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Unformatted text preview: ECE 329 Fall 2009 Homework5Solution
Due: Sep. 29, 2009 1.
a) According to Page 2 in Lecture 11, the mobility is dened as
qτ
q
=
.
m
mν mobility = Although mν is not explicitly given in the problem, it is related to Ne and σ by (see Page 3 in
Lecture 11)
σ= or Ne q 2
,
mν mν = Ne q 2
,
σ q
mν = therefore,
mobility = q
Ne q 2
σ = σ
Ne q 5.8 × 107
8.45 × 1028 × 1.6 × 10−19
= 4.3 × 10−3 m2 /(s · V ) ,
= where the unit is derived from v =
b)
R= qτ
m E. l
400
=
= 0.98 (Ω) .
2
7 × π × (1.5 × 10−3 )2
σ · πr
5.8 × 10 c) On the one hand,
J= I
.
πr2 On the other hand,
J = σE.
J
I
3
=
= 7.3 × 10−3 (V/m) .
E= =
2
σ
σ · πr
5.8 × 107 × π × (1.5 × 10−3 )2 ∴ d) The velocity of an electron is
v= qτ
E = mobility · E = 4.3 × 10−3 × 7.3 × 10−3 = 3.1 × 10−5 (m/s) ,
m so the time to drift from one end to the other is
t= l
400
=
= 1.3 × 107 (s) .
v
3.1 × 10−5 2. 1 ECE 329 Fall 2009 a) Since V (0) − V (z ) = ´z
0 E (z ) dz and V (0) = 0, we have ˆz
V (z ) = − E (z ) dz =
0 3
8 0 2z
3
8 0 [ 2d + 1 (z for 0
for d − d)] z
z d
(V) .
1 ∂
b) Inside the two regions, we have 2 V (z ) = ∂z2 V (z ) = 0, which means that Laplace's equation
is satised inside the two regions. However, at the interface z = d, the derivative of V (z ) goes
to innity, so Laplace's equation is not satised.
c) Method (1):
2 V (z = 1) = 3
[ 2d +
80 1 (1 − d)] (V) , ρs (z = 1) = (−z ) · D (z = 1) = (−z ) · [ 2 E (z = 1)] = (−z ) ·
ˆ
ˆ
ˆ
C= Q
ρs (z = 1) A
=
=
V
V (z = 1) − V (z = 0) − 2 3
80 1 2A
2d + 1 (1 − d) ˆ
1z = (C/V) . Method (2): the capacitance for the two slabs are given by
A
1d
A
2 1−d C1 =
C2 = 0
d z
z d
1 therefore, the total capacitance is
−
−
C = C1 1 + C2 1 −1 = d
1A 2 1
+ 1−d
2A = 1 2A
2d + 1 (1 − d) (C/V) . 312
C/m2 ,
80 ...
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This note was uploaded on 05/05/2010 for the course ECE 329 taught by Professor Franke during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 FRANKE

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