Hots questions based on Semi conductors.
These questions are important for class 12th board Examination.
Q. 1. What are different types of compound semiconductors? Give two examples of each.
Ans- The compound semiconductors are of following types
- Inorganic semiconductors: Cds, GaAs, CdSe
- Organic semiconductors: Anthracene, pthalocynines
- Organic polymers: Polyparrole, polyaniline, polythiophene.
Q. 2. Which characteristic of a semiconductors make them useful in fabrication of electronic devices?
Ans- The number of charge carriers and their direction of flow can be controlled within the solid semiconductor.
Q. 3. Give three advantages of semiconductor devices over conventional vacuum tube devices.
Ans- 1. The charge carriers can be manipulated within the semiconductor unlike the vacuum tubes where the electrons were manipulated in vacuum.
2. The size of semiconductors devices is very small because they do not require vacuum for manipulation of the charge carriers.
3. The semiconductor devices consume lesser energy because unlike vacuum tube, they do not require heating for electron emission.
Q. 4. What are the characteristics to be taken care of while doping a semiconductor? Justify your answer.
Ans-(i) The number of dopant atoms should be small and restricted to just a few parts per million (ppm) of the pure semiconductor atoms.
(ii) The size of the dopant atom used should be nearly equal to that of the semiconductor atom.
The above characteristics are essential so that the doping process does not cause any major distortion in the original pure semiconductor lattice. The dopant atom just substitute some of the semiconductor atoms in their lattice.
Q. 5. Give example of the dopants used to make
(i) n-type semiconductor (ii) p-type semiconductor.
Ans- n-type semiconductor are formed by using pentavalent impurity e.g. Arsenic (As); Antimony (Sb), Phosphorous (P) p-type semiconductors are formed by adding trivalent impurity to pure semiconductor e.g., Indium (In), Boron (B), Aluminium (Al).
Q. 6. What is the energy required to make electrons free for conduction in pure (i) germanium (ii) silicon? How does the energy value change in a doped semiconductor?
Ans- In a pure semiconductor, the energy required to make electrons available for conduction is large. It is about 0.72 eV for germanium and about 1.1 eV for silicon. In doped germanium, the energy required becomes nearly 0.01 eV and in doped silicon it is about 0.05 eV.
Q. 7. What is the net charge on a p-type extrinsic semiconductor? Justify.
Ans- Zero, the crystal maintains its overall neutrality because the charge on additional charge carriers is just equal and opposite to that of the ionised cores in the lattice.
Q.8. What should be the value of the reverse breakdown voltage of the diode used in a half-wave rectifier and why?
Ans- The reverse breakdown voltage of the pn-junction diode used in half-wave rectifier should be higher than the peak voltage of the a.c. voltage at the secondary of the transformer. The condition is necessary to protect the pn-junction diode from reverse breakdown.
Q.9. How is a sample of n-type semiconductor electrically neutral though it has an excess of negative charge carriers?
Ans- n-type semiconductors are fabricated by adding group 15 impurity atom which are neutral and having one extra proton for every conduction electron. So the sample in neutral.
Q.10. A student has to study the forward and reverse characteristics of a pn-junction. What kind of circuit arrangement should he use?
Draw the typical shape of input and output characteristic curves likely to be obtained.
Mark the knee voltage and reverse breakdown voltage on the curve. How can the static and the dynamic resistance be determined from the curves?
In which of the regions does the pn-junction remain when used as (i) full-wave rectifier (ii) stabiliser?
Q.11. What do you mean by host atoms in semiconductors?
Ans- In impure semiconductors, the semiconductors atom (e.g. Ge or Si) are called host atoms.
Q.12. What are optoelectronic devices? Give any two example of such devices.
Ans- The semiconductor junction diodes in which the charge carriers are generated by photons are called photoelectronic devices. The examples of such devices are
- Light emitting diodes
- Photovoltaic or solar cells.
Q.13. Two semiconductor pieces X and Y of same length and equal area or cross-section are provided. One of the materials is pure semiconductor and the other is doped.
(i) How will you identify the impure semiconductor from the two give pieces X and Y?
(ii) For a given rise in temperature what will be the relative charge in the behaviour of the two materials?
Ans- (i) Measure the electrical resistance offered by X and Y. The piece offering lower resistance will be impure semiconductor.
(ii) Both X and Y will have negative temperature coefficient of resistance. However, the resistance of pure semiconductor will decrease sharply with rise in temperature as compared to the impure semiconductor.
Q.14. A photodiode is operated is reverse bias region only although the forward current is known to be more than the current in reverse bias. Give reason.
Ans- A photodiode is sensitive to light and is used to measure/detect intensity of light incident on it. In a pn-junction, photodiode; consider n-type region. The electrons in this region are majority carriers and the holes are minority charge carriers
؞ ne >> nh.
When the pn-junction is illuminated the bonds between same atoms breaks up due to photo-excitation leading to increase in ne and nh say by ∆ne and ∆nh
We have ∆ne = ∆nh
؞ ∆nh/nh >> ∆ne/ne
i.e. The fractional increase in minority charge carriers is very large as compared to that in minority charge carriers.
The current through pn-junction in reverse bias is due to drift of the minority charge carriers. The number density of minority carriers being sensitive to light intensity; the reverse current show a rapid change with intensity of incident light.
So a photodiode is always operated in reverse bias.
Q.15. What is the minimum frequency that can be detected by a photodiode?
Ans- The minimum frequency (vm) that can be detected by a photodiode depends on the energy gap Eg of the semiconductor and is given by
H vm = Eg
؞ vm = Eg/h.
Q.16.What are the factors taken into consideration while selecting material suitable for fabrication of solar cell?
Ans-The following factors are taken into consideration while fabricating solar cell:
- Band gap is 1.0 eV to 1.8 eV
- High optical absorption
- Reasonable electrical conductivity
- Availability of raw material