all about may ecx5243 lesson

1. Name: T.M.M.Chanaka
Registration Number: 712256083
Centre: Kandy
PHYSICAL
ELECTRONICS
ECX5239
PRESENTATION 1

2.  QUESTION 01
Conductivity
I. Why is the conductivity of insulators,
when compared to that of a
semiconductor negligible?
Ability of a substance to conduct an electrical current
 q
 - Number of charge carriers per unit volume (free electrons)
q - Electron charge
- Charge mobility

3. Conductivity of insulators
The band-gap is large, the valance
band is full, and the conduction band
is empty.
 Interatomic bonding is ionic or
strongly covalent .The valence
electrons are tightly bonded.
In insulators there are no free
electrons to move throughout the
material.
According to the following equation
 q
Conductivity of insulator is nearly zero

4. Conductivity of semiconductors
The band-gap is smaller, the
valance band is full, and the
conduction band is empty.
 In semiconductors, bonding is
predominantly covalent .( relatively
weak )
These electrons are more easily
removed by thermal excitation and
free to move.
•Hence semiconductors have conductivity
The Conductivity of insulator is negligible as compared to
the conductivity of semiconductor

5. II. Briefly define the phrase “thermal
equilibrium”
Two physical systems are in thermal equilibrium if
no heat flows between them when they are connected
by a path permeable to heat

6. The location of the Fermi Level with respect to Ec & Ev
• no = total concentration
of electrons
• Nc = concentration of
available electron states
• Pv = total concentration
of holes
• Nv = concentration of
available holes states
• k = Boltzmann constant
(8.62* 10^(-5) eV/K)
• T = available
Temperature
Using Following Equations;
equation (1)
equation (2);
Question 02

7. I. Determine the location of the Fermi level with respect to
E c and E v when ,when T = 300K
• Phosphorous is a n type semiconductor.
Hence no = cm^(-3) & No of si = 2.8×10^19 cm^(-3) & kT =
0.0259eV
Using equation (1);
Ec – Ef = 0.2 eV
• no = pv & Nv of si = 1.04×10^19 cm^(-3)
Hence using equation (2);
Ef – Ev = 0.18 eV
ii). If 10^15 boron atoms per cc replace the
phosphourous atoms.
• Boron is a p type semiconductor.
Hence pv = 10^15cm^(-3) & No = 2.8×10^19 cm^(-3) and kT =
0.0259eV
Using equation (2);
Ef – Ev = 0.2395 eV & at here also Ec – Ef = 0.2652 eV

8. iii). When T = 600K, Repeat (i) ;
Therefore kT = 0.05172 eV and Using equation (1);
Ec – Ef = 0.4 eV and also
Ef – Ev = 0.36 eV
iv). Comment and comparison of (i) and (iii) ;
T = 300K;
Ec - Ef = 0.2 eV
Ef - Ev = 0.18eV
T = 600K;
Ec - Ef = 0.4 eV
Ef - Ev = 0.36eV
2×(Ec - Ef)T=300K = (Ec - Ef) T= 600k
• Therefore (Ec – Ef) is proportional to the T
(and also same of Ef - Ev)

9.  Question 03
I. Upon what physical factors does mobility depend?

10. (ii) Why and how does the mobility depend on doping?
Explain.
• The Conductivity of a semiconductor can be
controllably increased by “doping”.

11. Question 04
I. Explain the movement of the energy bands when a diode is
forward-biased.
Fermi level -EF
conduction band edge-EC
valance band edge- EV

12. •Now, when a p-n junction is built, the Fermi energy EF attains a
constant value.
• In this case the p-sides conduction band edge. Similarly n–side
valance band edge will be at higher level than Ecn, n-sides
conduction band edge of p - side.
•This energy difference is known as barrier energy. The barrier
energy is EB = Ecp - Ecn = Evp - Evn

13. II. Briefly explain the difference between the i-v
relationships of a silicone and a
gallium arsenide diode.