I3600 Homework 2
3. A silicon sample is maintained at 300K and characterized by the energy band
diagram in the following figure. Do the following:
a) Sketch the electrostatic potential (V) inside the semiconductor as a function of
x.
b) Sketch the electric field (E) inside the semiconductor as a function of x.
c) The carrier pictured on the diagram moves back and forth between x=0 and
x=L without changing its total energy. Sketch the kinetic energy and potential
energy of the carrier as a function of position inside the semiconductor. Let E F
be the energy reference level.
d) Roughly sketch n and p versus x.
e) Is the semiconductor degenerate at any point? If so, where?
4. A silicon step junction diode with a cross-sectional area A  10 4 cm 2 has a
doping of N A  1017 cm 3 and N D  1015 cm 3 . Let  n  801cm 2 / V  s and
 n  0.1s on the p-side; and let  p  477cm 2 / V  s and  n  1s on the n-side.
a) Calculate the current through the diode at room temperature
( KT / q  0.026V ) if:
i) VA  50V
ii) VA  0.1V
iii) VA  0.2V
b) Assuming that the mobilities and lifetimes do not vary significantly with
temperature, repeat part (a) for T  500 K
c) Summarize in your own words what has been exhibited by this problem.
Solution:
The current is given by:
 aVa

I  I o  e KT  1 where


 N

D

D
P
I o  qA N n po  P p no   KTq A
n po 
p no 
 n

LP
p
 LN



at room temperature or T  300 K , we have that ni  1.18  1010 cm 3 where we
have used Einstein relationship and the definition of the diffusion length. Thus
the doping on the n-type and p-type side will be much more than the intrinsic
carrier concentration. Hence we have:
n po 
ni2
 3.2  10 3 cm 3
NA
and
p no 
ni2
 3.2  10 5 cm 3
ND
With these values and the values stated in the problem, the equation for the
current at T  300 K becomes becomes:

 As
I V   1.1  10 14 e 38.45V  1
For a temperature of T  500 K , we have that ni  3  1014 cm 3 . Hence we have:
n po
ni2

 9.0  1011 cm 3
NA
and
p no 
ni2
2
 8.3  1013 cm 3
ND
N 
  D   ni2
2
 2 
With these values and the values stated in the problem, the equation for the
current at T  500 K becomes becomes:

 As
I V   6.3  10 6 e 23.21V  1
Thus we have the values of the current for the three voltages and different
temperatures as:
Voltage
-50
-0.1
0.2
Current at T  300 K
15
 8.2  10
 8.0  10 15
1.9  10 11
Current at T  500 K
 6.3  10 6
 5.6  10 6
6.5  10 4
We see that there is a higher reverse saturation current for the higher temperature diode.
Also, with increasing temperature, the diode will start to behave more like an intrinsic
device with a less step turn on characteristic.