Schottky diodes and metal-semiconductor junctions: Question bank

Definition Every symbol used on this page (earn them before you meet them)

$ϕ_{M}$ = metal work function (energy to lift an electron from the metal's Fermi level up to the vacuum level).
$ϕ_{S}$ = semiconductor work function; $χ$ = electron affinity (vacuum level down to the conduction-band edge $E_{C}$ ).
$q$ = the elementary charge, the magnitude of the charge on a single electron ( $q \approx 1.6 \times 1 0^{- 19}$ C). It appears whenever we turn a voltage (energy per charge) into an energy: multiplying a potential $V$ by $q$ gives the energy $q V$ an electron gains or loses. That is why the barriers written as energies carry a $q$ (e.g. $q V_{bi}$ ) while the work functions $ϕ_{M}, ϕ_{S}, χ$ are already energies.
$ϕ_{B} = ϕ_{M} - χ$ is the fixed metal-side barrier; $q V_{bi} = ϕ_{M} - ϕ_{S}$ is the semiconductor-side built-in barrier, the one bias can move. See Work Function and Electron Affinity if any of these feel shaky.
$V$ = the applied voltage. Sign convention on this whole page: $V > 0$ (forward) means the metal is at the higher potential relative to the semiconductor (grounded semiconductor is our reference node); $V < 0$ is reverse. Get this backwards and every "which barrier moves" answer flips.
$I$ = the terminal current, and our sign convention is: $I > 0$ means conventional current flows into the metal terminal and out of the semiconductor — i.e. the forward direction, the same direction that $V > 0$ pushes. So in $I = I_{S} (e^{q V / nk T} - 1)$ , forward bias ( $V > 0$ ) gives a large positive $I$ , and reverse bias ( $V < 0$ ) drives $I$ toward the small negative value $- I_{S}$ . (Because current is carried by electrons, the electrons physically move the opposite way, semiconductor → metal, in forward bias.)
$I_{S}$ = saturation current, given by $I_{S} = A A^{*} T^{2} e^{- ϕ_{B} / k T}$ , where $A$ = the junction cross-sectional area (bigger contact ⇒ more current), $A^{*}$ = the Richardson constant (a material constant setting how readily electrons are thermally emitted — see Thermionic Emission and Richardson's Law), $T$ = absolute temperature, $k$ = Boltzmann's constant.
$n$ = the ideality factor, a dimensionless fudge in $I = I_{S} (e^{q V / nk T} - 1)$ that stretches the exponential; $n = 1$ is perfect thermionic emission, $n > 1$ warns of extra non-ideal effects (image-force lowering, tunnelling, recombination). Real Schottkys sit around $n \approx 1.0$ – $1.1$ .
$ε_{s}$ = the semiconductor permittivity (how strongly the material screens electric field), and $N_{D}$ = the donor doping density (dopant atoms per unit volume that supply the conduction electrons) — both appear in the depletion-width formula $W = 2 ε_{s} (V_{bi} - V) / (q N_{D})$ . See Depletion Region and Poisson's Equation.

True or false — justify

A Schottky diode conducts using both electrons and holes, like a p–n diode.

False. It is a majority-carrier device: on n-type Si only electrons move across the junction, which is exactly why there is no minority-charge storage. See PN Junction Diode for the two-carrier contrast.

Applying forward bias lowers the metal-side barrier

ϕ_{B}

, letting more electrons in.

Mostly false. The main

ϕ_{B} = ϕ_{M} - χ

is set by materials and is essentially pinned; forward bias lowers the semiconductor-side barrier from

q V_{bi}

q (V_{bi} - V)

. (A tiny second-order image-force lowering does shave a few tens of meV off

ϕ_{B}

— see the dedicated item below — but that is not the mechanism that turns the diode on.)

For a metal on n-type semiconductor,

ϕ_{M} > ϕ_{S}

gives a rectifier.

True. When the metal's Fermi level sits lower, electrons spill from semiconductor to metal, exposing positive donors and raising a depletion barrier — that asymmetry rectifies.

The same condition

ϕ_{M} > ϕ_{S}

makes a rectifier on p-type semiconductor.

False. For p-type the majority carriers are holes, so the rectifying condition ==flips to

ϕ_{M} < ϕ_{S}

==. Applying the n-type rule to p-type is a classic sign error.

Heavier doping always improves the Schottky rectifier.

False. Higher

N_{D}

shrinks the depletion width

W

until electrons tunnel straight through, destroying rectification — this is precisely how Ohmic Contacts are made on purpose.

The 0.3 V turn-on of a Schottky comes from a smaller energy gap in the material.

False. It comes from a smaller barrier

ϕ_{B}

, which makes

I_{S} = A A^{*} T^{2} e^{- ϕ_{B} / k T}

much larger, so the same current is reached at a lower

V

. The bandgap of the semiconductor need not change.

Schottky diodes switch faster than p–n diodes because they are physically smaller.

False. Speed comes from having no stored minority charge to sweep out, giving near-zero reverse-recovery time — a physics reason, not a size reason.

In reverse bias the depletion region gets narrower.

False. With

V < 0

the term

(V_{bi} - V)

increases, so

W = 2 ε_{s} (V_{bi} - V) / (q N_{D})

grows — reverse bias widens

W

. See Depletion Region and Poisson's Equation.

ϕ_{B}

and

q V_{bi}

are the same barrier seen two ways, so they are equal.

False. They differ by the bulk offset:

ϕ_{B} = q V_{bi} + (E_{C} - E_{F})_{bulk}

, so the metal-side barrier is a little taller than the semiconductor-side one (read it off the figure above).

An ideality factor of

n = 1

and

n = 2

describe the same quality of diode.

False.

n = 1

signals clean thermionic emission (the ideal Schottky);

n

drifting toward

2

flags non-ideal current paths, so a larger

n

means a worse, leakier diode.

Spot the error

"Forward bias means

+ V

on the semiconductor for an n-type Schottky."

Error. By our convention forward bias puts

+ V

on the metal (semiconductor is the reference/ground), which lowers the barrier n-type electrons in the semiconductor see and floods them into the metal.

I_{S} = A A^{*} T^{2} e^{- ϕ_{B} / k T}

, so at absolute zero the saturation current is largest."

Error. As

T \to 0

the

T^{2}

prefactor and the thermal Boltzmann tail both collapse; thermionic emission needs thermal energy, so

I_{S} \to 0

, not maximum. See Thermionic Emission and Richardson's Law.

"Since

ϕ_{B} = ϕ_{M} - χ

, using a higher-work-function metal lowers the turn-on voltage."

Error. Higher

ϕ_{M}

raises

ϕ_{B}

, which reduces

I_{S}

and raises turn-on. To lower turn-on you want a smaller

ϕ_{B}

"An ohmic contact has no barrier, so no depletion physics applies anywhere."

Error. The bulk semiconductor and the rest of the circuit still obey the usual electrostatics; the ohmic contact simply has an accumulation (or tunnel-thin) region so it conducts both ways without a rectifying wall.

"The diode equation

I = I_{S} (e^{q V / nk T} - 1)

for a Schottky comes from diffusion of minority carriers."

Error. For a Schottky the same-shaped equation arises from thermionic emission of majority carriers over the barrier, not minority-carrier diffusion as in a p–n junction.

"Because

ϕ_{M} < ϕ_{S}

on n-type gives an ohmic contact, the metal must inject holes."

Error. It injects/accumulates electrons at the surface (n-type majority carriers), producing a low-resistance electron path — no holes involved.

A^{*}

I_{S}

is just the junction area written differently."

Error.

A

is the geometric junction area;

A^{*}

is the Richardson constant, a material property fixing the rate of thermionic emission. They are two independent factors that happen to multiply together.

Why questions

Why does the metal-side barrier

ϕ_{B}

stay essentially fixed under bias while

q V_{bi}

moves?

The applied voltage drops across the depletion region inside the semiconductor, bending its bands; the metal's electrons still start at the metal Fermi level and see the (nearly) unchanged

ϕ_{M} - χ

climb.

Why is the reverse current of a Schottky small and roughly constant?

It is the metal→semiconductor flow over the fixed

ϕ_{B}

, giving a nearly bias-independent

\propto e^{- ϕ_{B} / k T}

term — nothing in reverse lowers that wall (only the weak image-force effect nudges it).

Why does lower barrier automatically mean faster forward turn-on?

A smaller

ϕ_{B}

multiplies

I_{S}

e^{- Δ ϕ_{B} / k T}

, so the exponential

I = I_{S} e^{q V / nk T}

reaches useful current at a much smaller

V

Why can the exact same

ϕ_{M} > ϕ_{S}

rule that builds a rectifier be defeated by doping?

The rule sets barrier height (

ϕ_{B}

); the doping

N_{D}

sets barrier width (

W

). Enough doping makes the wall thin enough to tunnel through, so height no longer blocks carriers.

Why does aligning Fermi levels force the bands to bend at all?

In equilibrium

E_{F}

must be flat everywhere; if the two materials' Fermi levels started at different heights, charge transfers until the bands bend to make them match, and that bending is the built-in potential.

Why does the effective barrier depend a little on bias even though

ϕ_{M} - χ

is a material constant?

An electron just outside the metal induces an image charge, and this image-force (Schottky) lowering slightly reduces the peak barrier; reverse bias raises the field and lowers

ϕ_{B}

a touch more, which is why real reverse leakage creeps up rather than being perfectly flat.

Why are Schottky diodes preferred in switching power supplies and high-frequency rectifiers?

Their near-zero reverse-recovery time and low forward drop cut switching and conduction losses — see Rectifiers and Switching Power Supplies.

Edge cases

What exactly do we mean by "turn-on voltage," quantitatively?

It is the forward

V

at which the exponential current reaches some agreed threshold (commonly ~1 mA/mm² of junction, i.e. a datasheet current) — not a sharp switch. For a Schottky that threshold is crossed near ~0.2–0.4 V; for a Si p–n diode near ~0.7 V, purely because

I_{S}

differs.

What happens at exactly

ϕ_{M} = ϕ_{S}

on n-type?

q V_{bi} = ϕ_{M} - ϕ_{S} = 0

, so there is no built-in barrier and no depletion bending — the boundary between rectifying and ohmic, effectively a neutral (near-ohmic) contact.

What is the built-in potential when the metal work function only barely exceeds

ϕ_{S}

q V_{bi} = ϕ_{M} - ϕ_{S}

is tiny, so the barrier is shallow, the diode leaks easily, and rectification is weak — a poor Schottky diode.

What does

W

do in the limit of very strong forward bias

V \to V_{bi}

W = 2 ε_{s} (V_{bi} - V) / (q N_{D}) \to 0

: the depletion region vanishes as the barrier flattens, which is the fully-on state.

What does

I = I_{S} (e^{q V / nk T} - 1)

give at

V = 0

Exactly

0

: the forward and reverse thermionic flows balance in equilibrium, as any equilibrium contact must — independent of

n

At very high temperature, what happens to the reverse leakage

\propto e^{- ϕ_{B} / k T}

It grows, because more electrons in the Boltzmann tail can hop the fixed

ϕ_{B}

— Schottky reverse leakage worsens noticeably with heat.

Recall One-line self-test

Cover everything: name the one barrier bias mainly moves, the one it does not, and the doping trap. ::: Bias moves $q (V_{bi} - V)$ (semiconductor side); it leaves $ϕ_{B} = ϕ_{M} - χ$ essentially fixed (only weak image-force lowering acts on it); too much doping ( $N_{D}$ ) thins $W$ and creates tunnelling → ohmic contact.