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

Definition Sign and axis conventions — pin these down once

Two choices fix every minus sign on this page; we make them explicit so no line is ambiguous.

Charge-density sign $ρ$ : for the usual $n$ -channel device the fixed dopant charge left after depletion is from ionised acceptors, which are negative: $ρ = - q N_{A} < 0$ (here $q > 0$ is the elementary charge and $N_{A} > 0$ the acceptor concentration). So the source term $- ρ / ε_{s i}$ in Poisson is positive. If you flip to a $p$ -channel device, flip the sign of $ρ$ and every dependent sign with it.
$y$ -axis orientation: $y$ runs across the body, zero on the centre line, increasing upward, so the top surface is at $y = + t_{s i} /2$ and the bottom at $y = - t_{s i} /2$ . The outward surface normal at the top points in $+ y$ ; at the bottom in $- y$ . With this choice, "the gate pulls the surface potential toward $ψ_{g a t e}$ " is a positive restoring term, which is what we build below.

Step 1 — What are we even looking at? The channel as a strip of "electron pressure"

WHAT: we lay the transistor flat and draw the potential as a curve above it. WHY: the whole "leakage" story is about the shape of this curve — a smooth valley means the transistor is off; a sagging valley means the drain has snuck across. PICTURE (what to look for): the red curve is $ψ (x)$ drawn above a black rectangle (the channel). It dips into a valley in the middle; the two ends are pinned at source ( $x = 0$ , left) and drain ( $x = L_{g}$ , right). A black arrow marks the gate pressing the valley floor down. Observe: deep valley floor = electrons can't cross = transistor OFF.

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

Step 2 — The trick that makes it solvable: assume a parabola across the body

WHAT: we write $ψ$ as its value on the centre line plus a gentle $y^{2}$ bulge. WHY: a parabola is the simplest curve that (a) can be flat in the middle and (b) get pulled by the gate at both surfaces — exactly the physical picture. PICTURE (what to look for): a vertical cut through the sheet, $y$ up the page (top $y = + t_{s i} /2$ , bottom $y = - t_{s i} /2$ , centre $y = 0$ dotted). The red curve is $ψ$ vs $y$ : flattest at the centre ( $ψ_{0}$ , labelled), bulging out toward both gate-controlled surfaces (dashed black lines). Observe: the whole body's potential is captured by one centre value plus one bulge.

Step 3 — Let the gate speak: the surface boundary condition

WHAT: we demand that field lines are continuous crossing the oxide — the field leaving the silicon surface equals the field the gate imposes across the oxide. WHY: this is where the gate's authority enters the maths. Without it, the channel wouldn't feel the gate at all. PICTURE (what to look for): three stacked black boxes — GATE (top), oxide of thickness $t_{o x}$ (middle, labelled), silicon body (bottom). Red arrows shoot downward from gate through the oxide into the silicon: those are the field lines the boundary condition matches. Observe: shorten the oxide box and the same voltage drop makes steeper (stronger) arrows = tighter control.

Step 4 — Collapse 2-D into 1-D: the equation for the centre line

Intuition Folding the bulge back in — the full algebra for

$c_{2}$ (no term silently dropped) Take the symmetric case ( $c_{1} = 0$ ). Two facts from the parabola:

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

Put both into the Step-3 boundary condition: $c_{2} t_{s i} = \frac{ε _{o x}}{ε _{s i}} \cdot \frac{ψ _{g a t e} - ψ _{0} - c _{2} ( t _{s i} /2 ) ^{2}}{t _{o x}} .$ Solve exactly (keep every term). Multiply out and collect $c_{2}$ : $c_{2} t_{s i} t_{s i} \frac{ε _{s i}}{ε _{o x}} = t_{s i} (ψ_{g a t e} - ψ_{0}) - c_{2} t_{s i} \frac{t _{s i}^{2}}{4} \Rightarrow c_{2} (\frac{ε _{s i} t _{o x} t _{s i}}{ε _{o x}} + \frac{t _{s i}^{2}}{4}) = ψ_{g a t e} - ψ_{0},$ giving the exact coefficient $c_{2} = \frac{ψ _{g a t e} - ψ _{0}}{\frac{ε _{s i} t _{o x} t _{s i}}{ε _{o x}} + \frac{t _{s i}^{2}}{4}} .$ Now quantify the approximation. The $t_{s i}^{2} /4$ term is negligible next to the first when $\frac{t _{s i}^{2} /4}{ε _{s i} t _{o x} t _{s i} / ε _{o x}} = \frac{ε _{o x} t _{s i}}{4 ε _{s i} t _{o x}} ≪ 1.$ For a typical thin sheet ( $t_{s i} \approx 6$ nm, $t_{o x} \approx 1$ nm, $ε_{o x} / ε_{s i} \approx 3.9/11.7 = 1/3$ ) this ratio is $\approx (1/3) (6) / (4 \cdot 1) \approx 0.5$ — not tiny, so the leading form below is an order-of-magnitude ansatz, and the exact $c_{2}$ above is what a careful model keeps. Dropping the small term gives the classic $c_{2} \approx \frac{ε _{o x}}{ε _{s i}} \cdot \frac{ψ _{g a t e} - ψ _{0}}{t _{o x} t _{s i}} .$ Now Poisson in $y$ : $\partial^{2} ψ / \partial y^{2} = 2 c_{2}$ , so $2 c_{2} = \frac{2 ε _{o x}}{ε _{s i} t _{o x} t _{s i}} (ψ_{g a t e} - ψ_{0}) .$ That coefficient of $(ψ_{g a t e} - ψ_{0})$ is $1/ λ^{2}$ . So the $(ψ_{0} - ψ_{g a t e}) / λ^{2}$ term is nothing mysterious — it is the across-body curvature the parabola forces, re-expressed as a restoring pull toward $ψ_{g a t e}$ .

WHAT: algebra eliminates $y$ ; we are left with a 1-D differential equation in $ψ_{0} (x)$ . WHY: a 1-D equation is one we can actually solve and read a length off of — the payoff of the parabola bet. PICTURE (what to look for): on the left, a grid of small black down-arrows = the 2-D field map inside the body. A black "reduce" arrow points right to a single red curve $ψ_{0} (x)$ . Observe: the 2-D detail is summarised by one along-channel curve whose curvature is a lump of constants — that lump is $1/ λ^{2}$ .

Step 5 — Reading the answer: $λ$ is a screening length

Intuition Which branch? — choosing the

decaying exponential, carefully The general solution of $ϕ^{''} = ϕ / λ^{2}$ is $ϕ (x) = A e^{+ x / λ} + B e^{- x / λ},$ two branches: one growing ( $e^{+ x / λ}$ ), one decaying ( $e^{- x / λ}$ ). Physics picks between them by boundary conditions, not taste:

Origin at the drain. Put $x = 0$ at the drain edge and let $x$ increase into the channel toward the source. The drain injects a disturbance $ϕ (0) = ϕ_{D}$ at $x = 0$ .
Far from the drain the channel must relax to equilibrium, i.e. $ϕ \to 0$ as $x \to$ (deep into a long channel). The growing branch $A e^{+ x / λ}$ blows up there, which is unphysical, so we must set $A = 0$ .
What survives is $ϕ (x) = ϕ_{D} e^{- x / λ}$ — the decaying branch, with $B = ϕ_{D}$ fixed by the drain value at $x = 0$ .

In a short channel of length $L_{g}$ you cannot demand $ϕ \to 0$ ; instead you fix $ϕ$ at both ends (source and drain values), keeping both branches — their sum is a $cosh / sinh$ profile that never fully decays, which is precisely why the mid-channel floor lifts and the device leaks. Either way, the decay constant is $λ$ .

WHAT: we solve the shifted equation and select the physical branch; the drain's intrusion is $ϕ = ϕ_{D} e^{- x / λ}$ (long channel) or a non-decaying $cosh$ -type profile (short channel). WHY: now "how far does the drain reach?" has a crisp answer — about $λ$ . Need $L_{g} ≳ 5 λ$ so the intrusion has decayed to near nothing at mid-channel. PICTURE (what to look for): two potential curves over the same channel — a black long-channel curve (drain dip fades, valley floor stays deep = OFF holds) and a red short-channel curve (the two end-dips overlap, valley floor lifts = leaks). A red arrow labels the fade distance $λ$ . Observe: same $λ$ , but when $L_{g}$ is only a few $λ$ the exponential tails haven't died → leakage.

Step 6 — Add gate sides: how wrapping shrinks $λ$

WHAT: repeat Step 3 for 1, then 3, then 4 gated surfaces and add the terms. WHY: this is the payoff line for the whole GAA idea — literally why wrapping the channel helps, as a term in an equation. PICTURE (what to look for): three square cross-sections side by side — planar (only the bottom edge red), FinFET (top + two side edges red = 3), GAA nanosheet (all four edges red = 4). Observe: count the red edges = count the pull-back terms; more red = shorter reach $λ$ .

Step 7 — The all-around result, and its edge cases

Formula Gate-all-around screening length — reading the composite numerator

Carrying the surrounding-gate boundary all the way through (cylindrical/square wrap, gate fixing the potential on the entire perimeter) yields $λ_{G AA} = \frac{2 ε _{s i} t _{o x} d _{s i} + ε _{o x} d _{s i}^{2}}{16 ε _{o x}}$

$2 ε_{s i} t_{o x} d_{s i}$ — the oxide-coupling piece: it is the surface-matching term of Step 3 (note the $t_{o x}$ and the $ε_{s i}$ handshake), scaled by the body size.
$ε_{o x} d_{s i}^{2}$ — the body's own piece: even a perfect gate can't screen faster than the body's own size allows; this term grows as $d_{s i}^{2}$ .
$16 ε_{o x}$ in the denominator — the wrapped-perimeter payoff: full all-around control (vs $\sim 4$ for a single gate) puts a large number under $λ^{2}$ , dividing the reach down hard. This is the concrete face of the " $\frac{1}{4}$ -style factor."

Intuition Cover ALL cases — sanity checks

Very thin body ( $d_{s i} \to 0$ ): every numerator term carries a factor of $d_{s i}$ , so $λ_{G AA} \to 0$ . Reach vanishes — a paper-thin channel is perfectly controlled. ✅ matches intuition.
Very fat body ( $d_{s i}$ large): the $ε_{o x} d_{s i}^{2}$ term dominates, $λ_{G AA} \approx d_{s i} /4$ — grows with the body, so a wide sheet drifts toward planar-like poor control. This is the drive-vs-control trade-off the parent warned about.
Zero oxide ( $t_{o x} \to 0$ ): the $2 ε_{s i} t_{o x} d_{s i}$ term drops but $λ$ stays finite ( $λ \to d_{s i} /4$ ) — even a perfect gate can't beat the body's own thickness. You still need a thin body, not just a good gate.
High-k oxide (big $ε_{o x}$ ): the $2 ε_{s i} t_{o x} d_{s i}$ term shrinks relative to the $ε_{o x} d_{s i}^{2}$ term → smaller $λ$ . This is why GAA is paired with a high-k metal gate and ALD-thin oxides.

PICTURE (what to look for): $λ_{G AA}$ (red) plotted against body size $d_{s i}$ — nearly flat and small near $d_{s i} = 0$ , then rising to hug the black dashed asymptote $d_{s i} /4$ for large $d_{s i}$ . Two labelled arrows mark the thin-body (best) and fat-body (planar-like) regimes.

The one-picture summary

WHAT this final figure does: it lays the whole chain on one line — Poisson's law → parabola bet → gate boundary → the 1-D equation that births $λ$ → exponential decay → the $1/ N_{s i d es}$ gate-count win. Read it left to right as a recap of Steps 1–6, with the red final box (the gate-count payoff) being the punchline: more gated sides → smaller $λ$ → shorter $L_{g}$ without leaking. Each box below corresponds to exactly one step you just worked through. PICTURE (what to look for): six small boxes joined left-to-right by black arrows, one per step; only the last box ("N sides, 1/sqrt(N)") is red, and a red caption underneath states the punchline. Observe the flow: charge law in, gate-count payoff out.

Recall Feynman retelling — the whole walkthrough in plain words

Picture the channel as a valley and the potential as the water level in it, measured from the calm bulk silicon far away. First we fix conventions: the leftover dopant charge is negative, and $y$ points up across the body. The gate presses the valley floor down to keep it dry (transistor OFF). The drain, at the far end, is a reservoir trying to flood across. Poisson's equation is the rulebook connecting "how charge sits" to "what the valley looks like" (Step 1) — and because the thin body is fully depleted, the charge is just a fixed constant. To solve it we bet the potential bends like a simple parabola across the thin body (Step 2), then we let the gate speak by matching field lines across the oxide — thinner oxide, firmer press — and we note the top and bottom oxides may differ (Step 3). Doing the algebra in full (keeping the $t_{s i}^{2} /4$ term and only then judging when it's small), the parabola's across-body curvature turns into a restoring pull $(ψ_{0} - ψ_{g a t e}) / λ^{2}$ , and everything folds into one equation along the channel whose only mystery constant is a length called $λ$ (Step 4). The leftover dopant constant is swept away by measuring from equilibrium, leaving the clean $ϕ^{''} = ϕ / λ^{2}$ ; that has a growing and a decaying branch, and physics (relax to equilibrium far from the drain) throws away the growing one, so the drain's flood fades exponentially with distance, and $λ$ is exactly how far it reaches — so you need the channel about $5 λ$ long to stay dry (Step 5). Finally, every extra gated side adds another press, its own term in the sum, stacking in the denominator so $λ$ shrinks like $1/ N_{s i d es}$ (Step 6). Planar has one side, FinFET three, GAA all four — the smallest reach, the tightest valley, the least leak. And the edge cases behave: thin body → tiny $λ$ , fat body → $λ \approx d_{s i} /4$ creeping back toward planar (Step 7). That last line is the trade-off knob and is why GAA nanosheets exist.

Recall Quick self-test

What physical quantity does $λ$ measure? ::: The distance the drain's field penetrates (screens into) the channel; the drain disturbance decays like $e^{- x / λ}$ . Why does more gate coverage shrink $λ$ ? ::: Each gated surface adds a pull-back term to the $1/ λ^{2}$ sum, so for identical faces $λ \propto 1/ N_{s i d es}$ . Where does the constant dopant source go when we reach $ϕ^{''} = ϕ / λ^{2}$ ? ::: It is absorbed by shifting variables (measuring from the far-from-drain equilibrium); it sets where "flat" is, not the reach $λ$ . How do we pick the decaying exponential out of the two branches? ::: Put $x = 0$ at the drain and demand $ϕ \to 0$ far into a long channel; the growing branch blows up there, so its coefficient must be zero, leaving $ϕ = ϕ_{D} e^{- x / λ}$ . What is $λ_{G AA}$ in the fat-body limit? ::: $λ_{G AA} \to d_{s i} /4$ , drifting back toward planar-like poor control — the drive-vs-control trade-off. Why must $L_{g} ≳ 5 λ$ ? ::: So the exponentially decaying drain intrusion has faded to near zero by mid-channel, keeping the OFF valley deep.

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