How MOSFET gate capacitances (Cgs, Cgd) are calculated?
Reading in the tea leaves here, I'd surmise that for the case of:


  your equation 2)\$C_{gs}=C_{gd}=\frac{1}{2}C_{ox}WL + C_{ov}\$


They are taking the gate to channel capacitance and dividing equally between the S &amp; D.

In the case of:


  your equation 1)\$C_{gs}=\frac{2}{3}C_{ox}WL + C_{ov}\$


It looks like they are lumping the pinched off part of the channel which is attached to the source.

In your equation #2, while this is not strictly wrong, it is the wrong way to look at it. It would be best to think in terms of gate to channel.

In your equation #1, that might only hold true in one particular channel condition.  Once the channel pinches off the drain doesn't under go massive capacitance changes.

I'd be suspicious.

From the book "Operation and modelling of the MOS transistor" by Yannis Tsividis (recommended reading !!) the following equations from section 8.3.2 (page 391 in the 2nd edition).  For strong inversion:

$$ C_{gs} = C_{ox} \dfrac{2(1+2\eta)}{3(1+\eta)^2}  $$

$$ C_{gd} = C_{ox} \dfrac{2(\eta^2 + 2\eta)}{3(1+\eta)^2}  $$

\$ \eta = \$ degree of non-saturation. With \$ \eta = 1\$ at \$V_{ds}=0\$.

So in the case with the channel fully pinched off \$ \eta = 0\$. We get the case of your equation #1.