Week5HW S15 Solutions  nanohub.orgWeek5HW_S15_Solutions.pdf · HW5)Solutions(continued):) % 3d)%...
Embed Size (px)
Transcript of Week5HW S15 Solutions  nanohub.orgWeek5HW_S15_Solutions.pdf · HW5)Solutions(continued):) % 3d)%...

Mark Lundstrom Spring 2015
ECE305 Spring 2015 1
SOLUTIONS: ECE 305 Homework: Week 5
Mark Lundstrom Purdue University
The following problems concern the Minority Carrier Diffusion Equation (MCDE) for electrons:
nt
= Dnd 2ndx2
n n
+GL
For all the following problems, assume ptype silicon at room temperature, uniformly doped with N A = 10
17 cm3, n = 300 cm2/V sec, n = 10
6 s. From these numbers, we find:
Dn =kBTq
n = 7.8 cm2 s Ln = Dn n = 27.9 m
Unless otherwise stated, these parameters apply to all of the problems below. 1) The sample is uniformly illuminated with light, resulting in an optical generation rate
GL = 1020 cm3 sec1. Find the steadystate excess minority carrier concentration and
the QFLs, Fn and Fp . Assume spatially uniform conditions, and approach the
problem as follows. Solution:
1a) Simplify the Minority Carrier Diffusion Equation for this problem.
Begin with: nt
= Dnd 2ndx2
n n
+GL
Simplify for steadystate: 0 = Dnd 2ndx2
n n
+GL
Simplify for spatially uniform conditions: 0 = 0 n n
+GL
So the simplified MCDE equation is:
n n
+GL = 0
1b) Specify the initial and boundary conditions, as appropriate for this problem. Solution:
Since there is no time dependence, there is no initial condition. Since there is no spatial dependence, there are no boundary conditions.

Mark Lundstrom Spring 2015
ECE305 Spring 2015 2
HW5 Solutions (continued):
1c) Solve the problem. Solution:
In this case the solution is straightforward: n = GL n = 1020 106 = 1014 cm3
Now compute the QFLs: Since we are doped ptype and in low level injection:
p p0 = N A = nieEiFp( ) kBT
Fp = Ei kBT ln
N Ani
= Ei 0.026ln
1017
1010
= Ei 0.41 eV
n n >> n0 = nieFnEi( ) kBT
Fn = Ei + kBT ln
nni
= Ei + 0.026ln
1014
1010
= Ei + 0.24 eV
Note that since we are using the MCDE, we are assuming lowlevel injection. We should check the assumption that n

Mark Lundstrom Spring 2015
ECE305 Spring 2015 3
HW5 Solutions (continued): 2) The sample has been uniformly illuminated with light for a long time. The optical
generation rate is GL = 1020 cm3 sec1. At t = 0, the light is switched off. Find the
excess minority carrier concentration and the QFLs vs. time. Assume spatially uniform conditions, and approach the problem as follows.
2a) Simplify the Minority Carrier Diffusion Equation for this problem.
Solution:
Begin with: nt
= Dnd 2ndx2
n n
+GL
Simplify for spatially uniform conditions with no generation: dndt
= n n
2b) Specify the initial and boundary conditions, as appropriate for this problem. Solution:
Because there is no spatial dependence, there is no need to specify boundary condition. The initial condition is (from prob. 1): n t = 0( ) = 1014 cm3
2c) Solve the problem.
Solution: dndt
= n n
The solution is: n t( ) = Ae t / n
Now use the initial condition: n t = 0( ) = 1014 = A n t( ) = 1014 e t / n

Mark Lundstrom Spring 2015
ECE305 Spring 2015 4
HW5 Solutions (continued): 2d) Provide a sketch of the solution, and explain it in words. Solution:
For the electron QFL: Fn t( ) = Ei + kBT lnn t( ) + n0
ni
Initially, n t( ) >> n0 and n t( ) = n 0( )e t / n , so
Fn t( ) initially drops linearly with time towards EF . 3) The sample is uniformly illuminated with light, resulting in an optical generation rate
GL = 1020 cm3 sec1. The minority carrier lifetime is 1 sec, except for a thin layer (10
nm wide near x = 0 where the lifetime is 0.1 nsec. Find the steady state excess minority carrier concentration and QFLs vs. position. You may assume that the sample extends to x = + . HINT: treat the thin layer at the surface as a boundary condition do not try to resolve n x( ) inside this thin layer. Approach the problem as follows.
3a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution:
Begin with: nt
= Dnd 2ndx2
n n
+GL
Simplify for steadystate conditions: 0 = Dnd 2ndx2
n n
+GL
The simplified MCDE equation is:
Dnd 2ndx2
n n
+GL = 0 d 2ndx2
nLn2 +
GLDn
= 0 Ln = Dn n

Mark Lundstrom Spring 2015
ECE305 Spring 2015 5
HW5 Solutions (continued): d 2ndx2
nLn2 +
GLDn
= 0 where Ln = Dn n is the minority carrier diffusion
length. 3b) Specify the initial and boundary conditions as appropriate for this problem. Solution:
Since this is a steadystate problem, there is no initial condition. As x , we have a uniform semiconductor with a uniform generation rate. In a uniform semiconductor under illumination, n = GL n (recall prob. 1)), so
n x ( ) = GL n In the thin layer at the surface, the total number of eh pairs recombining per cm2 per second is the recombination rate per cm3 per sec, which is
R = n 0( ) S cm3s1 , times the thickness of the thin layer at the surface, x in cm. If we multiply these two quantities, we get the total number of minority carriers recombining per cm2 per s in the surface layer, which we will call RS .
RS = R x( )
0
x
dx n 0( ) S
x cm2s1 .
Rearranging this equation, we can write
RS =
x S
n 0( ) = SFn 0( ) cm2s1 where:
SF
x S
cm/s
is a quantity with the units of velocity. In practice, we usually dont know the thickness of the lowlifetime layer at the surface or the lifetime in this layer, so instead, we just specify the front surface recombination velocity. Typically,
0 < SF 107 cm/s . For this specific problem,
SF =
x S
= 106 cm
1010 s= 104 cm/s
The surface recombination velocity is simply a way to specify the strength of the recombination rate (in cm2s1) at the surface:
RS =
x S
n 0( ) = SFn 0( ) cm2s1

Mark Lundstrom Spring 2015
ECE305 Spring 2015 6
HW5 Solutions (continued):
In steadystate, carriers must diffuse to the surface at the same rate that they are recombining there so that the excess minority carrier concentration at the surface stays constant with time. The diffusion current of electrons at the surface is
Jn = qDn
dndx
A/cm2 .
The flux of electrons in the +x direction is
Jnq
= Dndndx
cm2s1
The flux of electrons in the x direction (to the surface where they are recombining) is:
Jnq
= +Dn
dndx
cm2s1
In steadystate, this flux of electrons flows to the surface at exactly the same rate that they recombine at the surface, so the boundary condition is
+Dn
dndx x=0
= RS = SFn 0( ) Note: Specifying surface recombination by just giving the surface
recombination velocity not the lifetime and thickness of the thin layer at the surface, is common practice in semiconductor work.
3c) Solve the problem. Solution:
d 2ndx2
nLn2 +
GLDn
= 0
To solve the homogeneous problem first, we set GL = 0. d 2ndx2
nLn2 = 0 The solution is n x( )= Ae x/Ln + Be+ x/Ln
Now solve for a particular solution by letting x + where everything is uniform:
nLn2 +
GLDn
= 0 The solution is: n = Ln2
DnGL = GL n
Add the two solutions: n x( )= Ae x/Ln + Be+ x/Ln +GL n

Mark Lundstrom Spring 2015
ECE305 Spring 2015 7
HW5 Solutions (continued):
To satisfy the first boundary condition as x , B = 0. Now consider the boundary condition at x = 0 :
+Dn
dndx x=0
= DnLn
A = SFn 0( ) = SF A+GL n( )
A =
SFGL nDn Ln + SF
= GL n
1+ Dn Ln( ) SF
n x( ) = GL n 1 e x / Ln
1+ Dn Ln( ) SF
Check some limits. i) SF = 0 cm/s, which implies that there is no recombination at the surface. Then we find: n x( ) = GL n , which make sense, since we have spatial uniformity.
ii) SF . Strong recombination at the surface should force n x = 0( ) = 0 ,
but in the bulk we should still have n x( ) = GL n . The transition from 0 to a finite value in the bulk should take a diffusion length or two.
From the solution:
n x( ) = GL n 1 e
x / Ln
1+ Dn Ln( ) SF
For SF , we find
n x( ) GL n 1 e x/ Ln
which behaves as expected.

Mark Lundstrom Spring 2015
ECE305 Spring 2015 8
HW5 Solutions (continued):
3d) Provide a sketch of the solution, and explain it in words. Solution:
The concentration is GL n in the bulk, but less at the surface, because of surface recombination. The transition from the surface to the bulk takes place over a distance that is a few diffusion lengths long.
The hole QFL is constant and almost exactly where the equilibrium Fermi level was, because we are in low level injection (the hole concentration is very, very near its equilibrium value). But the electron QFL is much closer to the conduction band edge. It moves away from EC near the surface, because surface recombination reduces n x( ) near the surface. The variation with position is linear, because n x( ) varies exponentially with position.
Note: Typically, in semiconductor work, these kinds of problems are stated directly in terms of surface recombination velocities and not in terms of very thin layers with low lifetime. The way problem 3) would usually be stated is as follows:

Mark Lundstrom Spring 2015
ECE305 Spring 2015 9
HW5 Solutions (continued): 3) The sample is uniformly illuminated with light, resulting in an optical
generation rate GL = 1020 cm3 sec1. The minority carrier lifetime is 1 sec. At x
= 0 the surface recombination velocity is SF = 104 cm/s . Find the steady state
excess minority carrier concentration and QFLs vs. position. You may assume that the sample extends to x = + . Approach the problem as follows.
3a) Simplify the Minority Carrier Diffusion Equation for this problem. 3b) Specify the initial and boundary conditions, as appropriate for this problem. 3c) Solve the problem. 3d) Provide a sketch of the solution, and explain it in words.
4) The sample is in the dark, but the excess carrier concentration at x = 0 is held constant
at n 0( ) = 1012 cm3. Find the steady state excess minority carrier concentration and QFLs vs. position. You may assume that the sample extends to x = + . Make reasonable approximations, and approach the problem as follows.
4a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution:
Begin with: nt
= Dnd 2ndx2
n n
+GL
Simplify for steadystate: 0 = Dnd 2ndx2
n n
+GL
No generation: GL = 0 the simplified MCDE equation is:
Dnd 2ndx2
n n
= 0 d2ndx2
nDn n
= 0 d2ndx2
nLn2 = 0 Ln = Dn n
d 2ndx2
nLn2 = 0 where Ln = Dn n is the minority carrier diffusion length.
4b) Specify the initial and boundary conditions, as appropriate for this problem.
Solution: Since this is a steadystate problem, there is no initial condition. As x , we expect all of the minority carriers to have recombined, so:
n x +( ) = 0

Mark Lundstrom Spring 2015
ECE305 Spring 2015 10
HW5 Solutions (continued): At the surface, the excess electron concentration is held constant, so
n x = 0( ) = 1012 cm3
4c) Solve the problem.
Solution: d 2ndx2
nLn2 = 0 The solution is n x( )= Ae x/Ln + Be+ x/Ln
To satisfy the first boundary condition in 4b): B = 0. Now consider the second:
n 0( ) = 1012 cm3
n x( ) = n 0( )e x/ Ln = 1012( )e x/ Ln
4d) Provide a sketch of the solution, and explain it in words. Solution:

Mark Lundstrom Spring 2015
ECE305 Spring 2015 11
HW5 Solutions (continued): 5) The sample is in the dark, and the excess carrier concentration at x = 0 is held
constant at n 0( ) = 1012 cm3. Find the steady state excess minority carrier concentration and QFLs vs. position. Assume that the semiconductor is only 5 m long. You may also assume that there is an ideal ohmic contact at x = L = 5m, which enforces equilibrium conditions at all times. Make reasonable approximations, and approach the problem as follows.
5a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution:
Begin with: nt
= Dnd 2ndx2
n n
+GL
Simplify for steadystate: 0 = Dnd 2ndx2
n n
+GL
Generation is zero for this problem: GL = 0 ; The simplified MCDE equation is:
Dnd 2ndx2
n n
= 0
d 2ndx2
nLn2 = 0
Ln = Dn n Since the sample is much thinner than a diffusion length, we can ignore recombination, so d 2ndx2
= 0 .
5b) Specify the initial and boundary conditions, as appropriate for this problem. Solution: Since this is a steadystate problem, there is no initial condition. As x , we expect all of the minority carriers to have recombined, so:
n x = L( ) = 0
At the surface:
n 0( ) = 1012 cm2
5c) Solve the problem. Solution: d 2ndx2
= 0 The general solution is n x( )= Ax + B To satisfy the first boundary condition in 5b): n L( )= AL + B = 0 . A = B L n x( )= B x L + B = B 1 x L( )

Mark Lundstrom Spring 2015
ECE305 Spring 2015 12
HW5 Solutions (continued): Now consider the second boundary condition:
n 0( )= B = 1012 cm3 n x( ) = n 0( ) 1 x L( ) = 1012( ) 1 x L( )
5d) Provide a sketch of the solution, and explain it in words.
6) The sample is in the dark, and the excess carrier concentration at x = 0 is held
constant at n 0( ) = 1012 cm3. Find the steady state excess minority carrier concentration and QFLs vs. position. Assume that the semiconductor is 30 m long. You may also assume that there is an ideal ohmic contact at x = L = 30m, which enforces equilibrium conditions at all times. Make reasonable approximations, and approach the problem as follows.
6a) Simplify the Minority Carrier Diffusion Equation for this problem.
Solution:
Begin with: nt
= Dnd 2ndx2
n n
+GL
Simplify for steadystate and no generation:
Dnd 2ndx2
n n
= 0 d2ndx2
nDn n
= 0 d2ndx2
nLn2 = 0 Ln = Dn n
d 2ndx2
nLn2 = 0 where Ln = Dn n is the minority carrier diffusion length.

Mark Lundstrom Spring 2015
ECE305 Spring 2015 13
HW5 Solutions (continued): 6b) Specify the initial and boundary conditions, as appropriate for this problem.
Solution: Steady state, so no initial conditions are necessary. The boundary conditions are:
n 0( ) = 1012 cm3
n 30m( ) = 0
6c) Solve the problem.
Solution: n x( )= Ae x/Ln + Be+ x/Ln Because the region is about one diffusion length long, we need to retain both solutions. n 0( )= A + B n L = 30 m( )= AeL /Ln + Be+L /Ln = 0 Solve for A and B to find:
A = n 0( )e+L/Ln
eL/Ln e+L/Ln( )
B = n 0( )eL/Ln
eL/Ln e+L/Ln( )
So the solution is:
n x( )= n 0( )eL/Ln e+L/Ln( ) e
xL( )/Ln + e+ xL( )/Ln
n x( )= n 0( ) sinh x L( ) / Ln sinh L / Ln( )

Mark Lundstrom Spring 2015
ECE305 Spring 2015 14
HW5 Solutions (continued):
6d) Provide a sketch of the solution, and explain it in words.
Solution:
The short base result is linear, but in this case, the slope in a little steeper initially and a little shallower at the end. Since the diffusion current is proportional to the slope, this means that inflow greater than outflow. This occurs because some of the electrons that flow in, recombine in the structure, so the same number cannot flow out.