Chapter 14 Semiconductor Electronics Materials Devices And Simple Circuits Solutions
Question - 1 : - In an n-type silicon, which of the following statement istrue:
(a) Electrons are majority carriers and trivalent atomsare the dopants.
(b) Electrons are minority carriers and pentavalentatoms are the dopants.
(c) Holes are minority carriers and pentavalent atomsare the dopants.
(d) Holes are majority carriers and trivalent atoms arethe dopants.
Answer - 1 : -
The correct statement is (c).
In an n-type silicon, the electrons are the majoritycarriers, while the holes are the minority carriers. An n-type semiconductor isobtained when pentavalent atoms, such as phosphorus, are doped in siliconatoms.
Question - 2 : - Which of the statements given in Exercise 14.1 is true forp-type semiconductors.
Answer - 2 : -
The correct statement is (d).
In a p-type semiconductor, the holes are the majoritycarriers, while the electrons are the minority carriers. A p-type semiconductoris obtained when trivalent atoms, such as aluminium, are doped in siliconatoms.
Question - 3 : - Carbon, silicon and germanium have fourvalence electrons each. These are characterised by valence and conduction bandsseparated by energy band gap respectively equal to (Eg)C, (Eg)Si and (Eg)Ge. Whichof the following statements is true?
(a) (Eg)Si < (Eg)Ge <(Eg)C
(b) (Eg)C < (Eg)Ge >(Eg)Si
(c) (Eg)C > (Eg)Si >(Eg)Ge
(d) (Eg)C = (Eg)Si = (Eg)Ge
Answer - 3 : -
The correct statement is (c).
Of thethree given elements, the energy band gap of carbon is the maximum and that ofgermanium is the least.
The energy band gap of these elements are related as: (Eg)C > (Eg)Si >(Eg)Ge
Question - 4 : - In an unbiased p-n junction, holes diffuse from the p-regionto 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 ton-region.
(d) All the above.
Answer - 4 : -
The correct statement is (c).
The diffusion of charge carriers across a junction takesplace from the region of higher concentration to the region of lowerconcentration. In this case, the p-region has greater concentration of holesthan the n-region. Hence, in an unbiased p-n junction, holes diffuse from thep-region to the n-region.
Question - 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.
Answer - 5 : -
The correct statement is (c).
When a forward bias is applied to a p-n junction, itlowers the value of potential barrier. In the case of a forward bias, thepotential barrier opposes the applied voltage. Hence, the potential barrieracross the junction gets reduced.
Question - 6 : - For transistor action, which of the following statementsare correct:
(a) Base, emitter and collector regions should have similarsize and doping concentrations.
(b) The base region must be very thin and lightly doped.
(c) The emitter junction is forward biased and collectorjunction is reverse biased.
(d) Both the emitter junction as well as the collectorjunction are forward biased.
Answer - 6 : -
The correct statement is (b), (c).
For a transistor action, the junction must be lightlydoped so that the base region is very thin. Also, the emitter junction must beforward-biased and collector junction should be reverse-biased.
Question - 7 : - For a transistor amplifier, the voltage gain
(a) remains constant for all frequencies.
(b) is high at high and low frequencies and constant in themiddle frequency range.
(c) is low at high and low frequencies and constant at midfrequencies.
(d) None of the above.
Answer - 7 : -
The correct statement is (c).
The voltage gain of a transistor amplifier is constant atmid frequency range only. It is low at high and low frequencies.
Question - 8 : - In half-wave rectification, what is the output frequencyif the input frequency is 50 Hz. What is the output frequency of a full-waverectifier for the same input frequency.
Answer - 8 : -
Input frequency = 50 Hz
For ahalf-wave rectifier, the output frequency is equal to the input frequency.
∴Output frequency = 50 Hz
For afull-wave rectifier, the output frequency is twice the input frequency.
∴Outputfrequency = 2 × 50 = 100 Hz
Question - 9 : - For a CE-transistor amplifier, the audio signal voltageacross the collected resistance of 2 kΩ is 2 V. Suppose the current amplificationfactor of the transistor is 100, find the input signal voltage and basecurrent, if the base resistance is 1 kΩ.
Answer - 9 : -
Collector resistance, RC = 2 kΩ = 2000 Ω
Audio signal voltage across the collector resistance, V =2 V
Current amplification factor of the transistor, β =100
Base resistance, RB = 1kΩ = 1000 Ω
Input signal voltage = Vi
Base current = IB
We have theamplification relation as:
Voltage amplification 
Therefore, the input signal voltage of the amplifier is0.01 V.
Baseresistance is given by the relation:
Therefore, the base current of the amplifier is 10 μA.
Two amplifiers are connected one after the other in series(cascaded). The first amplifier has a voltage gain of 10 and the second has avoltage gain of 20. If the input signal is 0.01 volt, calculate the output acsignal.
Answer - 10 : -
Voltage gain of the first amplifier, V1 = 10
Voltage gain of the second amplifier, V2 = 20
Input signal voltage, Vi =0.01 V
Output AC signal voltage = Vo
The totalvoltage gain of a two-stage cascaded amplifier is given by the product ofvoltage gains of both the stages, i.e.,
V = V1 × V2
= 10 × 20 =200
We have the relation:V0 = V × Vi
= 200 ×0.01 = 2 V
Therefore, the output AC signal of the given amplifier is2 V.