Microsoft hasn't disclosed the full workings of this memory from the last time I checked so this is what I understood from what we know. When I last read up on this, it's not just the idea of placing voltage on the terminals but electrically changing the properties of the material of the selector and the bit, in particular their resistive properties.
Filling in the gaps:
Now my thought process was that the resistive properties were controlled via a current sent through the bit and a voltage across the selector. To this regard if you do not apply enough voltage across the selector it would not reduce it's resistance enough to allow the system to alter the bit. My assumption is the system would apply the voltage to activate the selector and then perhaps alter the current (depending on 1 or 0) to change the bit from a 0 or a 1.
I mean, isn't there the possibility of a current being released from these cells?
If you are referring to leakage then yes, there will always be a leakage of current but the question to ask is if it is enough to change the memory unit. Even NAND memory has leakage at the gate, but the large amount of charge and relatively low leakage means we can classify it as non-volatile if power is off. Since 3D Xpoint was also classified as non-volatile I would hope similar ideas extend here.
If you are referring to write/read level current then there must be some low resistance circuit, which is a topic I talk about next.
Or the cells are constructed (namely the electrodes) in such a way that
will never be sufficient to [activate] the selector of the cell?
Once again, the simple answer would be yes because it is designed so but while Intel hides behind its proverbial curtain, I propose the following two possibilities.
Selector Does Not Engage
- The selector material may be of the sort that returns to high resistivity and charge through out when returning to the OFF position. The bit material may instead hold the resistive properties when the voltage is removed from it. This configuration ensures there is no standing charge on the selector that could add to the floating voltage and activate the selector, and that the bit can store memory.
The resistance path to the line would involve multiple selectors that would limit the current from changing the bit even if the selectors were to activate. Consider that the selector to the left of the cell you labeled (1) is "ON". I have drawn the shortest electrical path to the word line from the bit line in the image below. Notice here that even if we assume all selectors in that sequence are on (which I would assume is very unlikely), this would mean the impedance would be at least three times as much as the one that is being written/read to. I would argue that the current through any such path, or even summed, would be much too small to make a bit change.