Semiconductor Electronics: Materials, Devices and Simple Circuits

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Question 1

A charge Q is uniformly distributed over the surface of non conducting disc of radius R. The disc rotates about an axis perpendicular to its plane and passing through its centre with an angular velocity ω. As a result of this rotation, a magnetic field of induction B is obtained at the centre of the disc. If we keep both the amount of charge placed on the disc and its angular velocity to be constant and vary the radius of the disc than the variation of the magnetic induction at the centre of the disc will be represented by the figure

Solution

A.


Consider ring like the element of the disc of radius r and thickness dr.
If σ is charge per unit area, then charge on the element
dq = σ(2πr dr)
current ‘i’ associated with rotating charge dq is
straight i space equals space fraction numerator left parenthesis dp right parenthesis straight w over denominator 2 straight pi end fraction space equals space straight sigma space straight w space straight r space dr
Magnetic field dB at center due to element

dB space equals space fraction numerator straight mu subscript 0 straight i over denominator 2 straight r end fraction space equals space fraction numerator straight mu subscript 0 σωdr over denominator 2 end fraction
straight B subscript net space equals space integral dB space equals space fraction numerator straight mu subscript straight o σω squared over denominator 2 end fraction space integral subscript 0 superscript straight R dr space equals space fraction numerator straight mu subscript straight o σωR over denominator 2 end fraction
rightwards double arrow space straight B subscript net space equals fraction numerator straight mu subscript straight o Qω over denominator 2 πR end fraction
So if Q and w are unchanged then
straight B subscript net space proportional to 1 over straight R

Question 2

A p–n junction (D) shown in the figure can act as a rectifier. An alternating current source (V) is connected in the circuit.

Solution

C.

Only +ve current passes though the diode. Given figure is half wave rectifier.

Question 3

A piece of copper and another of germanium are cooled from room temperature to 77 K, the resistance of

  • each of them increases

  • each of them decreases

  • copper decreases and germanium increases

  • copper increases and germanium decreases.

Solution

D.

copper increases and germanium decreases.

Copper is metallic conductor and germanium is semiconductor therefore as temperature decreases resistance of good conductor decreases while for semiconductor it increases.

Question 4

A solid which is transparent to visible light and whose conductivity increases with temperature is formed by

  • Metallic binding

  • Ionic binding

  • Covalent binding

  • Van der Waals binding

Solution

C.

Covalent binding

Question 5

A thermocouple is made from two metals, Antimony and Bismuth. If one junction of the couple is kept hot and the other is kept cold then, an electric current will

  • flow from Antimony to Bismuth at the cold junction

  • flow from Antimony to Bismuth at the hot junction

  • flow from Bismuth to Antimony at the cold junction

  • not flow through the thermocouple

Solution

A.

flow from Antimony to Bismuth at the cold junction

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