[SOLVED] The effect of unused CMOS inputs

[pastro]

Okay, so I've always heard you aren't ever supposed to let unused CMOS inputs float, but I'm curious if anyone can help me understand the nuances of what happens if you DO float an CMOS input.

Let's say I have a two input CMOS OR gate, with one input directly tied high, and the other floating. Is it ever possible that the floating input could cause the output to be driven low?

I've also heard people mention that a floating CMOS input can cause excessive power draw. Why is this? If I had to guess, I'd think it's because the high impedance of the CMOS input means small noise currents can induce large noise voltages, which means the input MOSFET is constantly shutting on and off, thereby increasing dynamic power consumption. Is this right? If not, what really happens?

 

[lostinxlation]

I've also heard people mention that a floating CMOS input can cause excessive power draw. Why is this?

It depends on the threshold voltage of the cell, but nowadays, the cell is designed in a way that both of P and N mos are on at the same time for a moment, while the input transitions from 1 to 0 , or 0 to 1. If input is floating and voltage hangs somewhere between GND and VDD, it creates a permanent DC path.

 

[eladla]

I think the problems are quiescent current and SNR. Lets take your OR gate as an example.
If one input is tied high, the other could either be high or low or anywhere in between.
Lets say that a PMOS is open and charges the output to high. If the NMOS of the floating input opens wide enough to build up a drain current, it might discharge the output, since NMOS has higher current drive than PMOS.
This being said, this is unlikly. Usually the partially open transistor will just degrade the logic level at the output and thus degrade the SNR for the next logic level.
This might be a big problem, or not. It all depends on what signal margin you had to begin with. But why not just avoid this and tie the input to something
But more importantly, this will cause a DC current to flow from VDD to GND, and this will discharge your battery very quickly.


Hope this helps.