Post by thread:What to do with PINs not used

Why does this come up every 6 months or so??? NO!!! Don't tie them to anything! Define them as OUTPUTS, and set them to default as low. Why? Because, these are I/O pins. Unless its a dedicated input pin, don't tie it because at somepoint two things might happen. First, an unprogrammed chip might be put into the circuit, and I don't recall if by default the pins are input or outputs. If tied either way, they could might drive Vcc to Vss, or the other way around. Second, if you need to use that pin for something else down the road, you have to then isolate it first and wire it up. Most programmable logic devices define unused pins as no connects, as they tristate or isolate the pins when they burn them. Just my .02 worth. Everyone has their own ideas on what to do....

Harrison, What do I do with the statement in the Microchip data sheet that says "with all I/O pins in hi-impedance state and tied to Vdd or Vss"?? David > Why does this come up every 6 months or so??? > > NO!!! Don't tie them to anything! Define them as OUTPUTS, and set them to > default as low. > > Why? Because, these are I/O pins. Unless its a dedicated input pin, don't > tie it because at somepoint two things might happen. First, an unprogrammed {Quote hidden}> chip might be put into the circuit, and I don't recall if by default the > pins are input or outputs. If tied either way, they could might drive Vcc > to Vss, or the other way around. > > Second, if you need to use that pin for something else down the road, you > have to then isolate it first and wire it up. > > Most programmable logic devices define unused pins as no connects, as they > tristate or isolate the pins when they burn them. > > Just my .02 worth. Everyone has their own ideas on what to do.... >

PIC12C5XX, page 80, note 5. I am working on a low power battery project, and it so happens this topic has peaked my interest.....so maybe, the "application" is what dictates how the I/O and pins are configured. Could this be the confusion? > I think its taken out of context....when they say that, its for testing > conditions I believe. For worst case power consumption's and such, or to > measure prop delays, they must state those conditions such that a designer > could duplicate it. > > What page of the data sheet did you read that on? > > {Original Message removed}

The safest thing to do is to tie all unused pins to either ground or the positiv e supply voltage using resistors, and leave them tri-stated. Anything from about 1K to about 10K will work fine. This way, if the pins get set to an unintended state (due to software bugs or electrical noise or whatever), the chip will not be damaged because the current will be limited by t he resistor. You could get rid of the resistors by just leaving the pins unconnected and set as outputs, but this leaves the possibility of the pins floating while the chip is initializing, or in a fault situation in which the pins are tristated unintentionally. The important thing is to avoid having a tri-stated (input) pin floating, and to avoid having an output pin driven against an external connection so that it exceeds its rated source/sink capability (or the combined source/sink ca pability of the chip). Pull up or pull down resistors eliminate both possibilities in all conditions. The worst than can happen is accidentally driv ing an output against a pull up/down and thus wasting power. > What should I do with the pins on the PIC that I'm not using? Should > I leave them open ro tie them to ground? > > Dave > --- Peace, William Kitchen spam_OUTbillTakeThisOuTiglobal.net The future is ours to create.

>From: "William J. Kitchen" <.....billKILLspam@spam@iglobal.net> > The safest thing to do is to tie all unused pins to either ground or the posit ive supply voltage using resistors William is right. That's the safest thing to do. In addition to the safety against misprogrammed TRIS bits, there are are two other advantages to setting them as inputs and tieing them low. First, at reset, the pin WILL be an input until you program it otherwise. I know, setting up TRIS is the first thing you do in your program, but how long is the chip in reset from Vdd on until the program runs? Power up timer, etc, adds up. During that time, the pin condition is undefined. If something (stray capacitance, dirty PCB, "cosmic rays", etc...) tends to pull the pin into the invalid voltage range, the input buffer will draw excessive current for some time. If you reset the chip a lot in your application, and or its a battery powered system, this could add up to shorter battery life. Plus, its icky :). Second, the current consumption for an enabled output driver even with no load, may be slightly higher than an input tied low. (Due to leakage in the large driver FETs, I think.) Emphasis on *slightly*. I'm not sure the difference is significant at room temperature, but might be more so across the temperature range. The Microchip specifications for shutdown power consumption are for the case where all I/Os are set as inputs and tied to Vdd or Vss. This gives the lowest possible quiescent current. For experimental boards, I leave unused pins unconnected and program then as outputs, set low, so its easy to prototype new hair- brained circuits. But in production boards, I set them as inputs and tie them low (via resistor packs) if possible. If board space or cost are so critical you can't afford the resistors, then leave them as outputs set low, subject to the (probably slight) disadvantages. ------------ Barry King, KA1NLH Engineering Manager NRG Systems "Measuring the Wind's Energy" Hinesburg, Vermont, USA http://www.nrgsystems.com ------- End of forwarded message -------

William J. Kitchen wrote: > You could get rid of the resistors by just leaving the pins > unconnected and set as outputs, but this leaves the possibility of the > pins floating while the chip is initializing, This is the prototypical example of muddled thinking. So the pins float for a millisecond during initialization. So what? We are talking about currents of perhaps hundreds of microamps (per input) *at most* here, if it really mattered more, then *all transitions* would be prohibited! That is absurd. The situation is: An input in the "switching" voltage range *whether "floating" or biassed deliberately by the application circuit* causes some conduction of both input FETs, resulting in a current from rail to rail (temporary class A operation - class A amplification always has a quiescent current). The chip is designed to tolerate this current, but it represents wasted current in battery-powered applications. There has been a suggestion that inputs "floating" in this range may cause faulty operation. Think carefully! If this were so, then *none* of those nifty time-constant circuits to measure analog voltages or capacitance would be possible. There is *one* caveat to this - it *may* cause problems if you are using the ADC. That's all. > or in a fault situation in which the pins are tristated > unintentionally. Oh dear, what a nasty possibility! *However*, nowhere near as nasty as the equally probable exact opposite - a pin accidentally turned into an output whilst tied directly to rail. Even a 10 kilohm resistor will I suspect, draw more "leakage" current than an input held at ¸Vcc. In summary, leave them open, initialize them as outputs, use shadow registers, rest easy. If the TRIS registers subsequently get scrambled despite all this, well, tough luck, just be glad they weren't tied to a rail. Here endeth the FAQ. -- Cheers, Paul B.

hahaha. How about EXPUNGE PUKELIST or better yet EXPECTORATE PUKELIST :-) Sean At 07:52 AM 9/24/99 +1000, you wrote: >Now *that's* an option I hadn't considered! >-- > Cheers, > Paul B.X-POP3-Rcpt: paulb@ecpport2 [SNIP] > >REMOVE > > | | Sean Breheny | Amateur Radio Callsign: KA3YXM | Electrical Engineering Student \--------------=---------------- Save lives, please look at http://www.all.org Personal page: http://www.people.cornell.edu/pages/shb7 shb7KILLspamcornell.edu ICQ #: 3329174