Table of contents 
Hall Effect 
Fermi Level in Intrinsic and Extrinsic Semiconductors 
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When a magnetic field is applied to a currentcarrying conductor in a direction perpendicular to that of the flow of current, a potential difference or transverse electric field is created across a conductor. This phenomenon is known as the Hall Effect.
According to this effect the statements are:
If a specimen (metal or semiconductor) carrying a current I is placed in a transverse magnetic field B, an electric field E is induced in the direction perpendicular to both I and B. This phenomenon, known as the Hall effect, is used to determine whether a semiconductor is n or ptype and to find the carrier concentration. Also, by simultaneously measuring the conductivity σ, the mobility μ can be calculated.
Consider the figure shown below. Here current l is in +xdirection, magnetic field B is in +z direction then induced electric field will be in negative ydirection.
Semiconductor Bar
Hence a force will be exerted in the negative ydirection on the current carriers.
The current l may be due to holes moving from left to right or to free electrons travelling from right to left in the semiconductor specimen. Hence, independently of whether the carriers are holes or electrons, they will be forced downward toward side 1 of above figure.
If the semiconductor is ntype material, so that the current is carried by the electrons, these electrons will accumulate on side 1, and this surface becomes negativity charged with respect to side 2. Hence a potential, called the Hall voltage, appears between surface 1 and 2.
Now under the equilibrium condition
qE = Bvq
But, E = V_{H}/d and j = vρ = I/wd
Combining these relationships, we find
V_{h} = Ed = Bvd = Bjd/ρ = BI/ρw
Where ρ is the charge density, w is the width of the specimen and d is the distance between surfaces 1 and 2.
It is customary to introduce the Hall coefficient R_{H} defined by
R_{H} = 1/ρ
Hence, R_{h} = V_{H}w/BI
By hall experiment mobility of charge carriers is given as
μ = 8/3π σR_{H}
or, μ ≅ σR_{H}
⇒ Hall coefficient, R_{H} Temperature coefficient of resistance of given specimen.
⇒ For metals, σ is larger, V_{H} is small.
⇒ For semiconductors, σ is small, V_{H} is large.
The expression for the Hall voltage is given by:
V_{H} = IB/qnd
[Note: Minority carrier mobility (μ) and diffusion coefficient(D) can be measured independently with the help of HaynesShockley experiment.]
Example 1: A n type Ge Sample has a donor density N_{D} = 10^{21} atoms/m^{3}. It is arranged in a Hall experiment having a magnetic field B = 0.2 Wb/m^{2} and current density J = 500 A/m^{2}. Find the hall voltage generated when the thickness of sample is 2 mm. Also calculate the field intensity induced in magnitude ?
Solution:
V_{H} = BJDR_{H}
V_{H }= 0.2 x 500 x 2 x 10^{3} x
⇒ V_{H }= 1.25mV
If E > EF then f(E) < ½
If E > EF then f(E) > ½
A closer examination of f(E) indicates that at 0 K the distribution takes the simple rectangular form shown in figure. At temperature higher than 0K, some probability exists for states above the Fermi level to be filled.
In intrinsic semiconductor Fermi level E_{F} is given by
where, N_{C} = density of states in conduction band
N_{v} = density of states in valence band
In pure Semiconductor at T = 0K, Fermi level lies in the middle of bandgap.
Fermi level in ntype semiconductor is given by
E_{F} = E_{c}  kT In (N_{c}/N_{d})
Where, N_{D} = doping concentration.
shift = kT
shift ≅ kT
Fermi level in ptype semiconductor is given by
E_{f} = E_{v} + kT in (N_{v}/N_{A})
shift = kT
shift ≅ kT
Example 1: If the Fermi energy in silicon is 0.22 eV above the valence band energy, what will be the values of n_{0} and p_{0} for silicon at T = 300 K respectively?
Solution: Given that Fermi energy in silicon is 0.22 eV above the valence band energy, i.e.
E_{F} – E_{v} = 0.22 eV
So, we obtain the hole concentration as
p_{0} = N_{v} exp
= 1.04 x 1019 exp
= 2.13 x 10^{15}cm^{3}
Now, the energy bandgap for silicon is 1.12 eV, i.e.
E_{g} = E_{c} – E_{v} = 1.12 eV
Therefore, we obtain
Hence, the hole concentration is
n_{0} = N_{c} exp
= 2.8 x 10^{19} exp
= 2.27 x 10^{4} cm^{3}
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