Unused CMOS Pins

Thread Starter

mcardoso

Joined May 19, 2020
116
Hi All,

I know that you have to tie unused CMOS input pins to GND or VCC to prevent the circuitry from switching due to static charge due to drawing extra power I am guessing? Do you make these connections as direct connections, or should there be a current limiting resistor in the mix?

As a parallel question, what about unused CMOS outputs? For example I have a 4 channel AND gate chip that I am only using 3 channels on. I'm assuming this output pin should be left unconnected (floating) since a connection to VCC or GND could damage the chip. Is this correct?

Thanks!
 

jpanhalt

Joined Jan 18, 2008
9,771
Inputs are high impedance. Direct or with resistor, it doesn't matter. I tie to ground to minimize any leakage current.

Outputs are low impedance. I do not tie them to anything. Thus, with an MCU, I set unused pins to output, when possible.
 

WBahn

Joined Mar 31, 2012
25,924
Hi All,

I know that you have to tie unused CMOS input pins to GND or VCC to prevent the circuitry from switching due to static charge due to drawing extra power I am guessing? Do you make these connections as direct connections, or should there be a current limiting resistor in the mix?

As a parallel question, what about unused CMOS outputs? For example I have a 4 channel AND gate chip that I am only using 3 channels on. I'm assuming this output pin should be left unconnected (floating) since a connection to VCC or GND could damage the chip. Is this correct?

Thanks!
Yes, except the notion that it is "due to static charge due to drawing extra power."

If they are unconnected, then it takes very little external electric field to cause them to switch because they are so high impedance. They can also float to an intermediate state in which the internal circuitry sees the input as being both HI and LO with the result that the outputs can both be turned on resulting in shoot-through current. It is the shoot-through current that results in it drawing excess power (and possibly destroying the chip).
 

Papabravo

Joined Feb 24, 2006
13,962
Yes, except the notion that it is "due to static charge due to drawing extra power."

If they are unconnected, then it takes very little external electric field to cause them to switch because they are so high impedance. They can also float to an intermediate state in which the internal circuitry sees the input as being both HI and LO with the result that the outputs can both be turned on resulting in shoot-through current. It is the shoot-through current that results in it drawing excess power (and possibly destroying the chip).
To add the the above you don't ever want to have actual signals that make slow transitions through the threshold regions. If you do have such a signal, you would use a gate with a Schmitt Trigger input to avoid this problem.

 

andrewmm

Joined Feb 25, 2011
403
Slightly broader,
In commercial situations, it can be company practise that all none drive inputs are pulled up / down by resistors, not direct .
This mainly historical, it gives the bed of nails test points to work with, some chips have special modes that can be used for test, which can be toggled by the test equipment if not tied direct to Vcc / gnd
and once upon a time, like 40 year sago, there were questions about IC in put pins, if an input is tied direct to Vcc, it possible , even if only for a very short time, that at start up, the input will be at a higher volts than the internal of the chip ,

So the habit of using a 10K pull, or 470 Ohm pull down on un used inputs came about.
 
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