1000 Solved Problems in Modern Physics

310 5 Solid State Physics

5.39 The number of silicon atoms/m^3

n=

N 0 d

A

=

6. 02 × 1026 × 2 , 420

28

= 0. 52 × 1029

Let x be the fraction of impurity atom (donor). The general expression for the

conductivity is

σ=nneμn+npeμp

wherennandnpare the densities of the negative and positive charge carriers.

Becausennnp,

σ∼=nneμn=xnpeμn

x=

σ

npeμn

=

1. 08

0. 52 × 1029 × 1. 6 × 10 −^19 × 0. 13

=

9. 985

1010

or 1 part in 10^9.

Note that in normal silicon the conductivity is of the order of 10−^4 (Ω−m)−^1.

A small fraction of doping (10−^9 ) has dramatically increased the value by four

orders of magnitude.

5.40 ne=(4.^83 ×^1021 )T^3 /^2 e−Eg/^2 kTe/m^3

nGe

nSi

=e(ESi−EGe)/^2 kT

kT=

1. 38 × 10 −^23 × 400

1. 6 × 10 −^19

= 0 .0345 eV

nGe

nSi

=e(1.^14 −^0 .7)/(2×^0 .0345)= 588

5.41 kT=

1. 38 × 10 −^23 × 300

1. 6 × 10 −^19

= 0 .0259 eV

nC

nGe

=e−(EGe−EC)/^2 KT=e−(5.^33 −07)/^0.^052 =e−^89 ≈ 2. 2 × 10 −^39

5.42 I=I 0 [exp(eV/kT)−1]
whereI 0 is the forward bias saturation current.

I= 8 × 10 −^11

[

exp

(

0. 5 × 1. 6 × 10 −^19

1. 38 × 10 −^23 × 300

)

− 1

]

= 19. 7 × 10 −^3 A= 19 .7mA

5.43 W=

[

2  0 r

e

(V 0 −Vb)

(

1

Na

+

1

Nd

)] 1 / 2

whereris the relative permittivity,Vbis the bias voltage applied to the junc-

tion (hereVb=0),NaandNdare carrier concentrations in n-type and p-type

respectively.

W=

[

2 × 8. 85 × 10 −^12 × 16 × 0. 8

1. 6 × 10 −^19

(

1

1 × 1023

+

1

2 × 1022

)] 1 / 2

= 0. 29 μm