Name: T.M.M.Chanaka

Registration Number: 712256083

Centre: Kandy

PHYSICAL ELECTRONICS

ECX5239PRESENTATION 1

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

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

qConductivity of insulator is nearly zero

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

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

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

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 = 0.0259eVUsing 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 = 0.0259eVUsing equation (2);Ef – Ev = 0.2395 eV & at here also Ec – Ef = 0.2652 eV

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)

Question 03

I. Upon what physical factors does mobility depend?

(ii) Why and how does the mobility depend on doping? Explain.

• The Conductivity of a semiconductor can be controllably increased by “doping”.

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

•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

II. Briefly explain the difference between the i-v relationships of a silicone and a

gallium arsenide diode.