ECE 315 Homework 3 Solution
Fall 2007
1.
(PN junction, continued from Prob. 7 of HW 2.
Please follow the numbers there. )
For
a pn junction well described by the depletion region approximation as shown below, assume
the electron mobility
μ
n
=1000cm
2
/Vs, the hole mobility
μ
p
=400cm
2
/Vs, and the minority
lifetime
τ
n
=
τ
p
=10
7
s
(a)
In equilibrium, which region(s) satisfy the charge neutrality condition? Explain briefly.
(4 pts)
Regions 1 and 4. In these regions, there is negligible voltage drop and is therefore
characteristically similar to bulk silicon. In regions 2 and 3, we have majority carrier
diffused across the junction, which leaves behind ionized immobile dopant charges whose
electric charge is no longer neutralized by their corresponding free charge carriers. Regions
2 and 3 are thus approximated with the “depletion regions”, which means the majority
carriers are depleted (noted that the minority carriers are actually increased to keep the np
product constant.)
(b)
In equilibrium, which region(s) satisfy the relation of
np=n
i
2
? Explain briefly. (4 pts)
All Regions of 14.
In equilibrium,
np=n
i
2
is always true everywhere.
Rigorous proof has
to come from statistical mechanics and the definition of the Fermi levels, but you can think
of the situation as the chemical equation of
e

+ h
+
↔
nil.
(c)
At
V
D
=0, Calculate the depletion capacitance in F/cm
2
. (4 pts)
We can use the calculation in Homework 2 and have
x
p
= x
n
= 216 nm.
Hence the depletion
capacitance, or the capacitance across the depletion region, is: C
jo
= ε
si
ε
o
/W = ε
si
ε
o
/(x
p
+x
n
=
24.0 nF/cm
2
.
(d)
Estimate the ideal saturation current
I
S
at 250K, 300K and 400K.
You can use the
n
i
values from Problem 2 of Homework 2. Assume
A
= 1cm
2
. (6 pts)
We will ignore the temperature dependence of mobility and recombination lifetime here.
They do have some temperature dependence, but in view of the exponential dependence of
n
i
, are not very important to include.
I
S
= J
S
× A
J
S
= qD
p
p
no
/L
p
+ qD
n
n
po
/L
n
p
no
= n
i
2
(T)/ n
no
≈ n
i
2
/N
D
n
po
= n
i
2
(T)/ p
po
≈ n
i
2
/N
A
L
p
= (D
p
τ
p
)
1/2
L
n
= (D
n
τ
n
)
1/2
J
S
= qD
p
n
i
2
/ N
D
L
p
+ qD
n
n
i
2
/N
A
L
n
Since we have
μ
n
=1,000 cm
2
/Vs
and
μ
p
=400 cm
2
/Vs,
as well as
τ
n
=
τ
p
=0.1
μ
s
Then,
D
n
= kT/q ×
μ
n
,
D
p
= kT/q ×
μ
p
Therefore,
I
S
= qAn
i
2
(T) × [(
μ
p
kT/qτ
p
)
1/2
× 1/N
D
+ (
μ
n
kT/qτ
n
)
1/2
× 1/N
A
]
1
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at
T = 250 K, n
i
= 2.13 × 10
8
cm
3
, I
S
= 17.4 fA (×10
15
A)
at
T = 300 K, n
i
= 1.5 × 10
10
cm
3
, I
S
= 94.5 pA (×10
12
A)
at
T = 400 K, n
i
= 3.06 × 10
12
cm
3
, I
S
= 4.54 μA (×10
6
A)
You can see that the direct dependence of
n
i
2
has caused the current to change over 8 orders
of magnitude!!!
Your computer “can” behave quite differently in Fargo and in Phoenix,
though there are often other feedback/stabilizing techniques applied to minimize this effect.
2.
(Generation and recombination in the depletion region)
For the pn junction in Prob. 1 at
300K, the voltage across the diode is
V
D
and the current is
I
D
.
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 Fall '07
 SPENCER
 Semiconductors, Microelectronics, Pn junction, Charge carrier, Charge carriers in semiconductors

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