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MOSFET ID, VGS, VDS and saturation check

Short answer

VG = 8.095 V; VGS = 8.095 – 0.75 ID. Using ID = 0.556(VGS – 3)² gives ID = 3.47 mA, VGS = 5.50 V, VDS = 20 – 3.47(2.2+0.75) = 9.78 V. Since VDS > VGS – Vt, the MOSFET operates in saturation.

Find the value of ID, VGS, VDS and check whether the MOSFET operated in the saturation region or not. Given: Vt= 3V, k= 0.556 mA/V² [Figure: enhancement MOSFET with 20 V supply; gate divider of 10 MΩ from 20 V to gate and 6.8 MΩ from gate to ground; 2.2 kΩ drain resistor from 20 V to drain; 750 Ω source resistor from source to ground]

The enhancement MOSFET is biased by a voltage divider (no gate current). In saturation ID = k (VGS - Vt)², with k in mA/V².

Given

VDD = 20 V, R1 = 10 MΩ, R2 = 6.8 MΩ, RD = 2.2 kΩ, RS = 0.75 kΩ (resistances in kΩ in the formulas), Vt = 3 V, k = 0.556 mA/V².

Step 1: gate voltage

VG = VDD R2/(R1 + R2) = 20 × 6800/(10000 + 6800) = 8.095 V
VGS = VG - ID RS = 8.095 - 0.75 ID      (ID in mA, RS in kΩ)

Step 2: solve with the square law

ID = k (VGS - Vt)² = 0.556 (5.095 - 0.75 ID)²

Expanding and collecting terms gives 0.3128 ID² - 5.2494 ID + 14.4346 = 0, so

ID = 13.320 mA   or   3.465 mA

For ID = 13.320 mA the gate-source voltage would be VGS = -1.89 V, which is below Vt (the MOSFET would be off), so this root is rejected.

Step 3: results

ID  = 3.465 mA
VGS = 8.095 - 0.75 × 3.465 = 5.496 V
VDS = VDD - ID (RD + RS) = 20 - 3.465 × (2.2 + 0.75) = 9.778 V

Saturation check

Saturation requires VDS > VGS - Vt:

VGS - Vt = 2.496 V ,   VDS = 9.778 V  >  2.496 V

So the MOSFET is in the saturation region and the assumption is valid.

Answer: ID ≈ 3.47 mA, VGS ≈ 5.50 V, VDS ≈ 9.78 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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