written 6.0 years ago by |
If a current carrying conductor or semiconductor is placed in a transverse magnetic field, a potential difference is developed across the specimen in a direction perpendicular to both the current and magnetic field. The phenomenon is called HALL EFFECT. As shown consider a rectangular plate of a p-type semiconductor of width ‘w’ and thickness ‘d’ placed along x-axis. When a potential difference is applied along its length ‘a’ current ‘I’ starts flowing through it in x direction.
As the holes are the majority carriers in this case the current is given by
$I=n_h Aev_d$ ……………………………(1)
where $n_h$ = density of holes
A = w × d = cross sectional area of the specimen
$v_d$ = drift velocity of the holes.
The current density is
J= I/A =$n_h ev_d $ ……………………..(2)
The magnetic field is applied transversely to the crystal surface in z direction. Hence the holes experience a magnetic force
$F_m=ev_d B$ …………………………….(3)
In a downward direction. As a result of this the holes are accumulated on the bottom surface of the specimen.
Due to this a corresponding equivalent negative charge is left on the top surface.
The separation of charge set up a transverse electric field across the specimen given by,
$E_H= V_H/d $ …………………………..(4)
Where $V_H$ is called the HALL VOLTAGE and $E_H$ the HALL FIELD.
In equilibrium condition the force due to the magnetic field B and the force due to the electric field $E_H$ acting on the charges are balanced. So the equation (3)
$eE_H=ev_d B$
$E_H=v_d B$ ……………………………….(5)
Using equation (4) in the equation (5)
$V_H=v_d B d $………………………….(6)
From equation (1) and (2), the drift velocity of holes is found as
$v_d= I/(en_h A)= J/(en_h ) $ ……………………..(7)
Hence hall voltage can be written as
$V_H= IBd/(en_h A) = (J_x Bd)/(en_h )$
An important parameter is the hall coefficient defined as the hall field per unit current density per unit magnetic induction.
$R_H= E_H/(J_x B) $