pn Junction (5/11/00)
Page 1-2.1
1.2 - DEPLETION REGION OF A PN JUNCTION
INTRODUCTION
Objective
• Characterize and model the pn junction
Outline
• Physical aspects of pn junctions
• Mathematical models of the pn junction
- Depletion capacitance
- Breakdown characteristics
ECE 4430 - Analog Integrated Circuits and Systems
P.E. Allen
pn Junction (5/11/00)
Page 1-2.2
PHYSICAL ASPECTS OF THE PN JUNCTION
Abrupt Junction
Metallurgical Junction
n-type semiconductor
p-type semiconductor
iD
+vD Depletion
region
n-type
semiconductor
p-type
semiconductor
iD
+ v-D W
-W1
0
W2
x
Fig. 1.2-1
1. Doped atoms near the metallurgical junction lose their free carriers by diffusion.
2. As these fixed atoms lose their free carriers, they build up an electric field which opposes the diffusion
mechanism.
3. Equilibrium conditions are reached when:
Current due to diffusion = Current due to electric field
ECE 4430 - Analog Integrated Circuits and Systems
P.E. Allen
pn Junction (5/11/00)
Page 1-2.3
MATHEMATICAL CHARACTERIZATION OF THE PN JUNCTION
Abrupt PN Junction
Impurity concentration (cm-3)
Cross section of an ideal pn junction:
Apply a
reverse
bias,
vD = -VR
xd
xp
xn
n-type
semiconductor
p-type
semiconductor
ND
x
0
-NA
Depletion charge concentration (cm-3)
qND
-W1
iD
0
+vD -
W2
x
-qNA
Symbol of the pn junction:
Electric Field (V/cm)
iD
x
+v
D
-
E0
iD
Potential (V)
+v D
Fig. 1.2-1
V2
ψ0+ VR
x
V1
xd
ECE 4430 - Analog Integrated Circuits and Systems
Fig. 1.2-2
P.E. Allen
pn Junction (5/11/00)
Page 1-2.4
Summary of the PN Junction Characterization
Barrier potentialkT NAND
 NAND
ψo =
ln

 = Vt ln

2
q  ni 
 ni2 
Depletion region widthsW1 =
2εsi(ψo-vD)ND
qND(NA+ND)
W2 =
2εsi(ψo-vD)NA
qND(NA+ND)



W ∝
1
N
Depletion capacitanceCj = A
εsiqNAND
2(NA+ND)
Cj
1
ψo-vD
=
Cj0
vD
1- ψ
o
Cj0
Fig. 1.2-3
ECE 4430 - Analog Integrated Circuits and Systems
0
ψ0 vD
P.E. Allen
pn Junction (5/11/00)
Page 1-2.5
Example 1
An abrupt pn junction in silicon has the doping densities of N A = 1015 atoms/cm3 and ND = 1016
atoms/cm3. calculate the junction built-in potential, the depletion-layer widths, the maximum field with
10V reverse bias and the depeletion capacitance with 10V reverse bias if Cj0 = 3pF.
Solution
At room temperature, kT/q = 26mV and the intrinsic concentration is ni = 1.5x1010 cm-3. Therefore, the
junction built-in potential is
 1015·1016 
ψo = 0.026 ln
 = 0.637V
2.25x1026
The depletion width on the p-side is,
2·1.04x10-12·10.64
= 3.55x10-4 cm = 3.55µm
1.6x10-19·1015·1.1
The depletion width on the n-side is,
W1 =
2·1.04x10-12·10.64
= 0.35x10-4 cm = 0.35µm
1.6x10-19·1016·11
The maximum field occurs for x = 0 and is
qNA
 -1.6x10-19·1015·3.5x10-4
Εmax = - ε W 1 = 
 = -5.38x104 V/cm
1.04x10-12


The depletion capacitance can be found as
3pF
= 0.65pF
Cj =
10
1 + 0.5
W2 =
ECE 4430 - Analog Integrated Circuits and Systems
P.E. Allen
pn Junction (5/11/00)
Page 1-2.6
Summary of the PN Junction Characterization - Continued
Breakdown voltageεsi(NA+ND)
1
2
∝
VR = 2qN
E
max
N
AND
2
where Emax is the maximum electric field before breakdown occurs (usually due to avalanche
breakdown).
Reverse leakage currentThe reverse current, IR, increases by a multiplication factor M as the reverse voltage increases and is
IRA = MI R
where
M =
ID (mA)
1
V R n
1 -  
 B V
3
2
1
BV = VR
-25
-20
-15 -10
5
-5
VD (V)
-1
Breakdown
-2
-3
ECE 4430 - Analog Integrated Circuits and Systems
Fig. 1.2-4
P.E. Allen
pn Junction (5/11/00)
Page 1-2.7
Example 2
An abrupt pn junction has doping densities of NA = 5x1015 atoms/cm3 and ND = 1016 atoms/cm3.
Calculate the breakdown voltage if Εcrit = 3x105 V/cm.
Solution
VR =
εsi(NA+ND)
1.04x10-12·15x1015
2
=
Emax
= 88V
2qNAND
2·1.6x10-19·5x1015·1016
ECE 4430 - Analog Integrated Circuits and Systems
P.E. Allen
pn Junction (5/11/00)
Page 1-2.8
Summary of PN Junction Characterization - Continued
Graded junction:
ND
0
-NA
Above expressions become:
Depletion region widths 2εsi(ψo-vD)ND m
W 1 =  qN (N +N ) 

D A D 
 2εsi(ψo-vD)NA m
W2 = 

 qND(NA+ND) 



x
Fig. 1.2-5
 1 m
W ∝ 
N
Depletion capacitanceCj0
1
 εsiqNAND  m
=
Cj = A2(N +N )
v D m

A D  ( φo-vD) m

1 ψo

where 0.33 ≤m ≤ 0.5.
ECE 4430 - Analog Integrated Circuits and Systems
P.E. Allen