--- title: "E-MOSFET IDQ and VDSQ with graphical Q point" subject: "Electronics Devices and Circuit" chapter: 3 key: "semester_3/electronics_devices_and_circuit:emosfet-q-point-22m-18m" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/emosfet-q-point-22m-18m site: BCT Engineering --- # E-MOSFET IDQ and VDSQ with graphical Q point Electronics Devices and Circuit · Chapter 3: Field-Effect Transistor - Asked in: 2075 Ashwin, 5+3 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/emosfet-q-point-22m-18m - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3 ## Question Find IDQ and VDSQ from the following circuit. Show Q point graphically. [Figure: enhancement MOSFET with +40 V supply; gate divider of 22 MΩ from +40 V to gate and 18 MΩ from gate to ground; 3 kΩ drain resistor; 0.82 kΩ source resistor to ground; VGS(TH) = 5 V, ID(on) = 3 mA, VGS(on) = 10 V] ## Short answer k = 0.12 mA/V²; VG = 18 V. VGS = 18 – 0.82ID. Solving ID = 0.12(VGS–5)² gives IDQ = 6.72 mA, VGSQ = 12.5 V. VDSQ = 40 – 6.72(3+0.82) = 14.3 V. VDS > VGS–Vt, confirming saturation. ## Answer **Given:** VDD = 40 V, R1 = 22 MΩ, R2 = 18 MΩ, RD = 3 kΩ, RS = 0.82 kΩ, VGS(Th) = 5 V, ID(on) = 3 mA at VGS(on) = 10 V. #### Step 1: Find k ID = k (VGS − VGS(Th))² k = ID(on) / (VGS(on) − VGS(Th))² = 3 mA / (10 − 5)² = **0.12 mA/V²** #### Step 2: Gate voltage (gate current is zero) VG = VDD · R2/(R1 + R2) = 40 × 18/40 = **18 V** #### Step 3: Bias equation VGS = VG − ID RS = 18 − 0.82 ID (ID in mA, RS in kΩ) ID = 0.12 (18 − 0.82 ID − 5)² = 0.12 (13 − 0.82 ID)² Expanding: 0.0807 ID² − 3.5584 ID + 20.28 = 0, that is ID² − 44.10 ID + 251.3 = 0 ID = 6.72 mA or 37.4 mA For ID = 37.4 mA, VGS = 18 − 30.7 = −12.6 V, which is below the threshold, so this root is rejected. **IDQ = 6.72 mA**, VGSQ = 18 − 0.82(6.72) = 12.49 V #### Step 4: VDS VDSQ = VDD − ID (RD + RS) = 40 − 6.72 × 3.82 = **14.3 V** Check: VGS − VGS(Th) = 7.49 V < VDS = 14.3 V, so the MOSFET is in saturation and the square law is valid. #### Graphical Q point **(a) Transfer characteristic.** Plot ID = 0.12 (VGS − 5)²: | VGS (V) | 5 | 6 | 8 | 10 | 12 | 14 | |---|---|---|---|---|---|---| | ID (mA) | 0 | 0.12 | 1.08 | 3.0 | 5.88 | 9.72 | Draw the bias line ID = (18 − VGS)/0.82 through (VGS = 18 V, ID = 0) and (VGS = 0, ID = 21.95 mA). It cuts the curve at **Q = (12.49 V, 6.72 mA)**. **(b) Output characteristic.** DC load line: VDS = 40 − 3.82 ID, through (VDS = 40 V, ID = 0) and (VDS = 0, ID = 10.47 mA). Q lies on the VGS = 12.49 V curve at VDS = 14.3 V, ID = 6.72 mA. ```text ID(mA) 10.5 * | * 6.7 |------Q (VGS=12.5V) | \ * | \ * +-----------14.3---*-----> VDS 40 V ``` **Answer: IDQ = 6.72 mA, VDSQ = 14.3 V, VGSQ = 12.5 V (saturation region).** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With ID(on) = 3 mA at VGS(on) = 10 V and VGS(Th) = 5 V, k is: A. 0.12 mA/V^2 B. 0.3 mA/V^2 C. 0.6 mA/V^2 D. 0.06 mA/V^2 Answer: A. k = 3/(10-5)^2 = 0.12 mA/V^2. (Key point: Finding k) 2. VG for 40 V supply, R1 = 22 MΩ, R2 = 18 MΩ is: A. 22 V B. 18 V C. 40 V D. 9 V Answer: B. VG = 40*18/40 = 18 V since IG = 0. (Key point: Gate voltage) 3. Which formula gives VDSQ? A. VDD - ID RD B. VDD - ID (RD + RS) C. VG - ID RS D. VDD/2 Answer: B. VDS = 40 - 6.72*(3+0.82) = 14.3 V. (Key point: VDSQ) 4. On the transfer plot, where is Q? A. Where the curve ID = 0.12(VGS-5)^2 meets the bias line ID = (18 - VGS)/0.82 B. At VGS = 5 V C. At ID = 3 mA D. On the VDS axis Answer: A. The intersection gives Q = (12.49 V, 6.72 mA). 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