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JFET ID and VDS (IDSS 15 mA, VP -4.5 V)

Short answer

VG = 4 V; VGS = 4 – 1.65 ID. From ID = 15(1 – VGS/(–4.5))², the valid root is ID = 3.78 mA. VDS = 20 – 3.78×(1.85 + 1.65) = 6.76 V.

The n-channel JFET in the figure below has IDSS= 15mA and VP = -4.5V. 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 = 15 mA, VP = -4.5 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 = 15 [1 - (4.000 - 1.65 ID)/(-4.5)]²
   = 15 (1.8889 - 0.3667 ID)²

Expanding gives the quadratic:

2.0167 ID² - 21.7778 ID + 53.5185 = 0
ID = 7.017 mA   or   3.782 mA

Step 3: choose the valid root

The root 7.017 mA gives VGS = -7.58 V, which is outside the range VP ≤ VGS ≤ 0 (the channel would be cut off), so it is rejected.

ID  = 3.782 mA
VGS = 4.000 - 1.65 × 3.782 = -2.240 V
VDS = VDD - ID (RD + RS) = 20 - 3.782 × (1.85 + 1.65) = 6.763 V

Answer: ID ≈ 3.78 mA, VGS ≈ -2.24 V, VDS ≈ 6.76 V.

✓ Answer checked by Claude Opus 5.5 (high effort)

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