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ECE 315 Prelim Exam Solution
Fall 2005
Name: ___________________________________ Student net ID: _______________________
For a semiconductor resistor with length of
l
and cross section of
A
, the resistance can be estimated as
(
29
p
n
qp
qn
A
l
A
l
R
μ
ρ
+
=
=
, where
n
and
p
are the electron and hole concentration, and
n
and
p
are
the electron and hole mobility.
Diode equations in case:
W
d
= x
n
+ x
p
bi
D
A
si
V
N
N
q
+
=
1
1
2
0
ε
, N
A
x
p
= N
D
x
n
,
=
2
ln
i
D
A
bi
n
N
N
q
kT
V
,
0
=8.85
×
10
14
F/cm,
si
=11.7 is the relative dielectric constant of silicon,
q
=1.6
×
10
19
coul is the elemental
charge,
k
B
the Boltzmann constant, and
T
the temperature (default=300K).
The thermal voltage
k
B
T/q
at
room temperature is 26mV.
For nMOSFET with the threshold voltage
V
th
, the drain current
I
D
in the linear and saturation regions
above threshold (
V
GS
> V
th
) are:
(
29
(
29
saturation
2
linear
2
2
2
th
GS
Dsat
DS
th
GS
ch
ox
D
th
GS
Dsat
DS
DS
DS
th
GS
ch
ox
D
V
V
V
V
V
V
C
L
W
I
V
V
V
V
V
V
V
V
C
L
W
I

=
≥

=

=
<


=
And in the saturation region, the quasistatic circuit model can be approximated as:
1.
For a semiconductor wire as shown below with a cross section of 2
μ
m
×
1
μ
m and length of 100
μ
m
contains a pn junction in the middle with
N
A
=10
16
cm
3
and
N
D
=10
16
cm
3
.
Assume
n
i
=10
10
cm
3
and
n
=1,000cm
2
/Vs and
p
=400cm
2
/Vs. Calculate the wire resistance in Ohm.
Notice that a real number
is required, instead of just formula.
Show your derivation and brief justification. (
Hint: the depletion
region has majority carriers much less than the doping level.
) (15 pts)
g
mn
v
GS
N
A
=10
16
cm
3
,
N
D
=0
N
D
=10
16
cm
3
,
N
A
=0
1
m
1
m
1
m
100
m
1
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View Full DocumentThe majority carrier is the main component in drift current (or Ohm’s law resistivity), but we
need to substrate out the area that is in the depletion region.
m
W
V
mV
n
N
N
q
kT
V
i
D
A
bi
μ
43
.
0
72
.
0
10
2
10
6
.
1
10
85
.
8
7
.
11
2
72
.
0
12
60
ln
16
19
14
2
=
⋅
⋅
×
×
×
×
=
=
×
=
=


Since
N
A
=N
D
,
W
is equally distributed on the two sides, which makes the two resistors in parallel
have an effective area of 1
μ
m by (1  0.43/2)
μ
m.
The resistance is then:
(
29
Ω
=
+
×
×
×
⋅
×
×
×
=




k
R
570
400
1000
10
10
6
.
1
1
10
784
.
0
10
10
100
16
19
4
4
4
.
2.
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 Microelectronics

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