.

    And thus if you tie the unused pin to Vss, you just have to be very careful when you set the data direction register for that port. To mitigate some of the risk you can also set the output state to low, so it will take a double failure to cause processor damage, no?

    Anyway, just a thought.


    cheers,
    Tom
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    RE: Unused Pins 2005/09/15 12:59:08 (permalink)
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    As I said, direct connection is NOT recommended, unless you have a thorough tested hardware/firmware. Setting the PORT bit value to the power rail to which it is connected is a good precaution.
    ORIGINAL: temberson

    ORIGINAL: j_doin
    If your code is stable, and you don't want to use pulling resistors, the best practice is to configure unused pins as outputs and leave them unconnected.


    One question, would that not cause power consumption to increase. Even though there is no load on the pin, turning on the drivers for the output gates wouldstill consume some power. IIRC correct I believe the reference manual suggests leaving pins as inputs to reduce power consumption.
    Actually, an input which is connected to either VDD or VSS can have more current drawn due to board parasitics leakage than an output which is not switching with no current path at all. The output transistor channel is just like a low-value resistor to VDD/VSS. The comparison must be the leakage through the external input pin impedances against the internal leakage of the output amplifiers. Maybe if the output port is in a checkerboard pattern the output amps can have an increased leakage. I really have never measured this scenario, but from previous experiences with supercap leakage fighting in RTC chips I learned that a pin connected is a pin leaking current. Today when I have unused output pins in ultralow current systems I leave the pins without a soldering pad underneath, and with solder resistor mask tented.

    But I can always be plain wrong.
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    RE: Unused Pins 2005/09/15 20:42:35 (permalink)
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    ORIGINAL: j_doin

    As I said, direct connection is NOT recommended, unless you have a thorough tested hardware/firmware.
    . . . And with thoroughly tested hardware/firmware the manufacturing test engineers should give you a thorough cussin' if you tie pins directly to Vdd or Vss: this makes the node essentially untestable.

    Direct connections to Vdd or Vss also force a PWB re-layout in the event that the pin is ever needed for an engineering change: a pullup/pulldown lets you simply tack-solder "white wire" to the pads for signal routing, while a direct connection forces you to cut a trace. Which may be on an internal layer. Or tied to a copper-pour plane. (I suspect this is why you emphasized "thorough tested hardware/firmware".)

    This subject was also discussed in prior threads " What's the best way to buffer PIC inputs?" and " maximum value for pull-up resistor?".
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    RE: Unused Pins 2005/09/16 03:58:39 (permalink)
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    Let's re-phrase:

    "Direct connection is NOT recommended unless you have a mature product that undergone a 100K to 1M turnaround in production AND is moved to the ROM-mask bulk purchase to improve on costs, AND need pullup/pulldown removed for cost reasons, AND have a long AND stable record of ATE stats for the current firmware WHICH will NOT be modified in ANY way when going to ROM mask production AND shows NO low-impedance pin-config glitches."

    Meaning, almost never. (We analog guys don't like digital answers. Let's say the recommendation to use direct rail connection approaches the YES asymptotically over time).
    < Message edited by j_doin -- Sep. 16, 2005 8:00:29 AM >
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