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Fundamentals of Microelectronics [Behzad Razavi]习题解答

Behzad

Fundamentals

习题答案

Solution

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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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

SolutionsManualtoAccompany

FundamentalsofMicroelectronics,1stEdition

BookISBN:978‐0‐471‐47846‐1

Razavi



AllmaterialscopyrightedandpublishedbyJohnWiley&Sons,Inc.

Notforduplicationordistribution

2.1 (a)

k = 8.617 × 10

−5

eV/K

(T = 300 K) = 1.66 ×10

(300 K)

3/2

exp



−

0.66 eV

2 (8.617 ×10

−5

eV/K) (300 K)



−3

2.465 ×10

−3

(T = 600 K) = 1.66 ×10

(600 K)

3/2

exp



−

0.66 eV

2 (8.617 ×10

−5

eV/K) (600 K)



−3

4.124 ×10

−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

1.08×10

= 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

1.54×10

26.8 times higher than that in Si.

(b) Since phosphorus is a Group V element, it is a donor, meaning N

= 5 × 10

−3

. For an

n-type material, we have:

n = N

5 × 10

−3

p(T = 300 K) =

(T = 300 K)]

1.215 × 10

−3

p(T = 600 K) =

(T = 600 K)]

3.401 × 10

−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.

tot

= I

drift

= q(nµ

+ pµ

)AE

n = 10

−3

p = n

/n = (1.08 × 10

)

/10

= 1.17 × 10

−3

= 1350 cm

/V ·s

= 480 c m

/V ·s

E = V/d =

1 V

0.1 µm

= 10

V/cm

A = 0.05 µm × 0.05 µm

= 2.5 × 10

−11

Since nµ

≫ pµ

, we can write

tot

≈ qnµ

54.1 µA

(b) All of the parameters are the same except n

, which means we must re-calculate p.

(T = 400 K) = 3.657 ×10

−3

p = n

/n = 1.3 37 × 1 0

−3

Since nµ

≫ pµ

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

tot

≈ qnµ

54.1 µA

2.4 (a) From Problem 1, we can c alculate n

for Ge.

(T = 300 K) = 2.465 × 10

−3

tot

= q(nµ

+ pµ

)AE

n = 10

−3

p = n

/n = 6.0 76 × 10

−3

= 3900 cm

/V ·s

= 1900 cm

/V ·s

E = V/d =

1 V

0.1 µm

= 10

V/cm

A = 0.05 µm × 0.05 µm

= 2.5 × 10

−11

Since nµ

≫ pµ

, we can write

tot

≈ qnµ

156 µA

(b) All of the parameters are the same except n

, which means we must re-calculate p.

(T = 400 K) = 9.230 × 10

−3

p = n

/n = 8.5 20 × 10

−3

Since nµ

≫ pµ

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

tot

≈ qnµ

156 µA

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