Leaving unused pins just sitting there "floating" in any design is a serious and common error. A floating pin can pick up surrounding RF or electrostatic signals and oscillate, which can draw a surprising amount of current, leading to unnecessary heating, power supply drain, and additional cross talk.
The safest thing to do is to tie all unused pins, using resistors, to either ground ("pulldown") or the positive supply voltage ("pullup"). Anything from about 1K to about 10K will work fine.
If they are programmable I/O pins, also leave them tri-stated.
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 the resistor.
When using pullup (or pulldown) resistors, the worst that can happen is accidentally driving an output against a pull up/down and thus wasting power. (See "power considerations" section below).
Paul B. Webster says:
You should pull down to ground [rather than to Vcc] as:
- a short is more likely to occur to ground.
- If you have to use the points to connect to an external device, you are more likely to want to reference it to ground.
Well, what else could you possibly do? Let's exhaustively list all the options: You could have an input-only pin, an output-only-pin, or a programmable I/O ("tristate") pin. And you could either connect that pin to a resistor (covered above), connect that pin directly to Vcc or ground, or leave it unconnected.
I hope this page covers all 9 possibilities ...
Bad options:
Floating is not a good idea with CMOS inputs because there are usually no inherent pull-up or pull-down resistors. The input can float to any voltage, depending upon leakages to ground and the supply. Somewhere (actually, mostly everywhere) in the middle, the CMOS inverter or gate to which the input is connected goes into an indefinite and high power dissipation mode, with both N- and P-channel outputs conducting. This causes extra power dissipation and noise inside the chip.
Tony Nixon says:
A floating input WILL cause erratic chip operation. Maybe not while you force it to, but at some stage it will. I've had chips do all sorts of wierd things, until I eliminated a floating pin (or pins). Some of these chips froze just by placing a finger near them, some won't start, some start but run 'funny'. The worst problem is spending mutitudes of time finding out 'why', when you haven't realised the floating pin is the culprit. As far as I know, the excess currents are caused by input FETS not knowing what to do and the 'hi' and 'lo' ones turning on at the same time causing minute rail to rail shorts, which could be at any frequency. Surely this causes wonderful things to happen inside the die, false triggering the brown out detect etc. - who knows?.My opinion, is that there is some credibility in using a moderate value resistor to 'babysit' unused pins. Besides, there is your pin pad for other uses if need be :-)
Potentially useful options:
But if they are programmable I/O pins, it leaves open the possibility of the pins floating while the chip is initializing, or in a fault situation in which the pins are tristated unintentionally.Accidental grounding or energizing of an output pin (via a dropped pen or paperclip, ill-formed trace, bent component lead, etc...) to the opposite rail may smoke the chip and is far more likely than any other serious error.
Many low power uC applications involve the chips power cycling, sometimes rapidly. The power loss due to rapid switching of a tri-stated input pin may be significant. Disclamer: I do not have hard data on this, but I believe I'm correct. Rapid switching will not fry the chip, but may decrease battery life when the intention was to extend it.
It also introduces the (low but real) possibility of the pin being shorted to the other supply rail and causing functional failure or damage to the chip or supply.
Many experienced engineers will use this method to (slightly) reduce the cost of a production project AFTER (or in some cases in spite of) verifying that the part
- will not be destroyed by a short,
- is not suceptable to lockup problems,
- can program the unused pins as outputs (not always possible) and
- will not operate irratically while booting to the part of the program that initiallizes the pin.
But if they are programmable I/O pins, it leaves open the possibility of a fault situation in which the pins are programmed unintentionally to be outputs, leading to the "shorted output" problems.Many processors can (sometimes accidentally) change inputs to outputs. ESD spikes, brownouts, lightning crashes, screwy things related to general susceptibility, etc...
TTL chips: Letting inputs float is not too bad with TTL inputs because they sortof have pullup resistors built in.
Some chips have "internal pull ups".
stops gate voltages rising faster than any internal supply rails for any devices that can latch-up and so prevent an occasional failure due to bond wires melting. Latch-up is a danger in chips that have parasitic thyristors as an artifact of the manufacturing process. Most modern chips do not have this problem but its best to be safe
Pull down to GND (versus pull up to VCC) consumes %15 less power on the PIC.
Pulling up to Vcc (versus pull down to GND) consumes less power on TTL chips.
David VanHorn http://dvanhorn.org/ says:
There are some things you should not do, and there are several options of what you can do.
Which technique is best is subject to some debate. Most microcontrollers, on reset, default their I/O pins to inputs, so that you can control what happens before your code starts running, by using resistors to ground or VCC as appropriate. However, leaving unused pins in the input state can cause problems, unless the pins are tied to some definite logic level. CMOS inputs are very high impedance, and left unconnected, usually float to about 1/2 VCC, where the input stage draws much more current than normal. This also results in the input pin, as seen by the micro, changing state rapidly between '1' and '0'. This may cause other problems, depending on the micro, and on which pin we're looking at.
- First, you should not leave the pins as inputs, unless you tie them to some definite logic level.
This is pretty easy to satisfy, simply connect every unused pin, through a resistor, to ground or VCC.
Why the resistor? Two reasons.One: You may want to use a pin for debugging later, so having it tied directly to ground or VCC would be inconvenient.
Two: If, for some reason, your code were to assert the pin as an output, it could draw a lot of current if the output state were opposite of the level that it's tied to.
- You could always set the unused pins as outputs. This is what I do. This way, there is no need for resistors to tie them to any given state.
For some reason, which state you should set them in seems to be a hot topic with some people. I don't really see much preference, as long as the hardware is built correctly.
The 'minus' of setting unused pins as outputs comes up if there is a wiring mistake, or soldering problem, and an output pin gets connected to something it shouldn't be connected to. However, if the hardware is built correctly, there's no problem. I see this as purely a hardware issue.- If you are going to use pullup (to VCC) or pulldown (to Ground) ressitors, then you have to pick a value that won't allow too much current to flow if the pin is set to an output. Remember, we want to be able to use those pins for diagnostics, so we need to be able to drive the pin to the opposite state.
Most micros can source/sink a couple mA on each output, so we can put a lower limit on the resistors at about 4.7k (1mA at 5V)
Making the resistors too large has problems too. At high resistances, the pins could be "pushed" by electric fields or RF energy. Experience tells me that 10k is a good upper limit, where these external influences are not a problem.
Mark Willis says:
Usually, on Reset, all I/O pins are set as inputs (Gotcha here - use a pullup or pulldown so your devices don't act in "unpleasant ways" when your chip resets, you want a pull-down if using a logic-level Power FET or an NPN power Darlington or SCR that fires on a logic high input, or a pull-up if using a PNP transistor to do something safety-critical, of course - floating pins are a BAD idea if safety's at stake. This isn't a problem for an indicator LED; if it flashes on power-up, you probably won't notice the 10 microsecond flash <G> On the other hand if you have your ejector seat fire just because of a power glitch in the ejector seat controller you're making, you'll NOT be well liked by the pilot.)
"Inputs circuits (including schmidt trigger) draw considerably more power from the supply when the voltage strays from VDD or GND. Using schmidt inverters with an RC to make an oscillator draws a surprising amount of current, because the input connected to a capacitor is always in the linear range.I once inherited a design where current consumption was intended to be extremely low (~=60uA), and draw varied by the amount of ambient light falling of the PCB! Ends up, an LED was tied to an open-drain output. When the LED was off (normal), light level hitting the LED would change the voltage across it to the OD output (which was also an input, not unusual in a uP), and take the pin through different areas of the linear region. The solution was to pull-up the pin, so it was at VDD when the LED was off (the LED didn't care)."
-- Bruce Walter
Q: Recently, I had some odd problems with our [PICs] not going into sleep mode--or so it appears. Instead of the usual 30-70 uA of current draw from the battery, I am seeing 1-2 mA. Or what is even more baffling, a drop to sleep followed by a single slow rise to about 600 uA followed by a fall back to normal sleep current.
A: Dwayne Reid (dwayner at planet.eon.net) of Trinity Electronics Systems Ltd, Edmonton, AB, CANADA (780) 489-3199 voice (780) 487-6397 fax says:
I've had similar problems.One of them turned out to be the oscillator pins floating. I added a 10M resistor from ocs1 to gnd - problem gone.
The other was an input pin floating - solution was to make it an output during sleep.
Use a high impedance probe and look around while the PIC is asleep - keep an eye on the quiescent current each time you touch something. Even a 10M scope probe will bring an input LO - you might not even see that it was floating if the pin / trace capacitance is really lo. But your quiescent current will plummet if you hit the problem pin.
It might be a good idea to put some code like the following at the beginning of all your projects. This VERY BRIEFLY drives the pin and tests the result of that by reading the pin back.
movlw TRISB,FSR ; Point at TRIS register
bcf PORTB,x ; Get ready to drive a low
bcf INDF,x ; Start the short in this instruction
movf PORTB,W ; Get the value
bsf INDF,x ; Clear the short in this instruction
Simon Nield [simon.nield at QUANTEL.COM] says:
Reading Outputs and Inputs:might be a good opportunity to make use of the usually irritating lack of a shadow register for the port states: ; pin0 of port D is either connected to ground, vcc, or open circuit ; if o/c or vcc pin0 is driven high, if connected to ground pin0 is driven low bsf PORTD, 0 ; try and pull the pin high clrw ; (and wait a bit to let the voltage rise) xorwf PORTD, F ; read state of port D and drive it that way ; if pin0 was pulled low it will now be driven low ; if pin0 was floating or high it will now be driven high...detecting the open circuit state...:
; if pin n port x is pulled high => result = 0xfe, pin is driven high ; if pin n port x is pulled low => result = 0x00, pin is driven low ; if pin n port x is open circuit => result = 0xff, pin is driven high ; i.e. bit0 set => open circuit, bits7..1 = pin state bsf PORTx, n clrf result ; h:00 o:00 l:00 btfsc PORTx, n decf result, f ; h:ff o:ff l:00 bcf PORTx, n decf result, f ; h:fe o:fe l:ff btfss PORTx, n incfsz result, f ; h:fe o:ff l:00 bsf PORTx, n ; set pin high to match state
See also:
Q2: Well, which is it ? Do I tie them to Hi ? Or do I tie them to Lo ?
A2a: ``Sometimes it does matter, such as a completely unused flip-flop, where you would not want to tie both the reset and set inputs to their active level ... which could cause ... upset the performance of the other FFs in the same package.'' -- Jon Elson
A2b: CMOS inputs: ``whatever may make the PCB layout easiest (annotate to SCH), or makes using the input later (with a wire?) easier.'' -- Bruce Walter
``For example, if I am routing traces for a 74HC14 hex inverter, and I don't use 5 of the inverters, I tie the inputs of the unused gates on the pin 1 side of the chip to GND, and the inputs of unused gates on the pin 14 side of the chip to VCC. The way I route my power buses, the GND track is under the pin 1 side of the chip, and the VCC is under the pin 14 side of the chip. Electrically, there is no reason I know of to choose VCC over GND for CMOS inputs. ... This is what I do for double-sided boards. Of course, this advice is moot for boards with VCC and GND planes, since VCC and GND are equally accessible.'' -- Ivan Baggett http://www.bagotronix.com
``Actually, I have sometimes daisy chained unused sections of a HC14 or HC04 since the pins in question are adjacent and the lengths of the nets are minimized. The first in the chain connects to the nearest of Vcc and GND. If I recall correctly, TTL gates required a series resistor (I don't know why), LSTTL did not but both had to be pulled high because when pulled low they sourced appreciable current. Since CMOS the input behaviour has been independent of the input logic state. If you want to be able to use a spare gate or inverter in modifying a board, you could perhaps just make sure that the track to the input can be cut readily. If you daisy chain sections you can readily pull off a non-inverting buffer from 2 spare sections by just cutting the track to the second last inverter in the chain.'' -- Robert Mitchell
A2c: TTL inputs: Tie high. Use a 1 KOhm pull-up resistor. ``so that if the +5 V power supply goes to an abnormally high voltage, the chip may survive up to 7 Volts or so. Without the resistor, the input protection network would pop at about 5.5 V.'' -- Jon Elson
Inside the chip, TTL inputs ``are biased high, so connecting to VCC will give slightly less current draw.'' (but still much more current than CMOS) -- Ivan Baggett
``A grounded TTL input will sink 1.6 mA out of each input! That can add up quick on battery powered or other low power equipment. Of course, they DON'T use that stuff in battery powered gear anymore.'' -- Jon Elson
From: David Cary ..."Mike Reagan" on 2001-03-29 06:07:13 AM mentioned >Look at the low level schematic provided by the Mfg for their integrated >circuit.This is always good advice.
Now that the top IC suppliers have put their databooks online, it's very easy to get this information.
Logic ICs
- Philips http://www-us.semiconductors.philips.com/handbook/handbook_45.html
- On Semi http://www.onsemi.com/pub/prod/0,1824,productsm_Documentation_DocType=DataBook_LevelName1=DL121,00.html
- On Semi http://www.onsemi.com/pub/prod/0,1824,productsm_Documentation_DocType=DataBook_LevelName1=DL129,00.html
Allow me to quote from http://www-us.semiconductors.philips.com/acrobat/various/HCT_USER_GUIDE.pdf (which has some really low-level detail -- the shapes of the ``n'' and ``p'' regions in the silicon, which pins are pulled high or low for testing, etc.)
7.4 Termination of unused inputs To prevent any possibility of linear operation of the input circuitry of an LSTTL device, it is good practice to terminate all unused LSTTL inputs to VCC via a 1.2 kOhm resistor. Inputs should not be connected directly to GND or VCC , and they should not be left floating. Unlike LSTTL inputs, the impedance of 74HC and 74HCT inputs is very high and unused inputs must be terminated to prevent the input circuitry floating into the linear mode of operation which would increase the power dissipation and could cause oscillation. Unused 74HC and 74HCT inputs should be connected to VCC or GND, either directly (a distinct advantage over LSTTL), or via resistors of between 1 kOhm and 1 MOhm. Since the resistors used to terminate the inputs of LSTTL devices are usually between 220 Ohm and 1.2 kOhm, it is often possible to directly replace LSTTL circuits with their 74HCT counterparts. Some of the bidirectional (transceiver) logic devices in the HCMOS family have common I/O pins. These pins cannot be connected directly to VCC or GND. Instead, when defined as inputs, they should be connected via a 10 kOhm resistor to VCC or GND.On the other hand, AN2022: integrated bus hold circuits http://www-us.semiconductors.philips.com/acrobat/applicationnotes/AN2022.pdf starts out with "The problem with floating or unused CMOS inputs ..." and explains how a few specialized ICs are designed with "bus hold circuits" so it's OK to let their inputs float. -- David Cary
Questions:
Should I pull-up or down unused output pins? Is it OK to pull-up or down the unused output pins if I use the large value resistor as a weak pull-up or down? Your recommendation on bidirectional pins?(like open collector/drain?) Thanks, Yoon
Comments:
> stops gate voltages rising faster than any internal supply rails for any devices that can latch-up
Pull-up or pull-down resistors ARE useful for bus and signal integrity, but should have NO EFFECT on the parasitic current when tied to one of the supply rails directly with low inductance connection. (Like a common ground plane)
> and so prevent an occasional failure due to bond wires melting.
Bond wires DO NOT MELT. I have tried it, the bonding wire (10um, golden) can withstand 0.5A for short time without being interrupted. What usually fails is the silicon, either there is a oxide breakdown or some PN junction is destroyed by high voltage. (All diodes have a maximum reverse voltage, when this is crossed the junction can be irreversibly damaged). So you can destroy functional I/O pin by just 2mA. (Well that happened to me with no latch-up)
>Latch-up is a danger in chips that have parasitic thyristors as an artifact of the manufacturing process.
All CMOS devices do.
>Most modern chips do not have this problem but its best to be safe.
Not true. ALL CMOS devices contain a parasitic thyristor structure. I have not checked the SOI devices (some high-tech processors like 64-bit IBMs) if they are less sensitive. Only energy required to cause the chip to latch up (parasitic current) varies between the chips and also the current that the device will sink during a latch-up condition. Sone FLASH memories have latch-up current of 150mA max., some PIC devices of about 1A @3.4V saturation voltage which will damage the chip. Military processors can withstand 0.5A for a second into any I/O pin without triggering the parasitic thyristor (SCR for dummies), but at the cost of increased chip size and price.
Mario
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