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JFET ID and VDS (VP -5.5 V, IDSS 12 mA)

Short answer

VG=3.246V, VGS=3.246−ID. Using ID=12(1−VGS/(−5.5))² gives ID=5.14mA, VDS=2.57V, but VDS< VGS−VP so not pinch-off. Ohmic region solution: ID≈5.07mA, VDS≈2.80V.

Find ID and VDS for the given circuit. Given data are: VP = -5.5V, IDSS = 12 mA and assume all capacitors are ideal and check whether transistor is operating in pinch off region or not? [Figure: n-channel JFET amplifier with VDD = 18 V; gate divider 1.5 MΩ from VDD to gate and 330 kΩ from gate to ground; 2 kΩ drain resistor; 1 kΩ source resistor with bypass capacitor C3; input coupling capacitor C1 and output coupling capacitor C2]

Given: VDD = 18 V, R1 = 1.5 MΩ, R2 = 330 kΩ, RD = 2 kΩ, RS = 1 kΩ, VP = −5.5 V, IDSS = 12 mA. Capacitors are open at dc.

Gate voltage

VG = 18 × 330/(1500 + 330) = 3.246 V

Step 1: assume pinch-off (saturation) (ID in mA, RS in kΩ)

VGS = VG − ID RS = 3.246 − ID

ID = 12 (1 − VGS/VP)² = 12 (1 + (3.246 − ID)/5.5)² = 12 (1.590 − 0.1818 ID)²

0.3967 ID² − 7.939 ID + 30.34 = 0

ID = 14.87 mA or 5.14 mA

ID = 14.87 mA is larger than IDSS (12 mA) and would give VGS = −11.6 V (beyond VP), which is impossible, so it is rejected.

ID = 5.14 mA, VGS = 3.246 − 5.14 = −1.90 V

VDS = VDD − ID (RD + RS) = 18 − 5.14 × 3 = 2.57 V

Step 2: check for the pinch-off region

For pinch-off (saturation) the JFET needs VDS ≥ VGS − VP:

VGS − VP = −1.90 + 5.5 = 3.60 V

VDS = 2.57 V < 3.60 V, so the condition is not satisfied. The JFET is not in the pinch-off region; it is in the ohmic (triode) region, and the saturation-equation values above are not exact.

Step 3: solve in the ohmic region

In the ohmic region the drain current is

ID = IDSS [2 (1 − VGS/VP)(VDS/|VP|) − (VDS/VP)²]

with VGS = 3.246 − ID and VDS = 18 − 3 ID. Solving this numerically (trial and error):

ID ≈ 5.07 mA, VGS ≈ −1.82 V, VDS = 18 − 3 × 5.07 ≈ 2.80 V

Check: VGS − VP = −1.82 + 5.5 = 3.68 V > VDS = 2.80 V, so the ohmic-region assumption is consistent. (As expected, the actual ID is slightly smaller, and VDS slightly larger, than the saturation-equation estimate.)

Answer: ID ≈ 5.07 mA, VDS ≈ 2.80 V (VGS ≈ −1.82 V). Since VDS < VGS − VP, the transistor is not in the pinch-off region; it operates in the ohmic region. (The saturation equation alone gives ID ≈ 5.14 mA, VDS ≈ 2.57 V, which fails this check.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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