--- title: "MOSFET Q point graphically" subject: "Electronics Devices and Circuit" chapter: 3 key: "semester_3/electronics_devices_and_circuit:mosfet-q-point-10m-2k" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/mosfet-q-point-10m-2k site: BCT Engineering --- # MOSFET Q point graphically Electronics Devices and Circuit · Chapter 3: Field-Effect Transistor - Asked in: 2079 Baisakh, 6+2 marks - Video: [50- MOSFET DC Analysis, Q point, Biasing and LoadLine](https://www.youtube.com/watch?v=WZpbbSU7u_c) — Learners - 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/mosfet-q-point-10m-2k - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3 ## Question Find Q point and show it graphically. [Figure: enhancement MOSFET with +12 V supply, 2 kΩ drain resistor, 10 MΩ feedback resistor from drain to gate, 1 μF coupling capacitors at input Vi and output Vo; given ID(on) = 6 mA, VGS(on) = 8 V, VGS(Th) = 3 V] ## Short answer Feedback gives VGS = VDS. DC load: VDS = 12 – 2 ID. With ID = 0.24(VGS – 3)², solving yields ID = 2.79 mA, VDS = VGS = 6.41 V. VDS > VGS – Vt, so device is saturated. Q-point: (2.79 mA, 6.41 V). ## Answer The circuit is a drain-feedback biased enhancement MOSFET: the 10 MΩ resistor connects the drain to the gate. Since IG = 0, there is no drop across it, so VGS = VDS. #### Step 1: find k Given ID(on) = 6 mA at VGS(on) = 8 V and VGS(Th) = 3 V: ```text ID = k (VGS - VT)² → k = ID(on)/(VGS(on) - VT)² = 6 mA/(8 - 3)² = 0.24 mA/V² ``` #### Step 2: bias equations ```text Drain loop (dc load line): VDS = VDD - ID RD = 12 - 2 ID (ID in mA) Feedback: VGS = VDS Device: ID = 0.24 (VGS - 3)² ``` Substituting VGS = 12 - 2 ID: ```text ID = 0.24 (12 - 2 ID - 3)² = 0.24 (9 - 2 ID)² = 0.24 (81 - 36 ID + 4 ID²) 0.96 ID² - 9.64 ID + 19.44 = 0 ID = [9.64 ± √(92.9 - 74.65)]/1.92 = [9.64 ± 4.27]/1.92 ID = 7.25 mA or 2.79 mA ``` The root 7.25 mA gives VGS = 12 - 14.5 < 0 (impossible: MOSFET off), so it is rejected. ```text IDQ = 2.79 mA VGSQ = VDSQ = 12 - 2 × 2.79 = 6.41 V ``` Check saturation: VDS = 6.41 V > VGS - VT = 3.41 V. Yes. #### Graphical solution Plot the transfer curve ID = 0.24 (VGS - 3)² and the bias line ID = (12 - VGS)/2 (since VGS = VDS): | VGS (V) | 3 | 4 | 5 | 6 | 7 | 8 | |---|---|---|---|---|---|---| | ID curve (mA) | 0 | 0.24 | 0.96 | 2.16 | 3.84 | 6.0 | The line passes through (12 V, 0) and (0 V, 6 mA), e.g. at VGS = 6 V it is 3 mA and at 7 V it is 2.5 mA. ```text ID(mA) 6 |* . | \ . transfer curve | \ . 3 | \ . | \ .*Q (6.41 V, 2.79 mA) | . \ +--+---+--+-----+---- VGS 3 6.4 12 ``` The intersection of the curve and the load/bias line is the Q-point. **Answer: Q-point: IDQ ≈ 2.79 mA, VDSQ = VGSQ ≈ 6.41 V.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is VGS = VDS in the drain-feedback circuit? A. The MOSFET is in cut-off B. No gate current flows, so no voltage drops across the 10 MΩ resistor C. The drain resistor is zero D. The supply is applied directly to the gate Answer: B. IG = 0 means the feedback resistor drops no voltage, so gate and drain are at the same potential. (Key point: Drain feedback gives VGS = VDS) 2. How is k found from ID(on) = 6 mA, VGS(on) = 8 V, VT = 3 V? A. k = 6/8 B. k = 6/(8-3)^2 = 0.24 mA/V^2 C. k = 6*(8-3)^2 D. k = 6/3 Answer: B. From ID = k(VGS-VT)^2, k = ID(on)/(VGS(on)-VT)^2 = 6/25 = 0.24. (Key point: Finding k) 3. Graphically, where is the Q-point? A. Where the transfer curve meets the VGS axis B. At the intersection of the transfer curve and the bias line ID = (12 - VGS)/2 C. At VGS = VT D. At ID = ID(on) Answer: B. Since VGS = VDS, the load line becomes ID = (12-VGS)/2, and it crosses the transfer curve at the Q-point (6.41 V, 2.79 mA). (Key point: Q-point as intersection) ## Related in Electronics Devices and Circuit - [MOSFET ID, VGS, VDS and saturation check](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Amosfet-saturation-check-10m-6m8) — 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 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× - [E-MOSFET IDQ and VDSQ with graphical Q point](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Aemosfet-q-point-22m-18m) — Electronics Devices and Circuit, chapter 3 - [JFET ID and VDS (VP -5.5 V, IDSS 10 mA), pinch-off check](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Ajfet-self-bias-vp-5p5) — Electronics Devices and Circuit, chapter 3, asked 3× - [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× ## Related in other subjects - [Thevenin: 4 kΩ resistor, 12 V, 10 mA](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Athevenin-12v-10ma) — Basic Electronics Engineering, chapter 1, asked 2× - [Log-antilog circuit: Vo in terms of Vi and VR](https://www.bctengineer.com/question.md?key=semester_4%2Fadvanced_electronics%3Alog-antilog-circuit-vo-vi) — Advanced Electronics, chapter 5, asked 3× - [Block diagram and Vo/Vi of L-R ladder network](https://www.bctengineer.com/question.md?key=semester_3%2Fcontrol_systems%3Abd-l1-r1-l2-r2) — Control Systems, chapter 2 - [N-channel enhancement MOSFET construction, working](https://www.bctengineer.com/question.md?key=semester_2%2Fbasic_electronics_engineering%3Amosfet-n-channel-enhancement) — Basic Electronics Engineering, chapter 3, asked 9×