Gate-all-around (GAA) nanosheet transistors: Visual walkthrough

4.3.20 · D2 · Hardware › Semiconductor Fabrication › Gate-all-around (GAA) nanosheet transistors

Definition Sign aur axis conventions — inhe ek baar pakad lo

Do choices is page ke har minus sign ko fix karti hain; hum inhe clearly state karte hain taaki koi line ambiguous na rahe.

Charge-density sign $ρ$ : aam $n$ -channel device ke liye, depletion ke baad bache fixed dopant charge ionised acceptors se aate hain, jo negative hote hain: $ρ = - q N_{A} < 0$ (yahan $q > 0$ elementary charge hai aur $N_{A} > 0$ acceptor concentration hai). Toh Poisson mein source term $- ρ / ε_{s i}$ positive hai. Agar $p$ -channel device ho toh $ρ$ ka sign palto aur usse dependent har sign bhi palto.
$y$ -axis orientation: $y$ body ke across chalta hai, centre line par zero, upar ki taraf badhta hai, toh top surface $y = + t_{s i} /2$ par hai aur bottom $y = - t_{s i} /2$ par. Top par outward surface normal $+ y$ direction mein hai; bottom par $- y$ direction mein. Iss choice se, "gate surface potential ko $ψ_{g a t e}$ ki taraf kheenchta hai" ek positive restoring term ban jaata hai, jo hum neeche build karte hain.

Step 1 — Hum dekh kya rahe hain? Channel ek "electron pressure" ki strip ke roop mein

Definition Potential

$ψ$ , seedhi baat mein Electrostatic potential $ψ (x)$ (Greek letter "psi", ise ek height ki tarah padho) batata hai ki ek electron kisi jagah se kitna jaana chahta hai. Ise channel ke saath paani ke level ki tarah socho: jahan $ψ$ zyada hai, electrons ikatte hote hain aur flow kar sakte hain; jahan $ψ$ kam hai, nahi kar sakte. $x$ channel ke saath saath position measure karta hai, source ( $x = 0$ ) se drain ( $x = L_{g}$ ) tak.

Hum har potential ko gate se door neutral bulk silicon ke relative measure karte hain — woh reference point hamara " $ψ = 0$ " hai. Yahan ek symbol hai jo hum Step 2 mein earn karenge — $ψ_{0}$ , body ki centre line par potential; aur do aur — $ψ_{g a t e}$ aur $ψ_{s u r f a ce}$ — hum Step 3 mein earn karenge. Abhi sirf ek idea pakdo: $ψ (x)$ ek paani ka level hai, aur is page par har voltage usi bulk zero ke against measure kiya gaya hai.

KYA: hum transistor ko flat rakhke potential ko uske upar ek curve ki tarah draw karte hain. KYUN: poori "leakage" kahani is curve ki shape ke baare mein hai — ek smooth valley matlab transistor off hai; ek sagging valley matlab drain ne chori se cross kar liya. PICTURE (kya dhundho): laal curve $ψ (x)$ ek kale rectangle (channel) ke upar draw hai. Beech mein ek valley mein dip karti hai; dono ends source ( $x = 0$ , left) aur drain ( $x = L_{g}$ , right) par pinned hain. Ek kala arrow gate ko valley floor ko neeche dabaate hue dikhata hai. Gaur karo: gehra valley floor = electrons cross nahi kar sakte = transistor OFF.

Figure — Gate-all-around (GAA) nanosheet transistors

Formula Woh ek law jis par sab kuch tikaa hai: Poisson's equation

$\frac{d ^{2} ψ}{d x ^{2}} = - \frac{ρ}{ε _{s i}}$ Term by term:

$\frac{d ^{2} ψ}{d x ^{2}}$ — paani-level curve ki curvature (kitni tezi se slope badalta hai). Upar curve karna = valley bottom.
$ρ$ — body mein charge density. Hamaari fully-depleted assumption ke under yeh (constant) fixed dopant charge hai, aur hamare convention se $ρ = - q N_{A} < 0$ ek $n$ -channel device ke liye — toh $- ρ / ε_{s i} > 0$ .
$ε_{s i}$ — silicon ki permittivity: silicon kitni aasaani se electric field ko apne se guzarne deta hai. Bada $ε$ = material charge ke effect ko "dilute" karta hai.

Yeh tool kyun aur koi nahi? Poisson's equation woh rule hai jo "charge kaisa arranged hai" ko "potential curve kaisi dikhti hai" se jodta hai. Hum curve ki shape chahte hain → yahan se hi shuruat karni hogi.

Step 2 — Ise solvable banane ki trick: body ke across ek parabola assume karo

KYA: hum $ψ$ ko centre line par uski value aur ek gentle $y^{2}$ bulge ke roop mein likhte hain. KYUN: ek parabola sabse simple curve hai jo (a) beech mein flat ho sake aur (b) dono surfaces par gate ke zariye kheenchi ja sake — bilkul wahi physical picture. PICTURE (kya dhundho): sheet ke through ek vertical cut, $y$ page mein upar (top $y = + t_{s i} /2$ , bottom $y = - t_{s i} /2$ , centre $y = 0$ dotted). Laal curve $ψ$ vs $y$ hai: centre par sabse flat ( $ψ_{0}$ , labelled), dono gate-controlled surfaces ki taraf bahar bulge karta hai (dashed kali lines). Gaur karo: poori body ka potential ek centre value aur ek bulge se capture ho jaata hai.

Step 3 — Gate ko bolne do: surface boundary condition

KYA: hum demand karte hain ki field lines oxide cross karte waqt continuous rahen — silicon surface se nikalti field, gate ke oxide ke across lagayi field ke barabar ho. KYUN: yahan gate ki authority maths mein enter hoti hai. Iske bina, channel gate ko feel hi nahi karta. PICTURE (kya dhundho): teen stacked kale boxes — GATE (top), oxide thickness $t_{o x}$ ka (beech mein, labelled), silicon body (bottom). Laal arrows gate se oxide ke through silicon mein neeche shoot karte hain: woh field lines hain jo boundary condition match karta hai. Gaur karo: oxide box ko chota karo aur same voltage drop zyada steep (stronger) arrows banata hai = tighter control.

Step 4 — 2-D ko 1-D mein compress karo: centre line ka equation

Intuition Bulge ko wapas fold karna —

$c_{2}$ ke liye poora algebra (koi term chhupaaye nahi) Symmetric case lo ( $c_{1} = 0$ ). Parabola se do facts:

Surface value: $ψ_{s u r f a ce} = ψ_{0} + c_{2} (t_{s i} /2)^{2}$ .
Surface slope (top): $\frac{d ψ}{d y}_{+ t_{s i} /2} = 2 c_{2} (t_{s i} /2) = c_{2} t_{s i}$ .

Dono ko Step-3 boundary condition mein daalo: $c_{2} t_{s i} = \frac{ε _{o x}}{ε _{s i}} \cdot \frac{ψ _{g a t e} - ψ _{0} - c _{2} ( t _{s i} /2 ) ^{2}}{t _{o x}} .$ Exactly solve karo (har term rakho). Multiply out karo aur $c_{2}$ collect karo: $c_{2} t_{s i} t_{s i} \frac{ε _{s i}}{ε _{o x}} = t_{s i} (ψ_{g a t e} - ψ_{0}) - c_{2} t_{s i} \frac{t _{s i}^{2}}{4} \Rightarrow c_{2} (\frac{ε _{s i} t _{o x} t _{s i}}{ε _{o x}} + \frac{t _{s i}^{2}}{4}) = ψ_{g a t e} - ψ_{0},$ jisse exact coefficient milta hai $c_{2} = \frac{ψ _{g a t e} - ψ _{0}}{\frac{ε _{s i} t _{o x} t _{s i}}{ε _{o x}} + \frac{t _{s i}^{2}}{4}} .$ Ab approximation quantify karo. $t_{s i}^{2} /4$ term pehle wale ke saath negligible hai jab $\frac{t _{s i}^{2} /4}{ε _{s i} t _{o x} t _{s i} / ε _{o x}} = \frac{ε _{o x} t _{s i}}{4 ε _{s i} t _{o x}} ≪ 1.$ Ek typical thin sheet ke liye ( $t_{s i} \approx 6$ nm, $t_{o x} \approx 1$ nm, $ε_{o x} / ε_{s i} \approx 3.9/11.7 = 1/3$ ) yeh ratio $\approx (1/3) (6) / (4 \cdot 1) \approx 0.5$ hai — itna tiny nahi, isliye neeche diya leading form ek order-of-magnitude ansatz hai, aur exact $c_{2}$ jo upar hai wahi ek careful model rakhta hai. Small term drop karne par classic milta hai $c_{2} \approx \frac{ε _{o x}}{ε _{s i}} \cdot \frac{ψ _{g a t e} - ψ _{0}}{t _{o x} t _{s i}} .$ Ab Poisson in $y$ : $\partial^{2} ψ / \partial y^{2} = 2 c_{2}$ , toh $2 c_{2} = \frac{2 ε _{o x}}{ε _{s i} t _{o x} t _{s i}} (ψ_{g a t e} - ψ_{0}) .$ $(ψ_{g a t e} - ψ_{0})$ ka woh coefficient hi $1/ λ^{2}$ hai. Toh $(ψ_{0} - ψ_{g a t e}) / λ^{2}$ term kuch mysterious nahi hai — yeh woh across-body curvature hai jo parabola force karta hai, $ψ_{g a t e}$ ki taraf restoring pull ke roop mein re-express kiya gaya.

KYA: algebra $y$ ko eliminate karta hai; hum ek 1-D differential equation mein $ψ_{0} (x)$ ke liye reh jaate hain. KYUN: ek 1-D equation actually solve karne layak hai aur usse ek length padhi ja sakti hai — parabola ki bet ka payoff. PICTURE (kya dhundho): left par, small kale neeche arrows ka ek grid = body ke andar 2-D field map. Ek kala "reduce" arrow right ki taraf point karta hai ek single laal curve $ψ_{0} (x)$ par. Gaur karo: 2-D detail ek along-channel curve mein summarise ho gayi jiska curvature constants ka ek lump hai — woh lump $1/ λ^{2}$ hai.

Step 5 — Answer padhna: $λ$ ek screening length hai

Intuition Kaunsi branch? —

decaying exponential dhyaan se choose karna $ϕ^{''} = ϕ / λ^{2}$ ka general solution hai $ϕ (x) = A e^{+ x / λ} + B e^{- x / λ},$ do branches: ek growing ( $e^{+ x / λ}$ ), ek decaying ( $e^{- x / λ}$ ). Physics inme se boundary conditions se choose karta hai, pasand se nahi:

Drain par origin. $x = 0$ ko drain edge par rakho aur $x$ channel mein source ki taraf badhne do. Drain $x = 0$ par ek disturbance $ϕ (0) = ϕ_{D}$ inject karta hai.
Drain se door channel equilibrium par relax hona chahiye, yaani $ϕ \to 0$ jab $x \to$ (ek lambe channel mein gehraai mein). Growing branch $A e^{+ x / λ}$ wahan blow up karta hai, jo unphysical hai, isliye $A = 0$ set karna hoga.
Jo bachta hai woh $ϕ (x) = ϕ_{D} e^{- x / λ}$ hai — decaying branch, $B = ϕ_{D}$ drain value se $x = 0$ par fixed.

Ek short channel mein length $L_{g}$ ke saath tum $ϕ \to 0$ demand nahi kar sakte; balki tum $ϕ$ dono ends par fix karte ho (source aur drain values), dono branches rakhte ho — unka sum ek $cosh / sinh$ profile hai jo kabhi poori tarah decay nahi karta, aur yahi precisely wajah hai ki mid-channel floor lift hota hai aur device leak karta hai. Kisi bhi tarah, decay constant $λ$ hai.

KYA: hum shifted equation solve karte hain aur physical branch select karte hain; drain ki intrusion $ϕ = ϕ_{D} e^{- x / λ}$ hai (long channel) ya ek non-decaying $cosh$ -type profile (short channel). KYUN: ab "drain kitni door tak pahunchta hai?" ka ek clear jawab hai — lagbhag $λ$ . $L_{g} ≳ 5 λ$ chahiye taaki mid-channel mein intrusion almost nothing ho. PICTURE (kya dhundho): same channel ke upar do potential curves — ek kali long-channel curve (drain dip fade hoti hai, valley floor gehri rehti hai = OFF holds) aur ek laal short-channel curve (dono end-dips overlap karti hain, valley floor lift hota hai = leaks). Ek laal arrow fade distance $λ$ label karta hai. Gaur karo: same $λ$ , lekin jab $L_{g}$ sirf kuch $λ$ hai toh exponential tails nahi mare → leakage.

Step 6 — Gate sides add karo: wrapping kaise $λ$ chhotaa karta hai

KYA: Step 3 ko 1, phir 3, phir 4 gated surfaces ke liye repeat karo aur terms add karo. KYUN: yeh poore GAA idea ka payoff line hai — literally kyun channel ko wrap karna help karta hai, ek equation mein ek term ke roop mein. PICTURE (kya dhundho): teen square cross-sections side by side — planar (sirf bottom edge laal), FinFET (top + two side edges laal = 3), GAA nanosheet (saari char edges laal = 4). Gaur karo: laal edges gino = pull-back terms gino; zyada laal = chhotaa reach $λ$ .

Step 7 — All-around result, aur uske edge cases

Formula Gate-all-around screening length — composite numerator padhna

Surrounding-gate boundary ko puri tarah carry karne par (cylindrical/square wrap, gate poore perimeter par potential fix karta hai) milta hai $λ_{G AA} = \frac{2 ε _{s i} t _{o x} d _{s i} + ε _{o x} d _{s i}^{2}}{16 ε _{o x}}$

$2 ε_{s i} t_{o x} d_{s i}$ — oxide-coupling piece: yeh Step 3 ka surface-matching term hai (note karo $t_{o x}$ aur $ε_{s i}$ handshake), body size se scaled.
$ε_{o x} d_{s i}^{2}$ — body ka apna piece: ek perfect gate bhi body ki apni size se zyada tezi se screen nahi kar sakta; yeh term $d_{s i}^{2}$ ke saath badhta hai.
Denominator mein $16 ε_{o x}$ — wrapped-perimeter payoff: full all-around control (ek single gate ke $\sim 4$ ke comparison mein) ek bada number $λ^{2}$ ke neeche rakhta hai, reach ko hard divide karta hai. Yeh " $\frac{1}{4}$ -style factor" ka concrete face hai.

Intuition SAARE cases cover karo — sanity checks

Bahut thin body ( $d_{s i} \to 0$ ): numerator ka har term $d_{s i}$ ka factor carry karta hai, toh $λ_{G AA} \to 0$ . Reach vanish ho jaati hai — paper-thin channel perfectly controlled hai. ✅ intuition se match karta hai.
Bahut fat body ( $d_{s i}$ bada): $ε_{o x} d_{s i}^{2}$ term dominate karta hai, $λ_{G AA} \approx d_{s i} /4$ — body ke saath badhta hai, toh wide sheet planar-jaisi poor control ki taraf drift karti hai. Yeh woh drive-vs-control trade-off hai jiske baare mein parent ne warn kiya tha.
Zero oxide ( $t_{o x} \to 0$ ): $2 ε_{s i} t_{o x} d_{s i}$ term drop hota hai lekin $λ$ finite rehta hai ( $λ \to d_{s i} /4$ ) — ek perfect gate bhi body ki apni thickness se nahi jeet sakta. Tumhe phir bhi thin body chahiye, sirf achha gate nahi.
High-k oxide (bada $ε_{o x}$ ): $2 ε_{s i} t_{o x} d_{s i}$ term $ε_{o x} d_{s i}^{2}$ term ke relative shrink karta hai → chhotaa $λ$ . Isi liye GAA ko high-k metal gate aur ALD-thin oxides ke saath pair kiya jaata hai.

PICTURE (kya dhundho): $λ_{G AA}$ (laal) body size $d_{s i}$ ke against plot ki gayi — $d_{s i} = 0$ ke paas nearly flat aur small, phir bade $d_{s i}$ ke liye kale dashed asymptote $d_{s i} /4$ hug karne ke liye badhti hai. Do labelled arrows thin-body (best) aur fat-body (planar-like) regimes mark karte hain.

Ek-picture summary

YEH final figure kya karta hai: yeh poori chain ek line par rakhta hai — Poisson's law → parabola bet → gate boundary → 1-D equation jo $λ$ paida karta hai → exponential decay → $1/ N_{s i d es}$ gate-count win. Ise Steps 1–6 ke recap ke roop mein left se right padho, laal final box (gate-count payoff) ke saath jo punchline hai: zyada gated sides → chhotaa $λ$ → chhotaa $L_{g}$ bina leaking ke. Neeche har box exactly ek step se correspond karta hai jise tum abhi work through kar chuke ho. PICTURE (kya dhundho): chhe small boxes kale arrows se left-to-right joined, ek per step; sirf aakhri box ("N sides, 1/sqrt(N)") laal hai, aur neeche ek laal caption punchline state karta hai. Flow dekho: charge law in, gate-count payoff out.

Recall Feynman retelling — poora walkthrough seedhe shabdon mein

Channel ko ek valley imagine karo aur potential ko usmein paani ke level ki tarah, door ke calm bulk silicon se measure kiya gaya. Pehle hum conventions fix karte hain: bache hue dopant charge negative hai, aur $y$ body ke across upar point karta hai. Gate valley floor ko neeche dabaata hai taaki woh sukha rahe (transistor OFF). Drain, door ke end par, ek reservoir hai jo cross karne ki koshish karta hai. Poisson's equation woh rulebook hai jo "charge kaise baitta hai" ko "valley kaisi dikhti hai" se jodta hai (Step 1) — aur kyunki thin body fully depleted hai, charge sirf ek fixed constant hai. Ise solve karne ke liye hum bet lagate hain ki potential thin body ke across ek simple parabola ki tarah muda hai (Step 2), phir hum gate ko bolne dete hain oxide ke across field lines match karke — thinner oxide, mazbooti se dabana — aur note karte hain ki top aur bottom oxides alag ho sakte hain (Step 3). Algebra poori tarah karte hue ( $t_{s i}^{2} /4$ term rakho aur phir judge karo kab woh small hai), parabola ki across-body curvature ek restoring pull $(ψ_{0} - ψ_{g a t e}) / λ^{2}$ ban jaati hai, aur sab kuch channel ke saath ek equation mein fold ho jaata hai jiska single mystery constant ek length $λ$ kehlaata hai (Step 4). Bacha hua dopant constant equilibrium se measure karke sweep ho jaata hai, saaf $ϕ^{''} = ϕ / λ^{2}$ chhodke; uske paas ek growing aur ek decaying branch hai, aur physics (drain se door equilibrium par relax karo) growing wale ko throw away karta hai, toh drain ka flood exponentially fade hota hai distance ke saath, aur $λ$ exactly woh hai kitni door tak pahunchta hai — toh channel ko $5 λ$ lambe ki zaroorat hai taaki sukha rahe (Step 5). Finally, har extra gated side ek aur press add karta hai, apna term sum mein, denominator mein stack karke taaki $λ$ $1/ N_{s i d es}$ ki tarah shrink kare (Step 6). Planar mein ek side hai, FinFET mein teen, GAA mein saari char — sabse chhotaa reach, sabse tight valley, sabse kam leak. Aur edge cases sahi behave karte hain: thin body → tiny $λ$ , fat body → $λ \approx d_{s i} /4$ planar ki taraf wapas creep karta hai (Step 7). Woh aakhri line hi trade-off knob hai aur isi liye GAA nanosheets exist karte hain.

Recall Quick self-test

$λ$ kaun si physical quantity measure karta hai? ::: Woh distance jitni door drain ka field channel mein penetrate (screen into) karta hai; drain disturbance $e^{- x / λ}$ ki tarah decay karta hai. Zyada gate coverage $λ$ kyun chhotaa karta hai? ::: Har gated surface $1/ λ^{2}$ sum mein ek pull-back term add karti hai, toh identical faces ke liye $λ \propto 1/ N_{s i d es}$ . Constant dopant source kahan jaata hai jab hum $ϕ^{''} = ϕ / λ^{2}$ tak pahunchte hain? ::: Yeh variable shift karke absorb ho jaata hai (far-from-drain equilibrium se measure karke); yeh set karta hai ki "flat" kahan hai, reach $λ$ nahi. Do branches mein se decaying exponential kaise choose karte hain? ::: $x = 0$ drain par rakho aur demand karo ki $ϕ \to 0$ ek long channel mein gehraai mein; growing branch wahan blow up karta hai, toh uska coefficient zero hona chahiye, $ϕ = ϕ_{D} e^{- x / λ}$ bachta hai. Fat-body limit mein $λ_{G AA}$ kya hai? ::: $λ_{G AA} \to d_{s i} /4$ , planar-jaisi poor control ki taraf wapas drift — drive-vs-control trade-off. $L_{g} ≳ 5 λ$ kyun zaroori hai? ::: Taaki exponentially decaying drain intrusion mid-channel tak almost zero ho jaaye, OFF valley ko gehri rakhte hue.

Related: Subthreshold slope and gate electrostatics · Epitaxy and SiGe superlattices · CFET (complementary FET) · back to the parent topic.