Why use resistors to tie a logic input to ground?
Manvinder Bhullar
I'm looking at this circuit from Anchor Chips that shows unused inputs
of a 74LCX14 tied to ground with 10K resistors.
Can someone please advise what the rationale is behind this? I've
always tied unused inputs directly to ground. What role do the
resistors play?
Thanks
Manvinder S. Bhullar
Tom Faloon
There is no 'electrical' reason why you cannot tie unused inputs of this
device directly to ground. I can't tell you why inputs are tied low with
resistors on your board, but I can tell you that I often do this, and why I
do it.
I often tie inputs of unused logic elements to ground via a resistor,
especially if the board is a prototype, or a low volume product. My reason
is that it makes the input easily accessible if you ever want to use it. (If
you have to do a cut and strap mod on the PCB, just throw away the resistor,
and you have a pad to solder to.)
If you don't have that resistor, and the PCB layout person autoroutes the
board, the unused input will be tied to the nearest ground. This may be a
heavy ground plane, or on an inner layer, or to a via located underneath the
chip! Try cutting the ground track! Often the only solution is to lift the
pin! That is not always acceptable to the customer.
It may seem easier to connect unused pins to ground through a standard
track, and make sure it is easily accessible should it ever have to be cut.
But try telling that to a draftsman. Nine times out of ten it won't happen.
This practice may not be acceptable on a high volume product where cost is
important. If you do implement it, use a resistor value which is used
elsewhere on the circuit, to reduce component count. Anything from 0 Ohms to
several hundred K is fine for most MOS logic families. Purchasing, accounts,
stores, and the person who loads the pick and place machine will all be
pleased. (Well they might be pleased if they knew. Like the rest of us, they
usually see only the things that bother them.)
Tom Faloon
http://www.faloon.co.uk
A E
Manvinder Bhullar wrote:
If it's on a PCB it's easier to jump the resistor to use the gate than to
cut traces that go straight to a ground plane.
10K is low enough to pull down, and high enough to parallel easily.
Looking at the spec sheet for the LCX14 shows that there is 5uA max input
leakage current for Vcc 2.3v to 3.6v, so that at worst you get 50mV across
the 10K. Not too bad.
chris
Hi, it helps protect the chips inputs from excessive current, and in
the event of an ip short it limits the damage.
Sir Charles W. Shults III
First, the resistor limits current that may flow into or out of the input
terminal. Second, if you are debugging, that input terminal can be forced to a
different state and the resistor will allow it, where a direct connection to the
rail will not. The resistors saves you from having to desolder the terminal to
test the device.
Cheers!
Chip Shults
My robotics, space and CGI web page - http://home.cfl.rr.com/aichip
Tim Shoppa
If you tie them directly to ground or Vcc, *and* you have substantial
ground bounce (most of a volt), then you can violate the "Absolute
Maximum Ratings" for input voltage and current. For typical TTL-like
logic these say that inputs cannot go below -0.5V or above Vcc+0.5V
unless you ensure the current is limited.
If you put a resistor in series, then you do not violate the "Absolute
Maximum Ratings" because you stay under the input diode current rating.
Typical ground bounce in a small, well bussed system will not be above
half a volt and you can hook the input directly to Vcc or Ground. Even
if you have severe ground bounce problems you can typically get away
with hooking genuine TTL inputs directly to Vcc or Ground because the
ground bounce is a transient effect - but you *are* violating the specs.
Tim.
Dave VanHorn
> Maximum Ratings" because you stay under the input diode current rating.
Of course such ground bounce is a problem in itself, and you really should
fix that!
:)
Pete
My guess is that the resistors are there to permit easy design
changes during debug. If you need one of the unused inputs, you
merely have to remove the resistor and solder a wire to the pad. If
the pin were connected directly to the ground plane, you have to lift
the leg of the chip and solder a wire to it directly.
Pete
Tom Bruhns
They're useful in automated test situations; the tester may "know"
about the expected behavior of lots of parts and if (all) the inputs
can be set by the tester, it makes testing easier. Of course, that's
at odds with making compact equipment...and at least in my case, I can
usually get the guy who writes the test software, after some
grumbling, to accept a design with grounded inputs. (Actually he's
nicer than that about it; he just points out that it limits the
testing he can program, though not necessarily in a way that really
matters, and I guess it is more work for him.)
Cheers,
Tom
GregS
>mbhu...@tm.net.my (Manvinder Bhullar) wrote in message
> news:<27c42ac5.02101...@posting.google.com>...
>> Hi All,
>> I'm looking at this circuit from Anchor Chips that shows unused inputs
>> of a 74LCX14 tied to ground with 10K resistors.
>>
>> Can someone please advise what the rationale is behind this? I've
>> always tied unused inputs directly to ground. What role do the
>> resistors play?
>>
>> Thanks
>> Manvinder S. Bhullar
>
>
> My guess is that the resistors are there to permit easy design
>changes during debug. If you need one of the unused inputs, you
>merely have to remove the resistor and solder a wire to the pad.
You could also just attach it to a logic source or switch. Apply
a high source, and it will go high.
greg
fred bartoli
Tom Bruhns <k7...@aol.com> a écrit dans le message :
3200347.02101...@posting.google.com...
I used to work for Alstom, building boards that travel in trains and such
"quiet" environments.
Hard tying inputs was strictly forbidden on the assumption that it could
lead to non detected future field failures that would have been otherwise :
even if you don't use it, a dead input or gate is a bad sign for the future
life of the part.
Regards,
Fred.
Keith R. Williams
1...@hotmail.com says...
Which is why I never tie inputs directly to planes. I always run a
back surface wire to a via tied to the plane. If I need the input I
can simply cut the trace and solder to the chip pad/via. Lifting the
leg of a BGA isn't easy. ;-)
Resistors aren't a bad idea, but take space and cost a bit.
----
Keith
Tom Faloon
Tim Shoppa <sho...@trailing-edge.com> wrote in message
news:bec993c8.02101...@posting.google.com...
Tim,
The question was about unused inputs. I don't see how ground bounce can
damage unused inputs if they are tied directly to 0V, and tied, as they
usually are, reasonably close to the chip's own ground pin. In this
situation both input and chip ground are at the same potential, and held at
the same potential by a PCB track. I can't think of a better way of
protecting the input, regardless of how much ground bounce there is! Using
a current limiting resistor is meaningless, since the voltage across it is
zero, or very nearly so, and no input current can flow anyway.
I would apply exactly the same argument to pull up resistors in most modern
MOS type devices.
I agree there could be a potential problem if there was horrendous ground
bounce, AND the input was connected to 'ground' at a point physically
distant from it's own chip ground. This situation would be unusual. Ground
bounce 'abounds' but unused inputs are usually grounded close by.
I do agree with using a pull down, (or pull up) resistor where possible,
because it allows access to the input for PCB mods, and allows the input
level of active chips to be forced, during debug or fault finding. This has
been discussed in other posts.
Tom Faloon
http://www.faloon.co.uk
nospam
>I used to work for Alstom, building boards that travel in trains and such
>"quiet" environments.
>Hard tying inputs was strictly forbidden on the assumption that it could
>lead to non detected future field failures that would have been otherwise :
>even if you don't use it, a dead input or gate is a bad sign for the future
>life of the part.
Only if you bother to test them. Resistive tie up/downs are required to
allow node forcing during board test.
I suggest the above argument is bullshit, just how many manufacturing test
failures where down to faulty chips and what tiny proportion of these were
faults only detected by stimulating unused pins? I suggest that the
additional components and solder joints for pull up/down resistors would
have a far worse impact on reliability than not testing unused pins.
I am reminded of an old story I read somewhere about a Japanese manufacture
being required to supply an American company some chips with a 99.9% AQL.
On delivery they supplied the 0.1% of faulty chips in a separate package
with a covering note.
I suspect this argument was used to justify forced node testing of
individual chips which means the test programmers can just verify
individual chips against existing models instead of having to devise a
proper functional test for the board.