Mark

 Lundstrom     10/17/2014  

ECE  305  Homework  SOLUTIONS:  Week  8  

 
Mark  Lundstrom  
Purdue  University  
 
 
1) A  silicon  diode  is  asymmetrically  doped  at   N D = 1019  cm-­‐3  and N A = 1016  cm-­‐3.    (Note  
that  at   N D = 1019 the  semiconductor  is  on  the  edge  of  degeneracy,  but  we  can  assume  
that  non-­‐degenerate  carrier  statistics  are  close  enough  for  this  problem.)    Answer  the  
following  questions  assuming  room  temperature.    Assume  that  the  minority  electron  
and  hole  lifetimes  are   ! n = ! p = 10"6  s.    The  lengths  of  the  N  and  P  regions  are  
L = 500 µm  and   L >> x p , xn .  Assume  an  “ideal  diode”  and  answer  the  following  
questions.  
 

 
 
1a)    Compute   J D = I D A ,  the  diode  current  density  at  a  forward  bias  of     V A = 0.5  V.  
 
Solution:  
Since  this  in  a  one-­‐sided  junction  with   N D >> N A  ,  essentially  all  of  the  current  is  
due  to  electrons  injected  into  the  P-­‐region.  
 

JD = q
N A Ln
(
ni2 Dn qVA kBT
e ) ( )
! 1 = J 0 eqVA kBT ! 1     (*)  

 
From  Fig.  3.5  on  p.  80  of  SDF,   µ n = 1248 cm 2 V-s  at   N A = 1016  .  
Using  the  Einstein  relation,  we  find  
 
kT
Dn = B µ n = 0.026 ! 1248 = 32.4 cm 2 /s    
q
The  diffusion  length  is:  
 
Ln = Dn! n = 32.4 " 10#6 = 57 µm    

ECE-­‐305     1   Fall  2014  

Mark  Lundstrom     10/17/2014  

HW8  solutions  (continued):  

 
Since   Ln << L  ,  this  is  indeed  a  long  base  diode  as  assumed  in  (*)  
Now,  putting  numbers  n  (*):  
n2 D 1020 32.4
J 0 = q i n = 1.6 ! 10"19 16 "4
= 9.1! 10"12 A/cm 2  
N A Ln 10 57 ! 10
(
J D = J 0 eqVA k BT
) ( ) (
! 1 = 9.1" 10!12 e0.5/0.026 ! 1 = 9.1" 10!12 2.25 " 108 ! 1 = 2.1" 10!3   )
 
J D ( 0.5 V ) = 2.1! 10"3 A/cm 2  
 
1b)    Compute   J D = I D A ,  the  diode  current  density  at  a  forward  bias  of     V A = 0.6  V.  
 
Solution:  
Under  modest  forward  bias  we  can  ignore  the  -­‐1:  
( )
J D = J 0 eqVA kBT ! 1 " J 0 eqVA kBT  

J D ( 0.6 V ) = J 0 eq0.6 kBT = J 0 eq0.5 kBT ! eq0.1 kBT = J D ( 0.5 V ) ! 46.8  

J D ( 0.6 V ) = J D ( 0.5 V ) ! eq0.1 kBT = 2.1! 10"3 ! e0.1 0.025 = 9.6 ! 10"2  
 
J D ( 0.6 V ) = 9.6 ! 10"2 A/cm 2  
 
1c)    Compute   J D = I D A ,  the  diode  current  density  at  a  reverse  bias  of     V A = !0.5  V.  
 
Solution:  
J D ( !0.5 V ) = J 0 e ( )(
q !0.5 k BT
) ( ) (
! 1 = J 0 e!0.5 0.025 ! 1 = J 0 4.5 " 10!9 ! 1 # ! J 0   )
 
J D ( !0.5 V ) = ! J 0 = !9.1" 10!12 A/cm 2  
 
1d)    Compute   J D = I D A ,  the  diode  current  density  at  a  reverse  bias  of     V A = !0.6  V.  
 
Solution:  
J D ( !0.6 V ) = J 0 e ( )(
q !0.6 k BT
) ( ) (
! 1 = J 0 e!0.6 0.025 ! 1 = J 0 9.5 " 10!11 ! 1 # ! J 0   )
J D ( !0.6 V ) = ! J 0 = !9.1" 10!12 A/cm 2
 
 
For  an  ideal  diode,  the  reverse  current  is  constant  at   J D = ! J 0  as  long  as  the  
reverse  bias  is  greater  than  a  few   k BT q .  

ECE-­‐305     2   Fall  2014  

Mark  Lundstrom     10/17/2014  

HW8  solutions  (continued):  

 
2) Consider  the  diode  in  problem  1)  above  and  answer  the  following  questions.  
 
2a)    If  the  diode  is  biased  such  that   J D = 10!6 A/cm 2 ,  what  is   V A  ?  
 
Solution:  
n2 D
( )
Begin  with   J D = q i n eqVA kBT ! 1 = J 0 eqVA kBT ! 1  
N A Ln
( )
 
In  moderate  forward  bias,  we  can  drop  the  -­‐1:  
 
qV k T
J D = J 0 e A B  
 
so  
k T !J $
V A = B ln # D &  
q "J %
0

 
Putting  in  numbers:  
# 10!6 &
V A = 0.026ln % = 0.302 V     V A = 0.302 V  
$ 9.1" 10 ('
!12

 
2b)    If  the  temperature  changes  from  300  K  to  301  K,  how  much  does   V A  change?  
 
Solution:  
 
Differentiate  the  expression  that  we  obtained  above:  
d ') k BT ! J D $ +) k B ! J D $ k BT J 0 d
dV A
dT
= (
dT *) q
ln # & , = ln # & +
" J 0 % -) q " J 0 % q J D dT
( )
J D J 0`.1  

 
dV A k B ! J D $ k BT ! 1 dJ 0 $
= ln # & '  
dT q "J %
0
q #" J dT &%
0

dJ 0 d ! n Dn $ 2
= q  
i
dT dT #" N A Ln &%
 
The  strongest  part  of  the  temperature-­‐dependence  comes  from  the  exponential  factor  
in   ni2 ,  so  we  can  ignore  the  temperature  dependence  of  the  diffusion  coefficient,  the  
diffusion  length,  and  the  effective  densities-­‐of-­‐states  and  write:  
 

ECE-­‐305     3   Fall  2014  

Mark  Lundstrom     10/17/2014  

HW8  solutions  (continued):  

 
J 0 = Ke! EG k BT    
then  
1 dJ 0 1 " E % " E %
= ! EG k BT
Ke! EG kBT $ G 2 ' = $ G 2 ' ,  
J 0 dT Ke # k BT & # k BT &
 
which  can  be  used  to  find  
 
dV A k B ! J D $ k BT ! 1 dJ 0 $ k B ! J D $ k BT ! EG $
= ln # & ' = ln '  
dT q " J0 % q #" J 0 dT &% q #" J 0 &% q #" k BT 2 &%
 
so  
 
dV A k B ! J D $ ! EG q $
= ln # & ' # .  
dT q " J 0 % " T &%
 
Putting  numbers  in:  
 
dV A 1.38 ! 10"23 # 10"6 & # 1.12 &
= ln "  
dT 1.6 ! 10"19 %$ 9.1! 10"12 (' %$ 300 ('
 
dV A
= 1.00 ! 10"3 " 3.73! 10"3 = "2.7 ! 10"3  
dT
 
dV A
! "3 mV/K  
dT
 
So  we  need  to  lower  the  applied  bias  about  3  millivolts  to  keep  the  current  constant  as  
the  temperature  increases  1  K  or  1  degree  C.  
 
PN  junctions  can  be  used  as  thermometers  because  the  diode  current  depends  
sensitively  on  temperature.  
 
A  more  careful  treatment  would  include  the  temperature  dependencies  of  the  
bandgap,  effective  densities-­‐of-­‐states,  diffusion  coefficient,  etc.,  but  the  result  would  
be  close  to  the  value  obtained  here.  
 
 
 
 

ECE-­‐305     4   Fall  2014  

Mark  Lundstrom     10/17/2014  

HW8  solutions  (continued):  

 
3) The  sketch  below  shows  the  carrier  concentrations  in  a  PN  junction  at  room  
temperature.    Answer  the  following  questions.  
 

 
 
3a)    Is  the  diode  forward  or  reverse  biased?    Explain  your  answer.  
 
Solution:  
Forward  biased  because  there  are  excess  electrons  on  the  P-­‐side  and  excess  
holes  on  the  N-­‐side.  
 
3b)    What  is  the  acceptor  concentration  on  the  P-­‐side?  
 
Solution:     N A = 1016 cm -3  
 
3c)    What  is  the  donor  concentration  on  the  N-­‐side?  
 
Solution:     N D = 1014 cm -3  
 
3d)    What  is  the  intrinsic  carrier  concentration?  
 
Solution:  
n0 p0 = ni2    
On  the  P-­‐side:     n0 p0 = 1016 ! 107 = 1023     ni = 1023 = 3.16 ! 1011 cm -3    
On  the  N-­‐side:     n0 p0 = 1014 ! 109 = 1023   ni = 1023 = 3.16 ! 1011 cm -3    
 
ni = 3.16 ! 1011 cm -3  

ECE-­‐305     5   Fall  2014  

Mark  Lundstrom     10/17/2014  

HW8  solutions  (continued):  

 
3e)    Do  low  level  injection  conditions  apply?  
 
Solution:  
YES.  
On  the  P-­‐side:     !n ( "xn ) = 1010 << p0 = 1016    

( )
On  the  N-­‐side:     !p x p = 1012 << n0 = 1014    
 
3f)    What  bias  (in  volts)  is  applied  to  the  diode?  
 
Solution:  
According  the  the  law  of  the  junction:    
ni2 qVA kBT
!n ( "xn ) = e = n p0 eqVA kBT  
NA
k BT !n ( "xn ) # 1010 &
VA = ln = 0.026ln % 7 ( = 0.18 V  
q n p0 $ 10 '
V A = 0.18 V  
 
3g)   Which  is  longer:    the  electron  diffusion  length  on  the  P-­‐side  of  the  junction  or  the  
hole  diffusion  length  on  the  N-­‐side  of  the  junction.  
 
Solution:  
The  electron  diffusion  length  on  the  P-­‐side.    (The  diffusion  length  is  the  distance  
it  takes  for  the  minority  carrier  concentration  to  return  to  its  value  in  the  bulk.  
 
 
4) This  problem  concerns  the  energy  band  diagram  shown  below  for  a  diode  under  bias.  
 

ECE-­‐305     6   Fall  2014  

Mark  Lundstrom     10/17/2014  

HW8  solutions  (continued):  

 
4a)     Is  the  diode  forward  or  reverse  biased?  
 
Solution:  
Forward  biased  because   Fn > Fp  .  
 
4b)     What  is  the  value  of  the  applied  bias?  
 
Solution:  
qV A = Fn ! Fp  
V A = +0.5 V  
 
4c)     What  is  the  bandgap  of  the  semiconductor?  
 
Solution:  
Reading  from  the  graph:  
EC ! EV = 1.25 eV  
 
4d)     What  is  the  built-­‐in  potential  of  the  junction.  
 
Solution:  
From  the  plot:     V j = Vbi ! V A = 0.25 V  
Since:   V A = +0.5 V  
Vbi = V j + V A = 0.75 V   Vbi = 0.75 V  
 
 
5) Consider  the  Si  diode  of  prob.  1)  and  make  one  change.    The  minority  carrier  lifetime  
increases  by  a  factor  of  1000,  so   ! n = ! p = 10"3  s.    The  lengths  of  the  N  and  P  regions  
are  still   L = 500 µm  and   L >> x p , xn  .  Assume  an  “ideal  diode”  and  compute  the  
forward-­‐biased  current  at   V A = 0.6  V.    
 
Solution:  
We  recognize  that  the  minority  carrier  diffusion  length  will  change.  
Ln = Dn! n = 32.4 " 10#3 = 1800 µm    
 
Now   Ln >> L  ,  so  the  long  base  diode  of  prob.  1)  has  become  a  short  base  diode.  
Instead  of  eqn.    (*)  in  prob.  1),  we  have  to  replace  the  minority  carrier  diffusion  
length  by  the  length  of  the  P-­‐region:  

ECE-­‐305     7   Fall  2014  

Mark  Lundstrom     10/17/2014  

HW8  solutions  (continued):  

n2 D
( )
J D = q i n eqVA kBT ! 1 = J 0 eqVA kBT ! 1
NA L
( )
 
The  saturation  current  density  will  be  smaller  by  a  factor  of  
 
L 500
= = 8.8
Ln 57    
 
So  we  find:  
 
n2 D 9.1! 10"16
J0 = q i n = = 1.04 ! 10"16 A/cm 2
NA L 8.8  
 
The  current  becomes:  
 
( ) ( ) (
J D = J 0 eqVA kBT ! 1 = 1.04 " 10!16 e0.6/0.026 ! 1 = 1.04 " 10!16 1.05 " 1010 ! 1 = 1.1" 10!6 )  
J D ( 0.6 V ) = 1.1! 10 A/cm "6 2

 
 
 
Note:    The  lifetime  is  longer,  so  the  current  must  be  smaller  –  this  is  a  sanity  check.  
 
Key  point  to  remember:  
 
Long  base  diode:   L >> Ln  

JD = q
N A Ln
(
ni2 Dn qVA kBT
e !1 )
 
 
Short  base  diode,   L << Ln ,  replace  the  diffusion  length  by  the  length  of  the  quasi-­‐
neutral  region.  
 
n2 D
(
J D = q i n eqVA kBT ! 1
NA L
)
 
If  the  diode  is  neither  long  nor  short,  then  we  get  more  complicated  expressions  
involving  hyperbolic  functions.  
 

ECE-­‐305     8   Fall  2014