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MOSFET ID and VDS (VDD 10 V, 6 kΩ, Vt=1 V)

Short answer

VG=5 V, VGS=5−6ID. Using ID=0.5(VGS−1)² yields ID=0.5 mA (rejecting larger root). Then VDS=10−0.5×(6+6)=4 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 VDD = 10 V; gate divider R1 = 10 MΩ from VDD to gate and R2 = 10 MΩ from gate to ground; drain resistor RD = 6 kΩ; source resistor RS = 6 kΩ 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 = 10 V, R1 = 10 MΩ, R2 = 10 MΩ, RD = 6 kΩ, RS = 6 kΩ (resistances in kΩ in the formulas), Vt = 1 V, k = 0.5 mA/V².

Step 1: gate voltage

VG = VDD R2/(R1 + R2) = 10 × 10000/(10000 + 10000) = 5.000 V
VGS = VG - ID RS = 5.000 - 6 ID      (ID in mA, RS in kΩ)

Step 2: solve with the square law

ID = k (VGS - Vt)² = 0.5 (4.000 - 6 ID)²

Expanding and collecting terms gives 18 ID² - 25 ID + 8 = 0, so

ID = 0.889 mA   or   0.500 mA

For ID = 0.889 mA the gate-source voltage would be VGS = -0.33 V, which is below Vt (the MOSFET would be off), so this root is rejected.

Step 3: results

ID  = 0.500 mA
VGS = 5.000 - 6 × 0.500 = 2.000 V
VDS = VDD - ID (RD + RS) = 10 - 0.500 × (6 + 6) = 4.000 V

Saturation check

Saturation requires VDS > VGS - Vt:

VGS - Vt = 1.000 V ,   VDS = 4.000 V  >  1.000 V

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

Answer: ID ≈ 0.50 mA, VGS ≈ 2.00 V, VDS ≈ 4.00 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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