Why Small-Signal Output Resistance is Independent of Device Width

mosfet

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.

Best Answer

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.

$$R_o=\dfrac{1}{\dfrac{1}{2}{\lambda}\text{K}\frac{W}{L}(V_{GS}-V_T)^2} $$

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