Why does the Pinch-off Point in the MOSFET move towards the source as we increase \$V_{ds}\$?
Although this answer is oriented at OP, who declared the understanding of pinchoff, semiconductor conductivity and admittedly MOS capacitor, for consistency of narration I need to reiterate the derivation of analytical expressions for the MOSFET IV characteristics using the gradual channel approximation model. Let us consider the NMOS transistor: for the PMOS transistor, the voltage signs and carrier types has to be changed accordingly.
The gradual channel approximation (GCA) is valid for long-channel MOSFET device, i.e., of which the channel length is much greater than the device thickness. With the long-channel geometry we can assume (and this assumption is supported by calculations) that the component of the electric field directed from the drain to source electrode (along the channel, axis x) is much smaller than the electric field component directed into the depth of the bulk (axis y) in the inversion layer region.
In a configuration with the source and the body shorted and no potential applied to the drain electrode, under the strong inversion condition (VGS &gt; VTH) the surface concentration ns(x) of mobile carriers at the oxide film interface is non-zero and constant, q·ns(x)=Cox·(VGS-VTH), q is the elementary charge.
For NMOS, mobile carriers are electrons. For PMOS, the bulk is an n-type semiconductor, and mobile carriers in the inversion layer are majority carriers in the source and drain electrodes and minority carriers in its n-channel, i.e., holes. Select the voltage signs accordingly to the transistor type as well.
When a positive potential is applied to the drain, the local gate potential VCh(x) becomes the function of a distance from the source electrode. With the drain potential sufficiently small, VGS-VCh(x) &gt; VTh in the entire channel. Later we will elaborate on this condition in more detail. Also, we will see that the VCh(x) variation is gradual along the channel, hence the calculation method is called GCA.
Let us re-write the expression for surface concentration of mobile carriers with VCh(x) gradually changing from zero to VDS when moving along the channel. For that purpose, we partition the channel into slices orthogonal to the drain-source axis and use the formula for the charge at the plates of elementary capacitors in these partitions, the plates are separated by the oxide film. The capacitors are charged to voltages VGS-VCh(x)-VTh according to their x positions; using the formula for the charge at the plates of elementary capacitors, we calculate the charge and divide the result by the partition width:
$$
\tag 1
q·n_s(x) = C_{ox}·(V_{GS}-V_{Ch}(x)-V_{TH}) 
$$
Under the assumption of gradually changing VCh(x) this formula, derived for a MOS capacitor, is still valid.
Now that we have a formula connecting carrier concentrations and voltages, we turn our attention to current in the channel, under conditions of the strong inversion and a limited drain voltage being applied. The density of the drift current in the channel is proportional to the x component of electric field:
$$
j_x(x,y) = n(x,y)·(-q)·μ_n·E_x(x,y)
$$
, where the concentration n(x,y) and the field Ex(x,y) are changing along the channel. But the drain current,
$$
I_D = W·\int_0^t{j_x(x,y)·dy}
$$
, is constant along the channel per the charge conservation law. In this formula, t is the inversion layer thickness.
The drift current does vary in the y direction (into the bulk depth), but carriers are mostly concentrated at the oxide/semiconductor interface. This fact is not an assumption of GCA: the thin inversion layer is a manifestation of the quantum surface state phenomenon. We can safely replace the integral over the inversion layer depth in the drain current formula with the current density at the surface times thickness t:
$$
\tag 2
I_D = W·j_x(x,0)·t = W·n(x,0)·t·(-q)·μ_n·E_x(x,0) = \\W·n_s(x)·(-q)·μ_n·E_x(x)
$$
where we define the surface concentration of carriers ns(x) = n(x,0)·t and the x-component of the electric field vector Ex(x) = Ex(x,0) at the oxide/semiconductor interface.
Using equation 1, we define ns(x) via potentials and substitute it into equation 2:
$$
I_D = W·n_s(x)·(-q)·μ_n·E_x(x) = -W·C_{ox}·(V_{GS}-V_{Ch}(x)-V_{TH})·μ_n·E_x(x)
$$
By definition Ex(x) = -d(VCh(x))/dx (remember VCh(x) is a total potential, including the drain potential contribution), and we arrive at
$$
\tag 3
I_D = W·μ_n·C_{ox}·(V_{GS}-V_{Ch}(x)-V_{TH})·{dV_{Ch}(x)\over{dx}}
$$
Solving this equation for VCh(x), we arrive at the formula
$$
\tag 4
V_{Ch}(x) = V_{GS} - V_{TH} - \sqrt{(V_{GS}-V_{TH})^2 - {2·I_D·x\over{W·μ_n·C_{ox}}}}
$$
The elementary math indicates that the formula cannot give a valid value when a square root argument is negative, that is, when
$$
I_D·x &gt; W·μ_n·C_{ox}·{(V_{GS}-V_{TH})^2\over2}
$$
At a glance we cannot see if the solution ever enters the forbidden range, because we don't know yet how to calculate the drain current for the given drain voltage. We only know that it is constant (independent of x), and we have to have a constant in the solution, because equation 3 is the first order ODE and its solution must have one integration constant. Per ODE theory, this constant is defined from initial conditions, which, in our problem, state that VCh(L) = VDS: the channel potential at the drain is the drain electrode potential, because there is no insulator separating the channel and the drain but the depletion layer.
To find out this &quot;integration constant&quot; from our formulas, we integrate equation 3 over the channel length from 0 to L:
$$
\int_0^L{I_Ddx} = Wμ_nC_{ox}\int_0^L{(V_{GS}-V_{Ch}(x)-V_{TH}){dV_{Ch}(x)\over{dx}}dx} 
\\
I_D·L = Wμ_nC_{ox}\int_0^{V_D}{(V_{GS}-V_{Ch}-V_{TH})·dV_{Ch}}
$$
and arrive at the formula for ID
$$
\tag 5
I_D = W/L·μ_n·C_{ox}·\left((V_{GS}-V_{TH})·V_D-V_D^2/2\right)
$$
When VD &lt;= VDsat = VGS-VTH, with this ID the solution for VCh holds, at least from the point of view of elementary math, throughout the entire channel.
When VD &gt; VDsat, there exists the channel region adjacent to the drain in which formula 4 is invalid. To understand what happens, we look closer at Ex, the electric field, which is the potential gradient:
$$
E_x = -{dV_{Ch}\over{dx}}=-{I_D\over{Wμ_nC_{ox}\sqrt{(V_{GS}-V_{TH})^2 - {2·I_D·x\over{Wμ_nC_{ox}}}}}}
$$
When approaching the point xcritical moving from the source along the channel toward the drain, at the point
$$
\tag 6
x_{critical} = {Wμ_nC_{ox}(V_{GS}-V_{TH})^2\over{2·I_D}}
$$
the x component Ex of electric field grows infinitely in our solution, that is, the GCA assumption, Ex &lt; Ey, becomes broken. Fortunately, Ex grows rather rapidly with x, and GCA works nearly all the way up to the critical point. It breaks only at that point and beyond it up to the drain, where Ex &gt; Ey. This results in no inversion layer being able to form in this region of the channel, and there exists a depletion region separating the drain from the channel, the region length is L-xcritical. Therefore, xcritical is a pinchoff point, xcritical=xpinchoff. To extend the numerical calculation to the saturation mode, we use GCA locally in the inversion layer region and sew this solution with the electrostatic solution for the depletion region near the drain, where GCA, although not valid, is not necessary, because the depletion region contains no significant concentrations of mobile carriers resulting in noticeable surface/volume charges.
The closed expression for xcritical can be found by numerically solving equations 3 with the initial conditions of saturated mode. At small positive VD-VDsat, the pinchoff gap grows linearly with VD.
The answer to OP's question: when VD &gt; VDsat, the pinchoff point is separated from the drain by a depletion region of a finite size, because in this region the strong tangential electric field of the drain disturbs surface states and prevents the inversion layer from forming at the oxide/channel interface, and the appearance of the finite-size depletion region can be proven with the GCA calculations, which also can give the size of a depletion gap near the drain.