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JFET Q-point (IDSS 8 mA, VP -6 V)
Short answer
VG=0, so VGS=−ID×1k. Shockley equation gives 8(1−ID/6)²=ID; solving yields IDQ=2.59 mA, VGSQ=−2.59 V.
Find IDQ and VGSQ from the following circuit. [Figure: n-channel JFET with +20 V supply; 3.3 kΩ drain resistor; 1 MΩ gate resistor to ground; 1 kΩ source resistor to ground; IDSS = 8 mA, VP = -6 V]
This is a self-bias JFET circuit: the gate is returned to ground through a 1 MΩ resistor. Since IG ≈ 0, no voltage drops across it and VG = 0.
Given
VDD = 20 V, RD = 3.3 kΩ, RG = 1 MΩ, RS = 1 kΩ, IDSS = 8 mA, VP = -6 V.
Bias line
VGS = VG - ID RS = -ID RS = -(1 kΩ) ID (ID in mA gives VGS in V)
Shockley equation
ID = IDSS (1 - VGS/VP)² = 8 (1 - ID/6)²
= 8 (1 - ID/3 + ID²/36)
= 8 - 2.667 ID + 0.2222 ID²
0.2222 ID² - 3.667 ID + 8 = 0
ID = [3.667 ± √(13.44 - 7.111)]/0.4444 = [3.667 ± 2.516]/0.4444
ID = 13.91 mA or 2.59 mA
The root 13.91 mA would give VGS = -13.9 V, which is below VP = -6 V (the JFET would be off), so it is rejected.
IDQ = 2.59 mA
VGSQ = -ID RS = -2.59 V
Also VDS = VDD - ID (RD + RS) = 20 - 2.59 × 4.3 = 8.87 V. Since VDS > VGS - VP = 3.41 V, the JFET is in the pinch-off (saturation) region.
Answer: IDQ ≈ 2.59 mA, VGSQ ≈ -2.59 V.
✓ Answer checked by Claude Opus 5.5 (high effort)
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