Field plate

Field plates , also known as Magnetic Dependent Resistor (MDR), are sensors made of semiconductors that react to magnetic fields by changing the electrical resistance , i.e. to a magnetoresistive effect .

construction

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  Field plate

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  Indium antimonide InSb, high purity 99.999%

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  Nickel antimonide NiSb, high purity 99.999%

Field plates are made of indium antimonide (InSb), which has high electron mobility. Short circuit bridges made of nickel antimonide (NiSb) are inserted into the material during production . NiSb has a much higher conductivity than InSb and serves to distribute the electrons homogeneously before they reach the next plate section. Otherwise you would accumulate on one side.

If no magnetic field acts on the field plate, there is a defined basic resistance; the electrons cross the plate in a correspondingly straight line when the voltage is applied. If a magnetic field now acts from the outside, the electrons are deflected and the path is lengthened and thus the resistance increases. This increase can be used as a signal to e.g. B. to determine the strength of the magnetic field, as a proximity switch or - with a pole wheel - as a tachometer.

Field plates are increasingly being replaced by AMR or GMR sensors, as these primarily have higher operating temperatures and better linearity. Field plates are therefore rarely available in series and are obsolete in most designs.

Derivation of the increase in resistance

The force that acts on a charge when it passes through electromagnetic fields is given by the Lorentz force :

${\displaystyle {\vec {F}}=q\cdot \left({\vec {E}}+{\vec {v}}\times {\vec {B}}\right)}$

For semiconductors, as for metals, Ohm's law applies , here in its microscopic formulation:

${\displaystyle {\vec {j}}=\sigma \cdot {\vec {E}}}$

Here is the current density , the conductivity and the electric field strength. The electric field strength is equal to the voltage applied to the field plate. ${\displaystyle {\vec {j}}}$ ${\displaystyle \sigma }$ ${\displaystyle {\vec {E}}}$

The conductivity can be represented like this:

${\displaystyle \sigma =q\cdot n\cdot \mu =\rho \cdot \mu }$

It is the charge of the particles, the particle density (even particle distribution), the mobility of the particles and thus the charge density (or charge distribution). ${\displaystyle q}$ ${\displaystyle n}$ ${\displaystyle \mu }$ ${\displaystyle \rho }$

This transforms Ohm's law into:

${\displaystyle {\frac {\vec {j}}{\rho }}=\mu \cdot {\vec {E}}\quad \Rightarrow \quad {\vec {v}}=\mu \cdot {\vec {E}}\qquad ,}$ because ${\displaystyle {\vec {j}}:=\rho \cdot {\vec {v}}}$

The result is the drift speed of the particles.

With the force acting on a charge due to a field, the drift speed is converted to: ${\displaystyle {\vec {F}}=q\cdot {\vec {E}}}$

${\displaystyle {\vec {v}}=\mu \cdot {\frac {\vec {F}}{q}}}$

In the case of semiconductors, the flowing current is actually given by the sum of the electron current and the hole current. Since the semiconductors for field plates (often InSb) only have a high mobility for one of the charge carriers, the other current can be neglected. InSb has about 70 times better mobility for electrons:

${\displaystyle {\vec {j}}={\vec {j}}_{n}+{\vec {j}}_{p}\approx {\vec {j}}_{n}\quad ,\mu _{n}\gg \mu _{p}}$

Let us now assume that the plate lies in the xy-plane (it is therefore thin), the E-field is in the direction of the x-axis and the B-field (more precisely the magnetic induction) in the direction of the z-axis .

From this it follows for the Lorentz force:

${\displaystyle {\vec {F}}=q\cdot \left[{\begin{pmatrix}E_{x}\\0\\0\end{pmatrix}}+{\begin{pmatrix}v_{x}\\v_{y}\\0\end{pmatrix}}\times {\begin{pmatrix}0\\0\\B_{z}\end{pmatrix}}\right]=q\cdot {\begin{pmatrix}E+v_{y}B\\-v_{x}B\\0\end{pmatrix}}\qquad ,\vert {\vec {E}}\vert =E_{x}=E,\ \vert {\vec {B}}\vert =B_{z}=B,\ \vert {\vec {v}}\vert ={\sqrt {v_{x}^{2}+v_{y}^{2}}}}$

If you plug this into the equation for the drift velocity, it follows:

${\displaystyle {\begin{pmatrix}v_{x}\\v_{y}\end{pmatrix}}=\mu {\begin{pmatrix}E+v_{y}B\\-v_{x}B\end{pmatrix}}\quad \Rightarrow \quad v_{x}=\mu E-v_{x}\mu ^{2}B^{2}\quad \Leftrightarrow \quad v_{x}={\frac {\mu E}{1+\left(\mu B\right)^{2}}}}$

The resistance of the plate certainly only depends on the current in the direction of the E-field ( ), so only the speed in this direction is of interest. As you can see, this decreases as the B-field increases. ${\displaystyle j_{x}=\rho v_{x}}$

If the speed of the charge carriers is reduced, the current flow must decrease:

${\displaystyle v_{x}\rightarrow {\frac {v_{x}}{\alpha }}\quad \Rightarrow \quad j_{x}\rightarrow {\frac {j_{x}}{\alpha }}\quad \Rightarrow \quad \sigma \rightarrow {\frac {\sigma }{\alpha }}\qquad ,{\vec {j}}=\rho \cdot {\vec {v}},\quad {\vec {j}}=\sigma \cdot {\vec {E}}}$

If the current flow decreases, the conductivity must have decreased while the E-field (voltage) has remained the same.

Since the resistance is inverse conductivity, the resistance increases with a lower conductivity:

${\displaystyle \sigma \rightarrow {\frac {\sigma }{\alpha }}\quad \Rightarrow \quad R\rightarrow R\alpha \qquad ,R=\sigma ^{-1}}$

This means that the resistance of the field plate increases as the B-field increases as follows:

${\displaystyle R(B)=R_{0}\left(1+\left(\mu B\right)^{2}\right)}$

See also

• Hall sensor
• Magnetometer