Homework chapter 8
PROBLEMS
(Note: Use the semiconductor parameters listed in Appendix B if they are not
specifically given in a problem. Assume T = 300 K.)
Section 8.1 Carrier Generation and Recombination
8.1 Consider a semiconductor in which n0 = 1015 cm-3 and ni = 1010 cm-3. Assume that
the excess carrier lifetime is 10-6 s. Determine the electron-hole recombination rate
if the excess hole concentration is p = 5  1013 cm-3.
Section 8.2 Analysis of Excess Carriers
8.3 Derive Equation (8.16) from Equations (8.7) and (8.9).
8.5 Consider a one-dimensional hole flux as shown in Figure 8.1. If the generation
rate of holes in this differential volume is gp = 0 cm-3-s-1 and the recombination
rate is 2  1019 cm-3-s-1, what must be the gradient in the particle current density
to maintain a steady-state hole concentration?
Section 8.3 Ambipolar Transport
8.7 A sample of Ge at T = 300 K has a uniform donor concentration of 2  1013
cm-3. The excess carrier lifetime is found to be p0 = 24 s. determine the
ambipolar diffusion coefficient and the ambipolar mobility. What are the
electron and hole lifetimes?
8.9 Light is incident on a silicon sample starting at t = 0 and generating excess
carriers uniformly throughout the silicon for t > 0. The generation rate is g = 5
 1021 cm-3s-1. The silicon (T = 300 K) is n type with Nd = 5  1016 cm-3 and Na =
0. Let ni = 1.5  1010 cm-3, n0 = 10-6 s, and p0 = 10-7 s. Also let n = 1000
cm2/V-s and p = 420 cm2/V-s. Determine the conductivity of the silicon as a
function of time for t  0.
8.11 A silicon sample at T = 300 K is n type with Nd = 5  1016 cm-3 and Na = 0. The
sample has a length of 0.1 cm and a cross-sectional area of 10-4 cm2. A voltage
of 5 V is applied between the ends of the sample. For t < 0, the sample has been
illuminated with light, producing an excess carrier generation rate of g = 5 
1021 cm-3-s-1 uniformly throughout the entire silicon. The minority-carrier
lifetime is p0 = 3  10-7 s. At t = 0, the light is turned off. Derive the expression
for the current in the sample as a function of time t  0. (Neglect surface effects.)
8.13 In a silicon semiconductor material at T = 300 K, the doping concentrations are
Nd = 1015 cm-3 and Na = 0. The equilibrium recombination rate is Rp0 = 1011
cm-3-s-1. A uniform generation rate produces an excess carrier concentration of
n = p = 1014 cm-3. (a) By what factor does the total recombination rate
increase? (b) What is the excess carrier lifetime?
8.15 A semiconductor has the following properties:
Dn = 25 cm2/s
n0 = 10-6s
Dp = 10 cm2/s
p0 = 10-7s
The semiconductor is a homogeneous, p-type (Na = 1017 cm-3) material in
thermal equilibrium for t  0. At t = 0, an external source is turned on, which
produces excess carriers uniformly at the rate of g = 1020 cm-3-s-1. At t = 2  10-6
s, the external source is turned off. (a) Derive the expression for the excess
electron concentration as a function of time for 0  t . (b) Determine the value
of the excess electron concentration at (i) t = 0, (ii) t = 2  10-6 s, and (iii) t = .
(c) Plot the excess electron concentration as a function of time.
8.17 The x = 0 end of an Na = 1  1014 cm-3 doped semi-infinite (x  0) bar of silicon
maintained at T = 300 K is attached to a “minority carrier digester” which makes
np = 0 at x = 0 (np is the minority-carrier electron concentration in a p-type
semiconductor). The electric field is zero. (a) Determine the thermal-equilibrium
values of np0 and pp0. (b) What is the excess minority-carrier concentration at x =
0? (c) Derive the expression for the steady-state excess minority-carrier
concentration as a function of x.
*8.19 Consider an n-type silicon sample. Excess carriers are generated at x = 0 such
as shown in Figure 8.5. A constant electric field 0 is applied in the +x
direction. Show that the steady-state excess carrier concentration is given by
p(x) = A exp(s_x)
x>0
and
p(x) = A exp(s+x)
x<0
where
s 

1
  1  2
Lp
and

 p L p 0
2D p

8.21 Consider the semiconductor described in Problem 8.16. Assume a constant
electric field 0 is applied in the +x direction. (a) Derive the expression for the
steady-state excess-electron concentration. (Assume the solution is of the form
eax.) (b) Plot n verzuz x for (i) 0 = 0 and (ii) 0 = 12 V/cm. (c) Explain the
general characteristics of the two curves plotted in part (b).
*8.23 Consider the n-type semiconductor shown in Figure P.8.23. Illumination
produces a constant exce4ss carrier generation rate, G0, in the region L < x <
+L. Assume that the minority-carrier lifetime is infinite and assume that the
excess minority-carrier hole concentration is zero at x = 3L and at x = +3L.
Find the steady-state excess minority-carrier concentration versus x, for the case
of low injection and for zero applied electric field.
Illumination
3L
L
0
L
+3L
x
Figure 8.23 Figure for Problem 8.23.
8.25 Consider the function f(x, t) = (4Dt)1/2exp(x2/4Dt). (a) Show that this
function is a solution to the differential equation D 2 f x 2   f t . (b) Show
that the integral of the function f(x, t) over x from  to + is unity for all
values of time. (c) Show that this function approaches a  function as t
approaches zero.
Section 8.4 Quasi-Fermi Energy Levels
8.27 An n-type silicon sample with Nd = 1016 cm-3 is steadily illuminated such that g
= 1021 cm-3-s-1. If n0 = p0 = 10-6 s, calculate the position of the quasi-Fermi
levels for electrons and holes with respect to the intrinsic level (assume that ni =
1.5  1010 cm-3). Plot these levels on an energy-band diagram.
8.29 Consider an n-type gallium arsenide semiconductor at T = 300 K doped at Nd = 5
 1016 cm-3. (a) Determine EFn  EF if the excess carrier concentration is 0.1 Nd.
(b) Determine EFi  EFp.
8.31 Consider p-type silicon at T = 300 K doped to Na = 5  1014 cm-3. Assume excess
carriers are present and assume that EF  EFp = (0.01)kT. (a) Does this condition
correspond to low infection? Why or why not? (b) Determine EFn  EFi.
8.33 For a p-type silicon material doped at Na = 1016 cm-3, plot EFn  EF versus n
over the range 0  n  1014 cm-3. Use a log scale for n.
Section 8.5 Excess carrier Lifetime
8.35 Again consider Equation (8.71) and the definitions of p0 and n0 given by
Equations (8.75) and (8.76). Let p0 = 107 s and n0 = 5  107 s. Also let n = p
= ni = 1010 cm-3. Assume very low injection so that n << ni. Calculate R/n for a
semiconductor which is (a) n type (n0 >> p0), (b) intrinsic (n0 = p0 = ni), and (c)
p type (p0 >> n0).