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5.1: Physics of the Hall Effect

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    18965
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    Hall effect devices are direct energy conversion devices that convert energy from a magnetic field to electricity. The physics behind these devices is described by the Lorentz force equation. This discussion follows references [3] and [9]. If we place a charge in an external electric field, it will feel a force parallel to the applied electric field. If we place a moving charge in an external magnetic field, it will feel a force perpendicular to the applied magnetic field. The Lorentz force equation \[\overrightarrow{F} = Q\left(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B} \right) \nonumber \]

    describes the forces on the moving charge due to the external electric and magnetic fields. In the above equation, \(\overrightarrow{F}\) represents force in newtons on a charge moving with velocity \(\overrightarrow{v}\) in units \(\frac{m}{s}\). The quantity \(\overrightarrow{E}\) represents the electric field intensity in units \(\frac{V}{m}\), and \(\overrightarrow{B}\) represents the magnetic flux density in units \(\frac{Wb}{m^2}\). Charge in coulombs is denoted by \(Q\). Notice that the force on the charge due to the electrical field points in the same direction as the electrical field while the force on the charge due to the magnetic field points perpendicularly to both the velocity of the charge and the direction of the magnetic field.

    The Hall effect occurs in both conductors and semiconductors. In conductors, electrons are the charge carriers responsible for the effect while in semiconductors, both electrons and holes are the charge carriers responsible for the effect [9]. A hole is the absence of an electron. Consider a piece of semiconductor oriented as shown in Fig. \(\PageIndex{1}\)a. Assume the length is specified by \(l\), the width is specified \(w\), and the thickness is specified by \(d_{thick}\). For a typical Hall effect device, these dimensions may be in the millimeter range. Furthermore, assume the semiconductor is p-type with hole concentration \(p\) in units \(m^{-3}\). The charge concentration represents the net, or excess, charge density above a neutral material. Materials with a net negative charge, excess valence electrons, will have a positive value for the electron concentration \(n\) and are called n-type. Materials with a net positive charge, an excess of holes, will have a positive value for the hole concentration \(p\) which represents the density of holes in the material and are called p-type. Overall charge density is related to \(n\) and \(p\) by \[\rho_{ch} = -qn + qp \nonumber \]

    where \(q\) is the magnitude of the charge of an electron.

    Assume the semiconductor is placed in an external magnetic field oriented in the \(\hat{a}_z\) direction, with magnetic flux density \[\overrightarrow{B} = B_z\hat{a}_z. \nonumber \]

    Also assume a current is supplied through the semiconductor in the \(\hat{a}_x\) direction. The positive charge carriers in the semiconductor, holes, move with velocity \(\overrightarrow{v} = v_x\hat{a}_x\) because current is the flow of charge per unit time. These measures are illustrated in Fig. \(\PageIndex{1}\)b. Hall effect devices are typically used as sensors as opposed to energy harvesting devices because power must be supplied from this external current and because the amount of electricity produced is typically quite small.

    The force on the charges can be found from the Lorentz force equation. The force due to the external magnetic field on a charge of magnitude \(q\) is given by

    \[q\overrightarrow{v} \times \overrightarrow{B} = qv_x\hat{a}_x \times B_z\hat{a}_z = -qB_z\hat{a}_y \nonumber \]

    and is oriented in the \(-\hat{a}_y\) direction. Positive charges accumulate on one side of the semiconductor as shown in Fig. \(\PageIndex{1}\)c. This charge build up causes an electric field oriented in the \(\hat{a}_y\) direction which opposes further charge build up. Charges accumulate until an equilibrium is reached when the forces on the charges in the \(\hat{a}_y\) direction are zero.

    \[\overrightarrow{F} = 0 = Q\left(\overrightarrow{E} + \overrightarrow{v} \times \overrightarrow{B} \right) \nonumber \]

    5.1.1.png
    Figure \(\PageIndex{1}\): Illustration of Hall effect.

    The electric field intensity can be expressed as a function of the voltage \(V_{AB}\) measured across the width of the device, in the \(\hat{a}_y\) direction.

    \[\overrightarrow{E} = \frac{V_{AB}}{w}\hat{a}_y \nonumber \]

    \[q\overrightarrow{E} = -q\overrightarrow{v} \times \overrightarrow{B} \nonumber \]

    \[\frac{V_{AB}}{w} = v_xB_z \label{5.1.8} \]

    While the magnitude of the velocity of the charges \(v_x\) is often not known, the applied current, \(I_x\), in units amperes, is known. The current density through a cross section of the device is the product of the charge concentration, the strength of the charges, and the velocity of the charges.

    \[\text{current density} = \frac{I_x}{w \cdot d_{thick}} = q \cdot v_x \cdot p \nonumber \]

    From the above expression, velocity can be expressed in terms of the current.

    \[v_x = \frac{I_x}{w \cdot d_{thick} \cdot q \cdot p} \label{5.1.10} \]

    Equations \ref{5.1.8} and \ref{5.1.10} can be combined.

    \[V_{AB} = \frac{w \cdot I_x \cdot B_z}{w \cdot d_{thick} \cdot q \cdot p} \nonumber \]

    A magnetometer is a device that measures magnetic field. To use a Hall effect device as a magnetometer, start with a piece of semiconductor of known dimensions and known charge concentration, and then apply a current. If the voltage perpendicular to the current is measured, the magnetic field can be calculated. The measured voltage is proportional to the strength of the external magnetic flux density.

    \[B_z = \frac{d_{thick} \cdot q \cdot p \cdot V_{AB}}{I_x} \nonumber \]

    Voltage is easily measured with a voltmeter, so no specialized tools are needed. To reliably measure this voltage, it is often amplified.

    Alternatively, if the strength of an external magnetic field is known, the Hall effect can be used to measure the concentrations of holes or electrons in a piece of semiconductor. With some algebra, we can write the hole concentration as a function of the dimensions of the semiconductor, the known magnetic field strength, the applied current, and the measured voltage.

    \[p = \frac{I_x \cdot B_z}{d_{thick} \cdot q \cdot V_{AB}} \label{5.1.13} \]

    An analogous expression can be found if electrons instead of holes are the dominant charge carrier. The sign of this measured voltage is also used to determine whether a piece of semiconductor is n-type or p-type [58].

    The Hall resistance \(R_H\) is a parameter inversely proportional to the charge concentration, and it has the units of ohms [9] [59]. For the assumptions above, the Hall resistance is defined as

    \[R_H = \frac{B_z}{qp} \cdot \frac{w}{l \cdot d_{thick}}. \label{5.1.14} \]

    By combining Equations \ref{5.1.13} and \ref{5.1.14}, it can be written in terms of the measured voltage and applied current.

    \[R_H = \frac{V_{AB}}{I_x} \cdot \frac{w}{l} \nonumber \]

    As an example, suppose that a piece of silicon with a hole concentration of \(p = 10^{17} cm^{-3}\) is used as a Hall effect device. The device has dimensions \(l = 1 cm\), \(w = 0.2 cm\), and \(d_{thick} = 0.2 cm\), and it is oriented as shown in Fig. \(\PageIndex{1}\). The material has a resistivity of \(\rho = 0.9 \Omega \cdot cm\). A current of \(I = 1 mA\) is applied in the \(\hat{a}_x\) direction. The device is in an external magnetic field of \(\overrightarrow{B} = 10^{-5}\hat{a}_z \frac{Wb}{cm^2}\). If a voltmeter is connected as shown in the figure, what voltage \(V_{AB}\) is measured?

    \[V_{AB} = \frac{I_x \cdot B_z}{q \cdot d_{thick} \cdot p} = \frac{1^{-3} \cdot 10^{-5}}{1.6 \cdot 10^{-19} \cdot 0.2 \cdot 10^{17}} = 3.1 \cdot 10^{-6} V \nonumber \]

    Signals in the millivolt range are easily detected with a standard voltmeter, yet signals in the microvolt range often can be measured with some amplification. What output power is generated by this device? We can calculate the resistance along the \(\hat{a}_y\) direction. The resistivity of the silicon was given in the problem, and resistance \(R\) and resistivity \(\rho\) are related by

    \[R = \frac{\rho \cdot \text{length}}{\text{area}}. \nonumber \]

    The resistance across the width of the device is

    \[R_{width} = \frac{\rho w}{ld_{thick}} = \frac{0.09 \cdot 0.2}{1 \cdot 0.2} = 0.09 \Omega \nonumber \]

    We can use this calculated resistance and the measured voltage to find the power converted from the magnetic field to electrical power of the device.

    \[P = \frac{V^2_{AB}}{R_{width}} = 1.1 \cdot 10^{-11} W \nonumber \]

    This amount of power is tiny. While this device can make a useful sensor, it will not make a useful energy harvesting device. It generates tens of picowatts of power, and a \(1 mA\) current must be supplied to generate the power.


    This page titled 5.1: Physics of the Hall Effect is shared under a CC BY-NC 4.0 license and was authored, remixed, and/or curated by Andrea M. Mitofsky via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.