Extrinsic carrier concentration refers to the number of free electrons or holes in a semiconductor after adding impurity atoms through doping. Unlike intrinsic semiconductors, where carriers are generated naturally, doping introduces extra energy levels that increase electrical conductivity and make the material useful for electronic devices.
When donor or acceptor impurities are added, the semiconductor becomes n-type or p-type, increasing majority charge carriers. The carrier concentration mainly depends on doping level and temperature, and it plays an important role in understanding semiconductor conductivity and device behavior.
Now, to understand extrinsic carrier concentration in detail, let us study it separately for n-type and p-type semiconductors, where the roles of majority and minority charge carriers are different.
Carrier Concentration in n-type Semiconductor
Let ND be the concentration of donor atoms in an n-type semiconductor. At 0 K, donor atoms are not ionized and lie at the donor energy level ED, which is very close to the bottom of the conduction band energy level EC. When the temperature increases above 0 K, donor atoms get ionized and free electrons appear in the conduction band.
Hence, the electron concentration in the conduction band is given by:
$$n = N_D^+ =N_D-N_D^0$$
where,
ND+= number of ionized donor atoms and
ND0= number of unionized donor atoms
The concentration of ionized donors is:
$$N_D^+ =N_D\left[ 1-f\left( E_D \right) \right] ———-(1)$$
And as per the Fermi–Dirac distribution function:
$$f\left( E_D \right) =\frac{1}{1+ exp\left( \frac{E_D-E_F}{kT} \right)}———–(2)$$
In an n-type semiconductor, the Fermi level lies between the donor level and the conduction band bottom level. Hence, it is negative.
Therefore:
$$1-f\left( E_D \right) =exp\left( \frac{E_D-E_F}{kT} \right)—————(3)$$
Substituting this in eq. (1), we get the following:
$$n=N_D^+ =N_D exp\left( \frac{E_D-E_F}{kT} \right)————–(4)$$
But electron concentration in the conduction band is also given by:
$$n= N_C exp\left[ -\left( \frac{E_C-E_F}{kT} \right) \right]————-(5)$$
Comparing the above equations (4) and (5), we get:
$$N_C exp\left[ -\left( \frac{E_C-E_F}{kT} \right) \right]=N_D exp\left( \frac{E_D-E_F}{kT} \right)$$
$$=>exp\left[ \left( \frac{2E_F-E_D-E_C}{kT} \right) \right]=\frac{N_D}{N_C}$$
Taking the logarithm and solving, the Fermi level is:
$$E_F=\frac{E_D+E_C}{2}+\frac{kT}{2}log_{e}\left( \frac{N_D}{N_C} \right)———-(6)$$
At T = 0 K:
$$E_F=\frac{E_D+E_C}{2}$$
Thus, the Fermi level lies midway between the donor level and the bottom of the conduction band.
Substituting the value of EF from equation (6) in equation (5), we get:
$$n= N_C exp\left[ -\left( \frac{E_C-\left( \frac{E_D+E_C}{2}+\frac{kT}{2}log_{e}\left( \frac{N_D}{N_C} \right) \right)}{kT} \right) \right]$$
$$n= N_Cexp\left( \frac{E_D-E_C}{2kT} \right)exp\left[ \frac{1}{2}log_{e}\left( \frac{N_D}{N_C} \right) \right]$$
$$n= \left(N_DN_C\right)^{1/2}exp\left( \frac{E_D-E_C}{2kT}\right)$$
The final expression for electron concentration is:
$$n= \left(2N_D\right)^{1/2}\left( \frac{2\pi m_e^*kT}{h^2} \right)^{3/4}exp\left( \frac{E_D-E_C}{2kT}\right)$$
Carrier Concentration in p-type Semiconductor
Let NA be the concentration of acceptor atoms in a p-type semiconductor. At 0 K, acceptor atoms are not ionized and lie at the acceptor energy level, which is very close to the valence band energy level. When the temperature increases, the acceptor starts ionizing, and correspondingly, an electron from the valence band jumps to the acceptor level; therefore, a hole is created in the valence band.
Hence, the hole concentration is given by:
$$p = N_A^-$$
where,
NA–= number of ionized acceptor atoms
The concentration of ionized acceptor is:
$$N_A^-=N_Af\left( E_A \right) ———-(7)$$
And as per the Fermi–Dirac distribution function:
$$f\left( E_A \right) =\frac{1}{1+ exp\left( \frac{E_A-E_F}{kT} \right)}———–(8)$$
Since the acceptor level lies above the Fermi level, (EA – EF) is positive. Therefore:
$$p=N_A exp\left( \frac{E_F-E_A}{kT} \right)————–(9)$$
But hole concentration in the valence band is also given by:
$$p= N_V exp\left[ -\left( \frac{E_F-E_V}{kT} \right) \right]————-(10)$$
Comparing the above equations (9) and (10), we get:
$$N_V exp\left[ -\left( \frac{E_F-E_V}{kT} \right) \right]=N_A exp\left( \frac{E_F-E_A}{kT} \right)$$
$$=>exp\left[ \left( \frac{2E_F-E_A-E_V}{kT} \right) \right]=\frac{N_V}{N_A}$$
Taking the logarithm and solving, the Fermi level is:
$$E_F=\frac{E_A+E_V}{2}+\frac{kT}{2}log_{e}\left( \frac{N_V}{N_A} \right)———-(11)$$
At T = 0 K:
$$E_F=\frac{E_A+E_V}{2}$$
Thus, the Fermi level lies midway between the acceptor level and the top of the valence band.
Substituting the value of EF from equation (11) in equation (9), we get:
$$p= N_A exp\left( \frac{\left( \frac{E_A+E_V}{2}+\frac{kT}{2}log_{e}\left( \frac{N_V}{N_A} \right)-E_A \right)}{kT} \right)$$
$$p= N_Aexp\left( \frac{E_V-E_A}{2kT} \right)exp\left[ \frac{1}{2}log_{e}\left( \frac{N_V}{N_A} \right) \right]$$
$$p= \left(N_AN_V\right)^{1/2}exp\left( \frac{E_V-E_A}{2kT}\right)$$
The final expression for hole concentration is:
$$p= \left(2N_A\right)^{1/2}\left( \frac{2\pi m_h^*kT}{h^2} \right)^{3/4}exp\left( \frac{E_V-E_A}{2kT}\right)$$
Important Questions
Conceptual / Theory Questions
- What is an extrinsic semiconductor?
- Define majority and minority carriers.
- Why does doping increase carrier concentration drastically?
- Why is electron concentration high in n-type semiconductors?
- Why is hole concentration high in p-type semiconductors?
- What is the role of donor and acceptor impurities in carrier concentration?
- Why does an extrinsic semiconductor remain electrically neutral after doping?
- How does temperature affect carrier concentration in extrinsic semiconductors?
Derivation-Based Questions
- Derive expressions for electron and hole concentrations in an n-type semiconductor.
- Derive expressions for carrier concentration in a p-type semiconductor.
- Show donor and acceptor energy levels and explain their role in carrier concentration.