1 SOLUTIONS: ECE 305 Homework: Week 5 Mark

Mark Lundstrom 09/29/2014 SOLUTIONS: ECE 305 Homework: Week 5 Mark Lundstrom Purdue University (10/10/14) The following problems concern the Minority Carrier Diffusion Equation (MCDE) for electrons: ! "n
d 2 "n "n
= Dn
#
+ GL
!t
dx 2
$n
For all the following problems, assume silicon at room temperature, uniformly doped with N A = 1017 cm-‐3, µ n = 300 cm2/V sec, ! n = 10"6 s. From these numbers, we find: kT
Dn = B µn = 7.8 cm 2 s Ln = Dn! n = 27.9 µm q
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 cm-‐3 sec-‐1. Find the steady-‐state excess minority carrier concentration and the QFL’s Fn and Fp . Assume spatially uniform conditions, and approach the problem as follows. 1a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ! "n
d 2 "n "n
Begin with: = Dn
#
+ GL !t
dx 2
$n
d 2 !n !n
Simplify for steady-‐state: 0 = Dn
"
+ GL dx 2
#n
Simplify for spatially uniform conditions: 0 = 0 !
"n
+ GL #n
So the simplified MCDE equation is: "n
!
+ GL = 0 #n
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. ECE-‐305 1 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): 1c) Solve the problem. Solution: In this case the solution is straightforward: !n = GL" n = 10 20 # 10 $6 = 1014 cm -3 Now compute the QFLs: Since we are doped p-‐type and in low level injection: (E "F )
p ! p0 = N A = ni e
i
p
k BT
"N %
" 1017 %
Fp = Ei ! k BT ln $ A ' = Ei ! 0.026ln $ 10 ' = Ei ! 0.41 eV # 10 &
# ni &
n ! "n >> n0 = ni e(
Fn # Ei ) k BT
" !n %
" 1014 %
Fn = Ei + k BT ln $ ' = Ei + 0.026ln $ 10 ' = Ei + 0.24 eV # 10 &
# ni &
1d) Provide a sketch of the solution, and explain it in words. Solution: The excess carrier density is constant, independent of position. So are the QFL’s, but they split because we are not in equilibrium. The hole QFL is essentially where the equilibrium Fermi level was, because the hole concentration is virtually unchanged (low-‐level injection). But the electron QFL is much closer to the conduction band because there are orders of magnitude more electrons. 2) The sample has been uniformly illuminated with light for a long time. The optical generation rate is GL = 1020 cm-‐3 sec-‐1. At t = 0, the light is switched off. Find the excess minority carrier concentration and the QFL’s vs. time. Assume spatially uniform conditions, and approach the problem as follows. ECE-‐305 2 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): 2a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ! "n
d 2 "n "n
Begin with: = Dn
#
+ GL !t
dx 2
$n
Simplify for spatially uniform conditions with no generation: d!n
!n
="
dt
#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 cm -3 2c) Solve the problem. Solution: d!n
!n
="
The solution is: !n ( t ) = Ae"t /# n dt
#n
Now use the initial condition: !n ( t = 0 ) = 1014 = A !n ( t ) = 1014 e"t /# n 2d) Provide a sketch of the solution, and explain it in words. Solution: ECE-‐305 3 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): " !n ( t ) + n0 %
For the electron QFL: Fn ( t ) = Ei + k BT ln $
'& ni
#
Initially, !n ( t ) >> n0 and !n ( t ) = !n ( 0 ) e"t /# n , so Fn t initially drops linearly with time towards E F . ()
3) The sample is uniformly illuminated with light, resulting in an optical generation rate GL = 1020 cm-‐3 sec-‐1. The minority carrier lifetime is 1 μsec, 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 QFL’s vs. position. You may assume that the sample extends to x = +! . HINT: Ignore generation in the thin layer at the surface. 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: ! "n
d 2 "n "n
Begin with: = Dn
#
+ GL !t
dx 2
$n
Simplify for steady-‐state conditions: 0 = Dn
d 2 !n !n
"
+ GL dx 2
#n
The simplified MCDE equation is: d 2 !n !n
d 2 !n !n GL
Dn
"
+ GL = 0 " 2 +
= 0 Ln = Dn! n dx 2 # n
dx 2
Ln Dn
d 2 !n !n GL
" 2 +
= 0 where Ln = Dn! n is the minority carrier “diffusion dx 2
Ln Dn
length.” 3b) Specify the initial and boundary conditions, as appropriate for this problem. Solution: Since this is a steady-‐state problem, there is no initial condition. As x ! " , we have a uniform semiconductor with a uniform generation rate. In a uniform semiconductor under illumination, !n = GL" n , so !n ( x " # ) = GL$ n ECE-‐305 4 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): In the thin layer at the surface, the total number of e-‐h pairs recombining per cm2 per second is the recombination rate per cm3 per sec, which is R = !n 0 " S cm -3s-1 , 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 . !x
RS =
"
0
R ( x )dx #
!n ( 0 )
$S
!x cm -2 -s-1 . Rearranging this equation, we can write !x
RS =
!n ( 0 ) = S F !n ( 0 ) cm -2 -s-1 "S
where: "x
SF !
cm/s #S
is a quantity with the units of velocity. In practice, we usually don’t know the thickness of the low-‐lifetime layer at the surface or the lifetime in this layer, so instead, we just specific the front surface recombination velocity. Typically, 0 < S F ! 107 cm/s . For this specific problem, !x 10#6 cm
=
= 104 cm/s #10
"S
10 s
The surface recombination velocity is simply a way to specify the strength of the recombination rate (in cm-‐2-‐s-‐1) at the surface: !x
RS =
!n ( 0 ) = S F !n ( 0 ) cm -2 -s-1 "S
SF =
ECE-‐305 In steady-‐state, 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 d!n
J n = qDn
A/cm 2 . dx
The flux of electrons in the +x direction is Jn
d"n
= ! Dn
cm -2 -s-1 !q
dx
The flux of electrons in the –x direction (to the surface where they are recombining) is 5 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): "J %
d(n
! $ n ' = + Dn
cm -2 -s-1 dx
# !q &
In steady-‐state, this flux of electrons flows to the surface at exactly the same rated that they recombine at the surface, so the boundary condition is + Dn
d!n
= RS = S F !n ( 0 ) dx x=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 2 !n !n GL
" 2 +
=0
dx 2
Ln Dn
Solve the homogeneous problem first GL = 0. d 2 !n !n
" 2 = 0 solution is !n ( x ) = Ae" x/Ln + Be+ x/Ln
2
dx
Ln
Now solve for a particular solution by letting x ! " where everything is uniform: "n GL
L2n
! 2 +
= 0 The solution is: !n = "
GL = GL# n Ln Dn
Dn
Add the two solutions: !n ( x ) = Ae" x/Ln + Be+ x/Ln + GL# n To satisfy the first boundary condition as x ! " , B = 0. Now consider the boundary condition at x = 0 : D
d!n
+ Dn
= " n A = S F !n ( 0 ) = S F ( A + GL# n ) dx x=0
Ln
A= !
S F GL" n
GL" n
=!
Dn Ln + S F
1+ Dn Ln
(
)
SF
# x/ L
$
e n
!n x = GL" n &1#
&% 1+ Dn Ln
()
ECE-‐305 (
)
'
) S F )(
6 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): 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) S F ! " . 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 on the bulk should take a diffusion length or two. From the solution: # x/ L
$
e n
!n x = GL" n &1#
&% 1+ Dn Ln
For S F ! " , we find ()
(
)
'
) S F )(
!n ( x ) " GL# n %&1$ e$ x/ Ln '(
which behaves as expected. 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. ECE-‐305 7 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): 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. 4) The sample is illuminated with light, resulting in an optical generation rate GL = 1024 cm-‐3 sec-‐1, but all of the photons are absorbed in a thin layer (10 nm wide near x = 0). Find the steady state excess minority carrier concentration and QFL’s 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. 4a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ! "n
d 2 "n "n
Begin with: = Dn
#
+ GL !t
dx 2
$n
Simplify for steady-‐state: 0 = Dn
d 2 !n !n
"
+ GL dx 2
#n
Let’s treat the generation in a thin surface layer as a boundary condition, GL = 0 for x > 0. The simplified MCDE equation is: d 2 !n !n
d 2 !n !n
d 2 !n !n
Ln = Dn! n Dn
"
= 0 2 "
= 0 " 2 = 0 dx 2
#n
dx
Dn# n
dx 2
Ln
d 2 !n !n
" 2 = 0 where Ln = Dn! n is the minority carrier diffusion length. dx 2
Ln
ECE-‐305 8 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): 4b) Specify the initial and boundary conditions, as appropriate for this problem. Solution: Since this is a steady-‐state problem, there is no initial condition. As x ! " , we expect all of the minority carriers to have recombined, so: !n ( x " # ) = 0 At the surface, the total number of e-‐h pairs generation per cm2 per second is GS = GL !x = 102410"6 = 1018 cm -2s-1 . In steady-‐state, these must diffuse away at the same rate that they are generated, so ! Dn
d"n
= GS dx x=0
4c) Solve the problem. Solution d 2 !n !n
" 2 = 0 solutions is !n ( x ) = Ae" x/Ln + Be+ x/Ln 2
dx
Ln
To satisfy the first boundary condition in 4b): B = 0. Now consider the second: D
GS
d"n
1018
! Dn
# + n A = GS # A =
=
= 3.6 $ 1014 cm -3 !4
dx x=0
Ln
Dn Ln
7.8 27.9 $ 10
(
!n ( x ) =
(
)
)
GS
e" x/ Ln = 3.6 # 1014 e" x/ Ln ( Dn Ln )
4d) Provide a sketch of the solution, and explain it in words. Solution: ECE-‐305 9 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): Electrons are generated at the surface and diffuse into the bulk, so the concentration is high at the surface and approaches zero several diffusion lengths into the bulk. Deep in the bulk, there are no excess carriers so Fn = Fp = E F . The electron QFL must get closer to the conduction band near the surface, because the excess electron concentration is larger there. The variation is linear with position because !n x ()
5) varies exponentially with position. The sample is illuminated with light, resulting in an optical generation rate GL = 1024 cm-‐3 sec-‐1, but all of the photons are absorbed in a thin layer (10 nm wide near x = 0). Find the steady state excess minority carrier concentration and QFL’s vs. position. Assume that the semiconductor is only 5 μm long. You may also assume that there is an “ideal ohmic contact” at x = L = 5 μm, which enforces equilibrium conditions at all times. Make reasonable approximations, and approach the problem as follows. HINT: treat the thin layer at the surface as a boundary condition – do not try to resolve !n ( x ) inside this thin layer. 5a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ! "n
d 2 "n "n
Begin with: = Dn
#
+ GL !t
dx 2
$n
d 2 !n !n
"
+ GL dx 2
#n
Let’s treat the generation in a thin surface layer as a boundary condition, so GL = 0 in the semiconductor. Simplify for steady-‐state: 0 = Dn
ECE-‐305 10 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): The simplified MCDE equation is: d 2 !n !n
d 2 !n !n
d 2 !n !n
Ln = Dn! n Dn
"
=
0
"
=
0
" 2 = 0 dx 2
#n
dx 2 Dn# n
dx 2
Ln
Since the sample is much thinner than a diffusion length, we can ignore recombination, so d 2 !n
= 0 . dx 2
5b) Specify the initial and boundary conditions, as appropriate for this problem. Solution: Since this is a steady-‐state problem, there is no initial condition. At x = L , the “ideal ohmic contact” maintains equilibrium, so: !n ( x = L ) = 0 At the surface, the total number of e-‐h pairs generation per cm2 per second is GS = GL Δx = 102410−6 = 1018 cm -2s-1 . In steady-‐state, these must diffuse away at the same rate that they are generated, so ! Dn
d"n
= GS dx x=0
5c) Solve the problem. Solution: d 2 !n
= 0 The general solution is !n ( x ) = Ax + B dx 2
To satisfy the first boundary condition in 5b): Δn ( L ) = AL + B = 0 . B = −AL , so Δn ( x ) = Ax − AL = A ( x − L ) Now consider the second boundary condition: G
dΔn
− Dn
= − Dn A = GS
A= − S
dx x=0
Dn
and the solution is GS ⎛
x⎞
Δn ( x ) = A ( x − L ) =
⎜⎝ 1 − ⎟⎠ ( Dn L ) L
ECE-‐305 11 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): Compare to problem 4) – the factor out front is the same except that the diffusion length has been replaced by the length of the sample. Now put in numbers: GS
1018 cm -2s-1
=
× 5 × 10 −4 = 6.4 × 1013 cm -4 2 -1
Dn L 7.8 cm s
()
Δn x =
(
⎛
x⎞
1− ⎟ = 6.4 × 1013
⎜
L⎠
Dn L ⎝
GS
)
(
) ⎛⎜⎝ 1− Lx ⎞⎟⎠ cm
-3
(Note that we are in low level injection everywhere.) 5d) Provide a sketch of the solution, and explain it in words. Solution: 6) Concentration increases towards surface, because generation occurs the. Is zero at x = L because of the boundary condition there. Variation is linear with position because there is no recombination. Electron QFL(x) follows from n ( x ) ! "n ( x ) = ni e#( Fn ( x )# Ei )/k BT . The sample is in the dark, but the excess carrier concentration at x = 0 is held constant at !n 0 = 1012 cm-‐3. Find the steady state excess minority carrier concentration and ()
QFL’s vs. position. You may assume that the sample extends to x = +! . Make reasonable approximations, and approach the problem as follows. ECE-‐305 12 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): 6a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ! "n
d 2 "n "n
= Dn
#
+ GL
2
!t
dx
$
n
Begin with: 0 = Dn
d 2 !n !n
"
+ GL
dx 2
#n
Simplify for steady-‐state: G = 0 ; No generation: L
the simplified MCDE equation is: d 2 !n !n
d 2 !n !n
Dn
"
= 0 "
= 0 dx 2
#n
dx 2 Dn# n
d 2 !n !n
" 2 = 0 dx 2
Ln
Ln = Dn! n d 2 !n !n
" 2 = 0 where Ln = Dn! n is the minority carrier diffusion length. dx 2
Ln
6b) Specify the initial and boundary conditions, as appropriate for this problem. Solution: Since this is a steady-‐state problem, there is no initial condition. As x ! " , we expect all of the minority carriers to have recombined, so: !n ( x " # ) = 0 At the surface, the excess electron concentration is held constant, so !n ( x = 0 ) = 1012 cm -3 6c) Solve the problem. Solution: d 2 !n !n
" 2 = 0 solutions is !n ( x ) = Ae" x/Ln + Be+ x/Ln dx 2
Ln
To satisfy the first boundary condition in 10b): B = 0. Now consider the second: !n 0 = 1012 cm -3 ()
!n ( x ) = !n ( 0 ) e
ECE-‐305 " x/ Ln
( )
= 1012 e" x/ Ln 13 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): 6d) Provide a sketch of the solution, and explain it in words. Solution: Looks just like the solutions for prob. 4). Only difference is that instead of creating !n 0 by generation at the surface, we just specify !n 0 directly. ()
7) ()
The sample is in the dark, and the excess carrier concentration at x = 0 is held constant at !n 0 = 1012 cm-‐3. Find the steady state excess minority carrier ()
concentration and QFL’s vs. position. Assume that the semiconductor is only 5 μm long. You may also assume that there is an “ideal ohmic contact” at x = L = 5 μm, which enforces equilibrium conditions at all times. Make reasonable approximations, and approach the problem as follows. 7a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ! "n
d 2 "n "n
Begin with: = Dn
#
+ GL !t
dx 2
$n
d 2 !n !n
Simplify for steady-‐state: 0 = Dn
"
+ GL dx 2
#n
Generation is zero for this problem: GL = 0 ; the simplified MCDE equation is: d 2 !n !n
d 2 !n !n
Ln = Dn! n Dn
"
=
0
" 2 = 0 dx 2
#n
dx 2
Ln
Since the sample is much thinner than a diffusion length, we can ignore recombination, so d 2 !n
= 0 . dx 2
7b) Specify the initial and boundary conditions, as appropriate for this problem. Solution: Since this is a steady-‐state 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 cm -2 ECE-‐305 14 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): 7c) Solve the problem. Solution: d 2 !n
= 0 The general solution is !n ( x ) = Ax + B dx 2
To satisfy the first boundary condition in 7b): !n ( L ) = AL + B = 0 . Δn ( x ) = −B x L + B = B (1 − x L ) A = −B L Now consider the second boundary condition: !n 0 = B = 1012 cm -3 ()
Δn ( x ) = Δn ( 0 ) (1− x L ) = (10 ) (1− x L ) 12
8) 7d) Provide a sketch of the solution, and explain it in words. Just like problem 5) The sample is in the dark, and the excess carrier concentration at x = 0 is held constant at !n 0 = 1012 cm-‐3. Find the steady state excess minority carrier ()
concentration and QFL’s vs. position. Assume that the semiconductor is 30 μm long. You may also assume that there is an “ideal ohmic contact” at x = L = 30 μm, which enforces equilibrium conditions at all times. Make reasonable approximations, and approach the problem as follows. 8a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ! "n
d 2 "n "n
Begin with: = Dn
#
+ GL !t
dx 2
$n
Simplify for steady-‐state and no generation: d 2 !n !n
d 2 !n !n
d 2 !n !n
Ln = Dn! n Dn
"
=
0
"
=
0
" 2 = 0 dx 2
#n
dx 2 Dn# n
dx 2
Ln
ECE-‐305 15 Fall 2014 Mark Lundstrom 09/29/2014 d 2 !n !n
" 2 = 0 where Ln = Dn! n is the minority carrier diffusion length. dx 2
Ln
HW5 Solutions (continued): 8b) 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 cm-‐3 !n ( 30 µm ) = 0 8c) 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 µm ) = Ae" L / Ln + Be+ L / Ln = 0 Solve for A and B to find: !"n ( 0 ) e+ L/Ln
A = ! L/Ln
( e ! e+ L/Ln ) B=
!n ( 0 ) e" L/Ln
( e" L/Ln " e+ L/Ln ) So the solution is: !n ( 0 )
# "e"( x"L )/Ln + e+( x"L )/Ln %&
!n ( x ) = " L/Ln
+ L/Ln $
e
"e
(
!n ( x ) = !n ( 0 )
)
sinh #$( x " L ) / Ln %&
sinh ( L / Ln )
ECE-‐305 16 Fall 2014 Mark Lundstrom 09/29/2014 HW5 Solutions (continued): 8d) Provide a sketch of the solution, and explain it in words. Solution: 9) 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. Consider a sample that extends from !5 " x " 200 μm. The sample is illuminated with light, resulting in an optical generation rate of GL = 1024 cm-‐3 sec-‐1, but all of the photons are absorbed in a very thin layer (10 nm wide centered about x = 0). Do not try to resolve the electron density in this very thin layer, just find !n x = 0 . You may (
)
also assume that there are “ideal ohmic contacts”, which enforce equilibrium conditions at all times located at x = !5 μm and at x = 200 μm. Find the steady state excess minority carrier concentration vs. position. Make reasonable approximations, and approach the problem as follows. 9a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ! "n
d 2 "n "n
= Dn
#
+ GL
!t
dx 2
$n
Steady-‐state, no generation (treat generation near x = 0 as am internal boundary condition). For x > 0: ECE-‐305 17 Fall 2014 Mark Lundstrom d 2 !n !n
" 2 =0
dx 2
Ln
09/29/2014 HW5 Solutions (continued): For x < 0, we could also solve this equation, but the region is thin for x < 0, so we can ignore recombination and solve: d 2 !n
= 0 dx 2
9b) Specify the initial and boundary conditions, as appropriate for this problem. Solution: Steady-‐state, so no initial condition. At the boundaries: !n ( x = "5 µm ) = 0
!n ( x = 200 µm ) = 0
At x = 0, we need to determine !n x = 0 : (
)
The flux of carriers diffusing to the left from x = 0 is "n 0
F ! = Dn
5 # 10!4
and the flux diffusing to the right is "n 0
d"n
F + = ! Dn
= Dn
dx x=0
Ln
()
()
(where we have assume a decaying exponential solution as discussed below). The total flux diffusing away from the thin layer at x = 0 is $1
1 '
F TOT = F + + F ! = "n 0 Dn % +
!4 (
& Ln 5 # 10 )
The total flux diffusing away from x = 0 must (in steady-‐state) be equal to the total number of carriers per cm2 generated in the thin layer at x = 0: F TOT = GL × 10 nm =1018 cm -2s-1 ()
1 ⎪⎫
⎪⎧ 1
Δn 0 Dn ⎨ +
=1018 cm -2s-1 −4 ⎬
L
5
×
10
⎩⎪ n
⎭⎪
()
ECE-‐305 18 Fall 2014 Mark Lundstrom ()
Δn 0 =
1018
⎧⎪ 1
1 ⎫⎪
Dn ⎨ +
−4 ⎬
⎩⎪ Ln 5 × 10 ⎭⎪
=
09/29/2014 1018
⎧
1
1 ⎫
7.8 ⎨
+
⎬
−4
5 × 10−4 ⎭
⎩ 27.9 × 10
= 5.4 × 1011 ()
Δn 0 = 5.4 × 1011 HW5 Solutions (continued): 9c) Solve the problem. Solution: For x < 0-‐, the solutions is: !n ( x ) = Ax + B (
)
(
)
At x = -‐ 5 micrometers: !n x = "5 # 10 "4 = 0 = A "5 # 10 "4 + B
B
A=
( 5 ! 10"4 ) At x = 0: !n ( x = 0 ) = B For x < 0, the solution is: x
$
'
!n ( x ) = !n ( 0 ) &
+ 1)
#4
% 5 " 10
( x
⎛
⎞
Δn ( x ) = 5.4 × 1011 ⎜
+ 1⎟ cm -3
−4
⎝ 5 × 10
⎠
For x > 0+: the solution is: !n ( x ) = Ae" x/Ln + Be+ x/Ln !n ( x ) = Ae" x/Ln Because for x > 0, we can treat the region as being infinitely long. !n ( x ) = !n ( 0 ) e" x/Ln = 5.4 # 1015 e" x/Ln Δn ( x ) = 5.4 × 1011 e− x/Ln cm -3 9d) Provide a sketch of the solution, and explain it in words. Solution: ECE-‐305 19 Fall 2014 Mark Lundstrom 09/29/2014 Carriers are generated at x = 0 and diffuse away to both sides. ECE-‐305 20 Fall 2014

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