chapter 5 pn junction electrostatics
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President University Erwin Sitompul SDP 5/1
Lecture 5
Semiconductor Device Physics
Dr.-Ing. Erwin SitompulPresident University
http://zitompul.wordpress.com
2 0 1 3
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Chapter 5pn Junction Electrostatics
Semiconductor Device Physics
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Step junctionidealization
Metallurgical JunctionChapter 5 pn Junction Electrostatics
Doping profile
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Poisson’s EquationChapter 5 pn Junction Electrostatics
S 0K
E v DD E
S 0K
S 0x K
E
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.
In one-dimensional problems, Poisson’s equation simplifies to:
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Equilibrium Energy Band DiagramChapter 5 pn Junction Electrostatics
pn-Junction diode
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Band diagram
c ref
1( )V E E
q
Qualitative ElectrostaticsChapter 5 pn Junction Electrostatics
Equilibrium condition
Electrostatic potential
( )V x dx E
President University Erwin Sitompul SDP 5/7
Electric field
Qualitative ElectrostaticsChapter 5 pn Junction Electrostatics
Equilibrium condition
Charge density
dV
dxE
S 0K
Ex
S 0
( )x xK
E
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Formation of pn Junction and Charge DistributionChapter 5 pn Junction Electrostatics
D A( )q p n N N qNA– qND
+
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Formation of pn Junction and Charge DistributionChapter 5 pn Junction Electrostatics
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DF i n-side
i i
( ) ln lnNn
E E kT kTn n
bi F i n side i F p side( ) ( )qV E E E E
Ai F p-side
i i
( ) ln lnNp
E E kT kTn n
A Dbi 2
i
lnN N
qV kTn
Built-In Potential Vbi
Chapter 5 pn Junction Electrostatics
For non-degenerately doped material,
• Vbi for several materials:Ge ≤ 0.66 VSi ≤ 1.12 VGeAs ≤ 1.42 V
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A1
S
( )qN
x x c
E
A
S
qNd
dx
E
Dn
S
( ) ( )qN
x x x
E
Ap
S
( ) ( )qN
x x x
E
with boundary E(–xp) = 0
with boundary E(xn) = 0
The Depletion ApproximationChapter 5 pn Junction Electrostatics
On the p-side, ρ = –qNA
On the n-side, ρ = qND
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A p
D n
, 0 , 0 0,
qN x xqN x x
otherwise
Ap p
S
Dn n
S
( ), 0
( )( ), 0
qNx x x x
xqN
x x x x
E
2Ap p
S
2Dbi n n
S
( ) , 02
( )( ) , 0
2
qNx x x x
V xqN
V x x x x
Step Junction with VA = 0Chapter 5 pn Junction Electrostatics
Solution for ρ
Solution for E
Solution for V
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Relation between ρ(x), E(x), and V(x)Chapter 5 pn Junction Electrostatics
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
President University Erwin Sitompul SDP 5/14
2 2A Dp bi n
S S
( ) ( )2 2
qN qNx V x
A p D nN x N x
Step Junction with VA=0Chapter 5 pn Junction Electrostatics
At x = 0, expressions for p-side and n-side for the solutions of E and V must be equal:
President University Erwin Sitompul SDP 5/15
S An bi
D A D
2
( )
Nx V
q N N N
S Dp bi
A A D
2
( )
Nx V
q N N N
n px x W
Dn
A
Nx
N
Depletion Layer WidthChapter 5 pn Junction Electrostatics
Eliminating xp,
Eliminating xn,
Summing Sbi
A D
2 1 1V
q N N
Exact solution, try to derive
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S bi Dn p n
D A
2 , 0
V NW x x x
q N N
S bi2 VW
q N
S bi Ap n p
A D
2 , 0
V NW x x x
q N N
One-Sided JunctionsChapter 5 pn Junction Electrostatics
If NA >> ND as in a p+n junction,
If ND >> NA as in a n+p junction,
Simplifying,
where N denotes the lighter dopant density
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D Abi 2
i
lnN NkT
Vq n
S bi
D
2 VW
qN
n 0.115 mx W
Dp n
A
Nx x
N
Example: Depletion Layer WidthChapter 5 pn Junction Electrostatics
A p+n junction has NA 1020 cm–3 and ND 1017cm–3, at 300 K.
a) What isVbi?
b) What is W?
c) What is xn?
d) What is xp?
17 20
10 2
10 1025.86 mV ln 1.012 V
(10 )
1/ 214
19 17
2 11.9 8.854 10 1.0120.115 m
1.602 10 10
30.115 m 10 1.15 Å
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Step Junction with VA 0Chapter 5 pn Junction Electrostatics
• To ensure low-level injection conditions, reasonable current levels must be maintained VA should be small
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Step Junction with VA 0Chapter 5 pn Junction Electrostatics
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.
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A Dbi
i i
ln lnN NkT kT
Vq n q n
Sp n bi A
A D
2 1 1W x x V V
q N N
S Dp bi A
A A D
2,
Nx V V
q N N N
S An bi A
D A D
2 Nx V V
q N N N
Step Junction with VA 0Chapter 5 pn Junction Electrostatics
Built-in potential Vbi (non-degenerate doping):
A D2
i
lnN NkT
q n
Depletion width W :
,Dp
A D
Nx W
N N
W
NN
Nx
DA
An
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Effect of Bias on ElectrostaticsChapter 5 pn Junction Electrostatics
• If voltage drop â, then depletion width â• If voltage drop á, then depletion width á
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Linearly-Graded JunctionChapter 5 pn Junction Electrostatics
S
1dx
E V dx E
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1. (6.4)Consider a silicon pn junction at T = 300 K with a p-side doping concentration of NA = 1018 cm–3. Determine the n-side doping concentration such that the maximum electric field is |Emax| = 3×105 V/cm at a reverse bias voltage of VR = 25 V.
Chapter 5 pn Junction Electrostatics
Homework 4
Due: 17.09.2013.
2. (7.6)Problem 5.4Pierret’s “Semiconductor Device Fundamentals”.
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