### Document

```Semiconductor Device Physics
Lecture 6
Dr. Gaurav Trivedi,
EEE Department,
IIT Guwahati
Metallurgical Junction
 Doping profile
Poisson’s Equation
 Poisson’s equation is a well-known relationship in electricity and
magnetism.
 It is now used because it often contains the starting point in
obtaining quantitative solutions for the electrostatic variables.
 E 

K S 0
  D  v
D  E
  K S 0
 In one-dimensional problems, Poisson’s equation simplifies to:
E
x


K S 0
Equilibrium Energy Band Diagram
 pn-Junction diode
Qualitative Electrostatics
 Equilibrium condition
Band diagram
Electrostatic potential
V 
1
q
( E c  E ref )
V ( x )    E dx
Qualitative Electrostatics
 Equilibrium condition
Electric field
E 
dV
dx
E ( x) 
K

S
0
x
Charge density
E
x


K S 0
Formation of pn Junction and
Formation of pn Junction and Charge Distribution
Charge Distribution


  q( p  n  N D  N A )
qNA–
qND+
Formation of pn Junction and
Charge Distribution
Built-In Potential Vbi
• Vbi for several materials:
Ge
≤ 0.66 V
Si
≤ 1.12 V
GeAs ≤ 1.42 V
qV bi  ( E F  E i ) n  side  ( E i  E F ) p  side
 For non-degenerately doped material,
( E F  E i ) n-side
( E i  E F ) p-side
 n 
 ND 
 kT ln    kT ln 

n
n
 i 
 i 
 p 
 NA 
 kT ln    kT ln 

n
n
 i 
 i 
qV bi
 NAND 
 kT ln 

2
n
i


The Depletion Approximation
The Depletion Approximation
 On the p-side, ρ = –qNA
dE
 
dx
E ( x)  
E ( x)  
qN A
S
qN A
S
qN A
S
x  c1
( x  xp )
with boundary E(–xp)  0
 On the n-side, ρ = qND
E ( x)  
qN D
S
( xn  x )
with boundary E(xn)  0
 Solution for ρ
Step Junction with VA 0
  qN A ,  x p  x  0

   qN D ,
0  x  xn
 0,
o th e rw ise

 Solution for E



E ( x)  


 Solution for V
qN A
S
qN D
S
( x p  x ),
( x n  x ),
 xp  x  0
0  x  xn
 qN A
2
(
x

x
)
,
 xp  x  0
p
 2

S
V ( x)  
qN D
2
V bi 
( xn  x ) , 0  x  xn
2 S

Step Junction with VA 0
 At x = 0, expressions for p-side and n-side for the solutions of
E and V must be equal:
N A xp  N D xn
qN A
2 S
( x p )  V bi 
2
qN D
2 S
( xn )
2
Relation between ρ(x), E(x), and V(x)
1.Find the profile of the built-in potential Vbi
2.Use the depletion approximation  ρ(x)
 With depletion-layer widths xp, xn unknown
3.Integrate ρ(x) to find E(x)
 Boundary conditions E(–xp)  0, E(xn)0
4.Integrate E(x) to obtain V(x)
 Boundary conditions V(–xp)  0, V(xn)  Vbi
5.For E(x) to be continuous at x  0, NAxp  NDxn
 Solve for xp, xn
Depletion Layer Width
 Eliminating xp,
 Eliminating xn,
 Summing
xn 
xp 
2 S
NA
q
ND (NA  ND )
2 S
ND
q
NA (NA  ND )
xn  xp  W 
V bi
V bi
2 S  1
1


q  NA
ND
Exact solution,
try to derive


 V bi

ND
NA
xn
One-Sided Junctions
 If NA >> ND as in a p+n junction,
W  xn 
2  S V bi
q
ND
, xp  xn
ND
0
NA
 If ND >> NA as in a n+p junction,
W  xp 
2  S V bi
q
NA
, xn  xp
NA
0
ND
 Simplifying,
W 
2  S V bi
q
N
where N denotes the lighter dopant density
Step Junction with VA  0
• To ensure low-level injection conditions,
reasonable current levels must be
maintained  VA should be small
Step Junction with VA  0
 In the quasineutral, regions extending from the contacts to the
edges of the depletion region, minority carrier diffusion
equations can be applied since E ≈ 0.
 In the depletion region, the continuity equations are applied.
Step Junction with VA  0
 Built-in potential Vbi (non-degenerate doping):
V bi
 NAND 
 N A  kT
 ND 
kT
ln 

ln 
ln 


 
2
q
q
q
 ni

 ni 
 ni 
kT
 Depletion width W :
W  xp  xn 
xp 
2 S  1
1 


  V bi  V A 
q  NA
ND 
2 S
ND
q
NA NA  ND 
xp 
 V bi  V A  ,
ND
NA  ND
xn 
W , xn 
2 S
NA
q
ND NA  ND 
NA
NA  ND
W
 V bi  V A 
Effect of Bias on Electrostatics
• If voltage drop , then depletion width 
• If voltage drop , then depletion width 
Quasi-Fermi Levels
 Whenever Δn = Δp ≠ 0 then np ≠ ni2 and we are at nonequilibrium conditions.
 In this situation, now we would like to preserve and use the
relations:
n  ni e
( E F  E i ) kT
,
p  ni e
( E i  E F ) kT
 On the other hand, both equations imply np = ni2, which does
not apply anymore.
 The solution is to introduce to quasi-Fermi levels FN and FP such that:
n  ni e
( FN  E i ) kT
 n 
F N  E i  kT ln  
 ni 
p  ni e
( E i  FP ) kT
 p 
FP  E i  kT ln  
 ni 
• The quasi-Fermi levels is useful to describe the carrier
concentrations under non-equilibrium conditions
Example: Quasi-Fermi Levels
a) What are p and n?
2
n 0  N D  10 cm
17
3
, p0 
n  n 0   n  10 +10
3
31
n0
14
 10 cm
np  10  10 =10 cm
14
3
 10 cm
17
b) What is the np product?
17
 10 cm
14
17
p  p 0   p  10 +10
ni
3
14
3
3
3
Example: Quasi-Fermi Levels
 Consider a Si sample at 300 K with ND = 1017 cm–3 and
Δn = Δp = 1014 cm–3.
0.417 eV
c) Find FN and FP?
F N  E i  kT ln  n n i 
F N  E i  8.62  10
5

 300  ln 10
17
10
10
Ec
FN
Ei

FP
Ev
 0.417 eV
0.238 eV
FP  E i  kT ln  p n i 
E i  FP  8.62  10
5
 0.238 eV
np  n i e

 300  ln 10
14
10
10

 FN  E i 
kT
 ni e
 E i  FP 
0.417
kT
0.238
 10 e 0.02586  10 e 0.02586
10
10
 1.000257  10
 10 cm
31
3
31