10 Metal-Semiconductor Junctions

8/13/2019 10 Metal-Semiconductor Junctions

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Electronic DevicesMetal-Semiconductor junctions

prof. ing. Gianluca GiustolisiUniversit a degli studi di Catania

Academic Year 2012/2013(ver. January 16, 2013)
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The Schottky barrier diode
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A Schottky barrier contact is made between a metal and a

semiconductorIn general the semiconductor is n-typeThe vacuum level is used as reference level ( E vac )m and s are the metal and the semiconductor workfunctions, respectively ( m > s )

is the electron afnity of the semiconductor


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Work functions and Electron afnities

Work functionElement mAg, silver 4.26

Al, aluminium 4.28Au, gold 5.10Cr, chromium 4.50Mo, molybdenum 4.60Ni, nickel 5.15

Pd, palladium 5.12Pt, platinum 5.65Ti, titanium 4.33W, tungsten 4.55

Electron afnity

Element Ge, germanium 4.13

Si, silicon 4.01GaAs, gallium arsenide 4.07AlAs, aluminium arsenide 3.50
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Joining the two materials, electrons will ow into the metaland a space charge region is created in the semiconductorThe Fermi level becomes a constant through the system inthermal equilibriumTwo barriers ( B 0 and V bi ) are createdW is the space charge region width


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The Schottky barrier

B 0 is the Schottky barrier that represents the potentialbarrier seen by electrons in the metal trying to move intothe semiconductor

B 0 = m The barrier is independent of the semiconductor dopingconcentration


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The built-in potential

V bi is the built-in potential barrier, that is, the barrier seenby electrons in the semiconductor conduction band tryingto move into the metal

V bi = m s = B 0 n The barrier depends on the semiconductor doping

concentration as n = V t lnN c N d =

E g 2q V t ln

N d n i
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Reverse bias

Applying V R to the semiconductor with respect to themetal, the semiconductor-to-metal barrier increases whileB 0 remains constantThe current is constant and is supported by the smallnumber of electrons that ows across B 0
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Forward bias

Applying V a to the metal with respect to the semiconductor,the semiconductor-to-metal barrier decreasesElectrons can more easily ow from the semiconductor tothe metalNote that the current is supported by majority carriers
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Ideal junction properties

The electrostatic eld in the space charge region is due to

the semiconductor charge density made of positive ionsdE dx

= (x )

s =

qN d s

for 0 x x n Integrating the Poisson equation, we have

E = qN d s dx = qN d x s + C 1The electric eld is zero in x = x n , then

E = qN d

s (x x n )In the metal, a negative surface charge exists at themetal-semiconductor junction

(x ) = qN d x n (x )
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The space charge region width is

W = x n = 2 s qN d

(V bi + V R )1/ 2

A junction capacitance exists

C = dQ dV R

= qN d dx n dV R

= q s N d

2 (V bi + V R )

1/ 2

As for pn junctions we have

1C

2

= 2 (V bi + V R )

q s N d

which can be used to determine V bi , N d and, after few

manipulations, B 0
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Nonideal effects on the barrier height
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Nonideal effects on the barrier heightImage-force-induced lowering of the potential barrier (Schottky effect)

In vacuum, an electron q placed at a distance x from ametal layer will create an electric eldThe eld lines shall be perpendicular to the metal surfaceThe electron behaves as if no metal exists and a positivecharge + q is placed at the same distance from the metalsurface in x (image charge )In our case, a similar model exists by using the electroneffective mass and the semiconductor dielectric constant

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Due to the image charge, the force acting on the electron

q placed in x is

F ic = q 2

4 s (2x )2 = q E

The electric eld in x is

E ic = q

16 s x 2

And the electric potential results

ic (x ) =

x

+ E ( ) d =

= +

x

q 16 s 2

d =

=

q 16

s

1

+

x

= q

16 s x

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Given the potential

ic (x ) = q

16 s x the energy set by the image charge is

E ic (x ) = q ic (x ) = q 2

16 s x This contribution must be added to the energy set by theideal Schottky barrier theory

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The energy set by the ideal Schottky barrier theory

decreases almost linearly in proximity of themetal-semiconductor junction, hence

E sb (x ) E sb (0) + dE sb

dx x = 0 x = E sb (0) (q E sb )x

being E sb the (positive and constant) electric eldgenerated by the Schottky barrier 1 at x = 0The overall energy is

E (x ) = E ic (x ) + E sb (x ) = E sb (0) q 2

16 s x q E sb x 1Remember that the electric eld and the energy are related by

E (x ) = 1

q

dE (x )

dx

which denes E sb = 1

q

dE sb

dx x = 0> 0

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The overall energy E (x ) = E sb (0) q 2

16 s x q E sb x has amaximum atx m = q 16 s E sb

which results

E m = E sb (0)

q

q E sb 4 s



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The peak potential barrier is then lowered ( Schottky effect )With respect to the ideal case, the barrier lowering is set by

= E m E sb (0)

q = q E sb 4 s

The barrier height seen by electrons in the metal is now

Bn = B 0

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Nonideal effects on the barrier heightEffects of the surface states

Experimental measurements shows a monotonic relationbetween the work function and the barrier heightHowever the relationship is not linear as expected

Bn on GaAs

Metal Bn Au 0.87

Bn on Si

Metal Bn Al 0.55Au 0.81Pt 0.89W 0.68

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Nonideal effects on the barrier heightEffects of the surface states

Surface states may greatly inuence the potential barrieras wellSince the surface state density is not predictable with anydegree of certainty, the barrier height must be anexperimentally determined parameter
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Current-Voltage relationship

Current Voltage relationship
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The current across a metal-semiconductor junction ismainly due to majority carriersThree different mechanisms exists

Drift/diffusion of carriers from the semiconductor into themetalThermionic emission of carriers across the Schottky barrierQuantum-mechanical tunneling through the barrier

Typically, only one current mechanism dominates

Drift/diffusion current
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Electrons are pushed

from the semiconductorto the metal thanks to acombination ofdrift/diffusion processThe process takesplace in thesemiconductordepletion region [, x d ],with > 0 and 0

Actually, the depletionregion does not extendup to the point x = 0because of theSchottky effect

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Let us consider the total electron current

J dd = q n n E + D n dn dx

= qD n n V t

ddx

+ dn dx

Let us multiply both members by exp V t

J dd e

V t = qD n n V t

e

V t ddx

+ e

V t dn dx

= qD n d

dx n e

V t

Let us integrate it over the depletion region [, x d ]

J dd =qD n n e

V t

x d

x = x d

x = e

V t dx =

qD n n (x d )e

(x d )V t n ()e

( )

V t

x d x = e

V t dx

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We set the reference potential at x m 0, i.e., where E c hasthe maximum, E c (x m ). Hence (x m ) = 0In x = x d we have

n (x d ) = N d = N c e

E c (x d ) E Fn (x d )kT = N c e

Bn V bi

V t

(x d ) = V bi V a Hence we have

n (x d )e

(x d )V t = N c e

Bn V bi

V t e

V bi V a V t = N c e

Bn V t e

V a V t

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In x = we have

n () = N c e

E c ( ) E Fn ( )kT

() = E c () E c (x m )

q = Bn

E c () E Fm q

Hence we have

n ()e ( )

V t = N c eE Fn ( ) E c ( )

kT e q Bn + E c ( ) E Fm

kT = N c e Bn

V t eE Fn ( ) E Fm

kT

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The total electron current is

J dd =qD n n (x d )e

(x d )V t n ()e

( )V t

x d

e

V t dx =

= qD n N c e

Bn V t

e

V a V t

eE Fn ( ) E Fm

kT

x d

e

V t dx

The integral

x d

e

V t dx may be solved assuming that (x )is the solution of the Poisson equation d2 dx 2 = qN d s , that is

(x ) = qN d 2 s

(x x d )2+qN d x 2d

2 s =

qN d x 2d 2 s

1 1 x x d

2

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Since (x d ) = qN d x 2d

2 s = V bi V a we have(x ) = ( V bi V a ) 1 1

x x d

2

2(V bi V a )

x d x

where the quadratic term has been neglected to simplifythe integral evaluationMoreover, since

V bi V a = E max x d

2

we may write also(x ) = E max x

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The integral

x d

e

V t dx becomes

x d

e

V t dx

x d

0e

E max

V t x dx =

V t

E max1 e

E max x d

V t =

= V t

E max

1 e

2(V bi V a )V t

V t

E max

where we assumed 2 (V bi V a ) > 4V t and

E max = 2qN d (V bi V a )s The total electron current is

J dd = q n E max N c e

Bn V t e

V a V t e

E Fn ( ) E Fm kT

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The overall current may be written in the form

J dd = J dds eqV a kT

e

E Fn ( ) E Fm kT

where, the saturation current is dened as

J dds = q n E max N c exp q B 0kT

expq kT

Thermionic emission of carriers
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Only electrons with an

energy larger than thetop of the barrier maycross the barrierElectrons in the metalmust cross the xedbarrier q Bn (currentJ ms )Electrons in thesemiconductor must

cross the variablebarrier q (V bi V a )(current J sm )

J n = J sm

J ms

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The thermionic current must be evaluated at x = The ow coming from the semiconductor is made ofcarriers having energy higher than the barrier, E 0c = E c |x = 0

J s m =

E = E 0c qv x dn =

E = E 0c qv x dn dE

dE

The concentration of electrons between E and E + dE is

dn = g c (E )f F (E )dE

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Considering that

g c (E ) = 4(2m

n )3

2

h 3 E E c and f F (E ) exp E E Fn

kT

where E Fn = E Fn () and E c = E c (), we may write

dn dE dE =

4(2m n )3/ 2

h 3 E E c exp E

E

Fn kT dE



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Assuming a parabolic conduction band (with constant m n ),we may relate the carrier energy, E , to its velocity v , so that

E E c = 12

m n v 2 ; dE = m n v dv ; E E c = v m

n

2

The exponential term e E E Fn

kT may be written as

e E E Fn

kT = e E E c

kT e E c E

Fn

kT = e m n v

2

2kT e E c E

Fn

kT

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The term dn dE dE becomes

dn dE dE = 4(2m

n )3/ 2

h 3 v m

n 2 e

m n v

2

2kT e E c E

Fn

kT m n v dv

= 2 (m n / h )3 e

E c E Fn

kT e m n v

2

2kT 4v 2dv

Replacing v 2 = v 2x + v 2y + v 2z and observing that

4v 2dv = dv x dv y dv z we have

J s m = v 0x

v x = +

v y = +

v z = qv x 2m n h

3

e E c E

Fn

kT

e

m n v 2x

2kT e m n v

2y

2kT e m n v

2z

2kT dv x dv

y dv

z

For v y and v z the integrals are extended over any velocityin any verse ( + and signs)Along the x -axis, only negative velocities are consideredA minimum velocity, v 0x (i.e., energy) is required toovercome the barrier

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The term d n (v ) = f (v )dv represents the number ofelectrons in the innitesimal volume 4 v 2dv The number of electrons in the innitesimal volumedv x dv y dv z placed at a distance v from the center, i.e.,dn (v x , v y , v z ), is found dividing f (v )dv by the innitesimalvolume 4 v 2dv and multiplying the result by the

innitesimal volume d v x dv y dv z , that is

dn (v x , v y , v z ) = f (v )dv 4v 2dv

dv x dv y dv z = f (v )4v 2

dv x dv y dv z

Hence we may write

V f (v ) dv = V Z V Y V X f (v )4v 2 dv x dv y dv z where v 2 = v 2

x + v 2

y + v 2

z

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Manipulating the terms we have

J s m = 2m n h

3e

E c E Fn kT

v 0x

v x = qv x e

m n v 2x 2kT dv x

+

v y = e

m n v 2y

2kT dv y +

v z = e

m n v 2z

2kT dv z

Integrals over v y and v z may be solved considering that

+

e x 22 dx = 2

hence we have

+

e m n v

2y

2kT dv y = kT m n +

e m n v

2y

2kT d m

n kT v y = 2kT m n

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The integral over v x becomes

v 0x

qv x e

m n v 2x 2kT dv x =

q 2

v 0x

e m n v 2x

2kT dv 2x =

= q 2

2kT m n

v 0x

e m n v

2x

2kT d m

n 2kT v

2x =

= + qkT m n

e m n v 2x

2kT v 0x

= qkT m n

e m n v 20x

2kT

The term 12 m

n v 2ox is the minimum kinetic energy required toovercome the barrier and is equal to E 0

c E

c The integral over v x is

v 0x

qv x e

m n v 2x

2kT dv x = qkT

m n exp

E 0c E c kT

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Combining the terms we have

J s m = 2 m

n h

3

e E c

E

Fn

kT qkT m n e E

0c E

c

kT 2kT m n =

= 4qm n k 2

h 3 T 2e

E 0c E Fn

kT = 4qm n k 2

h 3 T 2e

E 0c E Fm E Fn + E Fm

kT =

= 4qm

n k 2

h 3 T 2e

q Bn kT e

E Fn E Fm kT

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The current of electrons owing from the semiconductor is

J s m = AT 2e

q Bn kT e

E Fn E Fm kT

where we have dened the effective Richardson constant

A = 4qm

n k 2

h 3

The current of electrons owing from the metal is constantsince the barrier height is almost constant

J m s is then equal to J

s m for V a = 0 and E Fn = E Fm

The overall current is then

J = J s m J m s = AT 2e

q Bn kT e

E Fn E Fm kT 1
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We have found that

J dd

= J dds e

qV a kT

eE Fn ( ) E Fm

kT

J = J s eE Fn ( ) E Fm

kT 1

Since J n = J dd = J , setting = eE Fn ( ) E Fm

kT , we have

J dds eqV a kT = J s ( 1)

Solving for yields

= J dds e

qV a kT + J s

J dds + J s Substituting into J , the nal current results

J n = J dds J s

J dds + J

s

eqV a kT

1

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Experimental and theoreticalreverse-bias currents in PtSi-Si

diode

Forward-bias current density J F versus V a for W-Si and W-GaAs

diodes
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Metal-semiconductor ohmic contacts

Metal-semiconductor ohmic contacts
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Contacts must be made between any semiconductordevice and the outside worldThese contacts are made via ohmic contacts

They are non-rectifying metal-semiconductor contactsWe have two types of non-rectifying contacts

Ideal non-rectifying barrierTunneling barrier

A specic contact resistance is used to characterize ohmiccontacts


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Ideal non-rectifying barriers


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When a positive voltage is applied to the metal, energy

bands in the semiconductor bend and electrons may easilyow downhill into the metalWhen a positive voltage is applied to the semiconductor,electrons can easily ow over the barrier into thesemiconductorThis junction performs an ohmic contact


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Tunneling barrier


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The space charge width in rectifying metal-semiconductorcontact is inversely proportional to the square root of the

semiconductor doping

x d = 2 s V bi qN d As the doping concentration increases, the probability oftunneling through the barrier increases

Tunneling barrier
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Let us analyze the case of a particle entering a potentialbarrier in x

[0 , L] with an energy, E , lower than the

potential energy, V (x )The Schr odinger equation is

d2

dx 2(x )

2m n 2

[V (x ) E ](x ) = 0If V (x ) = V 0 = constant, the solution for a particle movingalong the positive axis takes the form

(x ) = (x 0)e K (x x 0 ) with K =

2m n (V 0 E )

If V (x ) does not change in the interval [x , x + dx ], we maywrite

(x + dx ) = (x )e K (x )dx with K =2m n [V (x ) E ]

Tunneling barrier
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Manipulating the solution in [x , x + dx ], we have

ln (x + dx )(x ) = K (x )dx Expanding (x + dx ) in Taylor series, we have

(x + dx ) = (x ) + ddx dx

Substituting in the solution leads to

ln 1 + 1

(x )

d

dx dx =

K (x )dx

Approximating ln (1 + x ) x we have1

(x )ddx

dx = K (x )dx

Tunneling barrier
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Integrating from x = 0 to L

L

0

1(x )

ddx

dx = (L)

(0)

d(x )

= ln(L)(0)

= L

0K (x ) dx

Which leads to the nal solution

(L) = (0) exp L

0K (x ) dx

The transmission coefcient is

= (L)(L)(0)(0)

= exp 2 L

0K (x ) dx

Tunneling barrier
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Assume the barrier has a triangular form of height q Bn and extension L

V (x )E = q Bn 1 x L K (x ) = 2m

n q Bn 1 x L

The integral is

L

0K (x ) dx =

2m n q

Bn L

0 1 x L dx =

23 L

2m n q

Bn

Tunneling barrier
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The transmission coefcient results

= exp 43 L 2m n q Bn

with L = 2 s V bi qN d The tunneling current takes the form

J T = qv R n

where n is the density of available electrons and v R may beassumed equal to the Richardson velocity

v R = kT 2m n The tunneling probability is a strong function of the doping

concentration

Specic contact resistance
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A gure of merit of ohmic contacts is the specic contactresistance, R c This is dened as the reciprocal of the derivative of thecurrent density with respect to voltage at zero bias

R c = J

V

1

V = 0

cm 2

For a rectifying contact where the thermionic emissiondominates we have

J = A

T 2

exp q Bn kT exp

qV kT 1

and

R c =kT q exp +

q Bn kT

A

T2

Specic contact resistance
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For a metal-semiconductor

junction with high impuritydoping concentration thetunneling process willdominate and the currentdensity is

J exp 1

N d The specic contactresistance is

R c exp + 1

N d
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10 Metal-Semiconductor Junctions