--- title: "n-channel depletion MOSFET operation" subject: "Fundamental of Electrical and Electronics Engineering" chapter: 5 key: "semester_1/fundamental_of_electrical_and_electronics_engineering:depletion-mosfet-operation" source: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/depletion-mosfet-operation site: BCT Engineering --- # n-channel depletion MOSFET operation Fundamental of Electrical and Electronics Engineering · Chapter 5: Transistor - Asked in: 2081 Chaitra (new course), 6 marks - Video: [MOSFET- Depletion Type MOSFET Explained (Construction, working and Characteristics Explained)](https://www.youtube.com/watch?v=XqGBNyhImV4) — ALL ABOUT ELECTRONICS - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_1/fundamental_of_electrical_and_electronics_engineering/depletion-mosfet-operation - Whole chapter: https://www.bctengineer.com/md/semester_1/fundamental_of_electrical_and_electronics_engineering/chapter-5 ## Question Explain the operation of n-channel depletion MOSFET with the necessary diagrams. ## Short answer An n‑channel D‑MOSFET has a built‑in n‑channel. At V_GS = 0, I_D = I_DSS. Negative V_GS depletes the channel (I_D decreases; cut‑off at V_P). Positive V_GS enhances it (I_D > I_DSS). I_D = I_DSS (1 − V_GS/V_P)^2. ## Answer An **n-channel depletion MOSFET (D-MOSFET)** has a channel built in during manufacture, so drain current flows even at V_GS = 0. It can work in two modes: **depletion mode** (V_GS negative) and **enhancement mode** (V_GS positive). #### Construction ```text S G D | __|__ | ___|___ |metal| ___|___ | n+ ||=SiO2=|| n+ | |_______|________|_______| | diffused n-channel | | | | p-type substrate | |__________________________| | SS (body, tied to S) ``` - A lightly doped **p-type substrate** forms the body. - Two heavily doped **n+ regions** form the **source** and **drain**. - A thin **n-type channel** is diffused between them, linking source and drain. - A thin **SiO₂** layer insulates the metal **gate** from the channel, so gate current ≈ 0 and input resistance is very high. #### Operation **1. V_GS = 0:** with a positive V_DS, electrons flow from source to drain through the existing n-channel. The current at V_GS = 0 with the channel pinched off is **I_DSS**. **2. V_GS negative (depletion mode):** the negative gate repels electrons from the channel and attracts holes from the p-substrate, which recombine with electrons. The number of free electrons in the channel falls (the channel is depleted), so I_D decreases. At **V_GS = V_P** (pinch-off voltage, e.g. −4 V) the channel is fully depleted and I_D = 0. **3. V_GS positive (enhancement mode):** the positive gate attracts more electrons (minority carriers) from the substrate into the channel. The channel becomes richer in carriers and I_D rises above I_DSS. This is possible because the gate is insulated, so no gate current flows. (A JFET cannot do this.) **4. Effect of V_DS:** as V_DS rises, the channel narrows near the drain because the gate-to-drain voltage becomes more negative. At V_DS = V_GS − V_P the channel pinches off at the drain end, and I_D becomes nearly constant (saturation). #### Characteristics In saturation the transfer characteristic follows **Shockley's equation**: I_D = I_DSS (1 − V_GS/V_P)². ```text Transfer Drain I_D I_D | ______ +1 V (enh.) | / | /______ 0 V (I_DSS) | / | /_______ −1 V I_DSS * |/________ −2 V (depl.) | / +------------- V_DS |_/______ V_GS V_P 0 + ``` ```text Symbol (n-channel D-MOSFET): D | G ||--+ solid line = ||<-+-- SS built-in channel ||--+ S ``` #### Summary | V_GS | Mode | I_D | |---|---|---| | Positive | Enhancement | > I_DSS | | 0 | — | = I_DSS | | Negative (> V_P) | Depletion | < I_DSS | | = V_P | Cutoff | 0 | D-MOSFETs are used in RF amplifiers and as constant-current loads because they conduct at zero bias and accept both polarities of gate voltage. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. What distinguishes a depletion MOSFET from an enhancement MOSFET? A. It has a diffused channel, so I_D flows at V_GS = 0 B. It has no channel at V_GS = 0 C. It has no oxide D. It has a p-type channel always Answer: A. The built-in channel gives I_DSS at V_GS = 0. (Key point: Built-in channel) 2. What happens when V_GS is made negative in an n-channel D-MOSFET? A. I_D increases above I_DSS B. Electrons are repelled, channel is depleted, and I_D decreases until V_P C. The channel widens D. The oxide conducts Answer: B. At V_GS = V_P the channel is fully depleted and I_D = 0. (Key point: Negative V_GS) 3. Why can a D-MOSFET work with positive V_GS? A. The gate forward biases a pn junction B. It cannot work with positive V_GS C. The insulated gate draws no current, so more electrons enter the channel and I_D > I_DSS D. The substrate becomes n-type Answer: C. A JFET cannot do this because its gate junction would conduct. (Key point: Positive V_GS) 4. In saturation the transfer characteristic follows: A. I_D = k (V_GS - V_T)^2 B. I_D = I_DSS e^(V_GS/V_P) C. I_D = I_DSS (1 + V_GS/V_P) D. I_D = I_DSS (1 - V_GS/V_P)^2 Answer: D. Shockley's equation holds for the depletion device, with V_P negative for n-channel. 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