Download 971 Quiz 01

1001 Quiz 01
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1. (50%)The following figure shows a silicon bar that is subjected to electron
injection from the right.
2 1016 carriers
5 10 carriers
16
cm3
cm3
electrons
Silicon
0m n  1.45 1010 carriers
i
4 m
cm3
a. Determine the minority carrier concentration at x  0 .
b. Determine the minority carrier concentration at x  4 m .
c. Determine the electron diffusion current density J ndiff ( Dn  32cm 2 / s ).
d. Determine the hole diffusion current density J pdiff ( D p  12cm 2 / s ).
e. Determine the electric field E at x  4 m (  n  1350cm 2 / V  s
 p  480cm 2 / V  s ).
f. Determine the electron drift current density J ndrift .
g Determine the hole drift current density J pdrift .
h. Determine the total current density J .
i. Which type the silicon bar is?
2. (20%) True and false
(1) If we introduce some 3-valence atom into a Si-semiconductor, the minority
carriers of this impure semiconductor are electrons and it usual denoted by
p no .
(2) The depletion layer in a p-n junction contains null of free electron but few
numbers of free holes.
(3) There are two types of carriers in a semiconductor. The holes diffusion current
direction is opposite to that of the applied electric field. But the electrons
diffusion current is in the same direction of the applied electric field.
(4) The diffusion current in a semiconductor is relational to the concentration
gradient. The following electron concentration can drive the current in the x
direction.
ND
N D (x)
ni
x0
x  L2
xL
3.(30%) The following figure shows the minority carriers distribution of
a PN junction under the forward bias. Indicate whether each of the
following statements is true or false based on the information you
learn form the figure.
Minority concentration
n p (x )
Pn (x)
pno
n po
x2
x1
x
a. The doping concentration of P side is higher than N side.
b. The depletion width x2  x1 implies the majority carriers n no  p po .
c. The electrons diffusion current flow is in the direction of x .
d. The diode current in the direction of x .
e. The concentration n p ( x1 )  p n ( x2 ) , since the doping concentration at
right hand side higher than left hand side.