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For a MOSFET in deep triode region, we can approximate it as a resistor with the following:
$$ \frac{1}{{\mu}_n C_{ox}(W/L)(V_{GS} - V_{Th})} $$
However, in small-signal for a MOSFET in saturation, I know that \$R_{out} = \frac{1}{\lambda \cdot I_d}\$, where lambda is channel-length modulation factor and that is proportional to inverse of length. So increasing length increases \$R_{out}\$. But why isn't it dependent on width? Surely, if I make the transistor very wide, resistance should drop.
The small signal output resistance depends on W because \$I_D\$ depends on it.
After all, you have:
$$R_o=\dfrac{1}{{\lambda}I_D} $$
With \$I_D\$ being the current in the saturation region for the MOSFET. In saturation, $$I_D=\dfrac{1}{2}\text{K}\frac{W}{L}(V_{GS}-V_T)^2$$
So yeah, if you increase W, the \$R_o\$ does decrease because \$I_D\$ is in its denominator.
$$R_o=\dfrac{1}{\dfrac{1}{2}{\lambda}\text{K}\frac{W}{L}(V_{GS}-V_T)^2} $$
