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Lundstrom ECE 305 S15
ECE-305: Spring 2015
Metal-Semiconductor Junctions: II
Professor Mark Lundstrom
Electrical and Computer Engineering Purdue University, West Lafayette, IN USA
3/11/15
Pierret, Semiconductor Device Fundamentals (SDF) pp. 477-487
Question 1)
2
EC
EV
EFN
metal
EFM
E0
ΦM = 4.5 eV χ = 4.0 eV
EG = 1.0 eV
0.2 eV
V = 0 V
V = ? V
What is the voltage on the semiconductor?
Question 1)
3
What is the voltage on the semiconductor? a) +0.6 V b) -0.6 V c) +0.3 V d) -0.3 V e) +0.2 V
Question 1)
4
EC
EV
EFN
metal
EFM
E0
ΦM = 4.5 eV χ = 4.0 eV
V =Vbi =ΦM −ΦS
q
EG = 1.0 eV
0.2 eV
V = 0 V
V = +0.3V
= 4.5 − 4 + 0.2( ) = +0.3V
Question 1)
5
EC
EV
EFN
metal
EFM
E0
ΦM = 4.5 eV χ = 4.0 eV
EG = 1.0 eV
0.2 eV
V = 0 V
V = ? V
Draw the energy band diagram
Question 2)
6
EC
EVEFP
E0
χ = 4.0 eV
EG = 1.0 eV0.2 eV
metal
EFM
ΦM = 4.5 eV
V = 0 V
V = ? VWhat is the voltage on the metal?
Question 2)
7
What is the voltage on the metal? a) +0.6 V b) -0.6 V c) +0.3 V d) -0.3 V e) +0.2 V
Question 2)
8
EC
EVEFP
E0
χ = 4.0 eV
EG = 1.0 eV0.2 eV
metal
EFM
ΦM = 4.5 eV
V = 0 V
V = 0.3VV =Vbi =ΦS −ΦM
q= 4.0 + 1− 0.2( )− 4.5 = +0.3V
Question 2)
9
EC
EVEFP
E0
χ = 4.0 eV
EG = 1.0 eV0.2 eV
metal
EFM
ΦM = 4.5 eV
V = 0 V
V = 0.3VDraw the energy band diagram
Question 3)
10
EC
EV
EFP
E0
χ = 4.0 eV
EG = 1.0 eV0.2 eV
V = 0 V
EC
EV
EFN
χ = 4.0 eV
EG = 1.0 eV
0.2 eV
V = ? V
What is the voltage on the P-type semiconductor?
Question 2)
11
What is the voltage on the P-type semiconductorl? a) +0.6 V b) -0.6 V c) +0.3 V d) -0.3 V e) +0.2 V
Question 3)
12
EC
EV
EFP
E0
χ = 4.0 eV
EG = 1.0 eV0.2 eV
V = 0 V
EC
EV
EFN
χ = 4.0 eV
EG = 1.0 eV
0.2 eV
V = −0.6 V
Vbi =ΦP −ΦS
q= 4.0 + 1− 0.2( )− 4.0 + 0.2( ) = 0.6 V
V = −0.6 V
What is the voltage on the right contact?
13
N PV = 0 V = ?
V = +0.3 V = −0.3
V = 0
We can’t measure the built-in voltage of the NP junction.
14
E-band diagram
EF
EC
EV
outline
15
1) Electrostatics
2) (IV next lecture)
Lundstrom ECE 305 S15
A metal – P-type semiconductor junction
16
EC
EVEFS
Ei
metal
EFM
E0
ΦM = 4.5 ΦS
χ = 4.05
ΦBP
ΦM <ΦS
energy band diagram
17
EC
EVEF
Ei
metal
EF
ΦBP
qVbi
W
p x( ) << NA
ρ x( ) ≈ −qNA
electrostatics
18
ρ
x
metal P
W
ρ = −qNA
dEdx
=ρ x( )KSε0
dEdx
< 0
electric field
19
E x( )
x
metal P
W
ρ = −qNA
dEdx
=ρ x( )KSε0
E x( ) > 0
dEdx
< 0
electrostatic potential
20
E x( )
x
metal P
Vbi = E x( )∫ dx
W
final answers
21
E x( )
x
metal P
dEdx
= − qNA
KSε0→E 0( ) = qNA
KSε0W
12E 0( )W = Vbi
E 0( )
W
W = 2KSε0qNA
Vbi⎡
⎣⎢
⎤
⎦⎥
1/2
E 0( ) = 2qNAVbi
Ksε0
one-sided NP junction
22
W = 2KSε0q
NA + ND
NDNA
⎛⎝⎜
⎞⎠⎟Vbi
⎡
⎣⎢
⎤
⎦⎥
1/2
→ 2KSε0qNA
Vbi⎡
⎣⎢
⎤
⎦⎥
1/2
E 0( ) = 2Vbi
W
xn =NA
NA + ND
W ≈ 0
xp =ND
NA + ND
W ≈W
N P
ND = 1018
depletion region xp−xn0
E x( )
Vbi =kBTqln NDNA
ni2
⎛⎝⎜
⎞⎠⎟
⎛
⎝⎜⎞
⎠⎟
W NA = 1015
Vbi =ΦM −ΦS( )
q
summary
23
1) Built-in potential depends on workfunction differences.
2) Electrostatics is just like a one-sided NP junction. 3) Energy barrier in the e-band diagram we have
considered implies a rectifying characteristic.
equilibrium energy band diagram
24
EC
EVEF
Ei
metal
EF
ΦBPq Vbi −VA( )
W
P = e−ΔE kBT = e−q Vbi−VA( ) kBT ∝ eqVA kBT
Only majority carriers are involved!