--- title: "n-channel DMOSFET and JFET transconductance" subject: "Electronics Devices and Circuit" chapter: 3 key: "semester_3/electronics_devices_and_circuit:dmosfet-working-jfet-transconductance" source: https://www.bctengineer.com/q/semester_3/electronics_devices_and_circuit/dmosfet-working-jfet-transconductance site: BCT Engineering --- # n-channel DMOSFET and JFET transconductance Electronics Devices and Circuit · Chapter 3: Field-Effect Transistor - Asked in: 2079 Baisakh, 6+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/dmosfet-working-jfet-transconductance - Whole chapter: https://www.bctengineer.com/md/semester_3/electronics_devices_and_circuit/chapter-3 ## Question Explain the working of n channel DMOSFET with characteristics curves. Derive expression for JFET transconductance. ## Short answer An n-channel D-MOSFET has a built-in channel; VGS=0 gives IDSS. Negative VGS depletes the channel (depletion mode), positive VGS enhances it (enhancement mode). JFET transconductance: gm = dID/dVGS = (2IDSS/|VP|)(1–VGS/VP) = gm0√(ID/IDSS). ## Answer #### Working of n-channel D-MOSFET The n-channel depletion MOSFET has a p-type substrate, n⁺ source and drain, and a thin n-type channel built in between them, covered by SiO₂ and a metal gate. ```text S G D | +--||--+ | +--+---+-------+--+---+ | n+ n-channel n+ | | p-type substrate | +---------------------+ ``` - **VGS = 0:** electrons flow through the built-in channel when VDS is applied: ID = IDSS. - **VGS negative:** the gate repels electrons, the channel is depleted (narrower), and ID < IDSS. At VGS = VP the channel is cut off (ID ≈ 0). This is the **depletion mode**. - **VGS positive:** the gate attracts more electrons, the channel widens, and ID > IDSS. This is the **enhancement mode**. - At small VDS, ID varies linearly with VDS (ohmic); at VDS ≥ VGS - VP, the channel pinches off at the drain end and ID saturates: ID = IDSS (1 - VGS/VP)². - The insulated gate gives a very high input resistance (about 10¹⁴ Ω), and the gate may be driven positive or negative. **Characteristics:** ```text ID (drain) ID (transfer) | ____ VGS = +1 | / | _/____ VGS = 0 | _.' | / /____ VGS = -1 | IDSS .-' |/ /_____ VGS = -2 | .-' +----+---- VDS +--+------+---- VGS VP 0 ``` #### Transconductance of JFET (derivation) gm = ∂ID/∂VGS at constant VDS. In the saturation region: ```text ID = IDSS (1 - VGS/VP)² gm = dID/dVGS = 2 IDSS (1 - VGS/VP) · (-1/VP) = (2 IDSS/|VP|) (1 - VGS/VP) (VP is negative) gm = gm0 (1 - VGS/VP) , gm0 = 2 IDSS/|VP| ``` Using ID: gm = gm0 √(ID/IDSS). It is maximum at VGS = 0 and zero at VGS = VP. Example: IDSS = 10 mA, VP = -4 V: gm0 = 5 mA/V. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What makes a depletion MOSFET different from an enhancement one? A. It has a built-in n-channel between source and drain B. It has no gate C. It has no substrate D. It has a p-n junction gate Answer: A. The D-MOSFET has a physically built-in channel, so it conducts (ID = IDSS) at VGS = 0. (Key point: D-MOSFET structure) 2. What happens in an n-channel D-MOSFET when VGS is made positive? A. Channel narrows and ID falls below IDSS B. Channel widens and ID rises above IDSS (enhancement mode) C. Channel is cut off D. ID stays at IDSS Answer: B. A positive gate attracts more electrons, widening the channel so ID > IDSS. (Key point: Modes of operation) 3. Differentiating ID = IDSS(1 - VGS/VP)^2 w.r.t. VGS gives gm equal to: A. IDSS/VP B. (2 IDSS/|VP|)(1 - VGS/VP) C. IDSS (1 - VGS/VP)^2 D. 2 IDSS VP Answer: B. Chain rule gives 2 IDSS(1-VGS/VP)(-1/VP) = gm0(1 - VGS/VP) with gm0 = 2IDSS/|VP|. (Key point: Transconductance derivation) 4. Where is the JFET transconductance maximum? A. At VGS = VP B. At VGS = 0, where gm = gm0 C. At ID = 0 D. At VDS = 0 Answer: B. gm = gm0(1 - VGS/VP) is largest at VGS = 0 and zero at VGS = VP. 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