Taking logarithms on both sides, we get

……………… (29)

The number of free electrons per unit volume of semiconductor having energies in between E and E + dE is represented as  N(E) dE 

dE = width of  Energy band

Therefore, we have:

N(E) dE = ge(E) dE fe(E)   ……….(1)

ge(E) = The density of electron states per unit volume

fe(E)  = Fermi-Dirac distribution function i.e. probability that an electron occupies an electron state

The number of electrons present in the conduction band per unit volume of material ‘n’ is obtained by integrating N(E) dE between the limits Ec and Ect 

Where Ec = the bottom energy levels of conduction band

 Ect = the bottom and top energy levels of conduction band

n =    ……….(2)

can be written as

n = =

n = -   ……….(3)

we know that above Ect, there is no electrons.

Hence, Equation (3) becomes

n = -

n =     ……….(4)

The Fermi-Dirac distribution function fe(E) can be represented as:

(E) =      ……….(5)

Compared to the exponential value,  so the ‘1’ in the denominator can be neglected.

So  >>1

Hence, (E) = =   ……….(6)

The density of electron states ge(E) in the energy space from E = 0 to E can be written as:

(E) = [ ]3/2  (E - 0) ½    ……….(7)

where me* is the effective mass of an electron and h is Planck’s constant.

(E)dE = [ ]3/2  (E - 0) ½dE     ……….(8)

To evaluate n, the density of states is counted from Ec, since the minimum energy state in conduction band is Ec. so eq (8) can become

(E)dE = [ ]3/2  (E - Ec) ½dE     ……….(9)

Substituting Equations (6) and (9) in (4) gives

n = 3/2 (E-)1/2dE  

n = [ ]3/2 1/2dE  ………… (10)

The above equation can be simplified by the following substitution:

Put        ɛ = E − Ec     ………… (11)

So,      dɛ = dE

In Equation (11), Ec is constant, as we change the variable E to ε in Equation (10), the integral limits also change.

In Equation (11),  as E → Ec then ε → 0 and E → ∞, then ε also → ∞.

the exponential term in Equation (10) becomes:

= exp [] = exp []  

    ………… (12)

Substituting Equations (11) and (12) in (10), we get:

n = [ ]3/2 dE 

n = [ ]3/2 dE  ………… (13)

Above integral (I) can be simplified by substitution. Put ε = x2

so that        dɛ = 2x dx

I =2x dx

=dx

= =T)3/2    ………… (14)

Substituting Equation (14) in (13) gives:

n = [ ]3/2T)3/2   

 n  =

n =

n =    ………… (15)

The term   is almost a constant compared with the exponential term as the temperature changes. So, it is a pseudo constant and is given by the symbol Nc. So, we have

n =Nc     ………… (16)

Density of Holes in Valence Band

The number of holes per unit volume of semiconductor in the energy range E and E + dE in valence band is represented as P(E) dE.  Proceeding same way( as in case of electrons) we have 

Therefore, we have:

P(E) dE = gh(E) dE fh(E)      ……….(17)

dE = width of  Energy band

gh(E) = The density of holes states per unit volume

fh(E)  = Fermi-Dirac distribution function i.e. probability that an hole occupies an electron state

The number of electrons present in the conduction band per unit volume of material ‘n’ is obtained by integrating P(E) dE  between the limits Evb and EV 

where  EV = the bottom energy levels of valence band

 Evb  = the bottom and top energy levels of valence band

The total  number of holes present in the valence band per unit volume of material ‘p’ is obtained by integrating P(E) dE 

……….(18)

Equation (18) can be represented as:

 ……….(19)

 Now we know that below  Evb  no holes is present. Hence, Equation (19) becomes

    ……….(20)

We know hole can also be defined as absence of an electron.

presence of a hole = the absence of an electron

Hence, the Fermi-Dirac function of holes fh(E) in the valence band is:

Compared to exponential, the ‘1’ in the denominator is negligible,

Hence,   

         ……….(21) 

The density of hole states between E and E + dE in valence band can be written similar to Equation (8.9) for electrons.

   ……….(22) 

Where mh* is the effective mass of hole.

Substituting Equations (21) and (22) in (20),

……….(23) 

The above equation can be simplified by the substitution:

Put        ɛ = EV − E    ............. (24)

so        dɛ = − dE

In Equation (24), EV is constant, as we change the variable E to ε in Equation (23), the integral limits also change.

In Equation (24),

as E → EV then ε → 0

and E→ −∞, then ε → ∞

the exponential term in Equation (23) becomes:

............. (25)

Substituting Equations (24) and (25) in (23), we get:

………….(26)

From Equation (14), we know the integral value T)3/2 put here and we get

………….(27)

….The term   is almost constant compared with the exponential term as the temperature changes. So, it is a pseudo constant and is given by the symbol NV. So, we have

   …………… (28)

At equilibrium, the Lorentz force on a carrier  

 FL = Bevd      ……………..(1)

and the Hall force   

FH = eEH       ……………..(2)

Where EH is the Hall electric field due to accumulated charge.

At equilibrium,  FH = FL

eEH = Bevd

∴ EH = Bvd       ……………..(3)

If ‘d’ is the distance between the upper and lower surfaces of the slab, then the Hall field

EH =       ……………..(4)

In n-type material, Jx = –nevd

vd = -       ……………..(5)

Where n is free electron concentration, substituting (5) in (3), we have

 ∴ EH = -B        ……………..(6)

For a given semiconductor, the Hall field EH is proportional to the current density Jx and the intensity of magnetic field ‘B’ in the material.

i.e.  EH ∝ JxB

(or)  EH = RHJxB      ……………..(7)

Where RH = Hall coefficient

Equations (6) and (7) are the same so, we have

RHJxB  =-B   

RH = - = -       ……………..(8)

Where ρ is the charge density

 Similarly for p-type material

RH = =       ……………..(9)

Using Equations (8) and (9), carrier concentration can be determined.

Thus, the Hall coefficient is negative for n-type material. In n-type material, as the more negative charge is deposited at the bottom surface, so the top face acquires positive polarity and the Hall field is along negative Y-direction. The polarity at the top and bottom faces can be measured by applying probes.

Similarly, in the case of p-type material, a more positive charge is deposited at the bottom surface. So, the top face acquires negative polarity and the Hall field is along positive Y-direction. Thus, the sign of the Hall coefficient decides the nature of (n-type or p-type) material.

The Hall coefficient can be determined experimentally in the following way:

Multiplying Equation (7) with ‘d’, we have

EHd = VH = RHJxBd      ……………..(10)

From (Figure 15) we know the current density Jx

Jx =

Where W is the width of the box. Then, Equation (10) becomes

VH = RHBd = RH

RH =       ……………..(11)

Substituting the measured values of VH, Ix, B, and W in Equation(11), RH is obtained. The polarity of VH will be the opposite for n- and p-type semiconductors.

The mobility of charge carriers can be found by using the Hall effect, for example, the conductivity of electrons is

n = neμn

Or we can rewrite it as

μn = =n RH     ……………..(12)

by using equation (11)

μn = n       ……………..(13)

The solution to Schrödinger's equation for the Kronig-Penney potential is obtained by assuming that the solution is a Bloch function, namely a traveling wave solution of the form, eikx, multiplied with a periodic solution, u(x), which has the same periodicity as the periodic potential . The total wave function is therefore of the form:

where u(x) is the periodic function as defined by u(x) = u(x + a), and k(x) is the wave number. We now rewrite Schrödinger equation using this wave function considering first region I, between the barriers where V(x) = 0 and then region II, the barrier region where V(x) = V0:

In region I, Schrödinger's equation becomes:

          (1)

with

Since the potential, V(x), is finite everywhere, the solutions for uI(x) and uII(x) must be continuous as well as their first derivatives. Continuity at x = 0 results in:

and continuity at x = a-b combined with the requirement that u(x) be periodic:

Finally, continuity of the first derivative at x = a-b, again combined with the requirement that u(x) is periodic, results in:

 (15)

As a result we have four homogenous equations, (7), (9), (13), and (16), with four unknowns, A, B, C, and D, for which there will be a solution if the determinant of this set of equations is zero, or:

The first row of the determinant represents equation (7), the second row is obtained by combining (13) and (7), the third row represents equation (9) and the fourth row represents equation (16). This determinant can be rewritten as two determinants, each with three rows and columns, while replacing cosb(a-b) by bc, sinb(a-b) by bs, coshab eika by ac and sinhab eika by as, which results in:

 (18)

where eika + e-ika was replaced by 2coska. Since the result is independent of the sign of k, this equation also covers all solutions obtained when replacing e-ik by eik in equations (5) and (6).

A further simplification is obtained as the barrier width, b, is reduced to zero while the barrier height, V0, is increased to infinity in such manner that the product, bV0, remains constant and the potential becomes a delta function train at x = a and repeated with a period of a, namely bV0 (x - b - na) where n is an integer. As b approaches zero, sinh b approaches b. Equation (20) then reduces to:

The right hand side of equation (21) shows the relation between the energy (through ) and the wave-vector, k, and as you can see, since the left hand side of the equation can only range from −1 to 1 then there are some limits on the values that   (and thus, the energy) can take, that is, at some ranges of values of the energy, there is no solution according to these equation, and thus, the system will not have those energies: energy gaps. These are the so-called band-gaps, which can be shown to exist in any shape of periodic potential (not just delta or square barriers). This has been plotted in Figure.

Figure 4: Graph of the right hand side of equation (21) of P for P =2

Differentiating (1) w.r.t to T, we get

  =

=     is the temperature coefficient of resistance

Taking log of (1) and simplifying we get

=         …………..(2)

So

=      …………..(3)

A graph plotted between and in x-axis for NTC and PTC is shown below. The slope of the graph gives the value of

Figure 29: NTC - PTC Characteristics

The force is only due to electron cloud acting upon nucleus would only be due to the portion of the cloud enclosed by the sphere of radius x. portion outside the sphere of radius x does not apply any force on the nucleus.

 Volume of the sphere of radius x = (4/3)πx3  and

 Volume of the sphere of radius R = (4/3)πR3

 Now total negative charge of the electron cloud is -Ze  …..(2)

 Hence, the quantity of negative charge enclosed by the sphere of radius x is,

 [-Ze/(4/3)πR3] * (4/3)πx3  = -Ze (x3/ R3)   ……..(3)

 According to coulomb’s law = q1q2 /4 πR2
Here it becomes charge on electrons q1=-Ze (x3/ R3)

 Charge on nucleus q2= Ze

  So coulomb’s force = {-Ze (x3/ R3) * Ze}/4πƐox2 = Z2e2x/4πƐo R3  ……(4)

Note- magnitude is taken into account. Neglect negative sign.

i.e. At equilibrium Electrostatic force = Coulomb force .

EZe = Z2e2x/4πƐoR3   ……..(5)

Upon simplify

x = {4πƐoR3/Ze} E

Now dipole moment = either charge * separation between charges i.e. x

    = Ze *{4πƐoR3/Ze} E

 = 4πƐoR3E

Polarization is number of dipole moment per unit volume. Let us suppose N is the number of dipoles per unit volume so

Pe=4πƐoR3EN

Hence force F1 due to electric field is

F1=qE    ……(1)

Where q is the net charge on ion.

F2 restoring force due to binding force between ions

F2=kd  …..(2)

Where k is spring constant. Here we are considering bond as spring.

k may also expressed in terms of other constant which is related to the shape of  interatomic potential  as young’s modulus or modulus of rigidity as

k=Ydo      …..(3)

At Equilibrium F1 = F2 at equilibrium distance d

qE = kd 

qE = Ydod    using (3)

d = qE/Ydo

Now dipole moment p= qd = q2E/Ydo …..(4)

and polarization  Pi = Nq2E / Ydo ……(5)

This is called ionic polarization. This can also be expressed as

Pi =NαiE

Potential energy U of the dipole moment making an angle θ with the electric field. Where p0 is the dipole moment.

U=−p0⋅E=−p0Ecosθ

According to statistical mechanics, the number of dipoles or molecules making an angle with the electric field is proportional to

e−U/kT

Letting n(θ) be the number of molecules per unit solid angle at θ, we have

n(θ)=n0e+p0Ecosθ/kT

For normal temperatures and fields, the exponent is small, so we can approximate by expanding the exponential:

n(θ)= n0 (1+p0EcosθkT)

To find n0 , we integrate  over all angles and the result is equal to  N.

Where N is the total number of molecules per unit volume. The average value of cosθ over all angles is zero, so the integral is just n0 times the total solid angle 4π. We get

n0=N4π  or   N= n0 /4π

There exits molecules along the field (cosθ=1) than against the field (cosθ=−1). So in any small volume containing many molecules there will be a net dipole moment per unit volume.

 As dipole moment is present so polarization Pd is given

Pd =∑unitvolumep0cosθi

Evaluate Total Polarization over solid angle 2πsinθdθ. We have to integrate it

 Pd = 0∫π n(θ)p0cosθ2πsinθdθ

Substitute the value of n(θ)

Pd = −N/2 −1∫1 {1+(p0E/kT)cosθ)p0cosθd(cosθ)

By integrating we get,

 Pd =Np02E/ 3kT

The polarization is proportional to the field E and depends inversely on the temperature, because at higher temperatures there is more randomness by collisions.

This can also be expressed as

Pd =αdE