Lecture 13
OUTLINE
• pn Junction Diodes (cont’d)
– Charge control model
– Small-signal model
– Transient response: turn-off
Reading: Pierret 6.3.1, 7, 8.1; Hu 4.4, 4.10-4.11
Minority-Carrier Charge Storage
• Under forward bias (VA > 0), excess minority carriers are
stored in the quasi-neutral regions of a pn junction:
QN  qA

xp
n p ( x)dx

QP  qA pn ( x)dx
xn
 qApn ( xn ) LP
 qAn p ( x p ) LN
EE130/230M Spring 2013
Lecture 13, Slide 2
Derivation of Charge Control Model
Consider the n quasi-neutral region of a forward-biased pn junction:
•The minority carrier diffusion equation is (assuming GL=0):
pn
 2 pn pn
 DP

2
t
x
p
•Since the electric field is very small,
•Therefore
J P  qDP
(qpn )
J P qpn


t
x
p
EE130/230M Spring 2013
Lecture 13, Slide 3
p n
x
Derivation Assuming a Long Base
• Integrating over the n quasi-neutral region:
J P ()







1
qA  pn dx    A  dJ P  qA  pn dx 
t  x n
 p  xn


J p ( xn )
• Note that  A
J P ()
 dJ
P
  AJ P ()  AJ P ( xn )  AJ P ( xn )  I P ( xn )
J p ( xn )
dQP
QP
 I P ( xn ) 
• So
dt
p
EE130/230M Spring 2013
Lecture 13, Slide 4
Charge Control Model
We can calculate pn-junction current in 2 ways:
1. From slopes of np(-xp) and pn(xn)
2. From steady-state charges QN, QP stored in each excessminority-charge distribution:
dQP
QP
 I P ( xn ) 
0
dt
τp
QP
 I P ( xn ) 
τp
 QN
Similarly, I N ( x p ) 
τn
EE130/230M Spring 2013
Lecture 13, Slide 5
Charge Control Model for Narrow Base
•
For a narrow-base diode, replace p and/or n by
the minority-carrier transit time tr
–
time required for minority carrier to travel across the quasineutral region
– For holes in narrow n-side:
1
QP  qA pn ( x)dx  qA pn ( xn )WN
xn
2
dpn
pn ( xn )
I P  AJ P  qADP
 qADP
dx
WN
WN
QP WN 
 τ tr , p 

IP
2 DP
2

WP 
– Similarly, for electrons in narrow p-side: τ tr ,n 
2 DN
2
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Lecture 13, Slide 6
Charge Control Model Summary
• Under forward bias, minority-carrier charge is stored in the
quasi-neutral regions of a pn diode.
– Long base:


ni2 qVA / kT
QN  qA
e
 1 LN
NA


ni2 qVA / kT
QP  qA
e
 1 LP
ND
– Narrow base:


1 ni2 qVA / kT
QN  qA
e
 1 WP
2 NA


1 ni2 qVA / kT
QP  qA
e
 1 WN
2 ND
EE130/230M Spring 2013
Lecture 13, Slide 7
• The steady-state diode current can be viewed as the
charge supply required to compensate for charge loss
via recombination (for long base) or collection at the
contacts (for narrow base).
– Long base (both sides): I 
 QN QP

τn
τp
 QN QP
– Narrow base (both sides): I 

τtr ,n τtr , p
where
τ tr ,n
2

WP 

2 DN
and τ tr , p
2

WN 

2 DP
Note that
EE130/230M Spring 2013
Lecture 13, Slide 8
LN DN
L
D

and P  P
τn
LN
τ p LP
Small-Signal Model of the Diode
i
+
va
dva
i  C
R
dt
v

1 dI DC
d
d
qVA / kT


I 0 (e
 1) 
I 0 e qVA / kT
R dVA dVA
dVA
Small-signal
I DC
1
q
qVA / kT
conductance: G 

I 0e

R
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kT
Lecture 13, Slide 9
kT / q
Charge Storage in pn Junction Diode
EE130/230M Spring 2013
Lecture 13, Slide 10
pn Junction Small-Signal Capacitance
2 types of capacitance associated with a pn junction:
depletion capacitance CJ 
 due to variation of depletion charge
diffusion capacitance
dQdep
dVA
dQ
CD 
dVA
–due to variation of stored
minority charge in the quasi-neutral regions
For a one-sided p+n junction Q = QP + QN  QP so
τ p I DC
dQP
dI
CD 
 τp
 τ pG 
dVA
dVA
kT / q
EE130/230M Spring 2013
Lecture 13, Slide 11
Depletion Capacitance
CJ 
dQdep
dVA
A
s
W
What are three ways to reduce CJ?
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Lecture 13, Slide 12
Total pn-Junction Capacitance
C = CD + CJ


τI DC
CD 
 e qVA / kT  1
kT / q
CJ  A
s
W
•CD dominates at moderate to high forward biases
•CJ dominates at low forward biases, reverse biases
EE130/230M Spring 2013
Lecture 13, Slide 13
Using C-V Data to Determine Doping
2(Vbi  VA )
1
W2
 2 2  2
2
A q S N
CJ
A s
EE130/230M Spring 2013
Lecture 13, Slide 14
Example
If the slope of the (1/C)2 vs. VA characteristic is -2x1023 F-2 V-1,
the intercept is 0.84V, and A is 1 mm2, find the dopant
concentration Nl on the more lightly doped side and the
dopant concentration Nh on the more heavily doped side.
Solution: N l  2 /( slope  q s A2 )
 2 /( 2 10 1.6 10
23
19
10
12

 10
)
8 2
 6 1015 cm 3
2
qV
0.84
ni kTbi
10 20 0.026
kT N h Nl
18
3
Vbi 
ln

N

e

e

1
.
8

10
cm
h
2
q
Nl
6 1015
ni
EE130/230M Spring 2013
Lecture 13, Slide 15
Small-Signal Model Summary
C  C J  CD
I DC  I 0 (e qVA / kT  1)
A s
Depletion capacitance C J 
W
τI DC
Diffusion capacitance CD 
kT / q
EE130/230M Spring 2013
I DC
Conductance G 
kT / q
Lecture 13, Slide 16
Transient Response of pn Diode
• Suppose a pn-diode is forward biased, then suddenly turned
off at time t = 0. Because of CD, the voltage across the pn
junction depletion region cannot be changed instantaneously.
The time delay in switching between
the FORWARD-bias and REVERSE-bias
states is due to the time required to
change the amount of excess minority
carriers stored in the quasi-neutral regions.
EE130/230M Spring 2013
Lecture 13, Slide 17
Turn-Off Transient
• In order to turn the diode off, the excess minority
carriers must be removed by net carrier flow out of
the quasi-neutral regions and/or recombination
– Carrier flow is limited by the switching circuitry
EE130/230M Spring 2013
Lecture 13, Slide 18
Decay of Stored Charge
Consider a p+n diode (Qp >> Qn):
pn(x)
i(t)
ts
t
vA(t)
For t > 0:
dpn
dx
EE130/230M Spring 2013
x  xn
i

0
qAD p
Lecture 13, Slide 19
ts
t
Storage Delay Time, ts
• ts is the primary “figure of merit” used to characterize the
transient response of pn junction diodes

Qp 

i
  I R 


dt
τp
τ
p


dQ p
Qp
0  t  ts
• By separation of variables and integration from t = 0+ to t = ts,
noting that I F  Q p (t  0) / τ p
and making the approximation Q p (t  t s )  0
 IF 

We conclude that t s  τ p ln 1 
 IR 
EE130/230M Spring 2013
Lecture 13, Slide 20
Qualitative Examples
Illustrate how the turn-off transient response would change:
Increase IF
Decrease p
Increase IR
i(t)
i(t)
ts
EE130/230M Spring 2013
t
i(t)
ts
Lecture 13, Slide 21
t
ts
t