Certainly! Here is your content rewritten with the same technical information and mathematical expressions, but with slightly different wording and structure: --- **Part 1: NMOS Operation** - **Condition for NMOS to turn ON:** \[ V_{GS} \geq V_{th} \implies \text{NMOS is ON} \] where \( V_{th} \) is the threshold voltage for NMOS. - **Saturation Region (Active):** \[ \text{If} \quad V_{DS} \geq V_{GS} - V_{th} \implies \text{NMOS operates in saturation} \] \[ I_D = \frac{1}{2} \mu_n C_{ox} \left( \frac{W}{L} \right) (V_{GS} - V_{th})^2 \] - **Triode (Linear) Region:** \[ \text{If} \quad V_{DS} < V_{GS} - V_{th} \implies \text{NMOS operates in triode} \] \[ I_D = \mu_n C_{ox} \left( \frac{W}{L} \right) \left[ (V_{GS} - V_{th}) V_{DS} - \frac{V_{DS}^2}{2} \right] \] --- **Part 2: PMOS Operation** - **Condition for PMOS to turn ON:** \[ V_{SG} \geq |V_{TP}| \implies \text{PMOS is ON} \] where \( V_{TP} \) represents the threshold voltage of the PMOS transistor. - **Saturation Region:** \[ V_{SD} \geq V_{SG} - |V_{TP}| \implies \text{PMOS operates in saturation} \] \[ I_D = \frac{1}{2} \mu_p C_{ox} \left( \frac{W}{L} \right) (V_{SG} - |V_{TP}|)^2 \] - **Triode (Linear) Region:** \[ V_{SD} < V_{SG} - |V_{TP}| \implies \text{PMOS operates in triode} \] \[ I_{DP} = \mu_p C_{ox} \left( \frac{W}{L} \right) \left[ (V_{SG} - |V_{TP}|) V_{SD} - \frac{V_{SD}^2}{2} \right] \] ---