Carrier Concentration in Intrinsic Semiconductors

CARRIER CONCENTRATION IN INTRINSIC SEMICONDUCTORS

We know, at 0 K intrinsic pure semiconductor behaves as insulator. But as temperature increases some electrons move from valence band to conduction band as shown in Fig. 2.8. Therefore both electrons in conduction band and holes in valence band will contribute to electrical conductivity. Therefore the carrier concentration (or) density of electrons (n_e) and holes (n_h) has to be calculated.

Assume that electron in the conduction band is a free electron of mass m_C* and the hole in the valence band behaves as a free particle of mass m_h*.The electrons in the conduction band have energies lying from E_c to ∞ and holes in the valence band have energies from -∞ to E_v, as shown in Fig. 2.8. Here E_c represents the energy of the bottom (or) lowest level of conduction band and E_v represents the energy of the top (or) the highest level of the valence band.

DENSITY OF ELECTRONS IN CONDUCTION BAND

 …...(1)

From Fermi-dirac statistics we can write

 ……..(2)

Considering minimum energy of conduction band as E_c and the maximum energy can go upto we can write equation (2) as

 …….(3)

We know, Fermi function, probability of finding an electron in a given energy state is

 ………..(4)

Substituting equation (4) and (3) in equation (1) we have Density of electrons is conduction band within the limits E_c to ∞ as

………..(5)

Since to move an electron from valence band to conduction band the energy required is greater than 4K_B T. (i.e.,) E- E_F>> K_BT (or) (E- E_F)/K_BT>>1

(or) e ^{(E- E}_F^)/K_B^T >>1

1+e ^{(E- E}_F^)/K_B^T = e ^{(E- E}_F^)/K_B^T

Equation (5) becomes

 ………(6)

Let us assume that E-E_c = xK_BT

(or) E = E_C+ xK_BT

Differentiating we get dE = K_BT dx,

Limits: When E = E_C; x = 0

When E = ∞; x = ∞

Limits are 0 to ∞

Equation (6) can be written as

Density of electrons is conduction band is

 ……..(7)

DENSITY OF HOLES IN VALENCE BAND

We know, F (E) represents the probability of filled state. As the maximum probability will be 1, the probability of unfilled states will be [1-F (E)].

Example, If F (E) = 0.8 then 1 - F (E) = 0.2

i.e., 80% chance of finding an electron in valence band and 20% chance of finding a hole in valence band.

Let the maximum energy in valence band be E_v and the minimum energy be -∞. Therefore density of holes in valence band n_h is given by

 ..……(8)

 ……….(9)

Here E- E_F<< K_BT,

………..(10)

substituting equation (4) and (9) in (8), we get

 ……….(11)

Let us assume that E_v-E = xK_BT, (or) E = E_v-xK_BT

differentiating we get dE = -K_BT dx

Limits: When E = -∞

We have E_v-(-∞) = x

x = ∞

When E=E_v; x=0

Limits are ∞ to 0

Equation (11) becomes

To exclude the negative sign, the limits can be interchanged.

(or) 

density of holes in valence band is

……….(12)