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MOSFET drain current and VDS (Vt=1 V)

Short answer

VG=6 V, VGS=6−5ID. ID=0.5(VGS−1)² gives ID=0.754 mA (rejecting larger root). Then VDS=12−0.754×(5+5)=4.46 V.

Find the drain current (ID) and drain to source voltage (VDS) for the following circuit. Given parameters are: Vt=1 V and k=0.5 mA/V². [Figure: enhancement MOSFET with +12 V supply; gate divider of 8 MΩ from +12 V to gate and 8 MΩ from gate to ground; 5 kΩ drain resistor from +12 V to drain; 5 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 = 12 V, R1 = 8 MΩ, R2 = 8 MΩ, RD = 5 kΩ, RS = 5 kΩ (resistances in kΩ in the formulas), Vt = 1 V, k = 0.5 mA/V².

Step 1: gate voltage

VG = VDD R2/(R1 + R2) = 12 × 8000/(8000 + 8000) = 6.000 V
VGS = VG - ID RS = 6.000 - 5 ID      (ID in mA, RS in kΩ)

Step 2: solve with the square law

ID = k (VGS - Vt)² = 0.5 (5.000 - 5 ID)²

Expanding and collecting terms gives 12.5 ID² - 26 ID + 12.5 = 0, so

ID = 1.326 mA   or   0.754 mA

For ID = 1.326 mA the gate-source voltage would be VGS = -0.63 V, which is below Vt (the MOSFET would be off), so this root is rejected.

Step 3: results

ID  = 0.754 mA
VGS = 6.000 - 5 × 0.754 = 2.228 V
VDS = VDD - ID (RD + RS) = 12 - 0.754 × (5 + 5) = 4.457 V

Saturation check

Saturation requires VDS > VGS - Vt:

VGS - Vt = 1.228 V ,   VDS = 4.457 V  >  1.228 V

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

Answer: ID ≈ 0.75 mA, VGS ≈ 2.23 V, VDS ≈ 4.46 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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