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MOSFET ID and VDS (188 kΩ/47 kΩ, Vt=1 V)

Short answer

VG=4 V from divider; VGS=4−1.65ID. Solving ID=0.5(VGS−1)² yields ID=0.97 mA (rejecting larger root), VDS=20−0.97×(1.85+1.65)=16.6 V.

Find the drain current (ID) and drain to source voltage (VDS) for the following circuit. Given parameters are: Vt = 1 V and k = 0.5mA/V². [Figure: MOSFET with +20 V supply; gate divider of 188 kΩ from +20 V to gate and 47 kΩ from gate to ground; 1.85 kΩ drain resistor from +20 V to drain; 1.65 kΩ 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 = 188 kΩ, R2 = 47 kΩ, RD = 1.85 kΩ, RS = 1.65 kΩ (resistances in kΩ in the formulas), Vt = 1 V, k = 0.5 mA/V².

Step 1: gate voltage

VG = VDD R2/(R1 + R2) = 20 × 47/(188 + 47) = 4.000 V
VGS = VG - ID RS = 4.000 - 1.65 ID      (ID in mA, RS in kΩ)

Step 2: solve with the square law

ID = k (VGS - Vt)² = 0.5 (3.000 - 1.65 ID)²

Expanding and collecting terms gives 1.3612 ID² - 5.95 ID + 4.5 = 0, so

ID = 3.398 mA   or   0.973 mA

For ID = 3.398 mA the gate-source voltage would be VGS = -1.61 V, which is below Vt (the MOSFET would be off), so this root is rejected.

Step 3: results

ID  = 0.973 mA
VGS = 4.000 - 1.65 × 0.973 = 2.395 V
VDS = VDD - ID (RD + RS) = 20 - 0.973 × (1.85 + 1.65) = 16.595 V

Saturation check

Saturation requires VDS > VGS - Vt:

VGS - Vt = 1.395 V ,   VDS = 16.595 V  >  1.395 V

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

Answer: ID ≈ 0.97 mA, VGS ≈ 2.39 V, VDS ≈ 16.60 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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