What is the output of a CMOS inverter if the input is unconnected?

Would you tell someone analyzing such a circuit to just assume that the output is some particular value, or would you tell them that the output is indeterminate and that there is more than one possible value of the output that is consistent with the circuit?

 

What is the output of a CMOS inverter if the input is unconnected?

Would you tell someone analyzing such a circuit to just assume that the output is some particular value, or would you tell them that the output is indeterminate and that there is more than one possible value of the output that is consistent with the circuit?

Hi,

I think we are getting to the heart of this discussion now. It is interesting i think.

I am not sure the CMOS gate example is the same thing though, because that is a special case in itself. But if we dont apply any power to the gate, the output must be zero. So what we can say is that before we apply any power whatsoever, the output is zero. In a gate with only input capacitance to ground, if we turn the power on then the output should be high because the input is low, but then that starts to depend on the actual construction of the input circuit. Also, the output will be known if we apply a known input.

The dependent source is different i believe but if we apply a known input we also get a known output. Thus the output depends on the input. That means if we FORCE the input to be a certain value, then the output will assume a certain value. Now we can look at an example where we have a CCCS with gain of 5 and various inputs.
If we input 10 amps, we get 50 amps out.
If we input 5 amps we get 25 amps out.
If we input 1 amp we get 5 amps out.
If we input 0 amps we get 0 amps out.

So the output is not unknown, as long as we know what the input is.

Now in the circuit in question (i guess either one with the 2 ohm resistor that is changed to 0.25 ohms for this problem) if we input 10v across that 1 ohm resistor, the solution is 10v across that 1 ohm resistor. If we input 5v, the solution is 5v. If we input 1v the solution is 1v, and if we input 0v the solution is 0v. The solution always rests on what we make it. It can not really be anything, it will follow whatever we make it, even though we can make it anything we want.

Now with the power off, the solution must be 0v. When we turn the power on the 10amp current source comes to life but it's into a direct and perfect short, so we still have 0v across that 1 ohm resistor.

Now if we want to say that the output of the CCCS can be anything (and hence the voltage solution can be anything) then we at least have to show HOW the input of the dependent source was able to change from the state it was in just before power up.

For the CMOS gate there is a mechanism that could explain how the input might change, such as leakage current that charges the input cap and so the output goes low. But in the circuit with the 0.25 ohm resistor, we dont get any leakage current or any offset voltage, so there's no mechanism that would cause the dependent source to change.