MOSFET as a switch

2.4.14 · Hardware › Transistors: BJT & FET

KYA ho raha hai (mental model)

Ek N-channel enhancement MOSFET ke liye (jo sabse common switch hai), teen terminals matter karte hain:

Gate (G) — control knob.
Drain (D) — usually load/high side.
Source (S) — reference (low-side switch mein ground se connected).

Switch state sirf $V_{GS}$ se decide hoti hai, drain voltage se nahi:

$V_{GS} < V_{T H}$ → cut-off → switch OPEN.
$V_{GS} ≫ V_{T H}$ (fully driven) → triode / ohmic region → switch CLOSED, ek chhote resistor $R_{D S (o n)}$ jaisa behave karta hai.

KAISE: ON-resistance aur power loss ko first principles se derive karna

Step 1 — Triode region mein drain current

Yahan se kyun shuru karein? Kyunki "closed switch" = deep triode, aur humein jaanna hai ki yeh kitna resistive hai.

Gradual-channel model se, triode-region current hai:

$I_{D} = k [(V_{GS} - V_{T H}) V_{D S} - \frac{V _{D S}^{2}}{2}], k = μ_{n} C_{o x} \frac{W}{L}$

Yeh step kyun? $μ_{n} C_{o x}$ charge-mobility per area hai, aur $W / L$ scale karta hai ki channel kitna wide/short hai — bada $W / L$ = zyada current = better switch.

Step 2 — Small- $V_{D S}$ approximation

Jab switch closed hota hai, $V_{D S}$ chhota hota hai, isliye $V_{D S}^{2} /2$ term negligible hai:

$I_{D} \approx k (V_{GS} - V_{T H}) V_{D S}$

Yeh step kyun? Ek switch jo real current carry kare uske across sirf millivolts drop hone chahiye; quadratic term second order hai.

Step 3 — ON-resistance read off karo

$V_{D S} / I_{D}$ rearrange karo:

$R_{D S (o n)} = \frac{V _{D S}}{I _{D}} = \frac{1}{k ( V _{GS} - V _{T H} )}$

Step 4 — Conduction power loss

Jab ON ho aur load current $I_{D}$ carry kar raha ho:

$P_{co n d} = I_{D}^{2} R_{D S (o n)}$

Yeh step kyun? Yeh sirf on-resistance mein Joule heating hai — fundamental reason ki low $R_{D S (o n)}$ kyun matter karta hai.

Step 5 — Low-side switch ka output voltage

Load resistor $R_{L}$ ke saath jo $V_{D D}$ se drain tak hai, aur source grounded hai, drain node output ki tarah act karta hai:

$V_{o u t} = V_{D D} \frac{R _{D S (o n)}}{R _{D S (o n)} + R _{L}}$

OFF: $R_{D S (o n)} \to \infty \Rightarrow V_{o u t} \approx V_{D D}$ (HIGH).
ON: $R_{D S (o n)} ≪ R_{L} \Rightarrow V_{o u t} \approx 0$ (LOW).

Yeh step kyun? Dikhata hai ki is topology mein switch logic invert karta hai aur isliye hum $R_{D S (o n)} ≪ R_{L}$ chahte hain.

Common mistakes (steel-manned)

Recall

Recall Active recall — answers cover karo

Kaun sa variable ON/OFF decide karta hai? ⇒ == $V_{GS}$ relative to $V_{T H}$ ==
Kaun si region = "closed switch"? ⇒ triode/ohmic
On-resistance ka formula? ⇒ $R_{D S (o n)} = \frac{1}{k ( V _{GS} - V _{T H} )}$
Conduction power loss? ⇒ $P = I_{D}^{2} R_{D S (o n)}$
Better switch kaise banayein? ⇒ ==overdrive $V_{GS} - V_{T H}$ badhao== (aur $W / L$ )

Recall Feynman: 12-saal ke bachche ko samjhao

Socho ek garden hose hai jisme squeeze-valve hai. Gate tumhara haath hai. Zor se dabaao (high gate voltage) toh paani aasaani se rush karta hai — "pipe" wide open hai, switch CLOSED. Chhodon toh pipe pinch ho jaati hai, koi paani nahi — switch OPEN. Clever part yeh hai: pinching ke liye sirf ungliyon ka pressure chahiye (voltage), muscles ki pumping nahi (current). Isliye ek MOSFET switch ek tiny gentle signal se badi power control karta hai. Lekin fully-open pipe mein bhi kuch friction hoti hai (wahi hai $R_{D S (o n)}$ ), isliye jitna zyada dabaate ho, flow utna smooth hota hai.

Connections

Enhancement vs Depletion MOSFET
Triode vs Saturation regions
Threshold voltage VTH
BJT as a switch (compare: current-controlled vs voltage-controlled)
R_DS(on) and switching losses
Low-side vs High-side switching
Logic gates from MOSFETs (CMOS)

Kaun sa terminal voltage MOSFET switch ki ON/OFF state control karta hai?

V_{GS}

relative to

V_{T H}

Kaun si operating region ek CLOSED MOSFET switch ko correspond karti hai?

Triode (ohmic) region

Kaun si region OPEN switch ko correspond karti hai?

Cut-off (

V_{GS} < V_{T H}

)

Triode-region drain current equation likho.

I_{D} = k [(V_{GS} - V_{T H}) V_{D S} - V_{D S}^{2} /2]

Small-

V_{D S}

approximation se ON-resistance formula do.

R_{D S (o n)} = 1/ [k (V_{GS} - V_{T H})]

V_{GS}

badhane se

R_{D S (o n)}

pe kya asar padta hai?

Yeh kam ho jaata hai (bada overdrive → denser channel → lower resistance)

ON MOSFET mein conduction power loss kya hai?

P = I_{D}^{2} R_{D S (o n)}

Ek closed MOSFET perfect short kyun nahi hai?

Iska finite

R_{D S (o n)}

hai jo heat dissipate karta hai

Plain N-channel high-side switch gate ko

V_{D D}

pe rakh ke kyun ON nahi ho sakta?

Source ~

V_{D D}

tak rise karta hai, isliye

V_{GS} \to 0 < V_{T H}

; gate ko

V_{D D}

se upar boost karna padta hai

Low-side switch mein jab MOSFET OFF ho toh

V_{o u t}

kya hoga?

≈

V_{D D}

(koi current nahi, load pe koi drop nahi)