metallurgical junction
DESCRIPTION
Chapter 5. pn Junction Electrostatics. Metallurgical Junction. Doping profile. Step junction idealization. Chapter 5. pn Junction Electrostatics. Poisson’s Equation. Poisson’s equation is a well-known relationship in electricity and magnetism. - PowerPoint PPT PresentationTRANSCRIPT
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President University Erwin Sitompul SDP 7/1
Dr.-Ing. Erwin SitompulPresident University
Lecture 7
Semiconductor Device Physics
http://zitompul.wordpress.com
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President University Erwin Sitompul SDP 7/2
Step junctionidealization
Metallurgical JunctionChapter 5 pn Junction Electrostatics
Doping profile
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President University Erwin Sitompul SDP 7/3
Poisson’s EquationChapter 5 pn Junction Electrostatics
S 0K
E v DD E
S 0K
S 0K
Ex
Poisson’s equation is a well-known relationship in electricity and magnetism. It is now used because it often containes the starting point in
obtaining quantitative solutions for the electrostatic variables.
In one-dimensional problems, Poisson’s equation simplifies to:
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President University Erwin Sitompul SDP 7/4
Equilibrium Energy Band DiagramChapter 5 pn Junction Electrostatics
pn Junction diode
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President University Erwin Sitompul SDP 7/5
Band diagram
c ref
1( )V E E
q
Qualitative ElectrostaticsChapter 5 pn Junction Electrostatics
Equilibrium condition
Electrostatic potential
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President University Erwin Sitompul SDP 7/6
Electric field
Qualitative ElectrostaticsChapter 5 pn Junction Electrostatics
Equilibrium condition
Charge density
dV
dxE
S 0K
Ex
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President University Erwin Sitompul SDP 7/7
Formation of pn Junction and Charge DistributionChapter 5 pn Junction Electrostatics
D A( )q p n N N qNA– qND
+
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President University Erwin Sitompul SDP 7/8
Formation of pn Junction and Charge DistributionChapter 5 pn Junction Electrostatics
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President University Erwin Sitompul SDP 7/9
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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President University Erwin Sitompul SDP 7/10
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 E(–xp) 0
with 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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President University Erwin Sitompul SDP 7/11
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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President University Erwin Sitompul SDP 7/12
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 must be equal:
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President University Erwin Sitompul SDP 7/13
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
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President University Erwin Sitompul SDP 7/14
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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President University Erwin Sitompul SDP 7/15
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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President University Erwin Sitompul SDP 7/16
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 1025.86mV 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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President University Erwin Sitompul SDP 7/17
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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President University Erwin Sitompul SDP 7/18
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 :
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President University Erwin Sitompul SDP 7/19
Effect of Bias on ElectrostaticsChapter 5 pn Junction Electrostatics
• If voltage drop then depletion width • If voltage drop then depletion width
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President University Erwin Sitompul SDP 7/20
Linearly-Graded JunctionChapter 5 pn Junction Electrostatics
S
1dx
E V dx E
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President University Erwin Sitompul SDP 7/21
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 5
Deadline: 10 March 2011, at 07:30.
2. (7.6)Problem 5.4Pierret’s “Semiconductor Device Fundamentals”.