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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: ∂Δn t = D n d 2 Δn dx 2 Δn τ n + G L For all the following problems, assume ptype silicon at room temperature, uniformly doped with N A = 10 17 cm 3, μ n = 300 cm 2 /V sec, τ n = 10 6 s. From these numbers, we find: D n = k B T q μ n = 7.8 cm 2 s L n = D n τ n = 27.9 μm 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 G L = 10 20 cm 3 sec 1 . Find the steadystate excess minority carrier concentration and the QFL’s, F n and F p . Assume spatially uniform conditions, and approach the problem as follows. Solution: 1a) Simplify the Minority Carrier Diffusion Equation for this problem. Begin with: ∂Δn t = D n d 2 Δn dx 2 Δn τ n + G L Simplify for steadystate: 0 = D n d 2 Δn dx 2 Δn τ n + G L Simplify for spatially uniform conditions: 0 = 0 Δn τ n + G L So the simplified MCDE equation is: Δn τ n + G L = 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.

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

Page 1: Week5HW S15 Solutions - nanohub.orgWeek5HW_S15_Solutions.pdf · HW5)Solutions(continued):) % 3d)% Provideasketchofthesolution,andexplainitinwords. % % Solution:) % % Theconcentrationis

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:  

∂Δn∂t

= Dnd 2Δndx2

− Δnτ n

+GL

 For  all  the  following  problems,  assume  p-­‐type  silicon  at  room  temperature,  uniformly  doped  with   N A = 1017  cm-­‐3,   µn = 300  cm2/V  sec,   τ n = 10−6  s.    From  these  numbers,  we  find:  

Dn =kBTq

µn = 7.8 cm2 s   Ln = Dnτ n = 27.9 µm  

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.  

Begin  with:   ∂Δn∂t

= Dnd 2Δndx2

− Δnτ n

+GL    

Simplify  for  steady-­‐state:    0 = Dnd 2Δndx2

− Δnτ n

+GL  

Simplify  for  spatially  uniform  conditions:    0 = 0− Δnτ n

+GL  

So  the  simplified  MCDE  equation  is:    

− Δnτ 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.  

 

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Mark  Lundstrom     Spring  2015  

ECE-­‐305     Spring  2015  2  

HW5  Solutions  (continued):    

1c)    Solve  the  problem.    Solution:  

In  this  case  the  solution  is  straightforward:     Δn = GLτ n = 1020 ×10−6 = 1014 cm-3  Now  compute  the  QFLs:  Since  we  are  doped  p-­‐type  and  in  low  level  injection:  

p ≈ p0 = N A = nieEi−Fp( ) kBT        

Fp = Ei − kBT ln

N A

ni

⎝⎜⎞

⎠⎟= Ei − 0.026ln 1017

1010

⎛⎝⎜

⎞⎠⎟= Ei − 0.41 eV  

n ≈ Δn >> n0 = nieFn−Ei( ) kBT        

Fn = Ei + kBT ln Δn

ni

⎝⎜⎞

⎠⎟= 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   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.  

       

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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  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:  

Begin  with:   ∂Δn∂t

= Dnd 2Δndx2

− Δnτ n

+GL    

Simplify  for  spatially  uniform  conditions  with  no  generation:     dΔndt

=− Δnτ 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):  Δn t = 0( ) = 1014 cm-3  

 2c)    Solve  the  problem.    

Solution:  dΔndt

=− Δnτ n

  The  solution  is:  Δn t( ) = Ae− t /τ n  

Now  use  the  initial  condition:    Δn t = 0( ) = 1014 = A  Δn t( ) = 1014 e− t /τ n  

                       

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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 lnΔn t( ) + n0

ni

⎛⎝⎜

⎞⎠⎟  

Initially,  Δn t( ) >> n0  and  Δn t( ) = Δn 0( )e− t /τ 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  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:  

Begin  with:   ∂Δn∂t

= Dnd 2Δndx2

− Δnτ n

+GL    

Simplify  for  steady-­‐state  conditions:    0 = Dnd 2Δndx2

− Δnτ n

+GL  

The  simplified  MCDE  equation  is:  

Dnd 2Δndx2

− Δnτ n

+GL = 0     d 2Δndx2

− ΔnLn2 + GL

Dn

= 0   Ln = Dnτ n    

 

Page 5: Week5HW S15 Solutions - nanohub.orgWeek5HW_S15_Solutions.pdf · HW5)Solutions(continued):) % 3d)% Provideasketchofthesolution,andexplainitinwords. % % Solution:) % % Theconcentrationis

Mark  Lundstrom     Spring  2015  

ECE-­‐305     Spring  2015  5  

HW5  Solutions  (continued):    d 2Δndx2

− ΔnLn2 + GL

Dn

= 0  where   Ln = Dnτ 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,   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 .  

 

RS = R x( )

0

Δx

∫ dx ≈Δn 0( )τ S

Δx cm-2-s-1 .      

Rearranging  this  equation,  we  can  write  

RS =

Δxτ S

Δn 0( ) = SFΔn 0( ) cm-2-s-1  

where:  

SF ≡ Δx

τ S

 cm/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  specify  the  front  surface  recombination  velocity.    Typically,  

0 < SF ≤107 cm/s .    For  this  specific  problem,    

SF = Δx

τ S

= 10−6 cm10−10 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 =

Δxτ S

Δn 0( ) = SFΔn 0( ) cm-2-s-1

   

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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

dΔndx

A/cm2 .  

The  flux  of  electrons  in  the  +x  direction  is  

Jn

−q= −Dn

dΔndx

cm-2-s-1  

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

Jn

−q⎛⎝⎜

⎞⎠⎟= +Dn

dΔndx

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

dΔndx x=0

= RS = SFΔn 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 2Δndx2

− ΔnLn2 + GL

Dn

= 0  

To  solve  the  homogeneous  problem  first,  we  set    GL  =  0.  d 2Δndx2

− ΔnLn2 = 0    The  solution  is  Δn x( )= Ae− x/Ln + Be+ x/Ln

 Now  solve  for  a  particular  solution  by  letting   x → +∞  where  everything  is  uniform:  

− ΔnLn2 + GL

Dn

= 0    The  solution  is:  Δn = − Ln2

Dn

GL = GLτ n  

Add  the  two  solutions:    Δn x( )= Ae− x/Ln + Be+ x/Ln +GLτ n  

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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

dΔndx x=0

= −Dn

Ln

A = SFΔn 0( ) = SF A+GLτ n( )  

A = −

SFGLτ n

Dn Ln + SF

= −GLτ n

1+ Dn Ln( ) SF

 

Δn x( ) = GLτ 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:   Δn x( ) = GLτ n ,  which  make  sense,  since  we  have  spatial  uniformity.  

 ii)   SF →∞ .    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:  

Δn x( ) = GLτ n 1− e− x / Ln

1+ Dn Ln( ) SF

⎣⎢⎢

⎦⎥⎥  

For   SF →∞ ,  we  find  

Δn x( )→ GLτ n 1− e− x/ Ln⎡⎣ ⎤⎦

 which  behaves  as  expected.  

                       

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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   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  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:  

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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  μsec.  At  x  =  0  the  surface  recombination  velocity  is SF = 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.  

   4)   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:  

Begin  with:   ∂Δn∂t

= Dnd 2Δndx2

− Δnτ n

+GL    

Simplify  for  steady-­‐state:    0 = Dnd 2Δndx2

− Δnτ n

+GL  

No  generation:   GL = 0  the  simplified  MCDE  equation  is:  

Dnd 2Δndx2

− Δnτ n

= 0   d 2Δndx2

− ΔnDnτ n

= 0   d 2Δndx2

− ΔnLn2 = 0   Ln = Dnτ n    

d 2Δndx2

− ΔnLn2 = 0  where   Ln = Dnτ 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:  

Δn x → +∞( ) = 0    

Page 10: Week5HW S15 Solutions - nanohub.orgWeek5HW_S15_Solutions.pdf · HW5)Solutions(continued):) % 3d)% Provideasketchofthesolution,andexplainitinwords. % % Solution:) % % Theconcentrationis

Mark  Lundstrom     Spring  2015  

ECE-­‐305     Spring  2015  10  

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 2Δndx2

− ΔnLn2 = 0    The  solution  is  Δn x( )= Ae− x/Ln + Be+ x/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:  

 

 

Page 11: Week5HW S15 Solutions - nanohub.orgWeek5HW_S15_Solutions.pdf · HW5)Solutions(continued):) % 3d)% Provideasketchofthesolution,andexplainitinwords. % % Solution:) % % Theconcentrationis

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   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 = 5μm,  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:   ∂Δn∂t

= Dnd 2Δndx2

− Δnτ n

+GL    

Simplify  for  steady-­‐state:    0 = Dnd 2Δndx2

− Δnτ n

+GL  

Generation  is  zero  for  this  problem:   GL = 0 ;  The  simplified  MCDE  equation  is:  

Dnd 2Δndx2

− Δnτ n

= 0  

d 2Δndx2

− ΔnLn2 = 0

  Ln = Dnτ n    Since  the  sample  is  much  thinner  than  a  diffusion  length,  we  can  ignore  recombination,  so  d 2Δndx2

= 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:  

Δn x = L( ) = 0

   At  the  surface:    

Δn 0( ) = 1012 cm-2    

 5c)    Solve  the  problem.    Solution:  d 2Δndx2

= 0      The  general  solution  is  Δn x( )= Ax + B  To  satisfy  the  first  boundary  condition  in  5b):    Δn L( )= AL + B = 0 .    A = −B L     Δn x( )= −B x L + B = B 1− x L( )  

Page 12: Week5HW S15 Solutions - nanohub.orgWeek5HW_S15_Solutions.pdf · HW5)Solutions(continued):) % 3d)% Provideasketchofthesolution,andexplainitinwords. % % Solution:) % % Theconcentrationis

Mark  Lundstrom     Spring  2015  

ECE-­‐305     Spring  2015  12  

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 = 30μm,  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:   ∂Δn∂t

= Dnd 2Δndx2

− Δnτ n

+GL    

Simplify  for  steady-­‐state  and  no  generation:    

Dnd 2Δndx2

− Δnτ n

= 0  d2Δndx2

− ΔnDnτ n

= 0   d 2Δndx2

− ΔnLn2 = 0   Ln = Dnτ n    

 d 2Δndx2

− ΔnLn2 = 0  where   Ln = Dnτ n  is  the  minority  carrier  diffusion  length.  

 

Page 13: Week5HW S15 Solutions - nanohub.orgWeek5HW_S15_Solutions.pdf · HW5)Solutions(continued):) % 3d)% Provideasketchofthesolution,andexplainitinwords. % % Solution:) % % Theconcentrationis

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:  

Δn 0( ) = 1012    cm-­‐3  

Δn 30µm( ) = 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:  

A =−Δn 0( )e+L/Lne−L/Ln − e+L/Ln( )  

B =Δn 0( )e−L/Lne−L/Ln − e+L/Ln( )  

 So  the  solution  is:  

Δn x( )= Δn 0( )e−L/Ln − e+L/Ln( ) −e− x−L( )/Ln + e+ x−L( )/Ln⎡⎣ ⎤⎦  

Δn x( )= Δn 0( ) sinh x − L( ) / Ln⎡⎣ ⎤⎦sinh L / Ln( )  

                 

Page 14: Week5HW S15 Solutions - nanohub.orgWeek5HW_S15_Solutions.pdf · HW5)Solutions(continued):) % 3d)% Provideasketchofthesolution,andexplainitinwords. % % Solution:) % % Theconcentrationis

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.