5.3 Solutions
311
5.44 I = I 0 [exp(eV/kT ) − 1]
(1)
1
r e
=
dI
dV
=
eI 0
kT
exp(eV /kT )
( 2 )
But exp (eV/kT ) 1. Therefore
1
r e
=
eI
kT
or r e =
kT
eI
=
1.38 × 10
−23
× 300
1.6 × 10 −19 I
=
25.875 × 10
−3
I
If I is in milliamp.
r e ≈
26
I
(forward bias)
(3)
For the reversed bias we note from (2)
1
r e
=
dI
dV
=
e
kT
I 0 exp (eV/kT ) = 0
For V ≤ −4kT /e, r e → ∞. (reverse bias)
(4)
5.45 For a semiconductor in equilibrium the product of n(= N d ) and p(= N a ) is
equal to n
2
i , the square of the intrinsic concentration.
n × p = n
2
i
p =
n
2
i
n
=
(1.6 × 10
16 )
2
8 × 10 21 = 3.2 × 10
10 m
−3
5.3.5 Superconductor
5.46 λ =
1241
E(eV)
nm =
1, 241
2.73 × 10 −3 = 4.546 × 10
5 nm
5.47 E g =
1241
λ(nm)
=
1, 241
1.08 × 10 6 = 1.15 × 10
−3 eV
5.48 f =
2eV
h
=
2(1.602 × 10
−19 )(1.5 × 10
−6 )
6.626 × 10 −34
= 7.253×10
8 Hz = 0.7253 GHz
5.49 E g = 3.53kT c = (3.53)
(1.38 × 10
−23 )
1.6 × 10 −19 (3.4) = 1.035 × 10
−3 eV
5.50 T c (B) = T c
1 −
B
B c
1/2
2.0 = 7.19
1 −
0.074
B c
1/2
Solving for B c , we get B c = 0.079 T .