Majority and Minority carriers

Mobile charge carriers and immobile

ions
Conduction in Metals

Mobility:
Consider a metal to which a constant electric field E volts/m is applied. Due to the electric
field the electrons are accelerated from rest, acquires a linearly increasing velocity for a short
time. However due to collision the electrons lose energy which reduces its velocity to zero.
Again the electrons are accelerated and the process repeats itself. As a consequence ,electron
acquires a constant average drift velocity proportional to the electric field. Its direction is
opposite to that of the electric field.
Hence,
v∝ 𝐸
v=µ 𝐸.
Where µ is the mobility of the electron expressed in 𝑚2 /volt-sec
Current Density

Current Density
Consider a conductor of length L and cross sectional area A as shown in fig
Let the number of electrons contained in the length L is
equal to N. If the time taken for an electron to travel distance L meter is T
Then total number of electrons passing through any cross section of the conductor in unit time
is N/T.
Total charge ,Q=Nq L
𝑄 A(Area
Current is given by I = )
𝑇
𝑁𝑞
 = Metal or
𝑇
1 conducto
(N)r
electrons
Contd…..

𝐿
The drift velocity of an electron, v=𝑇
𝐿 2
therefore T= 𝑣

Substituting Eq 2 in 1
𝑁𝑞 𝑁𝑞𝑣
I= = 𝐿
𝐿/𝑣
𝐼 𝑁𝑞𝑣
The current density J= = where J is A/𝑚2
𝐴 𝐿𝐴
𝑁
is the electron concentration ‘n’ expressed as electron/ 𝑚3
𝐿𝐴
𝑁
Therefore n= and J=nqv
𝐿𝐴
And J=ρv where ρ=nq is the charge density in C/ 𝑚3

Hence we can write

J=nqµ𝐸
J=σ𝐸 [since v=µ 𝐸]

Where σ=nqµ
Conductivity in intrinsic
semiconductor
In intrinsic semiconductor, current distribution takes place by the charge carriers, electrons
and holes. Therefore the total current density is equal to sum of the electron and hole current
densities.
The total current density,
J=𝐽𝑛 + 𝐽𝑝 𝐽𝑛 =current density due to electrons
𝐽𝑝 =current density due to holes
Since J=nqµ𝐸
The electron current density , 𝐽𝑛 = nqµ𝑛 𝐸 where n=no. of electrons/ 𝑚3
Similarly, the hole current density=pqµ𝑝 𝐸 where p=no. of holes/ 𝑚3
Therefore, J= nqµ𝑛 𝐸 + nqµ𝑝 𝐸 =(n µ𝑛 +p µ𝑝 )q 𝐸 = σ𝐸
Where σ =(n µ𝑛 +p µ𝑝 )q is called conductivity of semiconductor
Contd…………

For an intrinsic semiconductor,

n=p= 𝑛𝑖 (𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛
= ℎ𝑜𝑙𝑒 𝑐𝑜𝑛𝑐𝑒𝑛𝑡𝑟𝑎𝑡𝑖𝑜𝑛)
Therefore, J= ( µ𝑛 + µ𝑝 )q 𝑛𝑖 𝐸
For n-type semiconductor, n>>p
σ = µ𝑛 qn
Therefore

And for p-type semiconductor, p>>n

Therefore σ = µ𝑝 qp
Numericals
1.Find the intrinsic carrier concentration of Germanium, if its intrinsic resistivity at 300K is
0.47 Ω-m. It is given that the electronic charge is 1.6x10−19 Coulombs and that electron and
hole mobilities at 300K are 0.39 and 0.19 𝑚2 /volt respt.
Sol: Given ρ= 0.47 Ω-m ;q= 1.6x10−19 Coulombs; µ𝑛 = 0.39 𝑚2 /volt ; µ𝑝 = 0.19 𝑚2 /volt
Let 𝑛𝑖 =intrinsic carrier concentration of germanium
1 1
Since conductivity σ= = =2.13 (Ω−m)−1
ρ 0.47
And conductivity in intrinsic semiconductor
σ =( µ𝑛 + µ𝑝 ) 𝑛𝑖 q
2.13=(.39+.19)x 1.6x10−19 x 𝑛𝑖
2.13
𝑛𝑖 = .93x10−19 =2.3x10−19 / 𝑚3 Ans
 Variation in semiconductor parameters

1.Intrinsic concentration
In an intrinsic semiconductor, the concentration (𝑛𝑖 ) increases with increase in
temperature.
The dependence on the temperature is given by the relation
𝑛𝑖 = 𝐴0 .𝑇 3 . 𝑒 − 𝐸𝐺0𝑘.𝑇
Where 𝐴0 =a constant independent of temperature
T=Temperature in K
𝐸𝐺0 =Energy gap at 0 K(in eV)
K=Boltzmann’s constant(in eV/K)
Contd………

2.Mobility:
The mobility (µ) of an intrinsic semiconductor decreases with increase in
temperature because at high temperature, the no of carriers are more and they
are more energetic. This causes an increased number of collisions with the atoms
and thus the mobility decreases.
3.Conductivity
The conductivity (σ) of an intrinsic semiconductor depends upon the number of
hole-electron pairs and mobility. The number of hole-electron pair increases with
increase in temperature, while its mobility decreases. Therefore the
conductivity of an intrinsic semiconductor increases with increase in
temperature.
Numericals

Q. Calculate the intrinsic conductivity of Si at room temp if 𝑛𝑖 = 1.4x1016 𝑚−3 , µ𝑛 =0.145 𝑚2 /Volt
sec, µ𝑝 =0.05 𝑚2 /Volt sec and q=1.6x 10−19 C. What are the individual contribution made by
electrons and holes.
Sol: Given 𝑛𝑖 = 1.4x1016 𝑚−3 ; µ𝑛 =0.145 𝑚2 /Volt sec; µ𝑝 =0.05 𝑚2 /Volt sec ;q=1.6x 10−19 C.
Conductivity of an intrinsic semiconductor
σ= ( µ𝑛 + µ𝑝 )q 𝑛𝑖
=q 𝑛𝑖 µ𝑛 + q 𝑛𝑖 µ𝑝
contribution by e- contribution by hole

= 1.6x 10−19 x1.4x1016 x 0.145 + 1.6x 10−19 x1.4x1016 x 0.05

=0.325x 10−3 + 0.112x 10−3 (Ω−m)−1
contribution by electron= 0.325x 10−3 (Ω−m)−1 and contribution by hole= 0.112x 10−3 (Ω−m)−1
 Mass action Law

Under thermal equilibrium, the product of the number of holes and the number
of electrons is constant and is independent of the amount of donor and acceptor
impurity doping. This relation is known as mass-action law and is given by the
relation

n.p= 𝑛𝑖 2

where n=free electron concentration

p=hole concentration
𝑛𝑖 = intrinsic carrier concentration
 Charge densities in N and P semiconductor

 Let 𝑁𝐷 =concentration of donor atoms (i.e the no. of positive charges/𝑚3

contributed by donor
 ions)
 𝑁𝐴 = concentration of acceptor atoms (i.e the no. of negative charges/𝑚3
contributed by
 acceptor ions)
 p=hole concentration (i.e the no of holes /𝑚3 )
 n=free electron concentration (i.e the no of free electrons /𝑚3 )
 Therefore,
 Total no. of positive charges /𝑚3 in a semiconductor= 𝑁𝐷 + p
 And total no. negative charges /𝑚3 in a semiconductor= 𝑁𝐴 +n
Contd…………

Since the semiconductor as a whole, is electrically neutral therefore magnitude

of total positive charge density is equal to total negative charge density i.e
𝑁𝐷 +p= 𝑁𝐴 +n
For N type semiconductor, the concentration of acceptor atoms is zero ( 𝑁𝐴 =0)
and also the number of free electrons is much larger than the number of holes
(i.e n>>p).Therefore
we can write as 𝑁𝐷 =n

𝑛𝑖 2 𝑛𝑖 2
And concentration of holes p= =
𝑛 𝑁𝐷
Contd…..

Similarly, in a P-type semiconductor

𝑁𝐷 =0
and p>>n
Therefore, we can write as
p= 𝑁𝐴

The concentration of free electrons in P-type semiconductor

𝑛𝑖 2 𝑛𝑖 2
 n= =
𝑝 𝑁𝐴
Numericals

Q. Calculate the donor concentration in N-type germanium having resistivity of

100 Ω-m.
Take q=1.6x10−19 C and µ𝑛 =.36𝑚2 /volt-sec.
1 1
Sol: Conductivity of Ge(σ)=ρ = 100 =0.01(Ω𝑚)−1

Again conductivity of germanium

0.01=qnµ𝑛
=1.6x10−19 x n x0.36
0.01
n= =1.74x 1017 atoms/ 𝑚3
0.36x1.6x 10−19

Since 𝑁𝐷 =n= 1.74x 1017 atoms/ 𝑚3 Ans

Find the resistivity of intrinsic Ge at 300K? If the donor type impurity is added
to the extent of 1 atom per 107 Ge atoms, what is the resistivity?

 Sol:For Ge, µ𝑛 = 3800 𝑐𝑚2 /V-sec

 µ𝑝 = 1800 𝑐𝑚2 /V-sec
n𝑖 = 2.5x 1013 𝑐𝑚−3
Conductivity σ= n𝑖 q(µ𝑛 + µ𝑝 )= 2.5x 1013 x1.6x 10−19 x(3800+1800)
=0.0224 (Ω𝑐𝑚)−1
1
Therefore resistivity ρ=σ =44.643 Ω𝑐𝑚

For each 107 Ge atoms, one impurity atom is added.

4.4𝑥1022
Therefore the no of atoms= 107 =4.4x 1015 atoms/ 𝑐𝑚3

Hence Number of Donor atoms N𝐷 = 4.4x 1015 atoms/ 𝑐𝑚3

 n= N𝐷 = 4.4x 1015 atoms/ 𝑐𝑚3
𝑛𝑖 2 2.5x 1013𝑥2.5x 1013
 p= = =1.42x 1011
𝑁𝐷 4.4x 1015
 Since p<<n
 Conductivity σ= nqµ𝑛
 =4.4x 1015 x1.6x 10−19 x3800
 =2.6752 (Ω𝑐𝑚)−1
1
 Resistivity ρ=σ = 0.3738 Ω𝑐𝑚 Ans
Numericals

 Q. Determine the concentration of the free electrons and holes in a sample of Ge at 300K
which has a concentration of donor atoms equal to 2x 1014 atoms/ c𝑚3 and a concentration
of acceptor ions equal to 3x 1014 atoms / c𝑚3 .

 Sol: Given 𝑁𝐷 = 2x 1014 atoms/ c𝑚3 ; 𝑁𝐴 =3x 1014 atoms / c𝑚3 ;

 n=number of free electrons and p=no of holes
 We know that number of free electrons
𝑛𝑖 2
 n= 𝑁𝐴

𝑛𝑖 =intrinsic concentration=2.3x 1019

2.3𝑥1019 𝑥2.3𝑥1019
n= =1.76 x 1024 electrons/ / c𝑚3 .
3𝑥1014
𝑛𝑖 2 2.3𝑥1019 𝑥2.3𝑥1019
And number of holes p= p= = =2.65x 1024 holes /c𝑚3 .
𝑁𝐷 2𝑥1014
Diffusion current in a semiconductor

 Consider a bar of P-type semiconductor having a non

uniform concentration of holes as shown in fig.
 Consider an imaginary surface Y-Y’ .The density of holes
on left side of YY’ is larger than that of the right side.
This means that there exist a concentration
gradient𝑑𝑝ൗ𝑑𝑥 whose value is given by
𝑑𝑝 𝑃𝑜−𝑃𝑥
 ൗ𝑑𝑥 =-
𝑥
 where Po=conc. Of holes at left end of the
semiconductor bar Px=conc. Of holes at the imaginary
surface YY’
 Because of the concentration gradient ,the holes from
left of the YY’ surface diffuse towards the right side.
This produces an electric current called diffusion
current .,
Contd……..

𝑑𝑝
 The magnitude of diffusion current per unit area is given by the 𝐽relation
𝑝 =-q 𝐷𝑝 . ൗ𝑑𝑥
Where q= charge of a hole in Coulomb; 𝐷𝑝 = Diffusion constant for holes in 𝑚2 /sec
𝑑𝑝
 ൗ𝑑𝑥=conc. gradient

 Similarly for N type semiconductor,

 𝐽𝑛 =-q 𝐷𝑛 . 𝑑𝑛Τ
𝑑𝑥
Where 𝐷𝑛 = diffusion constant for electrons in 𝑚2 /sec
 Total current in semiconductor is the sum of drift current and diffusion current.
 Thus for a P-type semiconductor, the total current per unit area
𝑑𝑝
 𝐽𝑝 = q. p.µ𝑝 . 𝐸 -q 𝐷𝑝 . ൗ𝑑𝑥
 Similarly, total current density for an N-type semiconductor
 𝐽𝑛 = q. n.µ𝑛 . 𝐸 -q 𝐷𝑛 . 𝑑𝑛Τ𝑑𝑥

 Einstein’s relationship for a semiconductor

 According to Einstein’s relationship for a semiconductor, the ratio of diffusion

constants (𝐷𝑝 𝑜𝑟𝐷𝑛 ) to the mobility (µ𝑛 or µ𝑝 ) of the charge carriers is
constant and is equal to the volt equivalent of the temperature(𝑉𝑇 ) .
 Mathematically,
𝐷𝑝 𝐷𝑛
=
µ𝑝 µ𝑛
Properties of intrinsic silicon and Germanium