THE (CLASSICAL) HALL EFFECT
So far we have not investigated the effect of a magnetic field on the transport properties of a medium and its effect on the distribution function f(k,r). The Hall effect, i.e. the effect of the application of a magnetic field on a medium, plays a central role in semiconductor technology since it is used both as a sensor and as a means of detecting the polarity of the carriers. We have left it to the end of this chapter to show that under well behaved conditions the particle description and the Boltzmann equation render the same result. However, this is not always true, there is also the Quantum Hall effect, not describable in particle terms. We will give both pictures, first the particle one (usually taught at undergraduate level) and then the one usingthe Boltzmann equation.
Consider a rectangular long wire of width w and depth d, as in figure 4.10, along the length of which a potential Vy is applied. A magnetic field Bz is also applied perpendicularly to its xy surface pointing upwards. Had there not been Bz, the electrons would have traveled along the positive у direction but because of Bz the carriers are curved along x, as shown in figure 4.10 by the magnetic force Fg =-evxB. At steady state, the accumulated electrons on one side of the width of the wire will create an electric field £ which will counterbalance the effect of Bz and we will then have zero force on the electrons.
FIGURE 4.10 Set-up for the measurement of the Hall effect; see text for the directions of the fields shown.
£ is negative since vyBz > 0. Since the charge carriers are electrons (i.e. we have an N - type semiconductor) their velocity is along the +y direction (towards the positive pole) and vxB points to the +x direction so that Fb = —evXВ points towards the -x direction and electron accumulation will occur on the left side of the specimen, producing an electrostatic voltage difference, the Hall voltage VH as shown. Had the carriers been holes the polarity of VH would have been the opposite. Hence the Hall effect may be used as a method of detecting the polarity of the carriers by just looking at VH. Equation 4.51 together with
£x= will give Now the current is Substituting for w we get
The critical reader must have seen the hidden assumption in deriving 4.54, the assumption that the induced charge is like a surface or sheet-like charge residing on the edges of the rectangular wire. We therefore proceed to give a more proper analysis based on the
Boltzmann equation. In section 4.3 we obtained a final form for this equation, equation 4.23. It is actually easier however if we work from its previous form, equation 4.21.
Assuming no temperature variation across the sample, we get from 4.21 that the deviation from equilibrium is
The first term of the RHS of equation 4.55 is exactly what one would get if there were only an electric field £ present. Guided by this, we try a solution of the form
where A is to be determined.
Assuming that we can write m*v(k) = Ш we substitute equation 4.56 in both the LHS and RHS of equation 4.55 and we get
Deleting j from both sides and rearranging we get
Solving 4.58 for A will give f'(k) and this will give in turn the currents of the Hall effect. However, we can follow a shorter path towards this goal.
We observe that expression 4.56 is of the same form as 4.25. Following the same steps as those of equations 4.25 to 4.31 (which we do not need to repeat here) we get
where a is the conductivity of the sample in the Hall effect assuming an isotropic medium. Substituting back in 4.58 we have
This equation describes fully the Hall effect. It tells us that at steady state the electric field
needed to create the current density J = Jy is not solely along у but has two components,
one £y = Jy/o and a perpendicular to this one £x = —-—B.Jy. It is left as a simple exercise
to the student to show that £x, £y are consistent with VH in equation 4.54 if q=-e. PROBLEMS
Using the general expression for the non-equilibrium distribution/(equation 4.41) obtain an expression for U in the form of equation 4.42.
4.4 Prove that the relationship between U and J is of the form
Find expressions for A,kg.
4.5 Prove the Wiedmannn-Franz law that the ratio of the electrical о to the thermal conductivity ke (defined above) is constant.