How to derive Hall's coefficient for a semiconductor?

Answer 1

''http://mems.caltech.edu/courses/EE40%20Web%20Files/Supplements/02_Hall_Effect_Derivation.pdf''

All the derivation is precisely explained.

You need to equate Newton second law to Lorentz equation, assuming the magnetic field is normal to current flow. You need do this separately for electrons and for holes. Both velocity and electric field have two components, transversal (y) and longitudinal (x) and the magnetic field is in z. It is more comfortable separate the vectorial equations into two equations, one for each component. So, you have four equations:
1) For time derivative of x-velocity of electrons
2) For time derivative of x-velocity of holes
3) For time derivative of y-velocity of electrons
4) For time derivative of y-velocity of holes


The set is complicate, but it is easier if you do the following approximation:
If magnetic field is small, you can assume that the second term in equations 1) and 2) is negligible. Yo cannot do the same with equations 3) and 4) because, since mobility of electrons and holes are not much different, this would give you as a result that transversal field does not exist, and you will not get anything.

You must bear in mind that velocities in your equations are individual velocities and you need average velocities. If you use the single model of a carrier starting from zero velocity and accelerating at a constant rate and suddenly stopped at a time tau, you can obtain the average velocities for the four equations multiplying the right side of each times tau/2. You can make sense of all this by plotting changing velocity as a saw tooth signal.

In agreement with mobility definition (charge times half-tau over effective mass) you can see that equations 1) and 2) are no more that the classic lineal relation between average velocity and electric field, with the mobility as a constant. Equations 3) and 4) have both a similar first term but the second term, including the magnetic field is more weird.

Now, you have average x-velocity for electrons, average x-velocity for holes, average y-velocity for electrons and average y-velocity for holes. If you multiply each velocity by charge and concentration you will get x-current density for electrons, x-current density for holes, y-current density for electrons and y-current density for holes. When this is done, you can add contributions of both types of carriers and obtain x-current density and y-current density.

What you will use for calculating the Hall´s coefficient is transversal current density which must be zero. Replace this zero in the formula for y-current density and you will obtain two terms equated to zero or rearranging, an equality between two terms. One of the terms is already in the shape we want, but the other still include longitudinal velocities for electrons and holes. You have expressions for them and must replace in the expression for y-current density. Now, you must identify expressions that can be replaced by electron and hole mobilities.

Now, you only have carrier concentrations, carrier mobilities, longitudinal and transversal fields and magnetic field magnitude. Solve for transversal electric field. For obtaining Hall´s coefficient you need to multiply both numerator and denominator for conductivity. Keep conductivity in the denominator in terms of concentrations and mobilities. This is not important for the numerator because you will must note that conductivity times longitudinal electric field is longitudinal density current. Now you have transversal electric field as a linear function of the product of current density (longitudinal, the only non-zero current density) times the magnetic field magnitude. The proportionality constant is Hall´s coefficient.

It is not easy!

Lydia Alvarez, Mexicali B.C. Mexico
lydia@iing.mxl.uabc.mx

Answer 2

R = Uh*d/(I*B),

where: R - Hall coefficient, Uh - Hall voltage, d - semiconductor thickness, I - current, B - magnitute of flux density.