You seem to be under the same miscomprehension as the TS regarding just what the CVD model for a diode is.

Not at all.
I know that the CVD model only indicates the forward drop when the diode is conducting.

To me, the purpose of the experiment is to show the the CVD model is only a rough approximation of an actual diode response when the diode is conducting for Vs = 1V or higher (we don't know how high the experiment is supposed to go).
I'm not just referring to a Vs of 0.5V.

But it certainly could be to show that the diode drop is not 0.7V when the supply is only 0.5V.

 

Hello again,

To state this as simple as possible, the CVD model diode is an open circuit for any voltage that is less than the 0.7 volt forward voltage even if that voltage source is impedance limited by a high impedance, and it is 0.7v for any impedance limited voltage source that is 0.7v or higher.
So do the math, it's all very simple from there.

Even simpler:
The diode has 0.7v across it for any applied voltage that is 0.7v or 'higher' (although it wont get any higher), and for any voltage less than 0.7v it will have whatever that voltage is across it.

There is no claim that this is somehow inferior to an exponential or log model, we know it is inferior, but the fact is it is the model chosen by the problem statement to be the one used in the question. Note there are a lot of things missing if we want to dig in deeper than the problem requires, for example they also do not include any reverse breakdown information.

So the ONLY two states for the diode are either 0.7v conducting, or OPEN circuit not conducting.