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A thin piece of semiconducting silicon will be used to fabricate an electrical device. This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. Electrical contacts are placed at opposite ends of its length. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.

Required:
a. If the application of 1.0 volt to the contacts results in a current of 0.019 amps, what is the resistivity in (ohm-cm) of the material?
b. If the material's conductivity is due to doping with aluminum to a level of [Al]= 1x10^17 atoms/cm^3, what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material (assuming that the aluminum is fully ionized - i.e. all Al atoms donated electrons).

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okpalawalter8

We have that for the Question "a)what is the resistivity in (ohm-cm) of the material? b) what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material"

Answer:

  • Resistivity = [tex]1.754 ohm-cm[/tex]
  • Conductivity = [tex]6.25*10^{25} cm^3/V-s[/tex]

From the question we are told

This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.

 

A) Resistivity is given as,

[tex]p = \frac{RA}{l}[/tex]

where,

[tex]R = \frac{V}{I}[/tex]

Therefore,

[tex]p = \frac{VA}{Il}\\\\p = \frac{1*(0.1*0.5)}{0.019*1.5}\\\\p = 1.754 ohm-cm[/tex]

B) Conductivity is given as,

[tex]U = \frac{\rho}{pe}\\\\U = \frac{10^{17}}{10^{10}*1.6*10^{-19}}\\\\U = 6.25*10^{25} cm^3/V-s[/tex]

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