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JFET ID and VDS (IDSS 18 mA, VP -5 V)

Short answer

VG=4 V from divider; VGS=4−1.65ID. Solving ID=18(1−VGS/(−5))² gives quadratic, valid root ID=4.02 mA. Then VDS=20−4.02×(1.85+1.65)=5.92 V.

The n-channel JFET in the figure below has IDSS = 18mA and Vp = -5V. Determine the values of ID and VDS. [Figure: JFET 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]

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 = 20 V, IDSS = 18 mA, VP = -5 V, RD = 1.85 kΩ, RS = 1.65 kΩ, R1 = 188 kΩ, R2 = 47 kΩ (in the formulas below, resistances are in kΩ)

Step 1: gate voltage and bias line

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

Step 2: Shockley equation

ID = IDSS (1 - VGS/VP)² = 18 [1 - (4.000 - 1.65 ID)/(-5)]²
   = 18 (1.8 - 0.33 ID)²

Expanding gives the quadratic:

1.9602 ID² - 22.384 ID + 58.32 = 0
ID = 7.397 mA   or   4.022 mA

Step 3: choose the valid root

The root 7.397 mA gives VGS = -8.21 V, which is outside the range VP ≤ VGS ≤ 0 (the channel would be cut off), so it is rejected.

ID  = 4.022 mA
VGS = 4.000 - 1.65 × 4.022 = -2.636 V
VDS = VDD - ID (RD + RS) = 20 - 4.022 × (1.85 + 1.65) = 5.923 V

Answer: ID ≈ 4.02 mA, VGS ≈ -2.64 V, VDS ≈ 5.92 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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