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

2.4.12 · Hardware › Transistors: BJT & FET

1. Teen cheezein jo zaroor honi chahiye (KYA)

Hum n-channel enhancement MOSFET (NMOS) study karte hain. Terminals hain: Gate (G), Source (S), Drain (D), Body/Bulk (source se tied).

Teen regions:

Cutoff: $V_{GS} < V_{t h}$ → koi channel nahi → $I_{D} = 0$ .
Triode (linear/ohmic): $V_{GS} > V_{t h}$ aur $V_{D S} < V_{o v}$ → continuous channel → resistor jaisa.
Saturation (active): $V_{GS} > V_{t h}$ aur $V_{D S} \geq V_{o v}$ → channel drain pe pinch off → current-source jaisa.

2. $I_{D}$ ko first principles se derive karna (KAISE)

Hum current ko scratch se banate hain. Kuch memorize nahi karna.

Step 1 — Channel mein kitna charge hai? Gate + oxide + channel ek parallel-plate capacitor banate hain jiska capacitance per unit area $C_{o x} = ε_{o x} / t_{o x}$ hai. Ek point pe jahan channel potential $V (x)$ hai, wahan mobile sheet charge (per area) sirf threshold ke upar ka charge hai: $Q_{n} (x) = - C_{o x} (V_{GS} - V (x) - V_{t h})$ Yeh step kyun? Sirf $V_{t h}$ se zyada gate voltage mobile electrons ko kheenchta hai; aur local channel voltage $V (x)$ effective gate-to-channel voltage ko reduce karta hai jaise hum drain ki taraf badhte hain.

Step 2 — Drift se current. Current $=$ charge density $\times$ velocity $\times$ width. Velocity $= μ_{n} E = - μ_{n} \frac{d V}{d x}$ : $I_{D} = W ∣ Q_{n} (x) ∣ μ_{n} \frac{d V}{d x} = W μ_{n} C_{o x} (V_{GS} - V_{t h} - V) \frac{d V}{d x}$ Yeh step kyun? $I_{D}$ channel ke saath-saath constant hai (charge conservation), isliye hum integrate kar sakte hain.

Step 3 — Channel ke saath integrate karein source se ( $x = 0, V = 0$ ) drain tak ( $x = L, V = V_{D S}$ ): $I_{D} \int_{0}^{L} d x = W μ_{n} C_{o x} \int_{0}^{V_{D S}} (V_{GS} - V_{t h} - V) d V$ $I_{D} L = W μ_{n} C_{o x} [(V_{GS} - V_{t h}) V_{D S} - \frac{V _{D S}^{2}}{2}]$

Step 4 — Saturation kahan se aati hai? Triode parabola peak karta hai jab $\frac{d I _{D}}{d V _{D S}} = 0$ : $(V_{GS} - V_{t h}) - V_{D S} = 0 \Rightarrow V_{D S, sat} = V_{o v}$ Is point pe drain end pe $Q_{n}$ $\to 0$ ho jaata hai (pinch-off). $V_{D S}$ aur badhao toh current continue nahi hoti badhna — extra voltage chhoti si pinched region mein drop ho jaati hai. Toh hum $I_{D}$ ko uske peak value pe freeze karte hain: triode formula mein $V_{D S} = V_{o v}$ substitute karein: $I_{D} = k_{n} [V_{o v} \cdot V_{o v} - \frac{V _{o v}^{2}}{2}] = \frac{k _{n}}{2} V_{o v}^{2}$

Dhyan do ki current ab $V_{GS}$ mein quadratic hai lekin (almost) $V_{D S}$ se independent hai — ek current source jo gate se control hota hai.

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

3. Worked examples

4. Common mistakes (Steel-man + fix)

5. Active recall

Recall Khud test karo (answers chhupa lo)

Triode↔saturation boundary kya decide karta hai? → $V_{D S} = V_{o v} = V_{GS} - V_{t h}$ .
Saturation mein current kyun badhna band ho jaati hai? → channel pinch off ho jaata hai; extra $V_{D S}$ pinched region mein drop ho jaata hai.
Deep triode mein $R_{o n}$ kya hai? → $1/ (k_{n} V_{o v})$ .

NMOS ke liye cutoff kaun si condition define karta hai?

V_{GS} < V_{t h}

, toh koi channel exist nahi karta aur

I_{D} \approx 0

Overdrive voltage

V_{o v}

define karo.

V_{o v} = V_{GS} - V_{t h}

; threshold se zyada gate voltage jo channel charge set karta hai.

Triode-region current equation batao.

I_{D} = k_{n} [(V_{GS} - V_{t h}) V_{D S} - V_{D S}^{2} /2]

, valid hai

V_{D S} \leq V_{o v}

ke liye.

Saturation-region current (ideal) batao.

I_{D} = \frac{1}{2} k_{n} (V_{GS} - V_{t h})^{2}

, valid hai

V_{D S} \geq V_{o v}

ke liye.

Triode aur saturation ke beech boundary kya hai?

V_{D S} = V_{o v} = V_{GS} - V_{t h}

(pinch-off condition).

Saturation ko "active" region kyun kehte hain?

Channel drain pe pinched hota hai;

I_{D}

almost

V_{D S}

se independent hota hai →

V_{GS}

-controlled current source ki tarah kaam karta hai (amplification ke liye use hota hai).

Channel-length modulation kya add karta hai?

Ek factor

(1 + λ V_{D S})

, jo finite output resistance

r_{o} \approx 1/ (λ I_{D})

deta hai.

V_{D S, s a t}

derive karo.

Triode mein

d I_{D} / d V_{D S} = 0

set karo:

(V_{GS} - V_{t h}) - V_{D S} = 0 \Rightarrow V_{D S} = V_{o v}

MOSFET switch (deep triode) ki on-resistance?

R_{D S (o n)} \approx 1/ (k_{n} V_{o v})

; zyada gate drive isse kam karta hai.

Channel charge equation

Q_{n} = - C_{o x} (V_{GS} - V - V_{t h})

kahan se aati hai?

Gate-oxide-channel parallel-plate capacitor; sirf threshold se upar aur local channel potential

V (x)

se upar ka voltage mobile charge hold karta hai.

Recall Feynman: 12 saal ke bachche ko explain karo

Ek garden hose imagine karo. Gate teri haath hai jo ek valve squeeze kar rahi hai: zyada press karo ( $V_{GS}$ upar) aur paani ke liye wider path khul jaata hai. Drain voltage woh hai jitna tum tap kholta ho. Jab tap thoda sa khula ho, zyada tap-turn = zyada paani (yeh triode hai, resistor ki tarah). Lekin jab path utna khul jaaye jitna teri squeeze allow karti hai, tab tap zyada turn karne se zyada paani nahi milta — flow teri squeeze se cap hota hai, tap se nahi (yeh saturation hai). Toh squeeze (gate) max flow set karta hai, aur transistor ek steady water source ban jaata hai.

Connections

BJT I-V curves and active region — MOSFET saturation ≈ BJT active region.
MOSFET as a switch (digital logic) — deep triode $R_{o n}$ aur cutoff.
Small-signal transconductance gm — $g_{m} = \partial I_{D} / \partial V_{GS} = k_{n} V_{o v}$ square law se.
Channel-length modulation and output resistance
CMOS inverter — NMOS aur PMOS dono in regions se guzarte hain.