ECE 584 Semiconductor Device Fundamentals
Chapter 4 PN Junction (2)
Qiliang Li
4.3 Reverse-Biased PN Junction
+
V
–
N
P
Wdep =
Ec
qφbi
Ec
Ef
Ev
Ef
Ev
(a) V = 0
2ε s (φi − Va )
2ε s ⋅ potential barrier
=
qN
qN
1
1
1
1
=
+
≈
N Nd Na lighter dopant density
Ec
qφbi - qVa
Ec
Efn
qV
Ev
(b) reverse-biased
Efp
Ev
Does the depletion layer widen or
shrink with increasing reverse bias?
Reverse-Biased PN Junction: Depletion Capacitance
Reverse biased PN junction is a capacitor.
Depletion capacitance
per unit area
N
Nd
Conductor
P
Na
Insulator
Wdep
Cdep =
εs
Wdep
Cdep =
Conductor
εs
Wdep
qε s
1/ 2
=[
]
1
1
2( +
)(φi − Va )
Na Nd
• Is Cdep a good thing?
• How to minimize junction capacitance?
Capacitance-Voltage Characteristics
(how to measure doping profile N(x)
1/C dep 2
1
Cdep
2
=
Wdep
εs
2
2
Capacitance data
2(φi − V )
=
qNε S
Slope = 2/qN εsA2
– φbi
Vr
Increasing reverse bias
• From this C-V data can Na and Nd be determined?
• How can we detect the doping profile N(x) of a Si bar?
P+N and N+P junction
EXAMPLE: A P+N junction has Na=1020 cm-3 and Nd
=1017cm-3. What is a) its built in potential, b)Wdep , c)xN ,
and d) xP ?
Solution:
a)
10 20 ×1017 cm −6
kT N d N a
φbi =
ln
= 0.026V ln
≈1V
2
20
−6
q
ni
10 cm
1/ 2
b) W ≈ 2ε sφbi =  2 ×12 × 8.85 ×10 ×1 
dep
 1.6 ×10 −19 ×1017 
qN d


−14
c) xN ≈ Wdep = 0.12 µm
d) xP = xN N d N a = 1.2 × 10 −4 µm = 1.2 Å ≈ 0
= 0.12 µm
P+N and N+P junction
P+
N
Metal/Semi. junction
Wdep
N+
Wdep
P
Wdep
N
M
P
M
Wdep
4.4 Junction Breakdown
Peak Electric Field
N+
N
0
a
Neutral Region
increasing
reverse bias
P
xp
E
Ep
 2qN

E p = E(x=0)= 
(φi + | Va |) 
 εs

increasing reverse bias
x
p
VB =
x
ε s Ecrit2
2qN
− φi
1/ 2
4.4.2 Avalanche Breakdown
E
c
original
electron
E
E fp
v
• impact ionization: an energetic
electron generating electron and
hole, which can also cause
impact ionization.
• Impact ionization + positive
feedbackavalanche breakdown
VB =
electron-hole
pair generation
E
Ec
fn
ε s Ecrit 2
2qN
1
1
1
VB ∝ =
+
N Na Nd
Lower doping High breakdown voltage
4.4.3 Tunneling Breakdown (Zener Breakdown)
Dominant if both sides of
a junction are very heavily
doped.
Filled States-
Empty States
Ec
Ev
B
qBL
J = G exp(- ) ≈ G exp()
ε
Eg
If assume: average electric field ε =Eg/qL
The field changes with applied voltage Va
I
V
Breakdown
To have TB, the field must be larger than
6
=
≈
10
V/cm
Ep Ecrit
4.5 Junction FET
Depletion of the channel to shut off the conduction
How about metal-semiconductor FET?