Solutions Chapter 6

6.1. Consider a base-collector junction of a silicon BJT (bipolar junction transistor)

like that if Figure 6.1. Assuming a linearly graded junction with a=1.2x1018 cm-3/um,
find Vbi

This can be done iteratively from Equation (6.11)

 1

2kT  a 12V bi 3 
Vbi  ln   
q 2ni  qa  

 


We choose a starting guess of 1V, and find:

Vbi cube root term logarithm new Vbi

1 0.000402474 10.0150146 5.21E-01

5.21E-01 0.00032381 9.79753926 5.09E-01

5.09E-01 0.000321449 9.7902212 5.09E-01

The built-in voltage is 0.51 V.

6.2. A silicon pn homojunction has a doping profile as indicated in Figure P6.1.

a) Find the value of the electric field in the bulk on the p side.

Adapting Equation 6.1 for the p side,

x  xo

p
N A  N A (0)e

Choosing the straight line on the p side,

x  0.2 m, N A  1018 cm-3 and x0  0.3 m, N A (0)  6 x1016 cm-3

  x  x0    0.2  0.3
or  p    3.6
NA 1018
ln ln
N A (0) 6 x1016

kT 1 -0.026V
Then Ep = - = = -0.72V / m m = -7.2kV / cm
q l 0.036 mm
b) Find the electric field in the bulk on the n side.

Similarly, on the n side,

x  0.6m, N D ( x )  1.8 x1017 cm 3 , x0  0.3 m, N D (0)  6 x1016 cm 3

n 
( x  x0 )

 0.6  0.3 m  0.27m
 N ( x)   6  1016 
ln  D  ln 1.8  1017 
 N D ( x0 )   
kT 1 -0.026V
En = - = = +0.096V / m m = 960V / cm
q l -0.27 m m

c) Plot NA-ND as a function of position

We first find the grading constant a. One way to obtain it is express the quantity
NA-ND as a function of x
 x  x 0 
(x  0.2)
p 18  0.036
N A (x)  N A (x 0 )e  10 e
x  x 0 
(x .3)
p 
N D (x )  N D (x 0 )e  6  1016 e 0.27

and the difference plotted below:

The slope is approximately –1018 cm-3/ m, or a=1018cm-3/ m (1022cm-4).

d) Find the built-in voltage.

This can be done iteratively from Equation (6.11)

  1

2kT  a 12V bi 3 
Vbi  ln   
q 2ni  qa  

 


Picking a starting voltage of 1V, this converges rapidly to 0.831V:

Vbi cube root term logarithm new Vbi

1 1.98517E-05 16.0336938 8.34E-01

8.34E-01 1.86843E-05 15.9730874 8.31E-01

8.31E-01 1.86608E-05 15.9718251 8.31E-01

e) Find the junction width at equilibrium

From Equation (6.7),

1

12Vbi  12 11.8  8.85  10 F / cm  0.83V 

1
14 3
3
w     1.9  105 cm  0.19m
 qa   1.6  10 19
C 10 22
cm 4
 
f) For the junction width you found in (e), comment on the validity of the
linear approximation used over this distance.

From the figure above in part (c) we find that over a distance of 0.2 m on either
side of the junction, the approximation is poor. 

6.3. In section 6.2.2, it was claimed that hyper-abrupt junctions exhibit a large
fractional change in junction capacitance with applied voltage. Explain physically
why we should expect this to be the case.

In a hyper-abrupt junction, the doping concentration decreases as one gets

further from the junction. Let us consider the reverse bias regime. As the bias
voltage gets larger, the junction width must get large to uncover the appropriate
number of ionized dopants. Since the dopants get scarcer away from the
junction, as the bias increases the junction width increases more and more
rapidly. This changes the distance between the capacitive plates more and more
rapidly, thus changing the capacitance more rapidly.

6.4. Figure P6.2 shows the equilibrium energy band diagram for a heterojunction
between n-type Semiconductor A (band gap 1.5 eV) and p-type Semiconductor B (1.0
eV) at equilibrium.
 a) Indicate the directions of:
Electron diffusion 
Electron diffusion current Jn(diff) 

Electron drift 
Electron drift current Jn(drift) 

Hole diffusion 
Hole diffusion current Jp(diff) 

Hole drift 
Hole drift current Jp(drift) 

Effective electric field for holes 

True electric field 

b) In the conduction band edge, there is a notch that looks like a quantum
well. The quantum well may have one or two discrete states.
 i) Indicate on the drawing where the potential well is for the electrons
Indicated above.

ii) If an electron is trapped inside the well, list the mechanisms by which it
can get out.
The electron can tunnel out to the left, it can be thermionically emitted over the

barrier to the left or to the right, or it can recombine with a hole.

6.5. Using the electron affinity model, draw (to scale) the energy band diagram for a
heterojunction between p-GaAs (Eg=1.43 eV, =4.07 eV) whose p=0.1 eV, and N-
Al0.3Ga0.7As (Eg=1.8eV, =3.74 eV) whose n=0.15 eV. Neglect interface states.
Under neutrality, we have:

At equilibrium, it becomes:
 The built-in voltage is Vbi=

Vbi    2  Eg 2   p    1   n    4.07  1.43  0.1   3.74  0.15  1.51V . . The

discontinuity in the conduction band edge is

EC   2  3   4.07  3.74  0.33eV

And the barrier for electrons is thus

EB electrons   Vbi  EC  1.51  0.33  1.18eV .

The discontinuity in EV is

EV   1  Eg1     2  Eg 2    3.74  1.8   (4.07  1.43)  0.04eV

and thus the barrier to holes is

EB ( holes )  Vbi  EV  1.51  0.04  1.55eV

6.6. Repeat the previous problem only now let the GaAs be n-type with n=0.1 eV and
let the AlGaAs be p-type with p=0.15 eV.
Setting the Fermi levels equal gives

The built-in voltage is Vbi=

Vbi   1  Eg1   p     2   n    3.74  1.8  0.15   4.07  0.1  1.22V The

discontinuity in the conduction band edge is

EC   2  1   4.07  3.74  0.33eV

And the barrier for electrons is thus

EBelectrons   Vbi  EC  1.22  0.33  1.55eV .

 The discontinuity in EV is

EV   1  Eg1     2  Eg 2    3.74  1.8   (4.07  1.43)  0.04eV

and thus the barrier to holes is

EB ( holes )  Vbi  EV  1.22  0.04  1.18eV

Now the glitch is in the valence band, and the barrier for holes is smaller than for

electrons.

6.7. Consider the Type I Np heterojunction of Figures 6.8 and 6.9 in which the net
doping N D' and N A' are uniform on the N and p sides respectively. Let n be the
permittivity on the N side and p that on the p-side. Solve Poisson's equation to find
the depletion widths wn and wp on the n and p sides and the total depletion width.
This is similar to the procedure used in Chapter 5 for homojunctions. (Hint: At the
interface, displacement eE is continuous.)
Poisson’s Equation:

QV
E (x) = ò dx
e
x
qN D' N D'
On the N side, En = ò
en
dx ' = q
en
( x - x ) for xn<x<x0
n
xn

xp
-qN A'
On the p side, E p = ò
x
ep
dx ' = q
N A'
ep
(x p )
- x for x0<x<xp

From continuity of displacement, we have

e nEn (x0 ) = e pEp (x0 )

Therefore


qN 'D x 0  x n   qN A' x p  x 0 or 
N D' w n  N A' w p

We integrate the field to find the voltage: V ( x )     ( x )dx  

 qN D'
N side: Vn (x )  Vn (x n )   x  x 0 2
2 n

qN 'A
 
2
p side: Vp (x )  Vp (x p )   x x
2 p p

On the N side the total built-in voltage is

qN D' '
V jn  x 0  x n 2  qN D w n2
2 n 2 n

qN 'A
   qN
2 '
and on the p side V jp  x  x0 A
w 2p
2 p p 2
p

The total junction voltage is

qN D' 2 qN 'A 2
Vbi  V jn  V jp  w  w
2 n n 2 p p

Substituting in N D w n  N A w p , we find
' '

2
qN D' 2 qN A' N D'  2
Vbi  w   '  w n or
2 n n 2 p N A 

Vbi  2  N ' 
wn    n p A' 
qN D' qN D'2  2  N 
  n p A
2 n 2 p N A'

2 n p N A' Vbi

qN D'   p N A'   n N D' 

Similarly

2 n  p N 'DV bi
wp 

qN A'  p N A'   n N D' 
And the total junction width is
 2 n p N A' Vbi 2 n p N D' Vbi
w  wn  wp  
qN D'   p N A'   n N D'  qN A'   p N A'   n N D' 

2 n p  N D'  N A'  Vbi

2


qN D' N A'   n N D'   p N A' 

6.8. Show that the junction capacitance per unit area for Problem 6.7 can be written as
dQV  q n p N A' N D' 
Cj=  A .
dVa
 
 2  n N D'   p N A' Vbi  Va  
 
From Equation (5.108), we have

dQ V
Cj 
dVa

In our case the charge is qND' wn  qN A' wp and we have

 d   d 
dQV
 A  qN D' wn    A   qN A' wp  
dVa  dVa   dVa 

d  ' 2 n p N A Vbi  Va  
'

 A  qN D 
dVa  qN D'   p N A'   n N D'  
 
 2 n p N A'  d 1
 A  qN D'  Vbi  Va  2
 qN D'   p N A'   n N D'   dVa
 
 q 2 n p N A' N D'   1  1
 

 A     V  V
 p N A'   n N D'    2 
2
bi a


 q n p N A' N D' 
 A 
 2   p N A'   n N D'  Vbi  Va  
 

6.9. For the Si:Ge Nn junction of Figure 6.17, sketch the energy band diagram that you
would expect using the simple electron affinity model (i.e., ignoring tunneling-
induced dipoles and interface states). Discuss the difference in current flow that
would result from the two energy band diagrams (simple model, and considering the
effects of the presence of dipoles). Let EC-Ef=0.1 eV for silicon and 0.15 eV for Ge.

Using the EAM, we would obtain the diagram shown below:

Current would flow easily in both directions in the ideal case since there is no
barrier to overcome as there is in Figure 6.17. Although that spike may be thin
enough to support tunneling, the current across the junction would not be as
great as in the ideal case.

If we add tunneling-induced dipoles to the case above, we would allow for

electrons from the valence band of the Ge to tunnel a short distance into the
silicon. This would place a negative charge near the surface of the Si and a
resulting positive charge near the surface of the germanium. Thus we would
expect band bending and a reduction of the barriers above.
6.10. We saw than in a heterojunction in which the spike and notch occurred in the
conduction band (as in Figure 6.12), it was possible for electrons in the valence band
of the narrow-gap material to tunnel a short distance into the forbidden band of the
wide-band-gap material. This tunneling induced a dipole that produced a
discontinuity in the bands at the junction. Consider a Pn heterojunction such as that
in Figure P6.3. Comment on the possibility of tunneling-induced dipoles in this case.

The possibility for electrons in the valence band to tunnel into the forbidden band
exists, and we can expect similar band alterations as a result.

6.11. Consider an n-GaAs Schottky barrier diode of area 10 µm2 and potential barrier
EB (0)  1eV .
a. Find the value of I0 from Equation (6.15).

qm* kT   E BkT(0)

2

J0  e
2 2 3
1.6  10 C0.067  9.1 1031 kg1.38  10 23 J/ K  300K 
19 2

exp 
1eV 

2 2 1.06  10 34 J s  0.026eV 
3

 7.1  109 1.98  10 17 

 1.4  10 7 A / cm 2  1.4  1011 A / cm2
where we used the conductivity effective mass for electrons in GaAs, because a
Schottky barrier is a majority carrier device.

b. Plot the forward I-V characteristic for a diode quality factor of n=1.3.

The equation for the current is Equation (6.14) (but note footnote in text):
  qV a

I  JA  AJ0 e nkT  1. With J0=1.410-11A/cm2 and n=1.3, we obtain the plot
 
below.

6.12. Consider the junction of Figure P6.4.

a) Draw its energy band diagram under forward and reverse bias.
 Forward bias

Reverse bias.

b) Will this junction be ohmic or rectifying? Why?

This is a metal-p-type semiconductor junction. Electrons from the metal can flow
easily to and from the valence band under either bias polarity.

6.13. Consider an n-Si:Al Schottky barrier diode for which N D' =1017cm-3 and the
measured built-in voltage is 0.70 V. Let the junction area be 10m2.
 a) Find the depletion width w. Note that this junction can be treated as a one-
sided junction.

We use Equation (5.40) for a p+n junction because the depletion region is entirely
in n-type material:
1

2V j  2(11.8)8.85  10 14 F / cm0.72

1
2
w   '      9.6  10 6 cm  0.096 m
 D  
qN  1.6  10 19
C10 17
cm 3



b) Find the junction capacitance at Va=-5 V.

First we find the junction width at Va=-5 V:

1

 
1
 2 Vbi  Va   2  2(11.8) 8.85 10 F / cm  0.7  (5)   2
14

w   
 qN D'    
1.6 1019 C 1017 cm3 
 2.7 105 cm  0.27  m

From Equation (6.16),

A 11.8 8.85 1014 F / C 10 108 cm2 

Cj   5
 3.9 1015 F
w 2.7 10 cm

6.14. A heterojunction is formed between n-type GaAs of N D' =1016 cm-3 and p-type
germanium of N A' =1017. The measured discontinuities in the band edges are
EC=0.27 eV and EV=0.49 eV.
a) Find the barrier to electrons in eV.

 N'   1016 
In the GaAs, EC  E f   n  kT ln  D   0.026eV ln  17 
 0.098eV
 NC   4.4 10 

 N'   1017 
In the Ge, E f  EV   p  kT ln  A   0,026eV ln  18 
 0.096eV
 NV   4 10 

The energy difference between the conduction band edge in the Ge and the
Fermi level is Eg-p=0.67-0.096=0.57eV. The barrier is this height, minus the
difference between the conduction band edge in the GaAs and the Fermi level, or
the barrier to electrons=0.57-0.098≈0.47 eV.
 b) Find the barrier to holes in eV.

The barrier to holes is equal to Eg(GaAs)-n -p=1.43-0.096-0.098=1.24eV.

c) Find the built-in voltage (measured at Evac) in eV.

From the figure, Vbi  GaAs  0.47  Ge  4.07  0.47  4.0  0.54eV

d). To what is the physical source of the discontinuity in the vacuum level
attributed?

The discontinuity is caused by the tunneling-induced dipoles.