Transcript 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
Linearly-Graded Junction
E
1
S
dx
V E dx
Linearly-Graded Junction
Homework Assignment
Derive the relations of electrostatic variables, charge density and
builtin voltage Vbi for the linearly graded junction.