Video: Hall Effect and a SIMPLE Derivation of The Hall Voltage

In the lesson on the Hall effect, you learned how Hall voltage is generated. In the following we want to derive a formula for the Hall voltage \( U_{\text H} \), which depends only on the quantities we can determine in the experiment.

We consider a Hall plate of width \( h \), thickness \( d \) and length \( L \). It can be made of a metal or a semiconductor.

Then we send an electric current \( I \) through the Hall plate. Electric current \( I \) means that in the material - positive or negative charge carriers move in a certain direction with a drift velocity \( v \). These charge carriers are either negatively charged (electrons with the charge \( q = -e\)) or positively charged (so-called holes with the charge \( q = +e\)).

Perpendicular to the direction of electron motion (\(v \perp \class{violet}{B}\)), a magnetic field with constant magnetic flux density \( \class{violet}{B} \) penetrates the Hall sample.

Electric and magnetic force inside the Hall sample

On the moving charge carriers always acts a magnetic force \( F_{\text m} \) (Lorentz force), which deflects the charge carriers perpendicular to the magnetic field and perpendicular to the direction of motion. In this case, the magnitude of the Lorentz force is:

Formula for magnetic force

Formula anchor$$ \begin{align} F_{\text m} ~=~ q \, v \, \class{violet}{B} \end{align} $$

Note! Lorentz force has a different direction, depending on whether you put for \( q \) negative elementary charge \( -e \) (for electrons) or positive elementary charge \( +e \) (for holes).

Due to the deflection of the charge carriers by the Lorentz force, there is a negative charge excess at one edge of the sample and a positive charge excess at the other edge. This charge difference generates an electric field \( E \), which exerts a electric force \( F_{\text e} \) on all subsequent charges. It acts against the magnetic force.

The top and bottom edge of the Hall sample can be considered as plates of a parallel plate capacitor. The electric force between the plates is given by:

Formula for electric force in a plate capacitor

Formula anchor$$ \begin{align} F_{\text e} ~=~ q \, E \end{align} $$

The electric force is not only directed in the opposite direction to the magnetic force; it also becomes larger the more the charge carriers have been deflected by the magnetic force. After a short time a equilibrium of forces between the magnetic and the electric force is established, which is why you may equate the formulas 1 and 2:

Electric force equated with the magnetic force

Formula anchor$$ \begin{align} q \, v \, \class{violet}{B} ~=~ q \, E \end{align} $$

With the equilibrium of forces, the electric field stabilizes at a certain value. At the edges of the sample this E-field can be measured as Hall voltage \( U_{\text H} \). Electric field is related to the voltage across the distance \( h \) between the edges of the sample: \(E = \frac{U_{\text H}}{h} \). Substitute the equation into the formula 3 for the E field. Charge \( q \) cancels out:

Formula anchor$$ \begin{align} U_{\text H} ~=~ h \, v \, \class{violet}{B} \end{align} $$

You cannot measure the average velocity \( v \) of the charge carriers directly, so substitute it using the formula for uniform motion. If a charge carrier travels the distance \( L \) (length of the sample) within the time \( t \), then we can write the velocity as follows:

Velocity is distance per time

Formula anchor$$ \begin{align} v ~=~ \frac{L}{t} \end{align} $$

The time \( t \) can then be determined using the formula for electric current:

Current is charge per time

Formula anchor$$ \begin{align} I ~=~ \frac{Q}{t} \end{align} $$

Rearrange Eq. 6 with respect to time:

Time equals charge per current

Formula anchor$$ \begin{align} t ~=~ \frac{Q}{I} \end{align} $$

Substitute equation 7 into equation 5:

Velocity equals distance times current divided by charge

Formula anchor$$ \begin{align} v ~=~ \frac{L \, I}{Q} \end{align} $$

Here, the amount of total charge \( Q \) moving through the sample is the product of the number \( N \) of moving charge carriers and their individual charge \( q \) (depending on the type of charge, \(q\) can be positive or negative):

Total charge is number of charge carriers miltiplied by single charge

Formula anchor$$ \begin{align} Q ~=~ N \, q \end{align} $$

Substitute Eq. 9 into Eq. 8 to get the following formula for drift velocity:

Speed is distance multiplied by current divided by charge carrier number multiplied by charge

Formula anchor$$ \begin{align} v ~=~ \frac{L \, I}{N \, q} \end{align} $$

Substituting the velocity 10 into the Hall voltage formula 4 yields:

Formula for Hall voltage via distance, current, thickness, magnetic field, number of charges and charge

Formula anchor$$ \begin{align} U_{\text H} ~=~ \frac{L\, I \, h \, \class{violet}{B}}{N \, q} \end{align} $$

We can express the number of charge carriers \( N \) contributing to the current with the charge carrier density \( n \). Charge carrier density is defined as the number of charge carriers \( N \) per volume \( V \) of the conductor (here: Hall sample):

Charge carrier density is number of charges per volume

Formula anchor$$ \begin{align} n ~=~ \frac{N}{V} \end{align} $$

Therefore, replace charge carrier number \( N \) in equation 11 using Eq. 12 by rearranging 12 for \(N\):

Formula for Hall voltage via distance, current, thickness, magnetic field, charge carrier density, volume and charge

Formula anchor$$ \begin{align} U_{\text H} ~=~ \frac{L\, I \, h \, \class{violet}{B}}{n \, V \, q} \end{align} $$

Now you can simplify the formula 13 a little bit. The volume \(V= h \, L \, d\) is the product of height \( h \), length \( L \) and thickness \( d \) of the Hall sample. Substitute it into Eq. 13 and then cancel \(L\) and \(h\):

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