Lecture #13
OUTLINE
• pn Junctions
– reverse bias current
– deviations from ideal behavior
– small-signal model
Reading: Chapters 6.2 & 7
Spring 2003
EE130 Lecture 13, Slide 1
Diode Current due to Generation
• If an electron-hole pair is generated (e.g. by
light) in the depletion region of a Schottky
diode or pn diode, the built-in electric field will
sweep the generated carriers out, resulting in
an additional component of current.
Spring 2003
EE130 Lecture 13, Slide 2
1
Review: Current Flow in Long-Base Diode
• Under forward bias (VA > 0):
– Holes are supplied (from the external circuit)
through the p-side ohmic contact
• Some of these recombine with injected
electrons; the rest are injected into the n-side
– Electrons are supplied (from the external
circuit) through the n-side ohmic contact
• Some of these recombine with injected holes;
the rest are injected into the p-side
Spring 2003
EE130 Lecture 13, Slide 3
Review: Current Flow in Short-Base Diode
• Under forward bias (VA > 0):
– Holes are supplied (from the external circuit)
through the p-side ohmic contact
• If the p-side is short (WP’ << Ln), ~all of the holes
are injected into the n-side, and recombine with
electrons at the n-side ohmic contact
– Electrons are supplied (from the external
circuit) through the n-side ohmic contact
• If the n-side is short (WN’ << Lp), ~all of the
electrons are injected into the p-side, and
recombine with holes at the p-side ohmic contact
Spring 2003
EE130 Lecture 13, Slide 4
2
Reverse Bias Current
• Consider a reverse-biased (VA<0) pn junction:
minority carrier concentration
x
-xp
xn
– Depletion of minority carriers at edges of depletion region
– The only current which flows is due to drift of minority carriers
across the junction. This current is fed by diffusion of minority
carriers toward junction (supplied by thermal generation).
Spring 2003
EE130 Lecture 13, Slide 5
Alternative Derivation of Formula for I0
“Depletion approximation”:
excess minority carrier concentration
x
-xp
xn
• I0 represents the rate at which carriers are generated
within a diffusion length of the depletion region
∆n
∂n
n / NA
=− p = i
∂t
τn
τn
2
-LN -x p ≤ x ≤ -x p
2
∂p
∆p
n / ND
=− n = i
∂t
τp
τp
Spring 2003
xn ≤ x ≤ xn + LP
 n 2 / ND 
 n 2 / NA 

 + qALP  i
I 0 = qALN  i

 τ

p
 τn 


EE130 Lecture 13, Slide 6
3
Deviations from the Ideal I-V Behavior
Spring 2003
EE130 Lecture 13, Slide 7
Effect of R-G in Depletion Region
• The net generation rate is given by
2
∂p ∂n
ni − np
=
=
∂t ∂t τ p (n + n1 ) + τ n ( p + p1 )
where n1 ≡ ni e ( ET − Ei ) / kT and p1 ≡ ni e ( Ei − ET ) / kT
ET = trap - state energy level
• R-G in the depletion region contributes an
additional component of diode current IR-G
∂p
− x p ∂t
I R −G = − qA∫
Spring 2003
xn
dx
R −G
EE130 Lecture 13, Slide 8
4
• For reverse bias greater than several kT/q,
I R −G = −
qAniW
1 n
p 
where τ 0 ≡  τ p 1 + τ n 1 
2τ 0
2  ni
ni 
I
n
Ip
Spring 2003
EE130 Lecture 13, Slide 9
• For forward biases,
I R −G ∝ qAniWe qV A / 2 kT
In
Ip
Spring 2003
EE130 Lecture 13, Slide 10
5
Effect of Series Resistance
Spring 2003
EE130 Lecture 13, Slide 11
High-Level Injection Effect
• As VA increases, the side of the junction
which is more lightly doped will eventually
reach HLI:
nn > nno
(p+n junction)
or
p p > p po
(n+p junction)
⇒ significant gradient in majority-carrier profile
Majority-carrier diffusion current reduces the diode
current from the ideal
Spring 2003
EE130 Lecture 13, Slide 12
6
Review: Charge Storage in pn-Diode
Spring 2003
EE130 Lecture 13, Slide 13
Small-Signal Model of the Diode
Small signal equivalent circuit:
i
v
R=1/G
C
i = Gv + C
dv
dt
Small-signal conductance :
G≡
1
dI
d
d
=
=
I 0 (e qVA / kT − 1) ≈
I 0e qVA / kT
R dVA dVA
dVA
G=
Spring 2003
q
kT
I 0 e qVA / kT ≅ I DC /
kT
q
EE130 Lecture 13, Slide 14
7
2 types of capacitance associated with a pn junction:
1. Cdep depletion capacitance
2. CD
diffusion capacitance (due to variation of
stored minority charge in the quasi-neutral
regions)
For a one-sided p+n junction, QP >> QN
so Q = QP + QN ≅ QP:
τ p I DC
dQ
dI
= τp
= τ pG =
dV A
dV A
kT / q
CD =
Spring 2003
EE130 Lecture 13, Slide 15
Depletion Capacitance
p
n
W
conductor
“insulator”
Cdep ≡
dQdep
dVA
=A
conductor
εs
W
What are three ways to reduce Cdep?
Spring 2003
EE130 Lecture 13, Slide 16
8
Total pn-Junction Capacitance
Spring 2003
EE130 Lecture 13, Slide 17
Cdep-vs.-VA (Reverse Bias)
1
Cdep
Spring 2003
2
W2
2(V − V )
= 2 2 ≅ 2bi A
A qε S N
A εs
EE130 Lecture 13, Slide 18
9
Example
If the slope of the (1/Cdep)2 vs. VA characteristic is 2x1023
F-2 V-1, the intercept is 0.84V, and A is 1 µm2, find the
lighter and heavier doping concentrations Nl and Nh.
Solution:
N l = 2 /( slope × qε s A2 )
= 2 /( 2 × 10 23 × 1.6 × 10 −19 × 12 × 8.85 × 10 −14 ×10 −8 cm 2 )
= 6 × 1015 cm −3
Vbi =
2
qV
0.84
kT N h N l
ni kTbi
10 20 0.026
ln
⇒
N
=
e
=
e
= 1.8 × 1018 cm −3
h
2
q
ni
Nl
6 ×1015
Spring 2003
EE130 Lecture 13, Slide 19
10