CBSE Class 12 Physics Semiconductor Devices And Communication System Worksheet

Question. In a p-type semiconductor the acceptor level is situated 60 meV above the valence band. The maximum wavelength of light required to produce a hole will be
(a) 0.207 × 10^–5 m
(b) 2.07 × 10^–5 m
(c) 20.7 × 10^–5 m
(d) 2075 × 10^–5 m

Answer: B

Question. Pure Si at 500K has equal number of electron (n_e) and hole (n_h) concentrations of 1.5 × 10¹⁶ m^–3. Doping by indium increases nh to 4.5 × 10²² m^–3. The doped semiconductor is of
(a) n–type with electron concentration ne = 5 × 10²² m^–3
(b) p–type with electron concentration ne = 2.5 ×10¹⁰ m^–3
(c) n–type with electron concentration ne = 2.5 × 10²³ m^–3
(d) p–type having electron concentration ne = 5 × 10⁹ m–3

Answer: D

Question. In insulator
(a) valence band is partially filled with electrons
(b) conduction band is partially filled with electrons
(c) conduction band is filled with electrons and valence band is empty
(d) conduction band is empty and valence band is filled with electrons.

Answer: D

Question. Symbolic representation of four logic gate are shown as 

Pick out which ones are for AND, NAND and NOT gates, respectively
(a) (ii), (iii) and (iv)
(b) (iii), (ii) and (i)
(c) (iii), (iii) and (iv)
(d) (ii), (iv) and (iii)

Answer: D

Question. Distance between body centred atom & a corner atom in sodium(a = 4.225 Å) is
(a) 3.66 Å
(b) 3.17 Å
(c) 2.99 Å
(c) 2.54 Å

Answer: A

Question. In a triode, g_m = 2 × 10^–3 ohm^–1 ; μ = 42; resistance of load, R = 50 kilo ohm. The voltage amplification obtained from this triode will be
(a) 30.42
(b) 29.57
(c) 28.18
(d) 27.15

Answer: B

Question. Zener diode is used for
(a) amplification
(b) rectification
(c) stabilisation
(d) all of the above

Answer: C

Question. Which one is the weakest type of bonding in solids ?
(a) Ionic
(b) Covalent
(c) Metallic
(d) Vander Wall’s

Answer: D

Question. A transistor has β = 40. A change in base current of 100 μ A, produces change in collector current
(a) 40 × 100 microampere
(b) (100 – 40) microampere
(c) (100 + 40) microampere
(d) 100/40 microampere

Answer: A

Question. In the energy band diagram of a material shown below, the open circles and filled circles denote holes and electrons respectively. The material is
(a) an insulator
(b) a metal
(c) an n-type semiconductor
(d) a p-type semiconductor

Answer: D

Question. Choose the only false statement from the following.
(a) In conductors, the valence and conduction bands may overlap.
(b) Substances with energy gap of the order of 10 eV are insulators.
(c) The resistivity of a semiconductor increases with increase in temperature.
(d) The conductivity of a semiconductor increases with increase in temperature.

Answer: C

Question. The current gain of a transistor in common base mode is 0.995. The current gain of the same transistor in common emitter mode is
(a) 197
(b) 201
(c) 198
(d) 199

Answer: D

Question. In forward biasing of the p–n junction
(a) the positive terminal of the battery is connected to p–side and the depletion region becomes thick
(b) the positive terminal of the battery is connected to n–side and the depletion region becomes thin
(c) the positive terminal of the battery is connected to n–side and the depletion region becomes thick
(d) the positive terminal of the battery is connected to p–side and the depletion region becomes thin

Answer: D

Question. If a small amount of antimony is added to germanium crystal
(a) it becomes a p–type semiconductor
(b) the antimony becomes an acceptor atom
(c) there will be more free electrons than holes in the semiconductor
(d) its resistance is increased

Answer: C

Question. By increasing the temperature, the specific resistance of a conductor and a semiconductor
(a) increases for both
(b) decreases for both
(c) increases, decreases
(d) decreases, increases

Answer: C

Question. At absolute zero, Si acts as
(a) non-metal
(b) metal
(c) insulator
(d) none of these

Answer: C

Question. On doping germanium with donor atoms of density 10¹⁷ cm^–3 its conductivity in mho/cm will be
[Given : μ_e = 3800 cm²/V–s and n_i = 2.5 × 10¹³ cm^–13]
(a) 30.4
(b) 60.8
(c) 91.2
(d) 121.6

Answer: B

Question. The energy band gap is maximum in
(a) metals
(b) superconductors
(c) insulators
(d) semiconductors.

Answer: C

Question. In a npn transistor 1010 electrons enter the emitter in 10–6 s. 4% of the electrons are lost in the base. The current transfer ratio will be
(a) 0.98
(b) 0.97
(c) 0.96
(d) 0.94

Answer: C

Question. A transistor has a base current of 1 mA and emitter current 90 mA. The collector current will be
(a) 90 mA
(b) 1 mA
(c) 89 mA
(d) 91 mA

Answer: C

ONE MARKS QUESTIONS

Question. Draw the truth table for a NOR gate.
Answer : 

Question. Which diode is used for voltage regulation? Give its symbol.
Answer : Zener diode (Image 1)

Question. Draw the logic symbol for a NAND gate.
Answer :

Question. Which biasing will make the resistance of p-n junction high?
Answer : Reverse biasing

Question. In the given diagram, is the diode D forward or reverse biased?
Answer : Reverse biased

TWO MARKS QUESTIONS

Question. Identify the equilent logic gate for the given logic gates. Write down the truth table for the final output of the combination.
Answer : NAND gate   

Question. Explain how the depletion region and barrier potential are formed in a p-n junction diode.
Answer : Formation of Depletion Layer: At the junction there is diffusion of electrons of n-region to p-region while holes of p-region diffuse into n-region. Some electrons combine with holes to neutralise each other. Thus near the junction there is an excess of positively charged ions in n-region and an excess of negatively charged ions in p-region. This sets up a potentialdifference and hence an internal electric field Ei (junction field) across the junctions. The field Eiis directed from n-region to p-region.

Question. Draw the circuit diagram of a n-p-n common emitter transistor as an amplifier with proper biasing.If a change of 0.4 mA in base current causes a change of 10mA in collector current in a common emitter amplifier, find a.c.current gain of the transistor.
Answer : ß= change in collector current/change in base current = 25
 

THREE MARKS QUESTIONS

Question. Draw energy band diagram for a (i) p- type extrinsic semiconductor (ii) n-type extrinsic semiconductor (iii) intrinsic semiconductors.
Answer : (Image 4) .(i) p-type semiconductor

Question. Define the terms ‘potential barrier’ and ‘depletion region’ for a p-n junction diode. State how the thickness of depletion region will change when the p.n junction diode is (i) forward biased. (ii) Reverse biased.
Answer : Potential barrier-The loss of electrons from the n-region and the gain of electron by the p-region cause a difference of potential across the junction of the two regions. Since this potential tends to prevent the movement of electron from the n region into the p region, it is often called a barrier potential.
Depletion region- The space-charge region on either side of the junction which is free of electrons and holes is known as depletion region.
(i) Thickness of depletion region decreases when diode is forward biased.
(ii) Thickness of depletion region increases when diode is reverse biased.

Question.With the help of labeled circuit diagram, explain the rectification action of a full wave rectifier.
Answer : Answer included in 10 years question answer.

Question.Distinguish between conductors, semiconductors and insulators on the basis of band theory of solids.
Answer : In metal either the conduction band is partially filled and the balanced band is partially empty or the conduction and valance bands overlap. When there is overlap electrons from valence band can easily move into the conduction band. This situation makes a large number of electrons available for electrical conduction. When the valence band is partially empty, electrons from its lower level can move to higher level making conduction possible (Image 4)
This situation is shown in Fig.(c). Here a finite but small band gap (Eg< 3 eV) exists. Because of the small band gap, at room temperature some electrons from valence band can acquire enough energy to cross the energy gap and enter the conduction band. These electrons (though small in numbers) can move in the conduction band. 
In this case, as shown in Fig.(b), a large band gap Egexists (Eg> 3 eV). There are no electrons in the conduction band, and therefore no electrical conduction is possible. Note that the energy gap is so large that electrons cannot be excited from the valence band to the conduction band by thermal excitation. This is the case of insulators.