Mark Lundstrom Spring 2015

ECE-305 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 p-type silicon at room temperature, uniformly doped with N A = 10

17 cm-3, 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 cm-3 sec-1. Find the steady-state 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 steady-state: 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

ECE-305 Spring 2015 2

HW5 Solutions (continued):

1c) Solve the problem. Solution:

In this case the solution is straightforward: n = GL n = 1020 106 = 1014 cm-3

Now compute the QFLs: Since we are doped p-type 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 low-level injection. We should check the assumption that n

Mark Lundstrom Spring 2015

ECE-305 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 cm-3 sec-1. 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 cm-3

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

ECE-305 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 cm-3 sec-1. 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 steady-state 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

ECE-305 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 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 (recall prob. 1)), so

n x ( ) = GL 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 = 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 .

RS = R x( )

0

x

dx n 0( ) S

x cm-2-s-1 .

Rearranging this equation, we can write

RS =

x S

n 0( ) = SFn 0( ) cm-2-s-1 where:

SF

x S

cm/s

is a quantity with the units of velocity. In practice, we usually dont 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 < 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 cm-2-s-1) at the surface:

RS =

x S

n 0( ) = SFn 0( ) cm-2-s-1

Mark Lundstrom Spring 2015

ECE-305 Spring 2015 6

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

Jn = qDn

dndx

A/cm2 .

The flux of electrons in the +x direction is

Jnq

= Dndndx

cm-2-s-1

The flux of electrons in the x direction (to the surface where they are recombining) is:

Jnq

= +Dn

dndx

cm-2-s-1

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

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

ECE-305 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

ECE-305 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

ECE-305 Spring 2015 9

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 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 cm-3. 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 steady-state: 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 steady-state 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

ECE-305 Spring 2015 10

HW5 Solutions (continued): At the surface, the excess electron concentration is held constant, so

n x = 0( ) = 1012 cm-3

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 cm-3

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

ECE-305 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 cm-3. 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 steady-state: 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 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

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( )

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ECE-305 Spring 2015 12

HW5 Solutions (continued): Now consider the second boundary condition:

n 0( )= B = 1012 cm-3 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 cm-3. 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 steady-state 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

ECE-305 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 cm-3

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

ECE-305 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.