Metodi Sperimentali della Fisica Moderna
P-N junctions
Short review of semiconductor properties
P-n junctions
References
S.M. Sze, Physics of semiconductor devices
(Wiley) Chap 1,2
Ashcroft-Mermin, Solid State Physics
(Saunders) Chap 4-5, 28-29
Grosso- Parravicini, Solid State Physics
Chap 13-14
Crystal lattice
Bravais lattice = set of all points such that
R = n1a1 + n2a2 + n3a3
a1 , a2 , a3 = primitive vectors
n1 , n2 , n3 = integers
Simple cubic (SC)
a1 = (-x+y+z)a/2
a2 = (x-y+z)a/2
a3 = (x+y-z)a/2
Body centered cubic (BCC)
Face centered cubic (FCC)
a1 = (y+z)a/2
a2 = (x+z)a/2
a3 = (x+y)a/2
Crystal lattice
a1 ; a2 ; a3 = primitive vectors are not unique
Primitive unit cell = volume of space that can fill up all
the available space with no overlapping (not unique)
Crystal lattice with basis
The crystal can be described by
different primitive vectors,
adding also the atoms within
the unit cell using a basis
Diamond lattice
Primitive
vectors
^
ax
^
ay
^
az
Basis
0
a ^ ^ ^
 x y z 
4

Lattice with
two point basis
Directions and planes
[n1, n2, n3] = crystal direction
(n1, n2, n3) = crystal plane
[1 1 0]
[0 1 0]
Miller indexes
[1 1 1]
Reciprocal lattice
Plane wave
e i kr
Bravais lattice
R = n1a1 + n2a2 + n3a3
The set of K yelding plane waves with the periodicity
of a Bravais lattice is the reciprocal lattice
e i K(r R)  e i Kr
e i KR  1
The vectors defining the reciprocal lattice are
b1  2
a2  a3
,
a1  a2  a3 
b2  2
a3  a1
a1  a2
, b3  2
a1  a2  a3 
a1  a2  a3 
bi  aj  2ij
If k is a linear combination
k  k1b1  k2b2  k3b3
k  R  2 k1n1  k2n2  k3n3 
To be a k of the reciprocal space the ki must be integers
Reciprocal lattice
Brillouin zone: region of points closer to a lattice point than to any
other lattice point in the reciprocal space.
The main symmetry directions are labelled
, K, X, W, L
L
X
W
bcc
What is the physical meaning of k?
Periodic potential U(r)
Crystal momentum
k 
fcc
p

Free electron
K
Fermi level
Consider N non-interacting electrons confined in a volume V = L3
Non interacting
1 e- Schroedinger
equation
The ground state is obtained from the single e- levels in V
and filling them up with the N electrons
2 2

  (r )   (r )
2m
 = Energy of the level
Represent the electron confinement to V with a boundary condition
 (x+L, y, z) =  (x, y, z)
 (x, y +L, z) =  (x, y, z)
 (x, y, z +L) =  (x, y, z)
This leads to running waves, while choosing as boundary condition for  to vanish
at the surface would give rise to standing waves……
STANDING WAVES
a) Neglect the boundary condition
With energies
  (r)
 ke i k r
i r
V
e i k r
1    k ( r )d r
V
2k 2
 (k ) 
2m
P 
The one-electron level k(r) is an
eigenstate of the momentum operator
P (r)  p (r)
1
 k (r ) 
The solution of the Schroedinger
equation is
Normalization condition
 

 
i r i
with eigenvalue
that is the electron
momentum
But k is also a wave vector of the plane wave
corresponding to the wavelength

2
k
p  k
b) Consider the boundary condition
The bc are satisfied by the plane wave only if
 k (r ) 
1
V
e
e i k( r L )
i kx L
e
i ky L
 e i kxz L  1
ez = 1 only for z = 2m
wavevector components
kx 
2
L
mx ; ky 
2
L
my ; kz 
2
L
mz
mx ; my ; mz integers
The allowed wavevectors in the k space are those
with components made by integer multiples of 2/L
the area per point
is (2/L)2
Fermi wavevector
How many allowed k are contained in a region 
of K space large compared to 2/L?
area per point
(2/L)2

V

3
2 / L  8 3
Volume occupied
by one point
The k space density of levels is
V
8 3
build the N-electron (non interacting) ground state by filling up the one-electron levels
2k 2
 (k ) 
2m
Each level can contain 2 electrons (one for each spin)
4kF3

3
For N large, we get a sphere
the sphere radius is defined kF
Number of allowed k within the sphere
volume
 4kF3  V  kF3


 3  8 3   6 2 V


kF3
kF3
N  2 2V 
V
6
3 2
For N non interacting e- in a volume V
The ground state is formed by occupying all single-particle levels
Occupied for k < kF
Unoccupied for k < kF
kF3
N
n 
V
3 2
FERMI WAVE VECTOR
Electronic structure of a solid
We start by assuming the positions of atoms R
The Hamiltonian describing the electronic structure is
N
pi2
Ze 2
1 N e2
H
 
 
2 i , j 1 ri  rj
i 1 2m
R i 1 ri  R
N
Electron
Kinetic energy
Electron-ion
interaction
Electron-electron
interaction
many-body problem
Impossible to calculate solutions
Hartree-Fock is not good for solids, it dumps the many body
problem in the correction term
1-D band theory: Tight-binding model
Semiconductors
Potential
V atomic
V between
Atomic planes
V along the ion line
N
VL (r)  Va (r  n a)
n 1
 
2

Va (r)  Ea (r) (r)  0
Atomic Schroedinger equation
(r) is the eigenstate of the isolated atom
The lattice potential is constructed from
a superposition of N free atoms
potentials Va(r) arranged on a chain
with lattice constant a
(but it is not the true lattice potential)
The Schroedinger equation for the crystal is
 
2
Va (r)  VL (r) Va (r) (r)  E (r)
The solution can be constructed using a linear combination of the type
 (r ) 
N
 cn (r  n a)
n 1
(r) are Wannier functions,
similar to the solutions  of the atomic
Schroedinger equations
cn : coefficient to be found
Insert the solution into the Schroedinger equation  matrix
l VL Va m  l ,m  l ,m 1
l ,m
Only the on-site and first neighbor coefficients are retained
thus defining the matrix elements ,
Recalling that the  must be of the type eikr
Eigenvector of two
generic sites
Reducing the matrix elements allows to obtain the expansion coefficients cn
Hence the eigen energy corresponding to a solution of the Schroedinger equation is
E  E (,  ,K )
Energy band
Wavevector along the
Brillouin directions
It is important how much the (r)
overlaps between neighboring sites.
The bandwidth depends on the overlap.
The potential breaks the degeneracy
Dispersion relation
for energy bands
The potential is also affecting the bands at the high symmetry points
of the brillouin zone giving rise to prohibited energies, i.e. energy gaps
Energy bands for electrons
Ge
Si
T dependence of energy gap
GaAs
Fermi level in metals
The ground state of N electrons in a solid is made similarly to the free electron case
The electron levels are identified by quantum numbers n and k,
and the structure of the solid gives rise to the electronic bands
When all levels are filled with electrons,
in a metal the last band is only partially filled
Fermi level is the energy below which the one-electron levels
are occupied and above which are unoccupied
Semiconductors
CB
VB
The semiconductors are characterized by a gap between
the last occupied band, defined as VALENCE BAND and the
first unoccupied band, defined as CONDUCTION BAND
Hence the definition of Fermi level as "the energy below which
the one-electron levels are occupied and above which are
unoccupied" is valid for all energies in the gap.
EF is not univocally determined
One has to define a number telling the energy position of the
electrons with the lowest binding energy
At T  0 there is a finite probability that some electrons will
be thermally excited across the gap, and there will be
conduction of ELECTRONS and HOLES
Nexc . el.  e
E G
2K BT
will see that
GaAs:
EG = 1.4 eV at T = 300 K, KBT  0.025 eV and N  7 1013
InSb:
EG = 0.16 eV at T = 300 K, KBT  0.025 eV and N  4 1016
Semiconductors
Thermally excited electrons will be mostly close to CBM and holes close to VBM
so the band dispersion can be approximated by a parabolic relation
2
 (k )   C 
2
k M  k


2
 (k )  V 
2
k M  k


1
1
0.28m
0.044m
The tensor M can be diagonalized so
0.49m
0.16m
 (k )   C
Si
 k12
k32 
k22

  



 2m1 2m2 2m3 
2
 k12
k32 
k22

 (k )  V   



 2m1 2m2 2m3 
2
Ge
Effective masses
Constant energy surfaces for e-
Constant energy surfaces for h
Carrier concentration in Semiconductors
To understand the equivalent of Fermi level in semiconductors,
we have to evaluate the number of carriers in thermal equilibrium
electrons
holes

1
C
 
KBT
nC (T )   d  gC ( )
e
V
1

 
KBT
pV (T )   d  gV ( )
e
g() = DOS
1


1


pV (T )   dgV ( )  1    




 e KBT  1 
V
1
d

g
(

)
V




V

e KBT  1
1
The carrier density depends on the chemical potential 
Suppose
C    KBT
  V  KBT
1
e
 
K BT
e
e
 
K BT

 
K BT
1
1
 
K BT

e
1
nC (T )  e
pV (T )  e

C  
KBT

 d gC ( )e

  C
KBT
C

  V
K BT

V

d gV ( )e

V  
K BT
Since only the carriers within KBT of the band edges contribute, the
effective mass approximation is good and the DOS g() is
gC ,V ( ) 
nC (T )  e
pV (T )  e
(mC)3,
(mV

C  
KBT
1  2mcKBT 


4  2 
 V
KBT
1  2mv KBT 


4  2 

)3
effective masses
3/2
3/2
2    C ,V
3
 
e
e

C  
KBT

 V
KBT
2
mC3,V/ 2
NC (T )
PV (T )
NC, PV
Effective density of states
integrated over all energies
n (T )  NC (T )e
p (T )  PV (T )e

C  
K BT
 V

K BT
Carrier density
Therefore it is the chemical potential that sets the density of carriers
n,p and also the energy position of the states within the gap.
This is usually referred to Fermi level in semiconductors but it is NOT a Fermi level
Carrier density
n (T )  NC (T )e
p (T )  PV (T )e
1  2m K T 
NC (T )   c 2B 
4  

1  2m K T 
PV (T )   v 2B 
4  

C  
K BT
 V
K BT
3/2
3/2
3/ 2
m
NC (T )  2.5  c 
m


3/ 2
 T 


 300K 
x 1019cm3
NC, PV effective DOS for carriers
Intrinsic semiconductors
At finite temperature thermal excitation of e- leaves an equal number of holes
in the VB so that
n = p = ni
intrinsic carrier density
nC (T ) pV (T )  n (T ) p (T )  e
NC (T )PV (T )e

 V  C  
K BT

C  
K BT
NC (T )e
 NC (T )PV (T )e

n (T ) p (T )  ni2 (T )  NC (T )PV (T )e
Mass action law
ni (T )  NC (T )PV (T )e

Eg
2K BT

 V
K BT
PV (T ) 
Eg
K BT

Eg
K BT
depends only on
the gap energy
Intrinsic semiconductors
nC (T )  pV (T )  ni (T )
What is the chemical potential for the intrinsic case i?
ni  nc  NC e
ni  pV  PV e
NC (T )e

 P
ln V
 NC
C  i
KBT


 C  i
KBT
i V
KBT
 PV (T )e

i V
KBT

 C  i
KBT

PV
e

i V  e

NC
e KBT
 C  i i V

KBT
KBT
e
 C 2 i V
KBT
 E g  V  2i  V
 E g  2V  2i
   C  2i  V




KBT
KBT
KBT

i  V 
Eg
2

KBT
 P
ln V
2
 NC




Eg
3KBT  mV 

i  EF  V 

ln

2
4
 mC 
Intrinsic semiconductors
i  EF  V 
Eg
2

3KBT  mV 

ln

4
 mC 
So for T  0, the chemical potential i,
is in the middle of the energy gap
Since (mV/mC)  1, the chemical potential i
does not change more than ~ KBT
C    KBT
  V  KBT
Are valid
Intrinsic semiconductors
3/ 4
m 
ni (T )  2.5  C 
m 
3/ 4
 mV 


m


3/2
 T 


300
K


e

EG
2KBT
x 1019cm 3
Considering (mV/m)  (mC/m)  1, at RT
ni (T )  2.5 e
Silicon: EG= 1.1 eV
GaAs: EG= 1.1 eV
InSb: EG= 0.17 eV

e
EG
2K BT

e
x 1019cm 3
EG
2KBT

 2.8x 10
EG
2KBT
e

ni (T )  7x 1010cm3
EG
10
 6.9x 1013
2KBT
 0.99
ni (T )  1.7x 107 cm3
ni (T )  2.4x 1019cm3
But mC and mV are 0.01
ni (T )  2.1x 1016cm3
Donors and acceptors
nC (T )  pV (T )  ni (T )
P: 5 valence eSi: 4 valence e-
B: 3 valence e-
n-type: a Si atom is replaced by P atom with an extra e-. The P atom is called donor
p-type: a Si atom is replaced by B atom with an extra hole. The B atom is called acceptor
Donors and acceptors introduce extra energy levels, whose energy can be estimated
using the H atom model
Donors and acceptors
Why hydrogen model?
1)
2)
Neglect the ion core of the inserted P atom
The P is represented by a Si atom with 1 hole (e+) fixed in the site + 1 eIsolated atom
Atom inside crystal
Charge field reduced by
the dielectric constant 
(13 for Si)
e- binding energy = 10.486 eV
(P ionization potential)
The e- moving in the lattice has energies
of the type E=E(k) (k =crystal wavevector)
The allowed energy levels has to be close
to conduction band minimum
parabolic bands with effective masses
The electron of the donor impurity is represented as a particle of charge –e and and mass m*
moving in the presence of attractive charge e/
H atom
a0 
2
me 2
;E 0 
me 4
2
e2

r0 
m
a
m* 0
m m *
E 
m* 1
E0
2
m 
e 
2
Extrinsic semiconductors
The density of donors or acceptors is above 1012/cm3
The level energies are very small compared to the energy gap
so it is very easy to excite an electron from a donor level or a hole from an acceptor level
Fermi level for intrinsic semiconductors
band diagram
density of states
Eg
KBT
Fermi-Dirac F(E)
P
i  EF  V 

ln V
2
2  NC




carrier concentrations
n  NC e
p  PV e
np  ni2


C  
K BT
 V
K BT
Fermi level for extrinsic semiconductors
Introduce impurities
not all dopants are necessarily ionized:
depends on the impurity energy level and T
DONOR
ND (cm-3) = donor concentration
D = energy of donor level
Probability of occupation
of donor levels
Probability of finding
ionized donor levels
1
P (ED ) 
1
1
g
e
g = ground state degeneracy = 2
The donor level are localized hence cannot
accommodate 2 electrons due to charge repulsion
 D EF
KBT
1  P (ED )



1

N

N
Density of ionized donors
D
D 1 
 D EF
 1  1 e KBT

g








ND 

ND
1  ge
EF  D
KBT
Fermi level for extrinsic semiconductors
ACCEPTOR
Introduce impurities
not all dopants are necessarily ionized:
depends on the impurity energy level and T
NA (cm-3) = acceptor concentration
A = energy of acceptor level
Probability of occupation
of acceptor levels
Probability of finding
ionized acceptor levels
Density of ionized acceptor
1
P (EA ) 
1
1
g
e
g = ground state degeneracy = 4
each acceptor impurity level can
accept one hole of either spin
The impurity level is doubly degenerate at K =0
 A EF
KBT
1  P (EA )



1
NA  ND 1 
 A EF
1
 1  e KBT

g








NA 
NA
1  ge
 A EF
KBT
Fermi level for extrinsic semiconductors
With impurity atoms introduced
preserve charge neutrality
Total negative charges
(electrons and ionized acceptors)
extrinsic
NA  n
Total positive charges
(holes and ionized donors)
intrinsic
ND  p
extrinsic
intrinsic
NA  n  ND  p
np  ni2
Add DONORS only
n  ND  p  ND


NC e

EC E F
K BT
 ND
1
1  2e
 D E F
K BT
Mass action law still valid
Fermi level for extrinsic semiconductors
DONORS
NC e

EC E F
K BT
 ND
1
1  2e
 D E F
K BT
Graphically solved to
determine EF
Plot for two
different values of D
Fermi level for extrinsic semiconductors
DONORS
band diagram
n-type semiconductor
density of states
Fermi-Dirac F(E)
carrier concentrations
Fermi level for extrinsic semiconductors
ACCEPTORS
band diagram
p-type semiconductor
density of states
Fermi-Dirac F(E)
carrier concentrations
NV e
EV EF
KBT
For a set of dopant concentrations
it is possible to estimate the Fermi level
 NA
1
1  2e
 A EF
KBT
NC e

EC E F
K BT
 ND
1
1  2e
 D E F
K BT
Carrier concentration temperature dependence
n  ND  p
Charge neutrality
D = energy level of the donor impurity
TDKB = D ionization impurity temperature
1) T<< TD = D/KB
freezing out region
D < EF < EC
NC e

EF 
EC EF
KBT
1 
 ND e
2
 D  EC
2

D EF
KBT
 N
ln D
2  2NC
KBT




Carrier density in the conduction band
n (T ) 
NC ND
2
e

D
2KBT
NC e

EC E F
K BT

ND 1
1  2e
 D E F
K BT
 NV e

E F EV
K BT
Carrier concentration temperature dependence
2) TD < T < EG/KB
NC e
saturation region

EC E F
K BT
ED < EF < EC
 ND
1
1
1 e
2
E F  D
K BT
 NV e
At RT almost all electron (donors) and holes (acceptors ) are excited = saturation condition
Intrinsic electrons negligible
n (T )  NC e
Minority carriers p(T)
ni2 (T ) ni2 (T )
p (T ) 

n (T )
ND
Intrinsic Silicon
N-type Silicon
3) TD < EG/KB < T
ni (T )  1010cm3
ND  1x 1014cm3
p(T )  1x 106cm3
intrinsic region

EC E F
K BT
 ND
Majority carriers
EV E F
K BT
Mobility
Why? We need to know what is happening to carriers for concentrations out of equilibrium
- Carrier transport for a semiconductor in the presence of an electric field E
- 2 parabolic bands (valence and conduction) and effective masses mv and mc
vdrift  E
Current density
= scattering time
n = carrier density
mobility
n 
e n
mc
ne 2 n
Jn 
 neE
mc
e p
p 
mv
ne 2 p
Jp 
 neE
mv
For e- and holes
  e (nn  p p )
 depends on carrier interaction
within the lattice and on temperature
Diffusion coefficient
Consider a concentration gradient in the solid
Suppose the flux goes from regions of high concentration
to regions of low concentration
with a magnitude that is proportional
to the concentration gradient
Jn  eDn n
Dn is the diffusion coefficient
For a doped semiconductor with a carrier concentration gradient and
applied electric field the total current density is
Jn  nenE  eDn n
Drift
Diffusion
Mobility-diffusion coefficient relation
The mobility p, and the diffusion coefficient D, are not independent
n
Jn  ne n   eDn
0
x
n
ne n   eDn
x
Consider an n-type semiconductor with nonuniform (n=n(x))
doping concentration and without an external applied field
Drift current balances the diffusion current
the nonuniform doping generates a potential (x)
that rigidly shifts the energy levels of the semiconductor
EC (x )  EC  (e )(x )
internal electric field
n  NC (T )e

EC E F
K BT
 n (x )
n
1 EC
e

n (x )  
n (x )
x
KBT x
KBT
nen   eDn
e
n
KBT
n kBT
Dn 
e

EC (x )
ex
The p-n junction
p-type semiconductor



1
ND  ND  1 
 D E F
 1  1 e  KBT

g

n-type semiconductor







NA 
NA
1  ge
 A EF
KBT
All donors and acceptors ionized (saturation condition)
ND  ND
NA  NA
ND,A (cm-3) = donor,acceptor concentration
The p-n junction
Bring together the two regions
p-type semiconductor
n-type semiconductor
1)
2)
Some e- move from n-type to p and recombine with h
Some h move from p-type to n and recombine with e-
3)
4)
In the n region close to x=0 remains non neutralized donors (+)
In the p region close to x=0 remains non neutralized acceptors (-)
A region depleted of majority carriers is formed at the interface (space charge region)
A strong electric field is built up opposing further diffusion of majority carriers, reaching equilibrium
At equilibrium and no external field E applied , no current flows so
n
Jn  nen   eDn
0
x
 kT n 
en 
0
 e x 
Dn 
n kBT
e
n  NC (T )e
E
n
n F
x
x

 C EF
KBT
E F
enn
0
x
The Fermi level must remain constant throughout the depletion layer and the sample
This means that the band in the two regions bend to adjust across the p-n junction
 eb
How much is the potential?
Depends on carrier concentration
eb  EFn  EFp
Contact potential
n (T )  ND  NC (T )e
p (T )  NA  PV (T )e


EC E F
KBT
E F EV
KBT


EC  E Fn
N
 ln D
KBT
NC
E Fp  EV
N
KBT
 ln
E Fp
A
PV

eb  EFn  EFp  EC  EV  KBT ln

eb  EG  KBT ln
ND
NC
P
 EV  KBT ln V
NA
E Fn  EC  KBT ln
NDNA
NC PV
ND
P 
 ln V 
NC
NA 
At equilibrium, free carrier concentrations (e- and holes)
depends on the position across the junction
For electrons
n x   NC T e

EC  x E F
K BT
EC (x )  EC  (e )(x )
Free electron density is high when Fermi level is close to bottom of CB
n (x )  NC T e
NC T e

n (x 1 )  n (x 2 )e
EC E F
K BT
e

EC ( x ) E F
KBT
e ( x )
KBT
e  ( x 1 )  ( x 2 ) 
K BT
 NC T e
 NDe
e ( x )
K BT

EC e ( x ) E F
KBT

At equilibrium, free carrier concentrations (e- and holes)
depends on the position across the junction
p (x )  NV e

E F EV ( x )
K BT
p (x )  PV T e

Free hole density is high when Fermi level is close to top of VB
EC ( x ) E F
K BT
 NAe
e ( x )
K BT
p (x1 )  p (x 2 )e
e  ( x1 )  ( x 2 ) 
K BT
Carrier concentrations
np 0
pp 0
nn 0
pn 0
x1
p (x1 )  p (x 2 )e
e  ( x1 )  ( x 2 ) 
K BT
x2
n (x 1 )  n (x 2 )e
x1 inside p
p (x1 )  pp 0
Equilibrium concentration of
holes in neutral bulk p material
x2 inside n
p (x1 )  pn 0
Equilibrium concentration of
holes in neutral bulk n material
pp 0  pn 0e
eb
K BT
e  ( x 1 )  ( x 2 ) 
K BT
x1 inside p
n (x1 )  np 0
Equilibrium concentration of
e- in neutral bulk p material
x2 inside n
n (x1 )  nn 0
Equilibrium concentration of
e- in neutral bulk n material
n p 0  nn 0e

eb
K BT
Space-charge region and internal electric field
Approaching the
depletion region
the number of holes
decreases and the layer
depleted of holes
is created.
The fixed charge
density is
p  NA
n  ND
dn
dp
Approaching the
depletion region the
number of edecreases and the
layer depleted of e- is
created.
The fixed charge
density is
  eND
  eNA
Depletion layer approximation
 eN A

  eN D
-d p  x  0
0  x  dn
The electrostatic potential  follows the Poisson equation
d 2  (x )


2
dx
r  0
p  NA
n  ND
Integrate the Poisson equation
Boundary conditions:
Boundary conditions:
Homogeneous semiconductor
at equilibrium,
no diffusion and Jd = 0
E=0
Homogeneous semiconductor
at equilibrium, no diffusion and Jd = 0
E=0
'( d p )  0
 ( d p )  0
E 
'(dn )  0
 (dn )  b
Choose the
Constant = 0
eNA
d

(x  d p )
dx
r  0
E 
Integrate the Poisson equation
–dp < x < 0
eNA
2

 (x ) 
x  dp 
2 0 r
d eND

(x  dn )
dx  r  0
0 < x < dp
 (x )  b 
eND
x  dn 2
2 0 r
2
 eNA
x

d
-d p  x  0
p

2



0 r
 (x )  
eND
b 
x  dn 2 0  x  dn

2 0 r



The continuity conditions for electric field at x = 0 gives
NAd p  NDdn
eND
eNA
dp 
d
r 0
r  0 n
Like charge neutrality
The continuity conditions for electric potential at x = 0 gives
NAd p2 
2 0 r
e
b  NDdn2
e

b 
NAd p2  NDdn2 
2 0 r
Solving for dp & dn
dp 
dn 
ND
2 0 r b
1
NA ND  NA
e
Depends on dopant concentration
2 0 r b
NA
1
ND ND  NA
e
The total width of the depletion layer is

N
w  d p  dn  dn 1  D
NA


 

w  d p  dn 
b  1V
NA  ND  1016cm 3  1022 m 3
 0 r  10
10
F /m
2 0 r b
e
 NA  ND

 NA
2
 NA
1

 ND NA  ND
NA  ND 2 0 r b
ND NA
e
w  5x 107 m  500nm
b
w
 2x 104 Vcm 1
AV
Current-voltage behavior of the p-n junction
Apply a voltage V < b to the junction
V is > 0 if the barrier is decreased
Forward bias
V < b
V is < 0 if the barrier is increased
Reverse bias
No V restriction
resistivity of space charge region >> bulk p,n regions
Total voltage drop is in the depletion layer
dp 
ND
2 0 r (b V )
1
NA ND  NA
e
dn 
2 0 r (b V )
NA
1
ND ND  NA
e
w  d p  dn 
NA  ND 2 0 r (b V )
ND NA
e
The total current through the junction is given by the changes in the minority carrier
concentration, since the majority carriers are absent in the depletion layer
Changes in minority carrier concentrations at boundaries of depletion layer
np 0
nn 0
Assume quasi-equilibrium conditions
n (x )  ND e

E C ( x ) E F
K BT
nn 0
minority carriers
np 0
n (dn )  ND e

E C E F
K BT
EC (x )  EC  (e )(x )
Equilibrium concentration
of e- in neutral bulk n material
Equilibrium concentration
of e- in neutral bulk p material
np 0
nn 0
Applying the potential + V
n ( d p )  n (dn )e
n ( d p )  nn 0e

Low injection conditions =
majority carrier conc. have negligible
change at the boundaries of depletion layer
e (b V )
K BT
e (b V )
KBT
n (dn )  nn 0
 np 0e
Equilibrium concentration
of e- in bulk n material
eV
KBT
Equilibrium concentration
of e- in bulk p material
(minority carriers)
The minority carrier concentration in –dp
(p-type region/depl. region boundary)
depends on V through the exponential
Changes in minority carrier concentrations at the ends of the depletion layer
np 0
n ( d p )  n p 0e

n ( d p )  n p 0  e


eV
K BT
eV
KBT
nn 0
p (dn )  pn 0e

 1


The e- (minority carrier) concentration in –dp
(p-type region/depl. region boundary)
depends on V through the exponential
eV
KBT
 KeVT

B
p (dn )  pn 0  e
 1




The hole (minority carrier) concentration in dp
(n-type region/depl. region boundary)
depends on V through the exponential
Changes in minority carrier concentrations
np 0
nn 0
The total electron current is the diffusion current
(drift due to minority carriers is negligible outside depletion region)
eV


K
T
Diff
B
( p )  n  ( p )  n p 0 
 1
n


J
d
eD n
d
eD n

e

The total hole current is the diffusion current (drift is negligible outside depletion region)
eV


K
T
Diff
B
J p (dn )  eDp p (dn )  eDp pn 0  e
 1




Narrow space-charge region:
neglect generation and recombination processes within it
the total current is given by the sum of the contributions

J  eDp pn 0  eDnnp 0
 KeVT

B
e
 1





Js  eDp pn 0  eDnnp 0 
reverse saturation current
Ideal diode equation
 KeVT

J  Js e B  1 




Js is made by minority carrier contribution only:
Holes are created thermally in the n-type region;
those close to the barrier region reach the
depletion layer by diffusion and are swept by the
internal electric field to the p-side
Same hold for e-
Ideal I-V characteristic
I

R V
1
Differential resistance
The impedance is low in forward bias
 1- 10 
The impedance is high in reverse bias
 1 105 
Real I-V characteristic
(a) Generation-recombination
(b) Diffusion-current
(c) High-injection
(d) Series-resistance effect
(e) Reverse leakage current due to generationrecombination and surface effects.
 KeVT

J  Js e B  1 




Junction breakdown
For high fields applied to a p-n junction, the junction
breaks down and conducts a very large current
only in reverse-bias
breakdown mechanisms
(1) thermal instability
(2) tunneling
(3) Avalanche multiplication
Thermal instability
reverse current at high reverse voltage  heat dissipation 
Increase in junction temperature  increases the reverse current Js  breakdown
Tunneling
Large fields can induce tunneling through the barriers
Junction breakdown
Avalanche multiplication
In the presence of high field, reverse carriers
can induce ionization within the barrier by scattering with
the material atoms
Breakdown voltage depends on impurity con. and gradient
3
VBD
N
 EG  2 

 60
  16
3 
 1.1eV   10 cm 

3
4
Zener diode = well controlled VBD
TRANSISTOR
Electronic devices
Light emitting diodes
Solar cells
Photodetectors
holes
e-
Emitter: higly doped p region ~ 1018 cm-3
Base: weakly doped n region ~ 1014 cm-3
Collector: medium doped p region ~ 1016 cm-3
Base region width << holes diffusion length in n type
holes
Bipolar: the device relies on the
behaviour of two types of carriers
Bipolar junction TRANSISTOR
1018 cm-3
1014 cm-3
Base-Emitter: forward bias
Setting VB to ground = 0
1016 cm-3
Base-Collector: reverse bias
VC << 0 < VE
BE in forward: the current is due to the hole injection in the base
(the current due to e injection in the emitter is small, since the
emitter is strongly doped, while the base is weakly doped).
base region width << holes diffusion length in n type
All holes are collected in C
TRAN(sfer)-(re)SISTOR
we expect IC to be only slightly smaller than IE
TRANSISTOR
VC << 0 < VE
Collector current
IC

 0.99
IE
Emitter current
 
Current transfer parameter
Base current
IC
IC



 100
IB IE  IC
1 
IB  IE  IC
Current gain
TRANSISTOR
eVC
E
 eV



JE  a11 e KT  1   a12 e KT  1 




eVC
E
 eV



JC  a21 e KT  1   a22 e KT  1 




For eVE >> KBT, VC << 0
JE
JC
 
DB NaE LE
DE NdB wB

IC
IB
E
 eV

KT

 a11  e
 1 


E
 eV

KT
 a21  e
 1 


aij  DB ,DE ,DC
JC
a21
 

JE  JC
a11  a21
NaE= acceptor impurity doping in the emitter
NdB= donor impurity doping in the base
For large , a small variation of IB gives large IC
SOLAR CELLS
Solar spectrum for different
relative atmospheric path lenghts
x
 incident on the front surface
Generation rate of e- hole pairs at x from the
semiconductor surface
G , x       1  R  e   x
() = absorption coefficient
() = photon flux
R() = reflected fraction
SOLAR CELLS
• Abrupt p-n junction solar cell
• Constant doping on each side
• ND >> NA Depletion region at the n-side can be
neglected
• No electric fields outside the depletion region
• Low injection condition
Assumed abrupt
doping profiles ND >> NA
Photogenerated carriers collection
n,p: diffusion
depletion region: drift
Jn , p  eDn , p n
Jn , p  nen , pE
 np  np 0  1 dJ n
 
Gn  
0
 n
 e dx
electrons in
p-type substrate
 pn  pn 0  1 dJ p

Gp  
0
 p
 e dx


holes in
n-type substrate
SOLAR CELLS
 dn p 


 dx 
 dp 
J p  epn  pE  eDp  n 
 dx 
Jn  enp n E  eDn 
Current density equations
n side
 d 2 pn
Dp 
2
 dx
 pn  pn 0 

x
0
   1  R e

 p




Solved with appropriate boundary conditions
Continuity equation
SOLAR CELLS
 dn p 


 dx 
 dp 
J p  epn  pE  eDp  n 
 dx 
Jn  enp n E  eDn 
Current density equations
p side
 d 2np
Dn 
 dx 2


 n  np 0 
   1  R e x   p
  0
 

n



Solved with appropriate boundary conditions
Continuity equation
SOLAR CELLS
Total photocurrent
JL    Jn    J p    Jdr  
Spectral response
Jn    J p    Jdr  
SR 
  1  R  
SR for different
recombination
velocities
Total current is obtained by integration on 
Equivalent circuit for an ideal pn diode solar cell