--- title: "MOSFET ID, VDS, region and small-signal circuit" subject: "Electronics Devices and Circuit" chapter: 3 key: "semester_3/electronics_devices_and_circuit:mosfet-id-vds-region-ac-equivalent" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/mosfet-id-vds-region-ac-equivalent site: BCT Engineering --- # MOSFET ID, VDS, region and small-signal circuit Electronics Devices and Circuit · Chapter 3: Field-Effect Transistor - Asked in: 2072 Chaitra, 3+3+2+2 marks - Video: [MOSFET Transconductance and MOSFET Small Signal Model Explained](https://www.youtube.com/watch?v=utBRWK4GmmE) — ALL ABOUT ELECTRONICS - 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-id-vds-region-ac-equivalent - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3 ## Question For the circuit given below, find ID and VDS. Also determine its region of operation and small signal ac equivalent circuit. Given data are: Vt = 1 V, k = 0.5 mA/V². [Figure: enhancement MOSFET with VDD = 10 V; gate divider R1 = 10 MΩ from VDD to gate and R2 = 10 MΩ from gate to ground; drain resistor RD = 6 kΩ; source resistor RS = 6 kΩ to ground] ## Short answer From VG=5V, VGS=5−6ID. Solve ID=0.5(VGS−1)², reject invalid root. ID=0.5mA, VDS=10−0.5×12=4V. Saturation since VDS=4V>VGS−Vt=1V. gm=2k(VGS−Vt)=1mS. ## Answer **Given:** VDD = 10 V, R1 = R2 = 10 MΩ, RD = RS = 6 kΩ, Vt = 1 V, k = 0.5 mA/V² (ID = k (VGS − Vt)²). #### 1. ID and VDS The gate current is zero: VG = 10 × 10/20 = 5 V VGS = VG − ID RS = 5 − 6 ID (mA, kΩ) ID = 0.5 (5 − 6 ID − 1)² = 0.5 (4 − 6 ID)² 18 ID² − 25 ID + 8 = 0 → ID = 0.889 mA or 0.5 mA For ID = 0.889 mA, VGS = 5 − 5.33 = −0.33 V < Vt, so the device would be off; this root is rejected. **ID = 0.5 mA**, VGS = 5 − 3 = 2 V **VDS** = VDD − ID (RD + RS) = 10 − 0.5 × 12 = **4 V** #### 2. Region of operation VGS − Vt = 2 − 1 = 1 V. VGS > Vt, so a channel exists, and VDS = 4 V > 1 V = VGS − Vt. Hence the MOSFET operates in the **saturation (pinch-off) region**. #### 3. Small-signal ac equivalent circuit gm = 2k (VGS − Vt) = 2 × 0.5 × 1 = **1 mA/V = 1 mS** The gate draws no current, so the input sees only the gate bias resistance RG = R1 ∥ R2 = 5 MΩ. Channel-length modulation is ignored (ro = ∞). Assuming the source is bypassed by a capacitor (common-source stage): ```text G o-----+ +----------+-----o D | + | | [RG] vgs gm·vgs (↓) [RD] 5 MΩ - | 6 kΩ | | | S o-----+-----------------+----------+-----o S (ground) ``` In words: RG = 5 MΩ is connected between gate and source; between drain and source there is the current source gm·vgs = 1 mS × vgs (flowing from drain to source) in parallel with RD = 6 kΩ (and RL, if connected). The gate and drain are separate nodes. Voltage gain: Av = −gm RD = −1 mS × 6 kΩ = −6 (or −gm (RD ∥ RL) with a load). If RS is not bypassed, RS appears in series with the source terminal and Av = −gm RD/(1 + gm RS) = −6/7 = −0.86. **Answer: ID = 0.5 mA, VDS = 4 V, saturation region, gm = 1 mS.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. With VG = 5 V, RS = 6 kΩ, which root of ID is rejected and why? A. 0.889 mA, as it gives VGS = -0.33 V < Vt (device off) B. 0.5 mA, as it is smaller C. Both are valid D. 0.889 mA since it exceeds 0.5 mA Answer: A. VGS must exceed Vt = 1 V; only ID = 0.5 mA (VGS = 2 V) satisfies this. (Key point: Solving for ID) 2. Why is the MOSFET in saturation here? A. VDS = 4 V > VGS - Vt = 1 V B. VGS < Vt C. VDS = 0 D. ID = IDSS Answer: A. VGS > Vt and VDS exceeds VGS - Vt, so it is in saturation. (Key point: Region) 3. What is gm = 2k(VGS - Vt) here? A. 1 mS B. 2 mS C. 0.5 mS D. 6 mS Answer: A. gm = 2*0.5*(2-1) = 1 mA/V. (Key point: Small-signal parameters) ## Close variants set in other exams - [E-MOSFET ID, VDS and region (k 0.12, Vt 5 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Aemosfet-id-vds-region-22m-18m) — Electronics Devices and Circuit, chapter 3, asked 2× ## Related in Electronics Devices and Circuit - [MOSFET ID and VDS (VDD 10 V, 6 kΩ, Vt=1 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Amosfet-id-vds-10m-6k) — Electronics Devices and Circuit, chapter 3, asked 7× - [MOSFET drain current and VDS (Vt=1 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Amosfet-bias-8m-5k-vt1) — Electronics Devices and Circuit, chapter 3, asked 7× - [MOSFET ID and VDS (188 kΩ/47 kΩ, Vt=1 V)](https://www.bctengineer.com/question.md?key=semester_3%2Felectronics_devices_and_circuit%3Amosfet-id-vds-188k-vt1) — Electronics Devices and Circuit, chapter 3, asked 7× - [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× - [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