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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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The Schottky barrier diode
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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The Schottky barrier diode
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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Ideal junction properties
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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Ideal junction properties
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
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)
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
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)
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
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)
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
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)
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
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)
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
Nonideal effects on the barrier height
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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
Nonideal effects on the barrier height
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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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Current-Voltage relationship
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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Drift/diffusion current
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
Drift/diffusion current
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Drift/diffusion current
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
Drift/diffusion current
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Drift/diffusion current
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
Drift/diffusion current
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Drift/diffusion current
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
Drift/diffusion current
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Drift/diffusion current
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
Drift/diffusion current
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Drift/diffusion current
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
Drift/diffusion current
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Drift/diffusion current
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
Drift/diffusion current
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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
Thermionic emission of carriers
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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
Thermionic emission of carriers
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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
Thermionic emission of carriers
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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
Thermionic emission of carriers
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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
Thermionic emission of carriers
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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
Thermionic emission of carriers
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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
Thermionic emission of carriers
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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
Thermionic emission of carriers
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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
Thermionic emission of carriers
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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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Current-Voltage relationship
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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
Current-Voltage relationship
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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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