Etching (wet vs dry - plasma): Question bank — Hardware deep dive

Definition The symbols used on this page (build these before answering)

Picture a rectangular block of film sitting on the wafer, with a mask hole opened on top. The etch chews downward AND sideways from that hole. Four quantities describe it — look at the figure below as you read each:

$t$ — the film thickness, the vertical distance from the top surface down to the layer we must NOT cut into. This is how deep the etch must reach.
$R_{v}$ — the vertical etch rate, how fast (say, nm/min) the etch front moves straight down.
$R_{l}$ — the lateral etch rate, how fast the etch front creeps sideways under the mask edge.
$u$ — the undercut, the horizontal distance the etch has crept sideways under the mask by the time it has cut all the way down. This is the "hidden" widening you cannot see from the top.
$A$ — the anisotropy, a single number scoring how vertical the cut is, defined as $A = 1 - \frac{R _{l}}{R _{v}}$ . If sideways is as fast as down, $R_{l} / R_{v} = 1$ and $A = 0$ (round, isotropic). If nothing goes sideways, $R_{l} = 0$ and $A = 1$ (perfectly straight walls).
$S$ — the selectivity, $S = \frac{R _{f i l m}}{R _{ma s k}}$ , the ratio of how fast the film etches to how fast the mask etches. High $S$ means the mask survives.

True or false — justify

TRUE or FALSE: A perfectly isotropic etch (

A = 0

) undercuts each side of the mask by one full film thickness.

TRUE. With

A = 0

the lateral and vertical rates are equal, so

u = (1 - A) t = t

— the moment the etch reaches the bottom it has also crept sideways by the same distance

t

under the resist.

TRUE or FALSE: Higher selectivity guarantees straighter sidewalls.

FALSE. Selectivity

S = R_{f i l m} / R_{ma s k}

compares etch rates of two different materials; anisotropy

A = 1 - R_{l} / R_{v}

compares vertical vs lateral rate in the same material. A wet etch can be extremely selective yet completely isotropic.

TRUE or FALSE: Dry/plasma etching is always faster than wet etching.

FALSE. You choose dry etching for directionality, not speed — its rate is often slower. Wet etching can strip thick layers very fast in a batch of many wafers at once.

TRUE or FALSE: RIE (reactive ion etching) is just physical sputtering by fast ions.

FALSE. RIE is a synergy of chemical radicals plus directional ion bombardment; the combination etches faster at the trench bottom than either mechanism alone would.

TRUE or FALSE: Undercut depends only on the etch chemistry, not on the film.

FALSE. Chemistry sets the ratio

R_{l} / R_{v}

, but

u = (1 - A) t

also scales with film thickness

t

— a thick film undercuts more even at identical anisotropy.

TRUE or FALSE: Wet etching leaves no ion-induced surface damage.

TRUE. It is purely chemical, so there is no energetic ion bombardment to displace atoms; the trade-off is that it cannot etch vertically.

TRUE or FALSE: Anisotropy of exactly

A = 1

means zero undercut.

TRUE.

A = 1

implies

R_{l} = 0

, so

u = (1 - 1) t = 0

— the walls are perfectly vertical and the mask width is transferred exactly.

TRUE or FALSE: Etching adds material to the wafer where the mask has holes.

FALSE. Etching is a subtractive step; it removes the exposed film. Adding material is Thin-Film Deposition, a different step.

Spot the error

"To make features smaller, just raise selectivity as high as possible." — what's wrong?

Selectivity protects the mask/underlayer but does nothing for lateral cutting. Small features need high anisotropy (

A \to 1

); a super-selective isotropic wet etch still undercuts by a full

t

per side and blurs the pattern.

"Since undercut is

u = (1 - A) t

, using a thinner film always fixes bad anisotropy." — what's wrong?

Thinning the film does shrink absolute undercut

u

, but the ratio of undercut to feature size can stay just as bad, and you may not have freedom to thin the film. The real cure for sub-micron features is raising

A

, i.e. switching to plasma/RIE.

"Plasma etches straight down because the reactive radicals travel vertically." — what's wrong?

Neutral radicals move in random directions like any gas. The directionality comes from the ions, which are accelerated vertically by the sheath electric field; radicals supply the chemistry, ions supply the direction.

"Two lines spaced

s = t

apart survive an isotropic wet etch fine." — what's wrong?

Isotropic etching undercuts

u = t

per side, so the gap erodes by

2 t

. You need

s > 2 t

or the two features merge — exactly why dense lines can't be wet-etched.

"Overetching is a mistake you should always avoid." — what's wrong?

Overetch is deliberate. Films aren't perfectly uniform, so you etch past the nominal clear-time to guarantee the whole wafer is cleared; high selectivity

S

is what lets the mask/underlayer survive that extra time.

"Sidewalls in RIE stay vertical because ions can't reach them." — mostly, but what's incomplete?

Geometry helps (vertical ions strike the bottom, not the walls), but the key extra ingredient is passivation — see the definition callout below; a thin protective film coats the walls and blocks chemical attack there, so only the ion-cleared bottom keeps etching.

"Selectivity

S < 1

is impossible." — what's wrong?

S = R_{f i l m} / R_{ma s k}

can be below 1 if the mask etches faster than the film — a disastrous but very real situation where your stencil dissolves before the film clears.

Why questions

Why does a liquid etchant produce rounded, undercut sidewalls?

A liquid molecule has no preferred direction; once it reaches the film through the mask hole it reacts equally sideways and downward (

R_{l} \approx R_{v}

), carving a curved profile that eats under the resist.

Why does wet etch rate rise sharply with temperature?

The surface reaction is chemically activated, so its rate follows the Arrhenius law

R \propto e^{- E_{a} / k T}

— warming the bath gives more molecules the activation energy

E_{a}

, exponentially speeding the etch.

Why can dry etching achieve near-vertical walls that wet etching cannot?

Ions are accelerated by the sheath field to arrive nearly vertical, so material is removed mainly downward (

R_{v} ≫ R_{l}

), pushing

A \to 1

while sidewalls stay protected by passivation.

Why is selectivity typically lower for dry etching than wet?

The physical ion-bombardment component removes atoms somewhat indiscriminately (by momentum, not chemistry), so it attacks mask and underlayer too — chemistry alone in wet etching can be tuned to ignore the mask almost completely.

Why must anisotropy and selectivity be treated as two independent knobs?

Anisotropy

A

is a within-one-material directionality ratio (

R_{l} / R_{v}

); selectivity

S

is a between-materials rate ratio (

R_{f i l m} / R_{ma s k}

). Changing gas pressure/bias can raise one while lowering the other, so a process must balance both.

Why does thicker film force you to accept more undercut?

Because

u = (1 - A) t

: the sideways creep accumulates over the whole time needed to reach depth

t

, so doubling the film doubles the undercut at fixed anisotropy.

Why does RIE etch faster at the trench bottom than on the sidewalls?

The vertical ion flux hits the bottom and drives the surface reaction there (ion-assisted etching), while sidewalls receive few ions and are shielded by the passivation film, so chemistry alone barely proceeds on them.

Edge cases

EDGE CASE: What is the undercut if

A = 0

and the film thickness is essentially zero (

t \to 0

u = (1 - 0) \cdot 0 = 0

. Even a fully isotropic etch causes negligible undercut on a vanishingly thin film — undercut is a product of poor anisotropy AND finite thickness

t

EDGE CASE: What does

A > 1

mean physically?

It would require

R_{l} < 0

, i.e. the sidewall growing outward — impossible for pure etching. Values above 1 signal deposition/passivation dominating, not etching, so

A

is bounded above by

1

for a real etch.

EDGE CASE: What does negative anisotropy (

A < 0

) mean, and when does it happen?

A < 0

requires

R_{l} > R_{v}

: the etch goes sideways faster than down. This happens in strongly radical-driven, low-bias plasmas (or aggressive wet etches with a slow-etching underlayer), producing a "notched" or barrel-shaped profile where the undercut

u = (1 - A) t

exceeds the film thickness

t

itself.

EDGE CASE: A mask with zero-width openings — what does etching do?

With no exposed film there is nothing to attack; the pattern is a blank field and no trench forms. The etch only acts where Photolithography opened the resist.

EDGE CASE: Infinite selectivity (

S \to \infty

) — is anisotropy then also perfect?

No.

S \to \infty

only means the mask/underlayer never erodes; the film could still be etched perfectly isotropically (

A = 0

). The two limits are unrelated.

EDGE CASE: Zero bias voltage in a plasma tool (ions not accelerated) — what happens to anisotropy?

Without vertical acceleration the ions lose their directionality and only chemical radicals act, so the etch becomes essentially isotropic (

A \to 0

) — you've turned an anisotropic tool into a dry chemical etch.

EDGE CASE: Etching a film already thinner than one monolayer of undercut tolerance — does the formula still apply?

The continuum formula

u = (1 - A) t

breaks down near atomic scales, where discrete atoms and roughness dominate; treat it as a smooth-film approximation valid only when

t

spans many atomic layers.

Recall One-line self-test

If someone says "my etch is very selective, so my 50 nm lines will be sharp," what's your rebuttal? ::: Selectivity $S$ says nothing about sidewall verticality; sharp small lines need high anisotropy $A$ (dry/RIE), because an isotropic etch undercuts by $(1 - A) t$ regardless of how selective it is.