Transcript 投影片 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
 2 p 
p 
p
p 
Dp
  p
 g 

2
x
x
 p0
t
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
d p 
p

t
 p0
(8.42)
EXAMPLE 8.1
 Solution
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
t /  p 0
pt   p0e
(8.43)
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
t /  p 0
nt   p0e
(8.44)
 Numerical calculation
Consider n-type gallium arsenide doped at Nd = 1016 cm-3. Assume that 1014 electronhole 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
14
 t / 10 8
-3
pt   10 e
cm
The excess hole and excess electron concentrations will decay to 1/e of their initial
value in 10 ns.
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.
EXAMPLE 8.2
OBJECTIVE
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
g 
p
p 

 p0
t
The solution to this differential equation is

pt   g  p 0 1  e t /
p0
(8.45)

(8.46)
 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

pt   510 10 1 e
21
7
t /107
 510 1 e
14
t /107
cm
-3
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.
1015
p(t) (cm3)
 = 2  10-7 s
0.5
 = 10-7 s
 = 0.5  10-7 s
0
1
2
3
4
5
Time (10-7s)
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.
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
 2 n 
n 
n
n 
Dn
  n
 g 

2
x
x
 n0
t
g
x=0
x
Figure 8.5 Steady-state generation rate at x = 0.
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
 2 n  n
Dn

0
2
x
 n0
(8.47)
Dividing by the diffusion coefficient, Equation (8.47) can be written as
 2 n 
n
d 2 n  n


 2 0
2
2
x
Dn n 0
dx
Ln
(8.48)
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
nx   Ae x / L  Bex / L
n
n
(8.49)
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
nx   n0e  x / L
n
x0
(8.50a)
EXAMPLE 8.3
 Solution
and
nx   n0e  x / L
n
x0
(8.50b)
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
Ln 
Dn n 0 
25 5 10 7   35.4m
Then for x  0, we have
nx  10 e
15
 Comment
 x / 35.410 4
cm-3
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.
Carrier
Concentration(cm-3)
(log scale)
p0+n(0)
51016
n0+n(0)
5.11016
p0
1015
n0
4.5103
35.4m
x=0
35.4m
x
Figure 8.6 steady-state electron and hole concentrations for the case when excess
electrons and holes are generated at x = 0.
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
 2 p 
p  p
p 
Dp
  p 0


2
x
x
 p0
t
(8.51)
The solution to this partial differential equation is of the from
px, t   px, t e
t /  p 0
(8.52)
By substituting Equation (8.52) into Equation (8.51), we are left with the partial
differential equation
 2 p x, t 
p x, t 
p x, t 
Dp
  p 0

2
x
x
t
(8.53)
EXAMPLE 8.4
 Solution
Equation (8.53) is normally solved using Laplace transform techniques. The solution,
without going through the mathematical details, is
p x, t  
1
4D pt 1/ 2
  x   p 0t 2 
exp

4Dpt




(8.54)
The total solution, from Equations (8.52) and (8.54), for the excess minority-carrier
hole concentration is
p x, t  
e
t / 
p0
4D t 
1/ 2
p
  x   p 0t 2 
exp

4Dpt




(8.55)
 Comment
We could show that Equation (8.55) is a solution by direct substitution back into the
partial differential equation, Equation (8.51).
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
d
or
 Comment


1 1.7 8.8 5 1 014 


 5.3 9  1 013 s

1.9 2
d = 0.539 ps
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.
EXAMPLE 8.6
OBJECTIVE
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
EF  E Fi
 n0
 kT ln
n
 i


  0.2982 eV

We can use Equation (8.70a) to determine the quasi-Fermi level for electrons in nonequilibrium.
We can write
EFn  EFi
 n0  n 

 kT ln
 0.2984 eV


ni


Equation (8.70b) can be used to calculate the quasi-Fermi level for holes in nonequilibrium. We
can write


EFi  EFp  kT ln


 Comment
p0  p

 0.179 eV

ni

We may note that the quasi-Fermi level for electrons is above EFi while the quasi-Fermi level for
holes is below EFi.
EXAMPLE 8.7
OBJECTIVE
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
R
np  n 
2
i
 p 0 n  n   n 0  p  p
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
2nin  n 
R
2ni  n  p 0   n 0 
2
If we also assume very low injection, so that n << 2ni, then we can write
R
n
n

 p0   n0

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.