1. EXAMPLE 8.1
OBJECTIVE
To determine the time behavior of excess carriers as a semiconductor returns to thermal equilibrium.
Consider an infinitely large, homogeneous n-type semiconductor with zero applied electric field. Assume that at time t = 0, a uniform concentration of excess carriers exists in the crystal, but assume that g = 0 for t > 0. If we assume that the concentration of excess carriers is much smaller than the thermal-equilibrium electron concentration, then the low-injection condition applies. Calculate the excess carrier concentration as a function of time for t  0.
Solution
For the n-type semiconductor, we need to consider the ambipolar transport equation for the minority-carrier holes, which was given by Equation (8.41). The equation is
We are assuming a uniform concentration of excess holes so that 2(p)/x2 =  (p)/x = 0. For t > 0, we are also assuming that g = 0. Equation (8.41) reduces to
(8.42)

2. EXAMPLE 8.1 Solution (8.43) Numerical calculation (8.44)
Since there is no spatial variation, the total time derivative may be used. At low injection, the minority-carrier hole lifetime, p0, is a constant. The solution to Equation (8.42) is
where p(0) is the uniform concentration of excess carriers that exists at time t = 0. The concentration of excess holes decays exponentially with time, with a time constant equal to the minority-carrier hole lifetime.
From the charge-neutrality condition, we have that n = p, so the excess electron concentration is given by
Numerical calculation
Consider n-type gallium arsenide doped at Nd = 1016 cm-3. Assume that 1014 electron-hole pairs per cm3 have been created at t = 0, and assume the minority-carrier hole lifetime is p0 = 10 ns.
We can note that p(0) << n0, so low injection applies. Then from Equation (8.43) we can write
The excess hole and excess electron concentrations will decay to 1/e of their initial value in 10 ns.
(8.43)
(8.44)

3. EXAMPLE 8.1
Comment
The excess electrons and holes recombine at the rate determined by the excess minority-carrier hole lifetime in the n-type semiconductor. This decay in the minority-carrier concentration is shown in Figure 8.3 for the 10-ns lifetime given in Example 8.1 (solid curve). Also shown are curves for lifetimes of 5 ns and 20 ns.

4. EXAMPLE 8.2 OBJECTIVE Numerical calculation (8.45) (8.46) Solution
To determine the time dependence of excess carriers in reaching a steady-state condition.
Again consider an infinitely large, homogeneous n-type semiconductor with a zero applied electric field. Assume that, for t > 0, the semiconductor is in thermal equilibrium and that, for t  0, a uniform generation rate exists in the crystal. Calculate the excess carrier concentration as a function of time assuming the condition of low injection.
Solution
The condition of a uniform generation rate and a homogeneous semiconductor again implies that 2(p)/x2 = (p)/x = 0 in Equation (8.41). The equation, for this case, reduces to
The solution to this differential equation is
Numerical calculation
Consider n-type silicon at T = 300 K doped at Nd = 2  1016 cm3. Assume that p0 = 10-7 s and g = 5  1021 cm3 s1. From Equation (8.46) we can write
(8.45)
(8.46)

5. EXAMPLE 8.2
Comment
We can note that for t  , we create a steady-state excess hole and electron concentration of 5  1014 cm-3. We can note that p << n0 so that low injection is valid. The exponential and steady-state behavior in the excess minority-carrier concentration is shown in Figure 8.4 for
the lifetime of 10-7 s (solid curve.) Also shown are curves for lifetimes of 0.5  10-7 s and 2  10-7 s. Note that, as the lifetime changes, the steady-state value of the excess carrier concentration and the time to teach steady state changes.

6. 0.5 1 3 2 4 5
1015
 = 2  10-7 s
 = 10-7 s
 = 0.5  10-7 s
Time (10-7s)
p(t) (cm3)
Figure 8.4 Exponential and steady-state behavior of the excess carrier concentration as
described in Example 8.2 for a lifetime of  = 10-7 s. Also shown, for comparison
purposes, are the curves for lifetimes of 0.5  10-7 s and 2  10-7 s.

7. EXAMPLE 8.3
OBJECTIVE
To determine the steady-state spatial dependence of the excess carrier concentration.
Consider a p-type semiconductor that is homogeneous and infinite in extent. Assume a zero applied electric field. For a one-dimensional crystal, assume that excess carriers are being generated at x = 0 only, as indicated in Figure 8.5. The excess carriers being generated at x = 0 will begin diffusing in both the +x and x directions. Calculate the steady-state excess carrier concentration as a function of x.
Solution
The ambipolar transport equation for excess minority-carrier electrons was given by Equation (8.40), and is written as g
x = 0 x
Figure 8.5 Steady-state generation rate at x = 0.

8. EXAMPLE 8.3
Solution
From our assumptions, we have  = 0, g = 0 for x  0, and (n)/t = 0 for steady state. Assuming a one-dimensional crystal, Equation (8.40) reduces to
Dividing by the diffusion coefficient, Equation (8.47) can be written as
where we have defined Ln2 = Dnn0. The parameter Ln has the unit of length and is called the minority-carrier electron diffusion length. The general solution to Equation (8.48) is
As the minority-carrier electrons diffuse away from x = 0, they will recombine with the majority-carrier holes. The minority-carrier electron concentration will then decay toward zero at both x = + and x = . These boundary conditions mean that B  0 for x > 0 and A  0 for x < 0. The solution to Equation (8.48) can then be written as
(8.47)
(8.48)
(8.49)
(8.50a)

9. EXAMPLE 8.3 Solution Numerical calculation and
Where n(0) is the value of the excess electron concentration at x = 0. the steady-state excess electron concentration decays exponentially with distance away from the source at x = 0.
Numerical calculation
Consider p-type silicon at T = 300 K doped at Na = 5  1016 cm-3. Assume that n0 = 5  10-7 s, Dn = 25 cm2/s, and n(0) = 1015 cm-3.
The minority-carrier diffusion length is
Then for x  0, we have
Comment
Figure 8.6 shows the results of this example. We can note that the steady-state excess concentration decays to 1/e of its value at x = Ln = 35.4 m. We can also note that the majority-carrier (holes) concentration barely changes under this low-injection condition. However, the minority-carrier (electron) concentration may change by orders of magnitude.
(8.50b)

10. 35.4m
35.4m
x = 0
4.5103
51016
5.11016
1015 n0 p0
Carrier
Concentration(cm-3)
(log scale)
p0+n(0)
n0+n(0) x
Figure 8.6 steady-state electron and hole concentrations for the case when excess
electrons and holes are generated at x = 0.

11. EXAMPLE 8.4
OBJECTIVE
To determine both the time dependence and spatial dependence of the excess carrier concentration.
Assume that a finite number of electron-hole pairs is generated instantaneously at time t = 0 and x = 0, but assume g = 0 for t > 0. Assume we have an n-type semiconductor with a constant applied electric field equal to 0, which is applied in the +x direction. Calculate the excess carrier concentration as a function of x and t.
Solution
The one-dimensional ambipolar transport equation for the minority-carrier holes can be written from Equation (8.41) as
The solution to this partial differential equation is of the from
By substituting Equation (8.52) into Equation (8.51), we are left with the partial differential equation
(8.51)
(8.52)
(8.53)

12. EXAMPLE 8.4 Solution Comment
Equation (8.53) is normally solved using Laplace transform techniques. The solution, without going through the mathematical details, is
The total solution, from Equations (8.52) and (8.54), for the excess minority-carrier hole concentration is
Comment
We could show that Equation (8.55) is a solution by direct substitution back into the partial differential equation, Equation (8.51).
(8.54)
(8.55)

13.   enNd = (1.6  1019)(1200)(1016) = 1.92 (-cm)-1
EXAMPLE 8.5
OBJECTIVE
To calculate the dielectric relaxation time constant for a particular semiconductor.
Consider n-type silicon at T = 300 K with an impurity concentration of Nd = 1016 cm-3.
Solution
The conductivity is found as
  enNd = (1.6  1019)(1200)(1016) = 1.92 (-cm)-1
where the value of mobility is the approximate value found from Figure 4.3. The permittivity of silicon is
 = r0 = (11.7)(8.85  1014) F/cm
The dielectric relaxation time constant is then or
d = ps
Comment
Equation (8.62) then predicts that in approximately four time constants, or in approximately 2 ps, the net charge density is essentially zero; that is , quasi-neutrality has been achieved. Since the continuity equation, Equation (8.58), used in this analysis does not contain any generation or recombination terms, the initial positive charge is then neutralized by pulling in electrons from the bulk n-type material to create excess electrons. This process occurs very quickly compared to the normal excess carrier lifetimes of approximately 0.1 s. The condition of quasi-charge neutrality is then justified.

14. EXAMPLE 8.6 Comment OBJECTIVE Solution
To calculate the quasi-Fermi energy levels.
Consider an n-type semiconductor at T = 300 K with carrier concentrations of n0 = 1015 cm-3, ni = 1010 cm-3, and p0 = 1015 cm-3. In nonequilibrium, assume that the excess carrier concentrations are n = p = 1016 cm-3.
Solution
The Fermi level for thermal equilibrium can be determined from Equation (8.69a). We have
We can use Equation (8.70a) to determine the quasi-Fermi level for electrons in nonequilibrium. We can write
Equation (8.70b) can be used to calculate the quasi-Fermi level for holes in nonequilibrium. We can write
Comment
We may note that the quasi-Fermi level for electrons is above EFi while the quasi-Fermi level for holes is below EFi.

15. EXAMPLE 8.7 Comment OBJECTIVE Solution
To determine the excess carrier lifetime in an intrinsic semiconductor.
If we substitute the definitions of excess carrier lifetimes from Equations (8.75) and (8.76) into Equation (8.71), the recombination rate can be written as
Consider an intrinsic semiconductor containing excess carriers. Then n = ni + n and p = ni + n. Also assume that n = p = ni.
Solution
Equation (8.77) now becomes
If we also assume very low injection, so that n << 2ni, then we can write
where  is the excess carrier lifetime. We see that  = p0 + n0 in the intrinsic material.
Comment
The excess carrier lifetime increases as we change from an extrinsic to an intrinsic semiconductor.