Final answer:
The NMOS transistor with the given parameters is in saturation mode, and the calculated drain current (ID) is approximately 246.2μA.
Step-by-step explanation:
To determine the mode of operation (saturation, linear, or cutoff) and the drain current (ID) for the given NMOS with kn=115μA/V², Vth=0.43V, VGS=2.5V, and VDS=2.6V, we first assess the thresholds. For an NMOS transistor:
- If VGS < Vth, the transistor is in cutoff.
- If VGS > Vth and VDS < (VGS - Vth), it's in the linear (or triode) region.
- If VGS > Vth and VDS >= (VGS - Vth), it's in saturation.
In our case, VGS (2.5V) is greater than Vth (0.43V), and VDS (2.6V) is greater than (VGS - Vth), which is 2.07V. Thus, the NMOS is in saturation.
To calculate the drain current (ID) in saturation for the NMOS, we use the equation:
ID = 0.5 × kn × (W/L) × (VGS - Vth)²
Substituting the given values:
ID = 0.5 × 115μA/V² × (2.5V - 0.43V)²
ID ≈ 0.5 × 115μA/V² × (2.07V)²
ID ≈ 0.5 × 115μA/V² × 4.2849V²
ID ≈ 0.5 × 115 × 4.2849 μA
ID ≈ 246.2μA
Therefore, the NMOS transistor is in saturation, and the drain current is approximately 246.2 μA.