Channel length and short-channel effects: Question bank

Before we start, the figure below fixes every symbol used on this page. Look at it as you read the definitions — each arrow in the picture is one of the symbols listed underneath.

Figure — Channel length and short-channel effects

Definition The three

directions (the three arrows in the figure above)

Lateral into the channel ( $x_{d S}, x_{d D}$ , the plum arrows pointing sideways): how far the source and drain depletion wedges reach toward each other. These set whether a device is short-channel: when $L ≲ x_{d S} + x_{d D}$ the wedges meet.
Vertical into the bulk ( $x_{d m}$ , the teal arrow pointing down): how deep the gate's own depletion region reaches downward. This shapes the trapezoid geometry of roll-off.
Along the current ( $L$ , the black double arrow across the bottom): source-to-drain spacing, the length carriers actually travel.
$x_{j}$ = junction depth, how deep the source/drain diffusions (the orange blocks) reach into the bulk. Deeper junctions make the lateral wedges reach further, worsening short-channel effects.

Mixing these up is the single most common error the topic invites — see Depletion region physics of pn junctions for why each junction carves its own region.

Definition The other symbols used below (define before you meet them)

$g \equiv \frac{x _{j}}{L} (1 + \frac{2 x _{d m}}{x _{j}} - 1)$ — the geometric factor in threshold roll-off. It is the fraction of bulk charge the source/drain corners steal from the gate; roll-off is $Δ V_{T} \propto - g$ , and since $g \propto x_{j} / L$ it grows as $L$ shrinks.
$v_{s a t}$ = saturation velocity, the top speed a carrier can reach in silicon; beyond it, more field gives no more speed. See Velocity saturation and carrier transport.
$E_{cr i t}$ = critical field, the lateral field at which velocity is about halfway to $v_{s a t}$ ; in the model $v = \frac{μ E}{1 + E / E _{cr i t}}$ it satisfies $v_{s a t} = μ E_{cr i t}$ . So $E_{cr i t}$ and $v_{s a t}$ are two views of the same speed limit.
$η \equiv - \frac{\partial V _{T}}{\partial V _{D S}}$ = DIBL coefficient (units mV/V): how many millivolts $V_{T}$ drops per volt of $V_{D S}$ . See Drain-Induced Barrier Lowering (DIBL).
$θ$ = mobility-degradation parameter (units $V^{- 1}$ ) in $μ_{e f f} = \frac{μ _{0}}{1 + θ ( V _{GS} - V _{T} )}$ : it measures how strongly the vertical field crushes carriers into the rough interface. See Surface scattering and effective mobility.

True or false — justify

A device is short-channel purely because

L

is small in nanometers

False — "short" is relative: the condition is

L ≲ x_{d S} + x_{d D}

. A 100 nm device on lightly-doped bulk (fat depletion wedges) can be short-channel while a 100 nm device on heavily-doped bulk (thin wedges) still behaves long-channel. See Depletion region physics of pn junctions.

Threshold roll-off makes

V_{T}

rise as

L

shrinks

False — it makes

V_{T}

fall. The source/drain wedges pre-deplete part of the bulk charge, so the gate has less charge

Q_{B}^{'}

to support and needs less voltage;

Δ V_{T} \propto - 1/ L

DIBL and threshold roll-off are the same effect

False — roll-off is

V_{T}

's dependence on $L$ at fixed

V_{D S}

(a geometry/charge-sharing effect); DIBL is

V_{T}

's dependence on $V_{D S}$ at fixed

L

(a drain-field-lowers-the-barrier effect). Both lower

V_{T}

but for different reasons. See Drain-Induced Barrier Lowering (DIBL).

Velocity saturation makes the drain current larger than the ideal square law predicts

False — it makes it smaller. Carriers hit a speed ceiling

v_{s a t}

, so current stops growing quadratically with overdrive and grows only linearly, capping the drive you'd naively expect. See Velocity saturation and carrier transport.

Higher

V_{GS}

always gives proportionally more current

False — vertical-field mobility degradation,

μ_{e f f} = μ_{0} / (1 + θ (V_{GS} - V_{T}))

, crushes carriers into the rough interface so drive rises sublinearly; velocity saturation caps it further. See Surface scattering and effective mobility.

Channel length modulation means the physical channel length actually gets shorter

Half-true — the physical

L

is fixed by lithography, but the electrically effective inverted channel shortens to

L - Δ L

as the drain-side pinch-off point retreats when

V_{D S}

rises. That is what gives finite output resistance.

In a long-channel device the gate is the sole controller of channel charge

True — the source/drain wedges are a negligible fraction of a long

L

, so essentially all of the bulk depletion charge

Q_{B}

sits under the gate's control. That is exactly why the long-channel model works.

Punch-through is just channel length modulation taken far enough

True in spirit — CLM grows the drain depletion into the channel; when

V_{D S}

pushes it all the way to touch the source depletion, the merged path is punch-through, a current the gate can no longer switch off.

DIBL only matters when the transistor is on

False — its worst damage is in the off state: a lower source-channel barrier at high

V_{D S}

multiplies subthreshold leakage even at

V_{GS} = 0

. See Subthreshold conduction and leakage.

Velocity saturation and vertical-field mobility degradation are two names for one phenomenon

False — velocity saturation comes from the lateral field

V_{D S} / L

(carriers hitting

v_{s a t}

); mobility degradation comes from the vertical field

\propto (V_{GS} - V_{T}) / t_{o x}

(surface scattering). Different fields, different directions, both present in short devices.

Spot the error

"To onset short-channel behaviour I plug the vertical gate depletion depth

x_{d m}

into

L ≲ x_{d S} + x_{d D}

Wrong symbol — the onset uses the lateral wedge widths

x_{d S}, x_{d D}

reaching sideways;

x_{d m}

is the vertical depth that only shapes the roll-off trapezoid. Never swap the two directions.

"Since

Δ V_{T} \propto x_{j} / L

, making junctions deeper reduces roll-off."

Backwards — deeper junctions (

x_{j}

, the junction depth, larger) make the removed triangular corners bigger, so

Δ V_{T}

grows. Shallow junctions suppress roll-off, which is why scaled devices use ultra-shallow junctions.

"At

L = 1 μ

m the geometric factor

g = \frac{x _{j}}{L} (1 + 2 x_{d m} / x_{j} - 1)

is huge, so roll-off dominates."

No —

g \propto x_{j} / L

, so a large

L

makes

g

tiny (e.g.

\approx 0.027

at 1 µm vs

\approx 0.55

at 50 nm). Roll-off is negligible for long channels.

"Because

I_{D, s a t} \propto (V_{GS} - V_{T})

in a short device, doubling

L

doubles the current."

Wrong reasoning — in the velocity-saturated regime

I_{D, s a t} \approx W C_{o x} v_{s a t} (V_{GS} - V_{T})

has no explicit $L$ ; the drain current is set by the saturation velocity

v_{s a t}

, not by how quickly

μ E

scales with

L

. Length dependence largely drops out.

"DIBL coefficient

η

is measured in volts, since it's a threshold shift."

Wrong units —

η = - \partial V_{T} / \partial V_{D S}

is a slope, volts of

V_{T}

per volt of

V_{D S}

, usually quoted in mV/V. It multiplies

V_{D S}

to give a shift.

"A high

E_{cr i t}

material is easier to velocity-saturate."

Backwards — the saturation condition is

E ≫ E_{cr i t}

, so a high critical field

E_{cr i t}

means you need a much larger field before carriers hit their speed limit, i.e. it resists velocity saturation.

"Since

μ_{e f f} = μ_{0} / (1 + θ (V_{GS} - V_{T}))

, at

V_{GS} = V_{T}

mobility is zero."

No — at

V_{GS} = V_{T}

the overdrive is zero, so the denominator is

1

and

μ_{e f f} = μ_{0}

, its maximum. Mobility falls only as overdrive grows.

Why questions

Why does a fixed corner charge hurt a short channel more than a long one?

Because the two triangular wedges are a fixed geometric size (set by junction depth

x_{j}

and gate depth

x_{d m}

), but they are a fraction

\propto 1/ L

of the total — the same corner eats a bigger slice of a shorter channel.

Why does raising

V_{D S}

act a little like raising

V_{GS}

in a short device?

The drain field reaches across the short channel to the source and lowers the injection barrier (DIBL) — the same barrier the gate lowers — so the drain partly does the gate's job.

Why does velocity saturation change the current from quadratic to linear in overdrive?

The square law comes from velocity scaling as

μ E

all along the channel; once carriers pin at the constant

v_{s a t}

, current

\propto W C_{o x} v_{s a t} (V_{GS} - V_{T})

becomes just the injected charge times a fixed speed — one power of overdrive.

Why do scaled devices suffer worse vertical-field mobility degradation?

Scaling thins the oxide

t_{o x}

and often keeps overdrive high, so the vertical field

\propto (V_{GS} - V_{T}) / t_{o x}

is enormous, pressing carriers hard against the rough Si–SiO₂ interface and amplifying surface scattering. See Surface scattering and effective mobility.

Why does channel length modulation give a MOSFET finite (not infinite) output resistance in saturation?

An ideal saturated device would have current independent of

V_{D S}

; but CLM shortens the effective channel as

V_{D S}

rises, so

I_{D}

keeps creeping up, giving a nonzero slope

\partial I_{D} / \partial V_{D S}

and hence finite

r_{o}

Why is

λ \propto 1/ L

The pinch-off retreat

Δ L

is a roughly fixed physical distance for a given

V_{D S}

; as a fraction

Δ L / L

of a shorter channel it is larger, so its current-boosting effect

λ

grows as

L

shrinks.

Why does DIBL worsen chip-level leakage power more than a single transistor suggests?

Millions of off transistors each leak, and because subthreshold current is exponential in the barrier, a modest DIBL-driven

V_{T}

drop multiplies every one of them — e.g. tens of mV can raise leakage by an order of magnitude. See Subthreshold conduction and leakage.

Edge cases

What happens to

Δ V_{T}

in the limit

L \to \infty

The geometric factor

g \propto x_{j} / L \to 0

, so

Δ V_{T} \to 0

and

V_{T} \to V_{T 0}

— the ideal long-channel threshold is recovered, exactly as it should be.

What if the junction depth

x_{j} \to 0

(perfectly shallow junctions)?

The factor

x_{j} (1 + 2 x_{d m} / x_{j} - 1)

tends to

2 x_{d m} x_{j} \to 0

, so roll-off vanishes — the physical reason ultra-shallow junctions are used to fight SCEs.

What is the current law in a device where

E = V_{D S} / L

is far below the critical field

E_{cr i t}

Carriers never reach

v_{s a t}

; velocity stays

v = μ E

(linear), so the classical square-law saturation current survives — this is the ordinary long-channel regime.

At exactly

E = E_{cr i t}

, what is the carrier velocity?

Plugging into

v = μ E / (1 + E / E_{cr i t})

gives

v = μ E_{cr i t} /2 = v_{s a t} /2

— the carrier is at half its ceiling, the crossover between linear and saturated transport. See Velocity saturation and carrier transport.

What happens when the drain depletion region just touches the source depletion region?

That is punch-through: a gate-independent conduction path opens beneath the channel, current shoots up, and the transistor stops working as a switch — the

L ≲ x_{d S} + x_{d D}

limit reached by high

V_{D S}

In the degenerate case

V_{GS} = V_{T}

exactly, what does

μ_{e f f}

equal?

With zero overdrive the degradation term

θ (V_{GS} - V_{T}) = 0

, so

μ_{e f f} = μ_{0}

— mobility is undegraded right at threshold and only falls as you drive harder.

What does Dennard scaling predict about SCEs if you scale

L

but forget to scale the junctions and doping?

SCEs blow up — the wedges stay fat relative to a shrunk

L

, roll-off and DIBL explode. Constant-field scaling requires shrinking

x_{j}

, thinning

t_{o x}

, and raising

N_{A}

together. See Scaling theory (Dennard scaling).

Recall Quick self-check before you close this page

Name the direction each symbol points ::: $L$ along the current, $x_{d S} / x_{d D}$ laterally into the channel, $x_{d m}$ vertically into the bulk, $x_{j}$ the junction depth downward. Roll-off vs DIBL — which variable? ::: Roll-off vs $L$ , DIBL vs $V_{D S}$ . Which field causes velocity saturation, which causes mobility degradation? ::: Lateral field ( $V_{D S} / L$ ) → velocity saturation; vertical field ( $(V_{GS} - V_{T}) / t_{o x}$ ) → mobility degradation. What links $E_{cr i t}$ and $v_{s a t}$ ? ::: $v_{s a t} = μ E_{cr i t}$ — the critical field and the saturation velocity are two views of the same speed limit.