physical electronic

1. Assignment #01
Name : D.A.T.P. Diddeniya
Reg No :110006073
Center : Colombo

2. 1.Introduction.
2.Energy Bands & Energy Band Diagrams.
3.The Conductivity Difference in Different Materials(Using
Energy Band Diagrams).
4.The Conductivity Variation in different semi-conductors
with temperature variation.
5.Answers for Q3.
6.Answers for Q6.

3. There are three categories of solids, based on
their conducting properties:
*Insulators
*conductors
*semiconductors
01.Introduction

4. In order of conductivity: conductors,
semiconductors, insulators
Conductors : material capable of carrying electric current, i.e.
material which has “mobile charge carriers”.
(electrons, ions,..)
E.g. :- metals, liquids with ions (water, molten ionic compounds),
plasma
Insulators : materials with no or very few free charge carriers.
E.g. :- quartz, most covalent and ionic solids, plastics
Semiconductors : materials with conductivity between that of
conductors and insulators
E.g. :- germanium(Ge) , silicon(Si) , GaAs, GaP, InP
Introduction (Cont..)

5. Electrical Conductivity
The electrical conductivity of solids,
depends upon the energy levels of
arbitrary electron, in general the valence
electron.
Introduction (Cont..)
Electrical Conductivity () may occur by
Electrons or IONS.
 qn
σ= niq(me + mh)

6. 02. Energy Bands And
Energy Band Diagrams
Valence Band
Energy Bands
(Shells)
Si has 14 Electrons
At T=0K, the
highest energy
band occupied by
an electron is
called the valence
band.
Silicon has 4 outer shell /
valence electrons

7. Electrons revolve around the nucleus in different energy levels or shells and
each shell is associated with definite energy. The energy of the K shell is the
least while those of L, M, N and O shells increases progressively. We also know
that any system that has least energy is the most stable.1st energy level is K
shell2nd energy level is L shell3rd energy level is M shell4th energy level is N shell
and so on
Energy Bands And Energy Band Diagrams (Cont..)

8. In solid materials, electron energy levels form bands of allowed energies,
separated by forbidden bands
Valence band : outermost (highest) band filled with electrons
(“filled” = all states occupied)
Conduction band : next highest band to valence band
(empty or partly filled)
Gap : energy difference between valence and
conduction bands, = width of the forbidden
band
The essential feature of the band theory is that the allowed energy states
for electrons are nearly continuous over certain ranges, called energy
bands, with forbidden energy gaps between the bands.
Energy Bands And Energy Band Diagrams (Cont..)

9. 03. The Conductivity Difference in Different
Materials (Using Energy Band Diagrams)

10. 04. The Conductivity Variation in different
semi-conductors with Temperature
Variation
Let's look at the factors that go into conductivity of a semiconductor and
consider how each of these are affected:
σ= ni q (me + mh)
Lastly, let's consider what will happen to ni for semiconductors as
temperature increases. The electrons in the valance band will gain energy
and go into the higher energy levels in the conduction band where they
become charge carriers! So this term will increase. Not only will it increase,
but it will increase exponentially! (Promoting electrons from the valance
band into the conduction band is a thermally activated process.)
Now consider mobility. The effect of an increase in temperature on
mobility is the same as it was for conductors. With the same reasoning, we
see that the drift velocity will decrease causing the mobility to decrease.
First let's consider q. As with conductors, as temperature increases,
the charge on each carrier will not change.

11. 05.Answers for Q3
σ =eNdµe
≈ (1.602 × 10-19 C)(1017 cm-3)(800 cm2 V-1 s-1)
= 12.8 Ω-1 cm-1
i) The conductivity of the sample is

12. In intrinsic Si, 𝐸𝑓 = 𝐸𝑓𝑖,
𝑁 𝑑
𝑛𝑖
= 𝑒−(
𝐸 𝑓𝑛−𝐸 𝑓𝑖
𝐾𝑇 )
𝑙𝑛(
𝑁 𝑑
𝑛𝑖
) = (
𝐸𝑓𝑛 − 𝐸𝑓𝑖
𝐾𝑇
)
𝐾𝑇 ∗ 𝑙𝑛
𝑁 𝑑
𝑛𝑖
= 𝐸𝑓𝑛 − 𝐸𝑓𝑖 = 𝐹𝑒𝑟𝑚𝑖 𝑙𝑒𝑣𝑒𝑙
𝐹𝑒𝑟𝑚𝑖 𝑙𝑒𝑣𝑒𝑙 = 𝐾𝑇 ∗ 𝑙𝑛
𝑁 𝑑
𝑛𝑖
∆𝐸 = (8.617 × 10-5 eV/K)(300 K)ln[(1017 cm-3)/ (1.45 × 1010 cm-3)]
= 0.407 eV

13. b)
The sample is further doped with Na = 9 × 1016 cm-3 = 0.9 × 1017cm-3 acceptors.
Due to compensation, the net effect is still an n-type semiconductor but with an
electron concentration given by,
n = Nd − Na = 1017 cm-3− 0.9 × 1017 cm-3
= 1 × 1016 cm-3 (>> ni)
σ = eNdµe
≈ (1.602 × 10-19 C)(1016 cm-3)(700 cm2 V-1 s-1)
= 1.12 Ω-1 cm-1

14. 𝐹𝑒𝑟𝑚𝑖 𝑙𝑒𝑣𝑒𝑙 = 𝐾𝑇 ∗ 𝑙𝑛
𝑁 𝑑
𝑛𝑖
=(0.02586 eV) * ln[(1016 cm-3) / (1.45 × 1010 cm-3)]
=0.348 eV

15. 06.Answers for Q6
Fraction of Conduction Band,
𝑛
𝑁
= 𝑒−(
𝐸 𝑔
2∗𝐾𝑇
)
Fraction of Conduction Band in Diamond,
𝑛
𝑁
= 𝑒−(
5.6𝑒𝑉
2∗0.026𝑒𝑉
)
=1.7x10-47 𝑒𝑉

16. Thank You……!