Threshold voltage (Vth) — Hinglish notes

2.4.11 · Hardware › Transistors: BJT & FET

$V_{t h}$ KYA hai?

Jo quantity actually current drive karti hai woh hai overdrive voltage: $V_{o v} = V_{GS} - V_{t h}$

Channel ko ek threshold ki ZAROORAT kyun padti hai?

Socho ek NMOS p-type body par (holes majority carriers hain).

Electrons ko conduct karaane ke liye, pehle aapko surface se saare holes door push karne padte hain (deplete karna). Iske liye kuch gate voltage lagti hai.
Phir aapko itne electrons kheenchne padte hain ki surface n-type ki tarah behave kare. Iske liye aur voltage lagti hai.

Toh gate voltage ka ek definite "budget" hota hai jo aapko koi bhi electron channel appear hone se pehle spend karna padta hai. Wahi budget $V_{t h}$ hai. Iske neeche aap sirf holes clear kar rahe ho; iske upar aap electrons stack kar rahe ho.

$V_{t h}$ ko first principles se kaise derive karein

Hum $V_{t h}$ ko un voltages ke sum ke roop mein build karte hain jo har physical stage par chahiye hoti hain.

Step 1 — Flat-band voltage $V_{F B}$ . Yeh step kyun? $V_{GS} = 0$ par bhi, gate metal aur semiconductor ke alag work functions aur trapped oxide charge bands ko bend kar dete hain. $V_{F B}$ woh voltage hai jo ise "undo" karke bands ko flat banata hai: $V_{F B} = ϕ_{m s} - \frac{Q _{o x}}{C _{o x}}$ jahan $ϕ_{m s}$ metal–semiconductor work-function difference hai aur $Q_{o x}$ fixed oxide charge hai.

Step 2 — Strong inversion tak band bending, $ϕ_{s} = 2 ϕ_{F}$ . $2 ϕ_{F}$ kyun? Strong inversion defined hai as: surface electron density = bulk hole density. Bulk Fermi level intrinsic se $ϕ_{F}$ neeche hota hai; invert karne ke liye, hum surface ko $ϕ_{F}$ se bend karke intrinsic tak pahunchen aur ek aur $ϕ_{F}$ aage jaayein taaki woh utna n-type ho jaaye jitna bulk p-type tha. Isliye: $ϕ_{s} = 2 ϕ_{F}, ϕ_{F} = \frac{k T}{q} ln \frac{N _{A}}{n _{i}}$

Step 3 — Depletion charge hold karne ke liye oxide ke across voltage, $V_{o x}$ . Yeh step kyun? Holes ko push karne se ionized acceptors ka ek depletion region expose hota hai jiska total charge $Q_{d e p}$ hai. Gate ko oxide capacitor $C_{o x}$ ke across equal opposite charge supply karni padti hai: $V_{o x} = \frac{Q _{d e p}}{C _{o x}}, Q_{d e p} = 2 q ε_{s} N_{A} (2 ϕ_{F})$ $Q_{d e p}$ Poisson's equation solve karke milta hai width $W$ ke fully depleted region ke liye: charge $= q N_{A} W$ aur $2 ϕ_{F} = \frac{q N _{A} W ^{2}}{2 ε _{s}}$ → $W$ eliminate karo.

Step 4 — Sab add karo.

Body effect (kyun $V_{t h}$ change ho sakta hai)

Agar source ko body se tie nahi kiya gaya hai, toh reverse body-source bias $V_{S B}$ depletion region ko widen kar deta hai, toh zyada charge support karna padta hai → $V_{t h}$ badhta hai: $V_{t h} = V_{t h 0} + γ (2 ϕ_{F} + V_{S B} - 2 ϕ_{F}), γ = \frac{2 q ε _{s} N _{A}}{C _{o x}}$ $γ$ body-effect coefficient hai. Square-root kyun? Kyunki $Q_{d e p} \propto band bending$ Step 3 se.

Worked examples

Common mistakes

80/20 — essentials

$V_{t h}$ = gate voltage jo strong inversion tak pahunchne ke liye chahiye; $V_{o v} = V_{GS} - V_{t h}$ current drive karta hai.
Teen physical costs: flat-band, $2 ϕ_{F}$ band bending, depletion charge over $C_{o x}$ .
Thinner oxide → lower $V_{t h}$ , higher doping → higher $V_{t h}$ , body bias → higher $V_{t h}$ .

Recall Feynman: ek 12-saal ke bacche ko explain karo

Socho ek garden hose hai jisme ek kink hai jo paani rok raha hai. Metal gate tumhara woh angootha hai jo plastic cover par kink ke upar press kar raha hai. Tumhe itna hard press karna padta hai ki pehle raaste se keechad push ho jaaye, phir paani ke liye ek saaf raasta khule. Jis point par paani sirf bahna shuru hota hai wahi threshold hai. Kam press karo: kuch nahi (achha, ek tiny drip). Zyada press karo: strong flow.

Active-recall flashcards

MOSFET mein

V_{t h}

physically kya mark karta hai?

Woh

V_{GS}

jis par surface strong inversion tak pahunchti hai, ek conducting channel banate hue.

Overdrive voltage define karo.

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

; yahi actually drain current control karta hai.

Threshold par band bending

2 ϕ_{F}

kyun hoti hai?

Kyunki strong inversion ka matlab hai surface electron density = bulk hole density — intrinsic tak pahunchne ke liye

ϕ_{F}

chahiye aur bulk ko mirror karne ke liye ek aur

ϕ_{F}

NMOS

V_{t h}

formula likho.

V_{t h} = V_{F B} + 2 ϕ_{F} + 2 q ε_{s} N_{A} (2 ϕ_{F}) / C_{o x}

Thinner oxide

V_{t h}

ko kaise affect karta hai?

C_{o x}

badhta hai, toh depletion-charge term chhota ho jaata hai →

V_{t h}

kam hota hai.

Higher substrate doping

N_{A}

V_{t h}

ko kaise affect karta hai?

Use badhata hai (dono

2 ϕ_{F}

aur

N_{A}

term badhte hain).

Body effect kya hai?

Reverse

V_{S B}

depletion ko widen karta hai → zyada charge →

V_{t h}

badhta hai:

V_{t h} = V_{t h 0} + γ (2 ϕ_{F} + V_{S B} - 2 ϕ_{F})

Kya

V_{t h}

se neeche drain current exactly zero hoti hai?

Nahi — chhota exponential subthreshold current phir bhi flow karta hai.

Flat-band voltage kya hai?

V_{F B} = ϕ_{m s} - Q_{o x} / C_{o x}

; woh gate voltage jo semiconductor bands ko zero applied field par flat banata hai.

C_{o x}

per unit area ka formula?

C_{o x} = ε_{o x} / t_{o x}

Connections

MOSFET operating regions — cut-off vs triode vs saturation $V_{t h}$ par split hote hain.
Overdrive voltage and drain current — $I_{D} \propto V_{o v}^{2}$ saturation mein.
CMOS scaling — kyun thin oxides & low $V_{t h}$ ne low-power chips enable ki.
Depletion region and Poisson's equation — $N_{A}$ term ka source.
Work function and flat-band voltage — $V_{F B}$ ka origin.
Subthreshold conduction — $V_{t h}$ se neeche leakage.