--- title: "JFET ID and VDS (VP -5.5 V, IDSS 10 mA), pinch-off check" subject: "Electronics Devices and Circuit" chapter: 3 key: "semester_3/electronics_devices_and_circuit:jfet-self-bias-vp-5p5" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/jfet-self-bias-vp-5p5 site: BCT Engineering --- # JFET ID and VDS (VP -5.5 V, IDSS 10 mA), pinch-off check Electronics Devices and Circuit · Chapter 3: Field-Effect Transistor - Asked in: 2079 Bhadra, 8 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/jfet-self-bias-vp-5p5 - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3 ## Question Find ID and VDS for the given circuit. Given data are VP = -5.5 V, IDSS = 10 mA and assume all the capacitors are ideal and check whether transistor is operating in pinch off region or not? [Figure: n-channel JFET amplifier with VDD = 20 V; gate divider of 1.5 MΩ from VDD to gate and 330 kΩ from gate to ground; drain resistor 2 kΩ; source resistor 1 kΩ with bypass capacitor C3; input coupling capacitor C1 and output coupling capacitor C2] ## Short answer DC bias: VG = 3.607 V, VGS = 3.607 – ID. Solving Shockley equation gives ID = 5.16 mA, VGS = –1.55 V. VDS = 20 – 5.16(2+1) = 4.53 V. Since VDS (4.53 V) > VGS–VP (3.95 V), the JFET operates in pinch-off. ## Answer The capacitors are open for dc, and C3 only bypasses RS for ac, so the dc analysis is that of a voltage-divider biased JFET. 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 = 10 mA, VP = -5.5 V, RD = 2 kΩ, RS = 1 kΩ, R1 = 1.5 MΩ, R2 = 330 kΩ (in the formulas below, resistances are in kΩ) #### Step 1: gate voltage and bias line ```text VG = VDD R2/(R1 + R2) = 20 × 330/(1500 + 330) = 3.607 V VGS = VG - ID RS = 3.607 - 1 ID (ID in mA) ``` #### Step 2: Shockley equation ```text ID = IDSS (1 - VGS/VP)² = 10 [1 - (3.607 - 1 ID)/(-5.5)]² = 10 (1.6557 - 0.1818 ID)² ``` Expanding gives the quadratic: ```text 0.3306 ID² - 7.0209 ID + 27.4147 = 0 ID = 16.081 mA or 5.157 mA ``` #### Step 3: choose the valid root The root 16.081 mA gives VGS = -12.47 V, which is outside the range VP ≤ VGS ≤ 0 (the channel would be cut off), so it is rejected. ```text ID = 5.157 mA VGS = 3.607 - 1 × 5.157 = -1.550 V VDS = VDD - ID (RD + RS) = 20 - 5.157 × (2 + 1) = 4.529 V ``` #### Pinch-off (saturation) check The JFET is in the pinch-off region if VDS > VGS - VP: ```text VGS - VP = -1.550 - (-5.5) = 3.950 V VDS = 4.529 V > 3.950 V ``` Yes, VDS is greater than VGS - VP, so the transistor operates in the pinch-off (saturation) region and the Shockley equation used is valid. **Answer: ID ≈ 5.16 mA, VGS ≈ -1.55 V, VDS ≈ 4.53 V.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why can the capacitors be ignored in finding the dc bias? A. They behave as open circuits for dc B. They are shorts for dc C. They supply the gate current D. They set the pinch-off voltage Answer: A. Capacitors block dc; C3 only bypasses RS for ac, so dc analysis is a plain voltage-divider bias. (Key point: DC analysis) 2. VG for VDD = 20 V, R1 = 1.5 MΩ, R2 = 330 kΩ is: A. 3.61 V B. 16.4 V C. 10 V D. 5.5 V Answer: A. VG = 20*330/1830 = 3.607 V. (Key point: Gate voltage) 3. Why is ID = 16.08 mA rejected? A. It gives VGS = -12.47 V, outside VP <= VGS <= 0 B. It is below IDSS C. It gives VDS > VDD D. Roots are always taken as the larger one Answer: A. VGS = 3.607 - 16.08 is far beyond VP = -5.5 V, which is impossible. (Key point: Rejecting a root) 4. VDS = 4.529 V and VGS - VP = 3.95 V. What follows? A. Not in pinch-off, since VDS > VGS - VP is false B. In pinch-off, since VDS > VGS - VP C. Cut off D. Ohmic region Answer: B. Pinch-off requires VDS > VGS - VP, which holds, so Shockley's equation is valid. (Key point: Pinch-off check) ## Close variants set in other exams - [JFET ID and VDS (VP -3.5 V, IDSS 10 mA), pinch-off check](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ajfet-self-bias-vp-3p5) — Electronics Devices and Circuit, chapter 3, asked 3× - [JFET ID and VDS (VP -5.5 V, IDSS 12 mA)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ajfet-self-bias-vp-5p5-idss12) — Electronics Devices and Circuit, chapter 3, asked 3× ## Related in Electronics Devices and Circuit - [JFET ID and VDS (IDSS 10 mA, VP -4 V, 25 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ajfet-divider-bias-idss10-vp4) — Electronics Devices and Circuit, chapter 3, asked 2× - [JFET ID and VDS (IDSS 18 mA, VP -5 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ajfet-divider-bias-idss18) — Electronics Devices and Circuit, chapter 3, asked 5× - [JFET ID and VDS (IDSS 15 mA, VP -4.5 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ajfet-divider-bias-idss15) — Electronics Devices and Circuit, chapter 3, asked 5× - [JFET DC operating point (IDSS 12 mA, VP -4 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ajfet-dc-operating-point-420k-180k) — Electronics Devices and Circuit, chapter 3 - [Depletion MOSFET Q-points (IDSS 10 mA, VP -4 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Adepletion-mosfet-qpoints) — Electronics Devices and Circuit, chapter 3 - [JFET divider bias ID, VGS, VDS (VP -2 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ajfet-voltage-divider-numeric-vp-2v) — Electronics Devices and Circuit, chapter 3