Q.1 : In a n- type silicon , which of the following statements is true :
(a) Electrons are majority carriers and trivalent atoms are the dopants.
(b) Electrons are minority carriers and pentavalent atoms are the dopants.
(c) Holes are minority carriers and pentavalent atoms are dopants।
(d) Holes are majority carriers and trivalent atoms are the dopants.
Solution : An n type semiconductor is obtained by doping a pure (intrinsic) semiconductor with a pentavalent impurity. In n-type semiconductor, electrons are the majority carriers and holes are the minority carriers. Hence option (c) is true.
Q.2. Which of the statement given in Q.1 is true for p-type semiconductor?
Solution : A p-type semiconductor is obtained by doping a pure (intrinsic) semiconductor with a trivalent impurity. In p-type semiconductor, holes are the majority carriers and electrons are the minority carriers. Hence option (d) is true.
Q.3. Carbon , silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to $(E_{g})_{C}$, $(E_{g})_{si}$ and $(E_{g})_{Ge}$ which of the following statements is true ?
(a)$(E_{g})_{si}<(E_{Ge})_{C}<(E_{g})_{C}$
(b) $(E_{g})_{C}<(E_{g})_{Ge}>(E_{g})_{si}$
(c) $(E_{g})_{C}>(E_{g})_{si}>(E_{g})_{Ge}$
(d) $(E_{g})_{C}=(E_{g})_{si}=(E_{g})_{Ge}$
solution : Out of the three given materials , the energy band gap is maximum for carbon , less for silicon and least for germanium . Hence (c) is true.
Q.4. In an unbiased p-n junction, holes diffuse from the p-region to n-region because
(a) free electrons in the n-region attract them.
(b) they move across the junction by the potential difference.
(c) hole concentration in p-region is more as compared to n-region.
(d) All the above.
Solution :
In an unbiased p-n junction , holes diffuse from the p-region to n-region. It is because hole centration in p-region is more as compared to n-region. Hence option (C) is correct.
Q.5 When a forward bias is applied to a p-n junction, it
(a) Raises the potential barrier
(b) reduces the majority carrier current to zero.
(c) lowers the potential barrier
(d) None of the above.
Solution :
When a forward bias is applied to a p-n junction, it opposes the potential barrier. As a result the potential barrier across the junction is lowered . Hence option (c) is correct.
Q.6. In half-wave rectification, what is the output frequency of the input frequency is 50 Hz ? What is the output frequency of a full-wave rectifier for the same input frequency?
Solution : In half-wave rectification , the output frequency is the same as the input frequency i.e 50Hz in this case . It is because here one output cycle is produced for every one input cycle.
In full wave rectification, the output frequency is twice the input frequency i.e 100 Hz in this case. It is because here two output cycles are produced for every one input cycle.