1. The RH of a specimen is 3.66x10-4 m3/C, its resistivity is 8.93x10-3 Ω-m. Find mobility and carrier concentration in a p-type semiconductor.
Given:
Hall coefficient, RH = 3.66x10-4 m3/C
Resistivity, ρ = 8.93x10-3 Ω-m
The mobility (\(\mu\)) is given by:
\[ \mu = \frac{R_H}{\rho} = \frac{3.66 \times 10^{-4}}{8.93 \times 10^{-3}} = 0.04 \, m^2/V \cdot s \]The carrier concentration (\( n \)) is given by:
\[ n = \frac{1}{R_H e} \] \[ = \frac{1}{3.66 \times 10^{-4} \times 1.6 \times 10^{-19}} \] \[ = 1.7 \times 10^{22} \, m^{-3} \]2. A semiconductor sample of width 0.5 mm is placed in a magnetic field of 500 Gauss. If the probe current and Hall voltage are 5.1 mA and 6.3 mV, respectively. Find the Hall coefficient, carrier concetration and type of semiconductor ?
Given:
Sample width = 0.5 mm = 5×10-4,
Magnetic field, B = 500 G = 500 ×10-4,
Current I= 5.1 mA = 5.1 ×10-3 A,
Voltage V = 6.3 mV = 6.3×10-3 V.
The Hall coefficient (\( R_H \)) is given by:
\[ R_H = \frac{V_H w}{IB} \] \[ R_H = \frac{6.3 \times 10^{-3} \times 5 \times 10^{-4}}{5.1 \times 10^{-3} \times 500 \times 10^{-4}} \] \[ R_H = 0.0123 m^3/C \]The carrier concentration (\( n \)) is given by:
\[ n = \frac{1}{R_H e} \] \[ n = \frac{1}{0.0123 \times 1.6 \times 10^{-19}} \] \[ n = 5.08 \times 10^{20} \, m^{-3} \]RH is positive, so the semiconductor is p-type.
3. Calculate the Hall voltage for a rectangular semiconductor bar when the magnetic field \( B = 0.5 \,\text{T} \), current \( I = 3 \,\text{A} \), width of the sample \( w \) = 2 mm and carrier concentration \( n = 5\times10^{21} \text{m}^{-3} \).
Given:
Magnetic field \( B = 0.5 \text{T} \)
Current \( I = 3 \text{A}\)
Width of the sample \( w = 2 mm =2\times 10^{-3} m \) and
Carrier concentration \( n = 5\times10^{21} \text{m}^{-3}\)
The Hall voltage across the width is
\[ V_H=\frac{B\,I}{n\,e\,w} \quad \Big( \because {R_H = \frac{1}{ne}} \Big) \]where \( e = 1.6\times10^{-19} \text{C} \) is the electronic charge.
Substitute the values:
\[ V_H=\frac{(0.5)(3)}{(5\times10^{21})(1.6\times10^{-19})(2\times10^{-3})} \] \[ V_H=\frac{(0.5)(3)}{5\times1.6\times2\times10^{\,21-19-3}} \] \[ V_H=\frac{1.5}{1.6}=0.9375\ \text{V}\ \approx\ \boxed{0.938\ \text{V}} \]For a comprehensive understanding, refer to a article on Hall effect.