Tutorial 7
Semiconductor
1
Formulae
1 2
KE= m v
2
I=ne v d A
1
P ( E )= ( E − EF )
kT
1+e
2
Numerical 1
Calculate the speed of conduction electron in copper having its kinetic
energy equal to Fermi energy of 7.0 eV. Also calculate the drift velocity if a
current of 5 A is flowing in a copper wire with a cross section of 0.5 mm 2.
28
Free electron density in copper is 8.5×10 /m3 and the charge on the electron
-19
is 1.6×10 C. Which velocity is greater? Why? Comment on the results
3
Solution
1 2 1 −19 1 −31 2
KE= m v ⇒ 7 eV = m v 2 ⇒ 7 ×1.6 × 10 J= × 9.1× 10 v
2 2 2
28 − 19 −6
I=ne v d A ⇒ 5=8.5 ×10 × 1.6 × 10 × v d × 0.5 × 10
4
Numerical 2
Find the probability with which an energy level 0.02 eV below Fermi level
will be occupied at room temperature of 300 K and 1000 K. What do the
results signify?
5
1
P ( E )= ( E − EF )
kT
1+e
As the energy level is below the Fermi level, ( E − E F ) =−0.02
− 23 J − 5 eV
k=1.38 × 10 =8.63 × 10
K K
Thus, at T = 300 K, At T = 1000 K ?
1
P ( E )= −0.02
=0.68=68 %
8.63 × 10− 5 × 300
1+e
6
Numerical 3
Find the probability of an electron occupying an energy level 0.02 eV
above the Fermi level at 300 K and 1000 K. What do the results signify?
( E − E F ) =+0.02
7
Numerical 4
At any given nonzero temperature, what is the probability of occupancy
for a state whose energy is equal to Fermi energy?
1
P ( E )= ( E − EF )
kT
1+e
If T ≠ 0 K and E=E F
1
P ( E=E F )= ( EF − EF )
=0.5=50 %
kT
1+e
8
Numerical 5
Find the temperature at which there is 1 % probability that a state with
energy 0.5 eV above Fermi energy will be occupied. What does the result
signify?
5793
1 1 1 1+eT
P ( E )= ( E − EF ) 0.01= ⇒ =
5793
kT T 0.01 1
1+e 1+e
5793
5793 5793
⇒ 99=e T ⇒ ln 99= ⇒ 4.595= ⇒ T =1261 K
T T
9
Numerical 6
Calculate the free electron density in copper, if each copper atom donates one
electron to the conduction band. (Properties of copper: Density = 8.96 gm/cc,
atomic weight 63.5 and Avogadro’s number = 6.02×1023 atoms/mole)
10
Numerical 7
Resistance of copper wire of diameter 1.03 mm is 6.51 ohm per 300 m. The
concentration of free electrons in copper is 8.4×1028 /m3. If the current is 2 A, find
the mobility of free electrons
11
Numerical 8
Calculate the conductivity of pure/intrinsic silicon if free electron
concentration is 1.5×1010 /cm3 and mobility of electrons and holes are 1500
cm2/V.s and 500 cm2/V.s respectively
σ i=e ni ( μ e + μh )
n i=1.5 × 1010 / cm3=1.5 ×1016 /m3
cm 2 m2 cm 2 m2
μ e =1500 =0.15 ∧ μh =500 =0.05
V .s V .s V .s V .s
σ i =1.6 × 10−19 ×1.5 ×1016 × ( 0.15+0.05 )=4.8 ×10− 4 mho/m
12
Numerical 9
Intrinsic silicon is doped with phosphorus with the atomic ratio of 10 8 (Si) : 1 (P).
Calculate the conductivity of N type of silicon thus formed
Properties of silicon: Density = 2.3290 g/cm3
Atomic weight =28.085
Avogadro’s number = 6.02×1023
conductivity = σN = neeμe
13
Numerical 10
A strip of copper having thickness 0.5 mm is placed in a magnetic field of
magnitude 0.75T. A current of 100 mA is sent through the strip. What is the Hall
potential difference that will appear across the width of the strip? The carrier
concentration of electrons in copper is 8.47×1028 electrons/m3
14
Numerical 11
A copper specimen having length 1m, width 1cm, and thickness 1mm is
conducting with 1A along its length and is applied with a magnetic field of 1T along
its thickness. It experiences Hall effect and a Hall voltage 0.074μV appears along
its thickness. Calculate the Hall coefficient, electron concentration and the mobility
of electrons in copper. Conductivity of copper is 5.8×107 (Ωm)−1.
15
Numerical 12
N type semiconductor having length 1cm, width 1mm and thickness 0.1mm is
made to conduct with 1mA current and is placed in the magnetic field acting along
it’s thickness. The Hall voltage is measured to be 3.44×10−7V. Calculate the
magnetic field, if the Hall coefficient of the specimen is -3.44×10 −8m3/C.
16
THANK YOU
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