# Is it wrong to use Thévenin's theorem on this memory circuit?

In my electricals beginner class I was given a paper about memory circuits. There was a small section about how writing data to a single memory cell worked.

Here is the sketch of the memory cell:

The paper is written in Norwegian and called "Resten av systemet" meaning "Rest of the system".

The thought process behind the circuit is that if Q has a certain logical value, it will keep this value since the voltage is inverted twice between the not-gates. But then the question arises about how to change the logical value Q. If Q has a initial voltage of 0 and x has a initial voltage of 1, what will the voltage above Q become?

The paper wants to explain that the sub-system with the smallest internal resistance will have the biggest impact on this voltage. This is done by replacing the upper not-gate with its Thévenin equivalent like this:

In this case $$\V_1\$$ is set to zero and $$\V_2\$$ is not zero. This is supposed to represent the case where you want to change Q from 0 to 1. The case where you want to change Q from 1 to 0 can be explained with the same logic. It then explains that there will go no current through the lower not-gate which means $$\R_1\$$ and $$\R_2\$$ are wired in series. The voltage above Q can then be written as:

$$\Q = \frac{R_1}{R_1 + R_2}V_2\$$

If $$\R_1\$$ has a much lower resistance then $$\R_2\$$, then the voltage above Q will be closer to zero. If $$\R_2\$$ has a much lower resistance than $$\R_1\$$, the voltage will be closer to the voltage level of $$\V_2\$$. This means you would be able to change the voltage above Q by making sure the internal resistance of the rest of the system is much lower than the internal resistance of the memory cell. So you could for example place a switch between Q and Y and close the switch whenever you want to change Q and Q will then keep this voltage after the switch is opened.

There are a couple of things I dont understand:

1. I thought Thévenin's theorem could only be used at linear electrical circuits consisting of voltage sources, current sources and resistances. But the not-gate is made up of transistors and is therefore non-linear. Would it not be wrong to replace the upper not-gate with a Thévenin equivalent then?
2. If you replace the upper not-gate with its Thévenin equivalent, you are saying that a current runs through it in this case. Would you not also have to replace the lower not-gate with its Thévenin equivalent in this case as well? This means the current going through Q would split between the two not-gates, right?