MOSFET I-V curves (triode and saturation) — Hardware notes

1. The three things that must happen (WHAT)

We study the n-channel enhancement MOSFET (NMOS). Terminals: Gate (G), Source (S), Drain (D), Body/Bulk (tied to source).

The three regions:

Cutoff: $V_{GS} < V_{th}$ → no channel → $I_D = 0$ .
Triode (linear/ohmic): $V_{GS} > V_{th}$ and $V_{DS} < V_{ov}$ → continuous channel → resistor-like.
Saturation (active): $V_{GS} > V_{th}$ and $V_{DS} \ge V_{ov}$ → channel pinched off at drain → current-source-like.

2. Deriving $I_D$ from first principles (HOW)

We build the current from scratch. No memorizing.

Step 1 — How much charge is in the channel? The gate + oxide + channel form a parallel-plate capacitor with capacitance per unit area $C_{ox} = \varepsilon_{ox}/t_{ox}$ . The mobile sheet charge (per area) at a point where the channel potential is $V(x)$ is only the charge above threshold: $Q_n(x) = -C_{ox}\big(V_{GS} - V(x) - V_{th}\big)$ Why this step? Only the gate voltage in excess of $V_{th}$ pulls in mobile electrons; and the local channel voltage $V(x)$ reduces the effective gate-to-channel voltage as we move toward the drain.

Step 2 — Current from drift. Current $=$ charge density $\times$ velocity $\times$ width. Velocity $=\mu_n E = -\mu_n \frac{dV}{dx}$ : $I_D = W\,|Q_n(x)|\,\mu_n \frac{dV}{dx} = W\mu_n C_{ox}\big(V_{GS}-V_{th}-V\big)\frac{dV}{dx}$ Why this step? $I_D$ is constant along the channel (charge conservation), so we can integrate.

Step 3 — Integrate along the channel from source ( $x=0,\,V=0$ ) to drain ( $x=L,\,V=V_{DS}$ ): $I_D \int_0^L dx = W\mu_n C_{ox}\int_0^{V_{DS}}\big(V_{GS}-V_{th}-V\big)\,dV$ $I_D\,L = W\mu_n C_{ox}\left[(V_{GS}-V_{th})V_{DS} - \frac{V_{DS}^2}{2}\right]$

Step 4 — Where does saturation come from? The triode parabola peaks when $\frac{dI_D}{dV_{DS}}=0$ : $(V_{GS}-V_{th}) - V_{DS} = 0 \;\Rightarrow\; V_{DS,\text{sat}} = V_{ov}$ At this point $Q_n$ at the drain end $\to 0$ (pinch-off). Push $V_{DS}$ higher and the current doesn't keep rising — the extra voltage drops across the tiny pinched region. So we freeze $I_D$ at its peak value: substitute $V_{DS}=V_{ov}$ into the triode formula: $I_D = k_n\left[V_{ov}\cdot V_{ov} - \frac{V_{ov}^2}{2}\right] = \frac{k_n}{2}V_{ov}^2$

Notice the current is now quadratic in $V_{GS}$ but (nearly) independent of $V_{DS}$ — a current source controlled by the gate.

Figure — MOSFET I-V curves (triode and saturation)

3. Worked examples

4. Common mistakes (Steel-man + fix)

5. Active recall

Recall Test yourself (hide the answers)

What determines the triode↔saturation boundary? → $V_{DS}=V_{ov}=V_{GS}-V_{th}$ .
Why does current stop rising in saturation? → channel pinches off; extra $V_{DS}$ drops across the pinched region.
What is $R_{on}$ in deep triode? → $1/(k_n V_{ov})$ .

What condition defines cutoff for an NMOS?

V_{GS} < V_{th}

, so no channel exists and

I_D \approx 0

Define the overdrive voltage

V_{ov}

V_{ov}=V_{GS}-V_{th}

; the gate voltage in excess of threshold that sets channel charge.

State the triode-region current equation.

I_D = k_n[(V_{GS}-V_{th})V_{DS} - V_{DS}^2/2]

, valid for

V_{DS}\le V_{ov}

State the saturation-region current (ideal).

I_D = \tfrac12 k_n (V_{GS}-V_{th})^2

, valid for

V_{DS}\ge V_{ov}

What is the boundary between triode and saturation?

V_{DS} = V_{ov} = V_{GS}-V_{th}

(pinch-off condition).

Why is saturation called the "active" region?

Channel is pinched at the drain;

I_D

is nearly independent of

V_{DS}

→ acts as a

V_{GS}

-controlled current source (used for amplification).

What does channel-length modulation add?

A factor

(1+\lambda V_{DS})

, giving finite output resistance

r_o\approx 1/(\lambda I_D)

Derive

V_{DS,sat}

Set

dI_D/dV_{DS}=0

in triode:

(V_{GS}-V_{th})-V_{DS}=0 \Rightarrow V_{DS}=V_{ov}

On-resistance of a MOSFET switch (deep triode)?

R_{DS(on)} \approx 1/(k_n V_{ov})

; larger gate drive lowers it.

Where does the channel charge equation

Q_n=-C_{ox}(V_{GS}-V-V_{th})

come from?

Gate-oxide-channel parallel-plate capacitor; only voltage above threshold and above local channel potential

V(x)

holds mobile charge.

Recall Feynman: explain to a 12-year-old

Imagine a garden hose. The gate is your hand squeezing a valve: press harder ( $V_{GS}$ up) and you open a wider path for water. The drain voltage is how hard you turn on the tap. When the tap is barely open, more tap-turn = more water (that's triode, like a resistor). But once the path is opened as wide as your squeeze allows, turning the tap harder doesn't give you more water — the flow is capped by your squeeze, not the tap (that's saturation). So the squeeze (gate) sets the max flow, and the transistor becomes a steady water source.

Connections

BJT I-V curves and active region — MOSFET saturation ≈ BJT active region.
MOSFET as a switch (digital logic) — deep triode $R_{on}$ and cutoff.
Small-signal transconductance gm — $g_m=\partial I_D/\partial V_{GS}=k_n V_{ov}$ from the square law.
Channel-length modulation and output resistance
CMOS inverter — both NMOS and PMOS traversing these regions.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, MOSFET ko samajhne ka sabse aasan tarika: ye ek voltage-controlled switch/resistor hai. Gate pe voltage $V_{GS}$ lagao — jab ye $V_{th}$ (threshold) se zyada hoga tabhi source aur drain ke beech ek channel banega jismein current flow karega. Is extra voltage ko hum overdrive kehte hain: $V_{ov}=V_{GS}-V_{th}$ . Yahi $V_{ov}$ har formula ka asli hero hai.

Ab do situations. Jab $V_{DS}$ chhota hai (specifically $V_{DS}<V_{ov}$ ), transistor ek resistor ki tarah behave karta hai — jitna $V_{DS}$ badhao utna current badhta hai. Isko triode region kehte hain. Lekin jaise hi $V_{DS}$ badha kar $V_{ov}$ ke barabar ho jaata hai, drain wale end pe channel pinch-off ho jaata hai, aur uske baad $V_{DS}$ aur badhao to bhi current nahi badhta — ab transistor ek constant current source ban gaya. Isko saturation (ya active) region kehte hain, aur yahi region amplifier banane ke kaam aata hai.

Ek common galti: log samajhte hain "saturation matlab transistor off" — galat! MOSFET mein saturation matlab useful, constant-current wala on-state. Off state to cutoff hai (jab $V_{GS}<V_{th}$ ). Isliye hamesha pehle $V_{ov}$ nikalo, phir $V_{DS}$ se compare karo: agar $V_{DS}<V_{ov}$ → triode formula, warna → square-law $\tfrac12 k_n V_{ov}^2$ . Boundary pe dono formule same answer dete hain — yahi continuity check karke apna confidence build karo.