UNIVERSITY OF CALIFORNIA
College of Engineering
Department of Electrical Engineering and Computer Sciences
EE 130/230M
Spring 2013
Prof. Liu & Dr. Xu
Solution to Homework Assignment #7
Problem 1: pn Diode Charge Control Model
Given: junction area A = 100 m2; minority-carrier lifetimes n = 10-6 s (p side) and p = 10-7 s (n side); T = 300K.
Since the minority carrier concentrations (np and pn) are enhanced within the quasi-neutral regions, the diode is
forward biased. The majority carrier concentrations (pp and nn) are not significantly enhanced, however, so lowlevel injection conditions prevail.
a) Since low-level injection conditions prevail, the “Law of the Junction” holds: within the depletion region and at
the edges of the depletion region, np=ni2  exp{qVA/kT}.
np and pn each are enhanced by a factor 1010 at the edges of the depletion region,
so 1010 = exp{qVA/kT}  VA = (kT/q)  ln(1010) = 10 × (kT/q)  ln(10) = 10 × (60 mV) = 0.6 V.
b) pp = NA = 1016 cm-3 and nn = ND = 1018 cm-3
c) np(-xp) = np(-xp) – np0(-xp) = 1014 – 104  1014 cm-3. pn(xn) = pn(xn) – pn0(xn) = 1012 – 102  1012 cm-3
The majority carrier concentrations (pp and nn) are not significantly enhanced within the quasi-neutral regions,
so low-level injection conditions prevail.
d) From Lecture 4, Slide 16 the electron mobility for NA =1016 cm-3 is n =1200 cm2/Vs and the hole mobility for
ND =1018 cm-3 is p =150 cm2/Vs.
The electron diffusion constant Dn= n (kT/q)=1200×0.026=31.2 cm2/s.
The hole diffusion constant, Dp= p (kT/q)=150×0.026=3.9 cm2/s.
The electron minority carrier diffusion length Ln = Dn t n = 31.2 ´10-6 = 5.5 ´10-3 cm = 55 m
And the hole minority carrier diffusion length Lp = D p t p = 3.9 ´10-7 = 6.24 ´10-4 cm= 0.624 m
e) Excess minority carrier charge is stored within the quasi-neutral regions:
QP = qApn(xn) Lp = 1.6×10-19×(100×10-8)×1012× 6.24×10-4 = 9.98×10-17 C (624 holes)
QN = qAnp(-xp) Ln = 1.6×10-19×(100×10-8)×1014× 5.5×10-3 = 8.8×10-14 C (550,000 electrons)
f) The diode current is found using the charge control model:
Ip(xn) = QP/p= 9.98×10-17/10-7 = 9.98×10-10 A
In(-xp) = QN/n = 8.8×10-14/10-6 = 8.8×10-8 A
I = Ip(xn) + In(-xp) = 8.9×10-8 A
The current is dominated by electron injection from the more heavily doped n side into the p side.
Problem 2: pn Junction Small-Signal Model
a) From Problem 1, IDC= 8.9×10-8 A. The small-signal resistance R = (kT/q)/ IDC=0.026/8.9×10-8 =2.9×105 .
Since the n-type side is degenerately doped (ND = 1018 cm-3), we should use the equation on Slide 20 of
Lecture 3 to find the reduction in band gap energy on the n side:
300
=35meV
EG  3.5  108 N1 3
T
The built-in potential is then
E  EG kT NA
Vbi  G

ln( ) = (1.12-0.035)/2 + 0.026×ln(1016/1010)=0.902V
2q
q
ni
The depletion width W 
2 s (Vbi  VA )

qNA
s A
2  1012  (0.902  0.6)
 0.194 m
1.6  1019  1016
1012  100  108
 5.15  1014 F
5
W
1.94  10
t
10-6
Diffusion capacitance CD = n =
= 0.39 ´10-11 F
5
R 2.95 ´10
Total capacitance C=Cj+ CD = 0.39×10-11 F.
Depletion capacitance C j 

A schematic of the small-signal model is shown below.
b) Under reverse bias, the stored minority carrier charge within the quasi-neutral regions is negligible and
so the depletion capacitance (CJ) is the dominant component of small-signal capacitance. From
Lecture 13 Slide 14,
2(Vbi -V A )
1 2(Vbi -V A )
1
=
=
= 1.25 ´103 ´ (Vbi -V A ) 2
2
2
-12
12
-19
16
Cj
es A qN A 10 ´10 ´1.6 ´10 ´10
F
The plot of 1/C2 vs. VA is shown below.
Extrapolating to zero, the x-intercept occurs at VA = Vbi = 0.902 V.
Problem 3: Transient Response of a pn Junction
a) From Lecture 13 Slide 20 the storage delay time is ts = t p ln[1+
IF
] = 10-6  ln(1+1) = 0.693 s
IR
b) Assuming that the diode turns on from i = 0 (QN = 0 and QP = 0) at 2 μs we can adapt the equation
kT  I F
t  / τ 
from Lecture 14 Slide 3 to obtain v A (t ) 
ln 1 
1 e p 
q  I0

where t' = t  2 μs.


Problem 4: Photodiode
a) hole diffusion equation within the quasi-neutral n-type region is
¶Dpn
¶2 Dpn Dpn
= Dp
+ GL
¶t
¶x 2
tp
In steady state Δpn/t = 0.
Far away from the junction (x  ∞) in the quasi-neutral region2Δpn/x2 = 0. Therefore
Dp (x ® ¥)
0=- n
+ GL Þ Dpn (x ® ¥) = GLt p
tp
b) Under steady state conditions, the hole diffusion equation within the quasi-neutral n-type region is
 2 p n p n
0  Dp

 GL
p
x 2
for which the general solution is
pn ( x ')  GL p  A1e
(x '
Lp
)
 A2e
(x'
Lp
)
where x' is defined to be 0 at the depletion region edge on the n side (ref. Lecture 10 Slide 12).
Assuming low-level injection so that the Law of the Junction (ref. Slide 8 of Lecture 10) holds, the
boundary conditions are
n 2 ( qVA )
Dpn (x ' = 0) = i [e kT -1]
ND
Dpn (x ' ® ¥) = GLt p
Because exp(x'/LP) → ∞ as x'→ ∞, the only way the second boundary condition can be satisfied is for
A2 to be zero.
With A2 = 0, the first boundary condition yields
pn ( x '  0)  GL p  A1 
ni 2 (qVA kT )
[e
 1]
ND
or
A1 
ni 2 (qVA kT )
[e
 1]  GL p
ND
and
pn ( x ')  GL p  [
( x ' )
ni 2 (qVA kT )
Lp
[e
 1]  GL p ]e
ND
The hole diffusion current density is then:
( x ' )
D n 2 (qVA )
d pn
Lp
J p ( x ')  qDp
 q p [ i [e kT  1]  GL p ]e
dx '
Lp ND
Similarly, under steady state conditions the electron diffusion equation within the quasi-neutral p-type
 2 n p n p
region is 0  Dn

 GL
n
x 2
for which the general solution is
n p ( x)  GL n  A3 e x / Lm  A4 e  x / Lm
where x'' is defined to be 0 at the depletion region edge on the p side and increases with distance into
the quasi-neutral p-type region, i.e. it is in the negative x direction (ref. Lecture 10 Slide 12).
2
Again assuming low-level injection, the boundary conditions are
n p ( x   0) 
qV
ni
( A )
[e kT  1]
NA
n p ( x   )  G L n
Following the same reasoning as for pn above, we obtain
 n2

n p ( x)  GL n   i e qVA / kT  1  GL n e  x / Lm
NA

The electron diffusion current density is then:
dn p

D  n2
J n ( x)  qDn
 q n  i e qVA / kT  1  GL n e  x / Lm
dx
Ln  N A





The total diode current I = AJ, where
D p  ni2 qVA / kT


Dn  ni2 qVA / kT
J  J p ( x  0)  J n ( x  0)  q
e
 1  GL p   q
e
 1  GL n 


Lp  N D
Ln  N A


(Note that there is a minus sign in front of Jn(x'') because x'' is in the negative x direction.)
 Dp
 D p p Dn n 
Dn  qVA / kT
qV / kT
 I  qAni2 

e
 1  qAGL 

  I 0 e A  1  qAGL L p  Ln 
L N

L
N
L
L

n
A 
n 
 p D
 p
2
(Note that Dpp = Lp and Dnn = Ln2.)








 I  I 0 e qVA / kT  1  I L where I L  qAGL L p  Ln 


c) The diode current is given by the ideal diode equation with an additional negative term due to
illumination when GL ≠0, i.e. the ideal I-V curve is shifted down by an amount equal to IL.
Since IL  GL, the shift downward increases proportionately with GL as shown in the plot below.