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Fundamentals of Microelectronics [Behzad Razavi]习题解答
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更新于2023-03-16
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《Fundamentals of Microelectronics[Behzad Razavi]》的课后习题答案手写版 不少学校用这本书作为模电的教材,但比较难找到答案,特此上传此资源以供参考学习使用 该资源为此书第一版的,第二版课后习题有部分增减但大体是相同的
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2.1 (a)
k = 8.617 × 10
−5
eV/K
n
i
(T = 300 K) = 1.66 ×10
15
(300 K)
3/2
exp
−
0.66 eV
2 (8.617 ×10
−5
eV/K) (300 K)
cm
−3
=
2.465 ×10
13
cm
−3
n
i
(T = 600 K) = 1.66 ×10
15
(600 K)
3/2
exp
−
0.66 eV
2 (8.617 ×10
−5
eV/K) (600 K)
cm
−3
=
4.124 ×10
16
cm
−3
Compared to the values obtained in Example 2.1, we can see that the intrinsic carrier concentration
in Ge at T = 300 K is
2.465×10
13
1.08×10
10
= 2282 times higher than the intrinsic carrier concentration in
Si at T = 300 K. Similarly, at T = 600 K, the intrinsic carrier concentration in Ge is
4.124×10
16
1.54×10
15
=
26.8 times higher than that in Si.
(b) Since phosphorus is a Group V element, it is a donor, meaning N
D
= 5 × 10
16
cm
−3
. For an
n-type material, we have:
n = N
D
=
5 × 10
16
cm
−3
p(T = 300 K) =
[n
i
(T = 300 K)]
2
n
=
1.215 × 10
10
cm
−3
p(T = 600 K) =
[n
i
(T = 600 K)]
2
n
=
3.401 × 10
16
cm
−3


2.3 (a) Since the doping is uniform, we have no diffusion current. Thus, the total current is due only to
the drift component.
I
tot
= I
drift
= q(nµ
n
+ pµ
p
)AE
n = 10
17
cm
−3
p = n
2
i
/n = (1.08 × 10
10
)
2
/10
17
= 1.17 × 10
3
cm
−3
µ
n
= 1350 cm
2
/V ·s
µ
p
= 480 c m
2
/V ·s
E = V/d =
1 V
0.1 µm
= 10
5
V/cm
A = 0.05 µm × 0.05 µm
= 2.5 × 10
−11
cm
2
Since nµ
n
≫ pµ
p
, we can write
I
tot
≈ qnµ
n
AE
=
54.1 µA
(b) All of the parameters are the same except n
i
, which means we must re-calculate p.
n
i
(T = 400 K) = 3.657 ×10
12
cm
−3
p = n
2
i
/n = 1.3 37 × 1 0
8
cm
−3
Since nµ
n
≫ pµ
p
still holds (note that n is 9 orders of magnitude larger than p), the hole
concentration once again drops out of the equation and we have
I
tot
≈ qnµ
n
AE
=
54.1 µA

2.4 (a) From Problem 1, we can c alculate n
i
for Ge.
n
i
(T = 300 K) = 2.465 × 10
13
cm
−3
I
tot
= q(nµ
n
+ pµ
p
)AE
n = 10
17
cm
−3
p = n
2
i
/n = 6.0 76 × 10
9
cm
−3
µ
n
= 3900 cm
2
/V ·s
µ
p
= 1900 cm
2
/V ·s
E = V/d =
1 V
0.1 µm
= 10
5
V/cm
A = 0.05 µm × 0.05 µm
= 2.5 × 10
−11
cm
2
Since nµ
n
≫ pµ
p
, we can write
I
tot
≈ qnµ
n
AE
=
156 µA
(b) All of the parameters are the same except n
i
, which means we must re-calculate p.
n
i
(T = 400 K) = 9.230 × 10
14
cm
−3
p = n
2
i
/n = 8.5 20 × 10
12
cm
−3
Since nµ
n
≫ pµ
p
still holds (note that n is 5 orders of magnitude larger than p), the hole
concentration once again drops out of the equation and we have
I
tot
≈ qnµ
n
AE
=
156 µA
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