MOSFET structure (gate, source, drain, body)

WHAT is a MOSFET made of?

For an n-channel MOSFET (NMOS):

Body is p-type
Source and drain are n⁺ (heavily doped n-type)
The channel, when it forms, carries electrons

For PMOS everything flips: n-type body, p⁺ source/drain, holes carry the current.

Figure — MOSFET structure (gate, source, drain, body)

WHY each terminal exists (Feynman-style)

HOW the channel forms — derivation from first principles

Step 1 — The MOS capacitor

The gate–oxide–body sandwich is a parallel-plate capacitor. Its capacitance per unit area:

$C_{ox} = \frac{\varepsilon_{ox}}{t_{ox}}$

Step 2 — Threshold: when does a channel appear?

Apply positive $V_{GS}$ on an NMOS. The gate's positive charge repels holes in the p-body directly under the oxide (depletion), then, once $V_{GS}$ is large enough, it attracts electrons to the surface. When enough electrons gather to form a conducting sheet of the opposite type to the body, we say inversion has occurred. The voltage at which this just happens is the threshold voltage $V_{th}$ .

$\boxed{\,V_{GS} > V_{th} \implies \text{channel exists}\,}$

Step 3 — Induced channel charge

Once $V_{GS} > V_{th}$ , the extra gate voltage beyond threshold ( $V_{GS}-V_{th}$ , called the overdrive $V_{OV}$ ) sets how much mobile charge is pulled in. Treating the channel like the capacitor's stored charge $Q = CV$ , per unit area the electron (negative) sheet charge is:

$\boxed{\,Q_{ch} = -\,C_{ox}\,(V_{GS} - V_{th})\,}$

The minus sign is physical: the inversion layer is made of electrons, so its charge density is negative. Its magnitude is $|Q_{ch}| = C_{ox}\,V_{OV}$ , and that magnitude is what determines the channel conductance and hence $I_D$ .

Formula Why

$|Q_{ch}| = C_{ox}V_{OV}$ (and why the minus sign)? Below $V_{th}$ , added gate charge mostly widens the depletion region (fixed acceptor ions, no mobile carriers). $V_{th}$ is defined to already include that depletion cost. Above threshold, the incremental gate charge is mirrored by incremental mobile electrons → magnitude $C_{ox}(V_{GS}-V_{th})$ . The sign is negative because electrons carry negative charge.

Approximation warning (steel-manned): This " $V_{th}$ absorbs all depletion charge, then a clean step to mobile charge" picture is the long-channel, constant-depletion-charge model. In reality, even in strong inversion the depletion region keeps widening slightly and a small extra depletion voltage drop develops, so the transition is smooth, not a perfect step. The formula is an excellent first-order approximation, not exact.

Step 4 — Why current then flows

Now apply a small $V_{DS}$ . The channel is a thin resistive sheet of electrons from source to drain. The horizontal field drives them: current flows. More overdrive ⇒ more mobile charge magnitude ⇒ lower resistance ⇒ more current. That is the entire operating principle in one sentence.

Worked examples

Worked example Example 3 — Naming source vs drain, and sign of channel charge

An NMOS ( $V_{th}=0.7\ \text{V}$ , $C_{ox}=8.6\times10^{-3}\ \text{F/m}^2$ ) has terminal P at $0\ \text{V}$ , terminal Q at $1\ \text{V}$ , gate at $2\ \text{V}$ . Electrons are the carriers, supplied from the lower potential → P is the Source, Q is the Drain. $V_{GS}=2-0=2\ \text{V}$ , so $V_{OV}=1.3\ \text{V}$ . Sheet charge: $Q_{ch}=-C_{ox}V_{OV}=-(8.6\times10^{-3})(1.3)\approx-1.1\times10^{-2}\ \text{C/m}^2$ (negative → electrons).

Why this step? Source is defined by carrier supply. For NMOS (electrons) → lowest-potential n⁺ well. The negative result confirms the inversion layer is electrons; using $V_{GS}$ measured to the drain would give the wrong overdrive.

Common mistakes (steel-manned)

Active recall

Recall Flashcards (reveal after answering)

#flashcards/hardware What do the letters MOS in MOSFET stand for? ::: Metal–Oxide–Semiconductor (the gate stack layers). Which layer makes gate current ≈ 0? ::: The thin oxide (SiO₂) insulator under the gate. For an NMOS, what type is the body and what type are S/D? ::: Body is p-type; source & drain are n⁺. Define the threshold voltage $V_{th}$ . ::: The $V_{GS}$ at which an inversion layer (channel) first forms, connecting S and D. Formula for oxide capacitance per unit area? ::: $C_{ox}=\varepsilon_{ox}/t_{ox}$ . Induced mobile channel (electron) charge per area above threshold? ::: $Q_{ch}=-C_{ox}(V_{GS}-V_{th})$ (negative; magnitude $C_{ox}V_{OV}$ ). Why is the sign of $Q_{ch}$ negative for NMOS? ::: The inversion layer is made of electrons, which carry negative charge. Is $Q_{ch}=C_{ox}V_{OV}$ exact? ::: No — it's the long-channel, constant-depletion-charge approximation; a small extra depletion drop keeps developing in strong inversion. How do you identify the source of an NMOS? ::: It's the n⁺ terminal at the lower potential (electron supply). What is the body effect? ::: A nonzero $V_{SB}$ increases the threshold voltage $V_{th}$ . Why is thinner oxide desirable? ::: Larger $C_{ox}$ → stronger gate control → more channel charge per volt. "Field Effect" refers to what? ::: The gate's electric field (not injected current) controls the channel.

Recall Feynman: explain to a 12-year-old

Imagine a garden hose (source to drain) with a squishy valve in the middle made of jelly. On top of the jelly is a glass sheet you can't push water through, and above that a metal lid. When you put a strong positive charge on the lid, it magically pulls a wire of tiny electrons together inside the jelly — even though nothing passes through the glass. That electron-wire lets water (current) flow from one end to the other. Take away the charge and the wire disappears, so no flow. You control everything from the lid, without ever touching the water. That lid is the gate, the glass is the oxide, and the electron-wire is the channel (and because it's made of electrons, its charge counts as negative).

Connections

BJT vs MOSFET comparison — BJT is current-controlled; MOSFET is voltage/field-controlled.
MOSFET operating regions (cutoff, triode, saturation) — uses $V_{OV}=V_{GS}-V_{th}$ from here.
Threshold voltage and body effect — deep dive on $V_{th}$ and $V_{SB}$ .
MOS capacitor and inversion — the physics of $C_{ox}$ and inversion layer.
CMOS logic — pairs NMOS + PMOS from this structure.
Channel-length modulation — refinement in saturation.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, MOSFET ek voltage-controlled switch hai. Iske char terminals hote hain: Gate, Source, Drain, aur Body. Sabse upar Gate hota hai (metal jaisa), uske neeche ek patli oxide (SiO2) ki insulator layer, aur uske neeche silicon (Body). Yeh oxide isliye important hai kyunki iske through koi current nahi jaata — Gate sirf ek capacitor ki plate ki tarah kaam karta hai. Jab tum Gate pe positive voltage dete ho (NMOS mein), toh iska electric field silicon ke andar electrons ko surface pe kheench leta hai, aur ek channel ban jaata hai jo Source aur Drain ko jodta hai.

Yeh channel tabhi banta hai jab $V_{GS}$ threshold voltage $V_{th}$ se zyada ho jaaye — isko inversion kehte hain. Jitna zyada overdrive $V_{OV}=V_{GS}-V_{th}$ , utni zyada charge magnitude, utni easily current flow karti hai. Formula: $Q_{ch}=-C_{ox}(V_{GS}-V_{th})$ — yahan minus sign zaroori hai kyunki channel electrons ka bana hota hai, aur electron ki charge negative hoti hai. Aur $C_{ox}=\varepsilon_{ox}/t_{ox}$ . Ek baat yaad rakho: yeh formula ek approximation hai (long-channel model); strong inversion mein bhi depletion region thoda-thoda badhta rehta hai, toh transition perfectly sharp step nahi hoti.

Source aur Drain naam voltage se decide hote hain, geometry se nahi. NMOS mein jahan potential kam hai wahan Source (electrons wahan se aate hain), zyada potential wala Drain. Aur Body ko mat bhoolna — agar $V_{SB}$ nonzero ho toh $V_{th}$ badh jaata hai, isko body effect kehte hain.

Sabse badi galti: sochte hain Gate mein current jaata hai. Nahi! Gate current lagbhag zero hota hai kyunki beech mein insulator hai. Yehi reason hai ki MOSFET ka static power kam hota hai aur yeh CMOS logic ke liye perfect hai.