半导体物理与器件第三版(尼曼)06章答案

+( g − R ) =
∂(δn )
µ n µ p ( p − n) µ nn + µ p p
∂(δn ) ∂t Q.E.D.
+ gp − and
τp
=
Then we have D ′∇ (δn ) + µ ′Ε • ∇(δn ) + ( g − R ) =
2
(2) Dn ∇ (δn ) + µ n Ε • ∇(δn ) + n∇ • Ε
n O p O n p n p
a
f
n = 3.6 x10 cm Also p= ni
2
In steady-state, δp = g ′τ So that
3.6 x10 n We have µ n = 3900 , µ p = 1900 Dn = 101 , D p = 49.2 Now D′ = Dn D p (n + p )
FG −t IJ Hτ K
pO
b
2
At t = 0 , δp = 0 so that 0 = g ′τ pO + A ⇒ A = − g ′τ pO Then
µ ′ = −868 cm / V − s
Now n p 3.6 x10
13
τn
=
τp
⇒
τn
=
1.6 x10
13
24 µs
which yields τ n = 54 µs 6.10 σ = eµ n n + eµ p p With excess carriers present n = nO + δn and p = pO + δp For an n-type semiconductor, we can write δn = δp ≡ δp
= or
g (101)b3.6 x10 g + (49.2)b1.6 x10 g
(101)(49.2) 3.6 x1013 + 1.6 x1013
13 13
Dn n + Dp p
b
(δp) H = A exp
FG −t IJ Hτ K
pO
and the particular solution is (δp) P = g ′τ pO so that the total solution is
19 −3 −1
G= (c)
2.25 x10 20 x10
4
−6
⇒ G = 1125 . x10 cm s
9 9 −3 −1
−3
−1
δp τ pO
=
5 x10 10
13
−6
R = G = 1125 . x10 cm s 6.4 (a) E = hν = or E = 3.15 x10 photon. Now hc
∂(δn ) ∂t
=
∂(δp ) ∂t
Also gn = g p ≡ g , p
Dp ∇ p − µ p (Ε • ∇p + p∇ • Ε )
2
τp
=
n
τn
≡R
+gp −
p
τp
=
∂p ∂t
Then we can write (1) Dp ∇ (δn) − µ p Ε • ∇(δn) + p∇ • Ε
+
bµ nD + µ pD g∇ (δn)
2 p p n
= g p − R p = 10 − 2 x10 ⇒
20 19
+ µ n µ p ( p − n )Ε • ∇(δn ) + µ n n + µ p p ( g − R) = µ n n + µ p p
dFp dx
+
= 8 x10 cm s
19
2
+g − R = 6.6 From Equation [6.18] ∂p p + = −∇ • Fp + g p − τp ∂t For steady-state, Then
+
∂(δn ) ∂t
and (2) Dn ∇ (δn ) + µ n Ε • ∇(δn ) + n∇ • Ε
2
∂p ∂t
=0
b2.4 x10 g =
13
∆σ = e µ n + µ p g ′τ pO
= 1.6 x10 cm
13 −3
b
gb g
6.11 n-type, so that minority carriers are holes. Uniform generation throughout the sample means we have ∂(δp ) δp = g′ − ∂t τ pO Homogeneous solution is of the form
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Chapter 6 Problem Solutions
Chapter 6
Problem Solutions
6.1 n-type semiconductor, low-injection so that R′ = or R ′ = 5 x10 cm s
δp = g ′τ pO 1 − exp
pO n O p O n p n O n p −19 16 −19 21 −7 pO
LM N
Then
67
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual
Then σ = eµ n nO + δp + eµ p pO + δp
n=
Nd 2
13
+
= 10 + or
13
F N I +n H2K b10 g + b2.4 x10 g
2 d 2 i 13 2 13 −3 13 2
2
a f or σ = eµ n + eµ p + eb µ + µ g(δp) so ∆σ = eb µ + µ g(δp)
6.2 (a) RnO = and nO = Then RnO = (b) Rn = so ∆Rn = Rn − RnO = 5 x10 − 5 x10 ⇒
18 10
λ
−19
= J
b6.625x10 gb3x10 g
−34 8
6300 x10
−10பைடு நூலகம்
nO
This is the energy of 1
τ nO
Chapter 6 Problem Solutions
∇ • Fp = µ p (Ε • ∇p + p∇ • Ε ) − Dp ∇ p
+ 2
By charge neutrality δn = δp ≡ δn ⇒ ∇(δn) = ∇(δp) and ∇ (δn ) = ∇ (δp ) and
2 2
Then ∂p p 2 = − µ p (Ε • ∇p + p∇ • Ε ) + D p ∇ p + g p − τp ∂t We can then write
Chapter 6 Problem Solutions
σ = 8 + 0.114 1 − exp
where τ pO = 10 6.12 n-type GaAs:
−7
LM N
FG −t IJ OP Hτ KQ
pO
6.14 (a) p-type GaAs, Dn ∇ (δn ) + µ n Ε • ∇(δn ) + g ′ −
n p n p n p
b
g
g
b
g ∂(∂δtn)
= −2 x10 cm s
19
−3
−1
6.8 We have the continuity equations (1) Dp ∇ (δp) − µ p Ε • ∇(δp) + p∇ • Ε
2
∂t Define µ n nD p + µ p pDn Dn Dp (n + p ) D′ = = µ nn + µ p p Dn n + D p p p ∂(δp ) ∂t and µ ′ =
2
+ gn −
n
τn
=
∂(δn ) ∂t
66
Semiconductor Physics and Devices: Basic Principles, 3rd edition Solutions Manual 6.9 For Ge: T = 300 K , ni = 2.4 x10 cm
13 −3
Chapter 6 Problem Solutions
ni
2
pO
b10 g =
10
2
10
16
= 10 cm
4
−3
1 W = 1 J / s ⇒ 3.17 x10
18
photons/s
+3
Volume = (1)(0.1) = 0.1 cm Then
10 −3 −1
10
4 −7
2 x10
⇒ RnO = 5 x10 cm s 10
12 −7
g=
317 . x10 0.1
−3
−1
b
Divide by µ n n + µ p p , then