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JFET ID and VDS (IDSS 10 mA, VP -4 V, 25 V)

Short answer

VG = 4.167 V from divider; VGS = 4.167 – ID·1kΩ. Solving ID = 10(1 – VGS/(–4))² yields ID = 5.26 mA. VDS = 25 – 5.26×(2.2k + 1k) = 8.15 V.

Find ID and VDS for the given circuit. [Figure: n-channel JFET with +25 V supply; gate divider of 1.5 MΩ from +25 V to gate and 300 kΩ from gate to ground; 2.2 kΩ drain resistor from +25 V to drain; 1 kΩ source resistor to ground; VP = -4 V, IDSS = 10 mA]

Method: find the gate voltage, write VGS in terms of ID, combine this with Shockley's equation, solve the quadratic and reject the root that is not physically possible. The gate current is zero.

Given

VDD = 25 V, IDSS = 10 mA, VP = -4 V, RD = 2.2 kΩ, RS = 1 kΩ, R1 = 1.5 MΩ, R2 = 300 kΩ (in the formulas below, resistances are in kΩ)

Step 1: gate voltage and bias line

VG = VDD R2/(R1 + R2) = 25 × 300/(1500 + 300) = 4.167 V
VGS = VG - ID RS = 4.167 - 1 ID      (ID in mA)

Step 2: Shockley equation

ID = IDSS (1 - VGS/VP)² = 10 [1 - (4.167 - 1 ID)/(-4)]²
   = 10 (2.0417 - 0.25 ID)²

Expanding gives the quadratic:

0.625 ID² - 11.2083 ID + 41.684 = 0
ID = 12.669 mA   or   5.264 mA

Step 3: choose the valid root

The root 12.669 mA gives VGS = -8.50 V, which is outside the range VP ≤ VGS ≤ 0 (the channel would be cut off), so it is rejected.

ID  = 5.264 mA
VGS = 4.167 - 1 × 5.264 = -1.098 V
VDS = VDD - ID (RD + RS) = 25 - 5.264 × (2.2 + 1) = 8.154 V

Answer: ID ≈ 5.26 mA, VGS ≈ -1.10 V, VDS ≈ 8.15 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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