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E-MOSFET ID, VDS and region (k 0.12, Vt 5 V)

Short answer

VG=18V, VGS=18−0.82ID. Solve ID=0.12(VGS−5)², reject larger root. ID=6.72mA, VDS=40−6.72(3+0.82)=14.3V. VGS−Vt=7.49V, VDS>7.49, operates in saturation.

Determine ID and VDS for the given circuit and find the region of its operation. Given: k = 0.12 mA/V² and Vt = 5 V. [Figure: enhancement MOSFET with VDD = 40 V; gate divider of 22 MΩ from VDD to gate and 18 MΩ from gate to ground; 3 kΩ drain resistor; 0.82 kΩ source resistor to ground]

Given: VDD = 40 V, R1 = 22 MΩ, R2 = 18 MΩ, RD = 3 kΩ, RS = 0.82 kΩ, k = 0.12 mA/V², Vt = 5 V (ID = k (VGS − Vt)²).

Step 1: Gate voltage

VG = 40 × 18/(22 + 18) = 18 V (IG = 0)

Step 2: Solve for ID

VGS = 18 − 0.82 ID

ID = 0.12 (18 − 0.82 ID − 5)² = 0.12 (13 − 0.82 ID)²

0.0807 ID² − 3.5584 ID + 20.28 = 0

RootID (mA)VGS = 18 − 0.82 IDValid?
137.4−12.6 VNo (VGS < Vt)
26.7212.49 VYes

ID = 6.72 mA, VGS = 12.49 V

Step 3: VDS

VDS = VDD − ID (RD + RS) = 40 − 6.72 × (3 + 0.82) = 40 − 25.7 = 14.3 V

Step 4: Region of operation

VGS − Vt = 12.49 − 5 = 7.49 V. Since VGS > Vt (a channel exists) and VDS = 14.3 V > 7.49 V, the MOSFET is in the saturation (pinch-off) region, so the square-law equation used is valid.

Answer: ID = 6.72 mA, VDS = 14.3 V, saturation region.

✓ Answer checked by Claude Opus 5.5 (high effort)

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