Carrier Concentration Semiconductors

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,

n

i : intrinsic electron concentration

p

i : intrinsic hole concentration

However,

n

i = pi

Simply,

n

i :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 10

10 cm-3

Gallium arsenide 1.8 x 10

6 cm-3

Germanium 2.4 x 10

13 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

n

0 : thermal-equilibrium concentration of electrons

p

0 : thermal-equilibrium concentration of holes

n

0p0 = 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,

n

0

: thermal-equilibrium concentration of electrons

p

0 : thermal-equilibrium concentration of holes

n

d : concentration of electrons in the donor energy state

p

a : concentration of holes in the acceptor energy state

N

d : concentration of donor atoms

N

a : concentration of acceptor atoms

N

d+ : concentration of positively charged donors (ionized donors)

N

a- : concentration of negatively charged acceptors (ionized acceptors)

By definition,

N

d+ = Nd - nd

Na

- = Na – pa

by the charge neutrality condition,

n

0 + Na- = p0 + Nd+

or

n

0 + (Na - pa) = p0 + (Nd – nd)

assume complete ionization,

p

a = nd = 0

then, eq # becomes,

n

0 + Na = p0 + Nd

by eq # and the Mass-Action law (

n0p0 = ni^2)

n

0 = 1/2.{(Nd - Na) + ((Nd - Na)2 + 4ni^2)^1/2}, where Nd > Na (n-type)

p

0 = 1/2.{(Na - Nd) + ((Na - Nd)2 + 4ni^2)^1/2}, where Na > Nd (p-type)

n

0 = p0 = ni, where Na = Nd (intrinsic)

If

Nd - Na >> ni,

then

n

0 = Nd - Na, p0 = ni^2 / (Nd - Na)

If

Na – Nd >> ni,

then

p

0 = Na – Nd, n0 = ni^2 / (Na – Nd)

Example 1)

Determine the thermal equilibrium electron and hole concentrations for a given doping

concentration.

Consider an n-type silicon semiconductor at

T = 300°K in which Nd = 1016 cm-3 and Na =

0. The intrinsic carrier concentration is assumed to be

ni = 1.5 x 1010 cm-3.

- Solution

The majority carrier electron concentration is

n

o = .{(Nd - Na) + ((Nd - Na)2 + 4ni^2)^1/2} ≅ 1016 cm-3

The minority carrier hole concentration is

p

0 = ni^2 / n0 = (1.5 x 1010)^2/1016 = 2.25 x 104 cm-3

- Comment

N

d >> ni, so that the thermal-equilibrium majority carrier electron concentration is

essentially equal to the donor impurity concentration. The thermal-equilibrium majority

and minority carrier concentrations can differ by many orders of magnitude.

Example 2)

Determine the thermal equilibrium electron and hole concentrations for a given doping

concentration.

Consider an germanium sample at

T = 300°K in which Nd = 5 x 10^13 cm^-3 and Na = 0.

Assume that

ni = 2.4 x 10^13 cm^-3.

- Solution

The majority carrier electron concentration is

n

o = 1/2.{(5 x 10^13) + ((5 x 10^13)^2 + 4(2.4 x 10^13)^2)^1/2} = 5.97 x 10^12 cm^-3

The minority carrier hole concentration is

p

0 = ni^2 / n0 = (2.4 x 10^13)^2/(5.97 x 10^12) = 9.65 x 10^12 cm^-3

- Comment

If the donor impurity concentration is not too different in magnitude from the intrinsic

carrier concentration, the thermal-equilibrium majority carrier electron concentration is

influenced by the intrinsic concentration.

Example 3)

Determine the thermal equilibrium electron and hole concentrations in a compensated ntype

semiconductor.

Consider a silicon semiconductor at

T = 300°K in which Nd = 10^16 cm^-3 and Na = 3 x 10^15

cm

-3. Assume that ni =1.5 x 10^10 cm^-3.

- Solution

The majority carrier electron concentration is

n

o = .{(10^16 – 3 x 10^15) + ((10^16 – 3 x 10^15)^2 + 4(1.5 x 10^10)^2)^1/2} ≅ 7 x 10^15 cm^-3

The minority carrier hole concentration is

p

0 = ni^2 / n0 = (1.5 x 10^10)^2/(7 x 10^15) = 3.21 x 10^4 cm^-3

- Comment

If we assume complete ionization and if

Nd - Na >> ni, the the majority carrier electron

concentration is, to a very good approximation, just the difference between the donor and

acceptor concentrations.

2. Carrier Transport

The net flow of the electrons and holes in a semiconductor will generate currents. The

process by which these charged particles move is called transport. The two basic

transport mechanisms in a semiconductor crystal:

- Drift: the movement of charge due to electric fields

- Diffusion: the flow of charge due to density gradients

a) Carrier Drift - Drift Current Density

Let,

J

dr : drift current density

ρ

: positive volume charge density

v

d : average drift velocity

then,

J

dr = ρvd

J

pdr = (qp)vdp (hole)

J

ndr = (-qn)vdn (electron)

J

dr = Jpdr + Jndr = (qp)vdp + (-qn) vdn

for low electric field,

v

dp = μpE (μp : proportionality factor, hole mobility)

v

dn = -μnE (μn : proportionality factor, electron mobility)

thus,

J

dr = Jpdr + Jndr = q(pμp + nμn)E

Significant drift current densities can be obtained in a semiconductor applying relatively

small electric fields. The drift current will be due primarily to the majority carrier in an

extrinsic semiconductor.
Anderson Jose Mariño Ortega
C.I. 17.456.750
E.E.S.
http://www-inst.eecs.berkeley.edu/~ee105/fa05/handouts/discussions/Discussion1.pdf