Science, asked by shreejabalaji16, 6 months ago

In a semiconductor, the concentration of free
electrons is 7 x 1015/cm and that of holes is 3 x
1021/cm?. The semiconductor is​

Answers

Answered by adityaabhisawane
2

1. Carrier Concentration

a) Intrinsic Semiconductors

- Pure single-crystal material

For an intrinsic semiconductor, the concentration of electrons in the conduction band is

equal to the concentration of holes in the valence band.

We may denote,

ni : intrinsic electron concentration

pi : intrinsic hole concentration

However,

ni = pi

Simply,

ni :intrinsic carrier concentration, which refers to either the intrinsic electron or hole

concentration

Commonly accepted values of ni at T = 300°K

Silicon 1.5 x 1010 cm-3

Gallium arsenide 1.8 x 106 cm-3

Germanium 2.4 x 1013 cm-3

b) Extrinsic Semiconductors

- Doped material

The doping process can greatly alter the electrical characteristics of the semiconductor.

This doped semiconductor is called an extrinsic material.

n-Type Semiconductors (negatively charged electron by adding donor)

p-Type Semiconductors (positively charged hole by adding acceptor)

c) Mass-Action Law

n0 : thermal-equilibrium concentration of electrons

p0 : thermal-equilibrium concentration of holes

n0p0 = ni

2 = f(T) (function of temperature)

The product of n0 and po is always a constant for a given semiconductor material at a

given temperature.

d) Equilibrium Electron and Hole Concentrations

Let,

n0 : thermal-equilibrium concentration of electrons

p0 : thermal-equilibrium concentration of holes

nd : concentration of electrons in the donor energy state

pa : concentration of holes in the acceptor energy state

Nd : concentration of donor atoms

Na : concentration of acceptor atoms

Nd

+ : concentration of positively charged donors (ionized donors)

Na

-

: concentration of negatively charged acceptors (ionized acceptors)

By definition,

Nd

+ =Nd - nd

Na- = Na – pa

by the charge neutrality condition,

n0 + Na- = p0 + Nd

+

or

n0 + (Na - pa) = p0 + (Nd – nd)

assume complete ionization,

pa = nd = 0

then, eq # becomes,

n0 + Na = p0 + Nd

by eq # and the Mass-Action law (n0p0 = ni

2

)

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