Mark Lundstrom Spring 2015 SOLUTIONS: ECE 305 Homework: Week 5 Mark Lundstrom Purdue University The following problems concern the Minority Carrier Diffusion Equation (MCDE) for electrons: ∂ Δn
d 2 Δn Δn
= Dn
−
+ GL
∂t
dx 2
τn
For all the following problems, assume p-­‐type silicon at room temperature, uniformly doped with N A = 1017 cm-­‐3, µ n = 300 cm2/V sec, τ n = 10−6 s. From these numbers, we find: kT
Dn = B µn = 7.8 cm 2 s Ln = Dnτ 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. Solution: 1a) Simplify the Minority Carrier Diffusion Equation for this problem. ∂ Δn
d 2 Δn Δn
Begin with: = Dn
−
+ GL ∂t
dx 2
τn
Simplify for steady-­‐state: 0 = Dn
d 2 Δn Δn
−
+ GL dx 2
τn
Simplify for spatially uniform conditions: 0 = 0 −
Δn
+ GL τn
So the simplified MCDE equation is: Δn
−
+ GL = 0 τn
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 Spring 2015 Mark Lundstrom Spring 2015 HW5 Solutions (continued): 1c) Solve the problem. Solution: In this case the solution is straightforward: Δn = GLτ 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 ≈ Δn >> n0 = ni e(
Fn − Ei ) k BT
⎛ Δn ⎞
⎛ 1014 ⎞
Fn = Ei + k BT ln ⎜ ⎟ = Ei + 0.026ln ⎜ 10 ⎟ = Ei + 0.24 eV
⎝ 10 ⎠
⎝ ni ⎠
Note that since we are using the MCDE, we are assuming low-­‐level injection. We should check the assumption that Δn << p . From the solution and the given p-­‐type doping density, we have: Δn = 1014 << p = 1017 . We are, indeed, in low level injection. 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 are 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. ECE-­‐305 2 Spring 2015 Mark Lundstrom Spring 2015 HW5 Solutions (continued): 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. 2a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ∂ Δn
d 2 Δn Δn
Begin with: = Dn
−
+ GL ∂t
dx 2
τn
Simplify for spatially uniform conditions with no generation: dΔn
Δn
=−
dt
τn
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): Δn ( t = 0 ) = 1014 cm -3 2c) Solve the problem. Solution: dΔn
Δn
The solution is: Δn ( t ) = Ae−t /τ n =−
dt
τn
Now use the initial condition: Δn ( t = 0 ) = 1014 = A Δn ( t ) = 1014 e−t /τ n ECE-­‐305 3 Spring 2015 Mark Lundstrom Spring 2015 HW5 Solutions (continued): 2d) Provide a sketch of the solution, and explain it in words. Solution: ⎛ Δn ( t ) + n0 ⎞
For the electron QFL: Fn ( t ) = Ei + kBT ln ⎜
⎟⎠ ni
⎝
Initially, Δn ( t ) >> n0 and Δn ( t ) = Δn ( 0 ) e−t /τ 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: treat the thin layer at the surface as a boundary condition – do not try to resolve Δn ( x ) inside this thin layer. Approach the problem as follows. 3a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ∂ Δn
d 2 Δn Δn
Begin with: = Dn
−
+ GL ∂t
dx 2
τn
Simplify for steady-­‐state conditions: 0 = Dn
d 2 Δn Δn
−
+ GL dx 2
τn
The simplified MCDE equation is: d 2 Δn Δn
d 2 Δn Δn GL
Dn
−
+
G
=
0
− 2 +
= 0 Ln = Dnτ n L
dx 2 τ n
dx 2
Ln Dn
ECE-­‐305 4 Spring 2015 Mark Lundstrom Spring 2015 HW5 Solutions (continued): d 2 Δn Δn GL
− 2 +
= 0 where Ln = Dnτ 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, Δn = GLτ n (recall prob. 1)), so Δn ( x → ∞ ) = GLτ n 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 = Δn ( 0 ) τ S cm -3s-1 , times the thickness of the thin layer at the surface, Δx 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 . Δx
RS =
∫ R ( x )dx ≈
0
Δn ( 0 )
τS
Δx cm -2 -s-1 . Rearranging this equation, we can write Δx
RS =
Δn ( 0 ) = S F Δn ( 0 ) cm -2 -s-1 τS
where: Δx
cm/s SF ≡
τS
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 specify the front surface recombination velocity. Typically, 0 < S F ≤ 107 cm/s . For this specific problem, Δx 10−6 cm
=
= 104 cm/s −10
τS
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: Δx
RS =
Δn ( 0 ) = S F Δn ( 0 ) cm -2 -s-1
τS
SF =
ECE-­‐305 5 Spring 2015 Mark Lundstrom Spring 2015 HW5 Solutions (continued): 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 dΔn
J n = qDn
A/cm 2 . dx
The flux of electrons in the +x direction is Jn
dΔn
= − Dn
cm -2 -s-1 −q
dx
The flux of electrons in the –x direction (to the surface where they are recombining) is: ⎛J ⎞
dΔn
− ⎜ n ⎟ = + Dn
cm -2 -s-1 −q
dx
⎝ ⎠
In steady-­‐state, 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
dΔn
= RS = S F Δn ( 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 Δn Δn GL
− 2 +
=0
dx 2
Ln Dn
To solve the homogeneous problem first, we set GL = 0. d 2 Δn Δn
− 2 = 0 The solution is Δn ( x ) = Ae− x/Ln + Be+ x/Ln
2
dx
Ln
Now solve for a particular solution by letting x → +∞ where everything is uniform: Δn G
L2
− 2 + L = 0 The solution is: Δn = − n GL = GLτ n Ln Dn
Dn
Add the two solutions: Δn ( x ) = Ae− x/Ln + Be+ x/Ln + GLτ n ECE-­‐305 6 Spring 2015 Mark Lundstrom Spring 2015 HW5 Solutions (continued): To satisfy the first boundary condition as x → ∞ , B = 0. Now consider the boundary condition at x = 0 : D
dΔn
+ Dn
= − n A = S F Δn ( 0 ) = S F ( A + GLτ n ) dx x=0
Ln
A= −
S F GLτ n
GLτ n
=−
Dn Ln + S F
1+ Dn Ln
(
− x/ L
⎡
e n
Δn x = GLτ n ⎢1−
⎢⎣ 1+ Dn Ln
()
(
)
)
SF
⎤
⎥ S F ⎥⎦
Check some limits. i) SF = 0 cm/s, which implies that there is no recombination at the surface. Then we find: Δn ( x ) = GLτ n , which make sense, since we have spatial uniformity. ii) (
)
S F → ∞ . Strong recombination at the surface should force Δn x = 0 = 0 , but in the bulk we should still have Δn ( x ) = GLτ n . The transition from 0 to a finite value in the bulk should take a diffusion length or two. From the solution: − x/ L
⎡
e n
Δn x = GLτ n ⎢1−
⎢⎣ 1+ Dn Ln
For S F → ∞ , we find ()
(
)
⎤
⎥ S F ⎥⎦
Δn ( x ) → GLτ n ⎡⎣1− e− x/ Ln ⎤⎦
ECE-­‐305 which behaves as expected. 7 Spring 2015 Mark Lundstrom Spring 2015 HW5 Solutions (continued): 3d) Provide a sketch of the solution, and explain it in words. Solution: The concentration is GLτ 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 Δn ( x ) near the surface. The variation with position is ECE-­‐305 linear, because Δn ( 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: 8 Spring 2015 Mark Lundstrom Spring 2015 HW5 Solutions (continued): 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. At x 4) = 0 the surface recombination velocity is S F = 104 cm/s . Find the steady state excess minority carrier concentration and QFL’s 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. The sample is in the dark, but the excess carrier concentration at x = 0 is held constant at Δn ( 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. 4a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ∂ Δn
d 2 Δn Δn
Begin with: = Dn
−
+ GL ∂t
dx 2
τn
Simplify for steady-­‐state: 0 = Dn
d 2 Δn Δn
−
+ GL dx 2
τn
No generation: GL = 0 the simplified MCDE equation is: d 2 Δn Δn
d 2 Δn Δn
Dn
−
= 0 −
= 0 dx 2
τn
dx 2 Dnτ n
d 2 Δn Δn
− 2 = 0 dx 2
Ln
Ln = Dnτ n d 2 Δn Δn
− 2 = 0 where Ln = Dnτ n is the minority carrier diffusion length. dx 2
Ln
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: Δn ( x → +∞ ) = 0 ECE-­‐305 9 Spring 2015 Mark Lundstrom Spring 2015 HW5 Solutions (continued): At the surface, the excess electron concentration is held constant, so Δn ( x = 0 ) = 1012 cm -3 4c) Solve the problem. Solution: d 2 Δn Δn
− 2 = 0 The solution is Δn ( x ) = Ae− x/Ln + Be+ x/Ln 2
dx
Ln
To satisfy the first boundary condition in 4b): B = 0. Now consider the second: Δn ( 0 ) = 1012 cm -3 ( )
Δn ( x ) = Δn ( 0 ) e− x/ Ln = 1012 e− x/ Ln 4d) Provide a sketch of the solution, and explain it in words. Solution: ECE-­‐305 10 Spring 2015 Mark Lundstrom Spring 2015 HW5 Solutions (continued): 5) The sample is in the dark, and the excess carrier concentration at x = 0 is held constant at Δn ( 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. 5a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ∂ Δn
d 2 Δn Δn
Begin with: = Dn
−
+ GL ∂t
dx 2
τn
d 2 Δn Δn
Simplify for steady-­‐state: 0 = Dn
−
+ GL dx 2
τn
Generation is zero for this problem: GL = 0 ; The simplified MCDE equation is: d 2 Δn Δn
d 2 Δn Δn
Dn
−
=
0
− 2 =0
2
Ln = Dnτ n
dx 2
τn
dx
Ln
Since the sample is much thinner than a diffusion length, we can ignore recombination, so d 2 Δn
=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. As x → ∞ , we expect all of the minority carriers to have recombined, so: Δn ( x = L ) = 0
At the surface: Δn ( 0 ) = 1012 cm -2 5c) Solve the problem. Solution: d 2 Δn
= 0 The general solution is Δn ( x ) = Ax + B dx 2
To satisfy the first boundary condition in 5b): Δn ( L ) = AL + B = 0 . A = −B L ECE-­‐305 Δn ( x ) = −B x L + B = B (1 − x L ) 11 Spring 2015 Mark Lundstrom Spring 2015 HW5 Solutions (continued): Now consider the second boundary condition: Δn ( 0 ) = B = 1012 cm -3
()
( )(
) (
)(
Δn x = Δn 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 Δn ( 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. 6a) Simplify the Minority Carrier Diffusion Equation for this problem. Solution: ∂ Δn
d 2 Δn Δn
Begin with: = Dn
−
+ GL ∂t
dx 2
τn
Simplify for steady-­‐state and no generation: d 2 Δn Δn
d 2 Δn Δn
d 2 Δn Δn
Ln = Dnτ n Dn
−
=
0
−
=
0
− 2 = 0 dx 2
τn
dx 2 Dnτ n
dx 2
Ln
d 2 Δn Δn
− 2 = 0 where Ln = Dnτ n is the minority carrier diffusion length. dx 2
Ln
ECE-­‐305 12 Spring 2015 Mark Lundstrom Spring 2015 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: Δn ( 0 ) = 1012 cm-­‐3 Δn ( 30 µm ) = 0 6c) Solve the problem. Solution: Δn ( x ) = Ae− x/Ln + Be+ x/Ln
Because the region is about one diffusion length long, we need to retain both solutions. Δn ( 0 ) = A + B Δn ( L = 30 µm ) = Ae− L / Ln + Be+ L / Ln = 0
Solve for A and B to find: −Δn ( 0 ) e+ L/Ln
A = − L/Ln
( e − e+ L/Ln ) B=
Δn ( 0 ) e− L/Ln
( e− L/Ln − e+ L/Ln ) So the solution is: Δn ( 0 )
⎡ −e−( x−L )/Ln + e+( x−L )/Ln ⎤
Δn ( x ) = − L/Ln
+ L/Ln ⎣
⎦ (e − e )
Δn ( x ) = Δn ( 0 )
sinh ⎡⎣( x − L ) / Ln ⎤⎦
sinh ( L / Ln )
ECE-­‐305 13 Spring 2015 Mark Lundstrom Spring 2015 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. ECE-­‐305 14 Spring 2015