Exercises Problems Answers Chapter 6

6.3 An n-type silicon sample contains a donor concentration of Nd = 1016 cm-3. The
minority carrier hole lifetime is found to be τp0 = 20 µs. (a) What is the lifetime of
the majority carrier electrons? (b) Determine the thermal-equilibrium generation
rate for electrons and holes in this material. (c) Determine the thermal-equilibrium
recombination rate for electrons and holes in this material.

6.4 (a) A sample of semiconductor has a cross-sectional area of 1 cm2 and a thickness
of 0.1 cm. Determine the number of electron–hole pairs that are generated per unit
volume per unit time by the uniform absorption of 1 watt of light at a wavelength of
6300 Å. Assume each photon creates one electron–hole pair. (b) If the excess minority
carrier lifetime is 10 µs, what is the steady-state excess carrier concentration?

TYU 6.1 Silicon at T = 300 K has been doped with boron atoms to a concentration of
Na = 5 × 1016 cm-3. Excess carriers have been generated in the uniformly doped
material to a concentration of 1015 cm-3. The minority carrier lifetime is 5 µs.
(a) What carrier type is the minority carrier? (b) Assuming g’ = E = 0 for t > 0,
determine the minority carrier concentration for t > 0.

TYU 6.2 Consider silicon with the same parameters as given in TYU 6.1. The material is
in thermal equilibrium for t < 0. At t = 0, a source generating excess carriers is
turned on, producing a generation rate of g’ = 1020 cm-3-s-1. (a) What carrier
type is the minority carrier? (b) Determine the minority carrier concentration for t > 0.
(c) What is the minority carrier concentration as t → ∞?
Exercise Solutions

6.3
(a) Recombination rates are equal
no po
=
τ nO τ pO
n o = N d = 1016 cm −3
n2
po = i =
(
1.5 × 1010 )2

= 2.25 × 10 4 cm −3
no 1016
Then
1016 2.25 × 10 4
=
τ nO 20 × 10 − 6
which yields
τ nO = 8.89 × 10 +6 s
(b) Generation rate = recombination rate
Then
2.25 × 10 4
G= = 1.125 × 10 9 cm −3 s −1
20 × 10 − 6
(c)
R = G = 1.125 × 10 9 cm −3 s −1

6.4

(a) E = hν =
hc
=
(6.625 ×10 )(3 ×10 )
−34 8

λ 6300 × 10 −10

or
E = 3.15 × 10 −19 J; energy of one photon
Now
1 W = 1 J/s ⇒ 3.17 × 1018 photons/s
Volume = (1)(0.1) = 0.1 cm 3
Then
3.17 × 1018
g=
0.1
= 3.17 × 1019 e-h pairs/cm 3 -s
(b)
δn = δp = gτ = (3.17 × 1019 )(10 × 10 −6 )
or
δn = δp = 3.17 × 1014 cm −3

TYU 6.1
(a) p-type; Minority carriers = electrons
 −t 
(b) δn(t ) = δn(0) exp 

 t no 
Then
  −t 
δn(t ) = 1015 exp −6
 cm −3
 5 × 10 
_______________________________________

TYU 6.2
(a) p-type; Minority carriers = electrons
  − t 
(b) δn(t ) = g ′t no 1 − exp 

  t no 

( )(   −t
= 10 20 5 × 10 − 6 1 − exp ) −6


  5 × 10 
  −t 
or δn(t ) = 5 × 1014 1 − exp −6

  5 × 10 
(c) As t → ∞ , δn(∞ ) = 5× 1014 cm −3
_______________________________________

6.10
For Ge: n i = 2.4 × 1013 cm −3
2
Nd N 
no = +  d  + n i2
2  2 
2
 4 × 1013 
=
4 × 1013
+  (
 + 2.4 × 1013
 )
2

2  2 
= 5.124 × 1013 cm −3

po =
n i2
=
(
2.4 × 1013 )
2

= 1.124 × 1013 cm −3
n o 5.124 × 1013
(a) We have:
µ n = 3900 cm 2 /V-s, D n = 101 cm 2 /s
µ p = 1900 cm 2 /V-s, D p = 49.2 cm 2 /s
For very, very low injection,
D n D p (n + p )
D′ =
Dn n + D p p

=
(101)(49.2)(5.124 ×1013 + 1.124 ×1013 )
(101)(5.124 ×1013 ) + (49.2)(1.124 ×1013 )
= 54.2 cm 2 /s
and
µ n µ p ( p − n)
µ′ =
µnn + µ p p

=
(3900)(1900)(1.124 ×1013 − 5.124 ×1013 )
(3900)(5.124 ×1013 ) + (1900)(1.124 ×1013 )
= −1340 cm 2 /V-s
(b) For holes, t pt = t p 0 = 2 × 10 −6 s
 For electrons,
n p
=
t nt t p0
5.124 × 1013 1.124 × 1013
=
t nt 2 × 10 − 6
⇒ t nt = 9.12 × 10 −6 s
_______________________________________

6.11
σ = eµ n n + eµ p p
With excess carriers
n = n o + δn and p = p o + δp
For an n-type semiconductor, we can write
δn = δp ≡ δp
Then
σ = eµ n (n o + δp ) + eµ p ( p o + δp )
or
σ = eµ n n o + eµ p p o + e(µ n + µ p )(δp )
so
(
∆σ = e µ n + µ p (δp ) )
In steady-state, δp = g ′τ pO
So that
∆σ = e(µ n + µ p )(g ′τ pO )
_______________________________________

6.12
(a) p o = N a = 1016 cm −3

no =
n i2
=
(
1.5 × 1010 ) 2

= 2.25 × 10 4 cm −3
po 1016
σ = eµ n (n o + δn ) + eµ p ( p o + δp )
≅ eµ p p o + e µ n + µ p δn ( )
Now δn = δp = g ′t n 0 (1 − e −t / t n 0 )
( )( )
)(
= 8 × 10 20 5 × 10 −7 1 − e −t / t n 0
= 4 × 10 ) cm
14
(1 − e −t / t n 0 −3

Then σ = (1.6 × 10 )(380)(10 ) −19 16

+ (1.6 × 10 )(900 + 380 ) −19

× (4 × 10 )(1 − e ) 14 −t / t n 0

σ = 0.608 + 0.0819(1 − e ) ( Ω -cm) −t / t n 0 −1

(b) (i) σ (0) = 0.608 ( Ω -cm) −1

(ii) σ (∞ ) = 0.690 ( Ω -cm) −1
_______________________________________

6.22
n-type, so we have
d 2 (dp ) d (dp ) dp
Dp − µ pΕo − =0
dx dx τ pO
Assume the solution is of the form
δp = A exp(sx )
Then
d (dp ) d 2 (dp )
= As exp(sx ) , = As 2 exp(sx )
dx dx 2
Substituting into the differential equation
D p As 2 exp(sx ) − µ p Ε o As exp(sx )
A exp(sx )
− =0
τ pO
or
1
Dp s2 − µ pΕo s − =0
τ pO
Dividing by D p , we have
µ pΕo 1
s2 − s− =0
Dp L2p
The solution for s is
 2 
1 µp  µp


 4 
s=  Εo ± Ε + 2 
2 Dp  Dp o  Lp
   
which can be rewritten as
 2 
1  µ p LpΕo  µ p LpΕo 
 + 1
s=  ±  
Lp 2D p  2D p 
   
Define
µ p LpΕo
β≡
2D p
Then
1 
s= β ± 1 + β 2 
L p  
In order that δp = 0 as x → +∞ , use the
minus sign for x > 0 and the plus sign for
x < 0 . Then the solution is
δp = A exp(s − x ) for x > 0
δp = A exp(s + x ) for x < 0
where
 1 
s± = β ± 1 + β 2 
L p  
_______________________________________

Ex 6.7
p o = N a − N d = 1016 − 3 × 1015
= 7 × 1015 cm −3

no =
n i2
=
(
1.5 × 1010 )2

= 3.214 × 10 4 cm −3
po 7 × 1015
(a) In thermal equilibrium,
p 
E Fi − E F = kT ln o 
 ni 
 7 × 1015 
= (0.0259 ) ln 

 1.5 × 10
10

= 0.33808 eV

(b) Quasi-Fermi levels,

 p + δp 
E Fi − E Fp = kT ln o 

 ni 
 7 × 1015 + 4 × 1014 
= (0.0259 ) ln 

 1.5 × 1010 
= 0.33952 eV
 n + δn 
E Fn − E Fi = kT ln o 

 ni 
 3.214 × 10 4 + 4 × 1014 
= (0.0259 ) ln 

 1.5 × 1010 
= 0.26395 eV
_______________________________________

Ex 6.8
n-type; n o = 1015 cm −3 , p o = 2.25 × 10 5 cm −3
δn = δp = 1014 cm −3 ,
τ no = τ po = 5 × 10 −7 s
We have

R=
[(n + δn )( p o + δp ) − n i2
o ]
τ po (n o + δn + n i ) + τ no ( p o + δp + n i )

[( )( ) (
≅ 1015 + 1014 1014 − 1.5 × 1010 )] 2

÷ {(5 × 10 )(10 + 10 ) + (5 × 10 )(10 )}

−7 15 14 −7 14

or
R = 1.83 × 10 20 cm −3 s −1
_______________________________________
6.41
(a) From Equation (6.56)
d 2 (dp ) dp
Dp + g′− =0
dx 2
τ pO
Solution is of the form
−x  
δp = g ′τ pO + A exp  + B exp + x 
 Lp   Lp 
   

At x = +∞ , δp = g ′τ pO so that B = 0 ,
Then
−x
δp = g ′τ pO + A exp 
 Lp 
 
We have
d (dp )
Dp = s(dp )
dx x =0 x =0

We can write
d (dp ) −A
= and (δp ) = g ′τ pO + A
dx x =0 Lp x =0

Then
− AD p
Lp
(
= s g ′τ pO + A )
Solving for A , we find
− sg ′τ pO
A=
Dp
+s
Lp
The excess concentration is then
 s  − x 
δp = g ′τ pO 1 − ⋅ exp 
 (D p Lp + s )  L p 
 
where
L p = D pτ pO = (10)(10 −7 ) = 10 −3 cm
Now
δp = (10 21 )(10 −7 )
  
× 1 −
s
⋅  − x 
( )
exp
−3
 10 10 + s  L p 
 

or
 s  − x 
δp = 1014 1 − ⋅ exp 
10 + s  L p 

4
 
(i) For s = 0 ,
δp = 1014 cm −3
 (ii) For s = 2000 cm/s,
  − x 
δp = 1014 1 − 0.167 exp 
  L p 
 
(iii) For s = ∞ ,
  − x 
δp = 1014 1 − exp 
 
  L p 

(b)
(i) For s = 0 ,
δp(0) = 1014 cm −3
(ii) For s = 2000 cm/s,
δp(0) = 0.833 × 1014 cm −3
(iii) For s = ∞ ,
δp(0) = 0
_______________________________________