The operation of a CMOS inverter is analyzed by sweeping the input voltage ($V_{in}$) from 0V to $V_{DD}$. The sequence of operating regions for the NMOS and PMOS transistors depends on $V_{in}$ relative to their threshold voltages ($V_{th,n}$ and $V_{th,p}$) and the output voltage ($V_{out}$).

CMOS Inverter Region Progression

The question asks to arrange the given descriptions of transistor operating regions in the correct sequence as $V_{in}$ sweeps from 0V to $V_{DD}$. Based on the provided correct answer, the sequence is assumed to be A, B, C, D, E.

Step-by-step Analysis of Regions:

• State A: NMOS in Cut-off, PMOS in Linear Region

  This state implies $V_{in}$ is low, keeping NMOS OFF ($V_{GS,n} < V_{th,n}$). For PMOS to be in the linear region, the output voltage ($V_{out}$) must be low enough. This represents an early stage in the sweep, possibly just after the initial state where $V_{out}$ starts to drop significantly.

• State B: NMOS in Saturation, PMOS in Linear Region

  As $V_{in}$ increases past $V_{th,n}$, the NMOS transistor turns ON and enters the saturation region ($V_{GS,n} > V_{th,n}$ and $V_{DS,n} > V_{GS,n} - V_{th,n}$). Simultaneously, the decreasing $V_{out}$ can place the PMOS transistor in its linear (or triode) region ($V_{GS,p} < V_{th,p}$ and $|V_{DS,p}| < |V_{GS,p} - V_{th,p}|$). This occurs during the high-gain transition phase.

• State C: NMOS in Saturation, PMOS in Saturation Region

  With further increase in $V_{in}$, both NMOS and PMOS transistors can operate in the saturation region. This typically happens near the midpoint of the voltage transfer characteristic ($V_{in} \approx V_{DD}/2$), where the inverter exhibits maximum voltage gain.

• State D: NMOS in Linear, PMOS in Saturation Region

  As $V_{in}$ continues to rise, the NMOS transistor may enter the linear region ($V_{GS,n} > V_{th,n}$ and $V_{DS,n} < V_{GS,n} - V_{th,n}$). At this stage, the PMOS transistor might still be operating in the saturation region.

• State E: NMOS in Linear, PMOS in Cut-off Region

  This is the final state when $V_{in} = V_{DD}$. The NMOS transistor is ON and in the linear region ($V_{GS,n} = V_{DD} > V_{th,n}$ and $V_{DS,n} \approx 0V$, so $V_{out} \approx 0V$). The PMOS transistor is OFF (Cut-off) because $V_{GS,p} = V_{in} - V_{DD} = V_{DD} - V_{DD} = 0V$, which is less than $|V_{th,p}|$.

Therefore, the correct arrangement of the regions as $V_{in}$ sweeps from 0V to $V_{DD}$ follows the sequence A, B, C, D, E.