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301.

For the given combination of gates, if the logic states of inputs `A,B,C,` are as follows `A=B=C=0` and `A=B=1, C=0` then the logic states of output `D` are A. 0, 0B. 0, 1C. 1, 0D. 1, 1

Answer» Correct Answer - D
The output D for the given combination
`D=bar((A*B)*C)=bar((A+B))+barC`
If A=B=C=0, then `D=bar((0+0))=bar0=bar0+bar0=1+1=1`
`If A=B=1`, then `D=bar((1+1))=bar0=bar1+bar0=0+1=1`
302.

How does potential barrier of a semiconductor vary with temperature?A. 0.6 VB. 0.8 VC. 0.9 VD. 0.4 V

Answer» Correct Answer - D
The potential barrier decreases on heating. Hence it can not be 0.6, 0.8 or 0.9 V.
303.

The potential barrier of a semiconductor is 0.6 Vat room temperature. What is the approximate value of its potential barrier, if the temperature is increased by `20^@C`?A. 0.7VB. 0.8VC. 1.00 VD. 0.5V

Answer» Correct Answer - D
The potential barrier is decreased if the temperature is increased. Hence it can be 0.5 V. It cannot be 0. 7 V, 0.8 V or 1 V.
304.

The number of silicon atoms per `m^(3) is 5xx10^(28)`. This is doped simultaneously with `5xx10^(22)` atoms per `m^(3)` of Arsenic and `5xx10^(20) per m^(3)` atoms of indium. Calculate the number of electrons and holes. Given that `n_(i)=1.5xx10^(16)m^(-3)`. Is the material n-type or p-type?A. `3.24xx10^(6)m^(-3)`B. `6.24xx10^(8)m^(-3)`C. `4.54xx10^(9)m^(-3)`D. None of these

Answer» Correct Answer - C
For each atom doped of arsenic, one free electron is received similarly of each atom doped of indium, a vecancy is created. So, number of free electrons introduced by pentavalent impurity is
`N_(AS) = 5 xx 10^(22)m^(-3)`
The number of holes introduced by trivalent impurity added is
`N_(I)=5xx10^(20)m^(-3)`
So, net number of electrons added is
`n_(e)=N_(AS)-N_(I)`
`=5 xx10^(22)-5xx10^(20)`
`=4.95xx10^(22)m^(-3)`
Now, by the law of mass action.
`n_(e)n_(h) =n_(i)^(2)`
So, `n_(h) =(n_(i)^(2))/(n_(e))=((1.5xx10^(16))^(2))/(4.95xx10^(22))`
`rArr n_(h)=4.54xx10^(9)m^(-3)`
As, `n_(e) gt n_(h)` ( number of hles ). So, the material is n-type semiconductor.
305.

What are donor and acceptor impurities?

Answer»

1. Every pentavalent dopant atom which donates one electron for conduction is called a donor impurity. 

2. Each trivalent atom which can accept an electron is called an acceptor impurity.

306.

What are some features of p-type semiconductors?

Answer»

1. These are materials doped with trivalent impurity atoms (acceptors).

2. Electrical conduction in these materials is due to majority charge carriers i.e., holes.

3. The acceptor atoms acquire electron and become negatively charged-ions.

4. Number of holes is very large compared to the number of free electrons. nh >> ne. Holes are majority charge carriers.

5. When energy is supplied externally, positively charged holes (majority charge carriers) and negatively charged free electrons (minority charge carriers) are available for conduction.

307.

What are some features of n-type semiconductor?

Answer»

1. These are materials doped with pentavalent impurity (donors) atoms. 

2. Electrical conduction in these materials is due to majority charge carriers i.e., electrons. 

3. The donor atom loses electrons and becomes positively charged ions. 

4. Number of free electrons is very large compared to the number of holes, ne >> nh. Electrons are majority charge carriers.

5. When energy is supplied externally, negatively charged free electrons (majority charges carries) and positively charged holes (minority charges carries) are available for conduction.

308.

What is extrinsic semiconductors?

Answer»

The semiconductor with impurity is called a doped semiconductor or an extrinsic semiconductor.

309.

State advantages of semiconductor devices.

Answer»

1. Electronic properties of semiconductors can be controlled to suit our requirement.

2. They are smaller in size and light weight.

3. They can operate at smaller voltages (of the order of few mV) and require less current (of the order of pA or mA), therefore, consume lesser power.

4. Almost no heating effects occur, therefore these devices are thermally stable.

5. Faster speed of operation due to smaller size. 

6. Fabrication of ICs is possible.

310.

Explain applications of semiconductors.

Answer»

i. Solar cell:

1. It converts light energy into electric energy. 

2. t is useful to produce electricity in remote areas and also for providing electricity for satellites, space probes and space stations.

ii. Photo resistor: It changes its resistance when light is incident on it.

iii. Bi-polar junction transistor:

1. These are devices with two junctions and three terminals.

2. A transistor can be a p-n-p or n-p-n transistor. 

3. Conduction takes place with holes and electrons. 

4. Many other types of transistors are designed and fabricated to suit specific requirements. 

5. They are used in almost all semiconductor devices.

iv. Photodiode: It conducts when illuminated with light.

v. LED (Light Emitting Diode): 

1. It emits light when current passes through it. 

2. House hold LED lamps use similar technology. 

3. They consume less power, are smaller in size and have a longer life and are cost effective.

vi. Solid State Laser: It is a special type of LED. It emits light of specific frequency. It is smaller in size and consumes less power.

vii. Integrated Circuits (ICs): A small device having hundreds of diodes and transistors performs the work of a large number of electronic circuits.

311.

Transistor as a switch is used for controlling high power devices inA. motorsB. rectifierC. carsD. buses

Answer» Correct Answer - A
312.

State disadvantages of semiconductor devices.

Answer»

1. They are sensitive to electrostatic charges. 

2. Not very useful for controlling high power. 

3. They are sensitive to radiation. 

4. They are sensitive to fluctuations in temperature. 

5. They need controlled conditions for their manufacturing. 

6. Very few materials are semiconductors.

313.

In the figure, potential difference between A and B is A. zeroB. 5 VC. 10 VD. 15 V

Answer» Correct Answer - C
Here p-n junction is forward biased. If p-n junction is ideal, its resistance is zero. The effective resistance across A and B
`=(10xx10)/(10+10) = 5KOmega`
Current in the circuit, `I=(V)/(R )=(30)/(15xx10^(3))=(2)/(10^(3))A`
Current in arm `AB=I=(2)/(10^(3))`
Potential difference across A and B
`=(2)/(10^(3))xx5xx10^(3)=10V`
314.

The Boolean expression for the output Y of the logic operation shown is-A. `overline((A+B))+C`B. `(A+B)+overline(C)`C. `overline(overline((A+B))+C)`D. `overline((A+B+C))`

Answer» Correct Answer - C
315.

What is the name of the gate obtained by the combination shown in figure ? A. NANDB. NORC. NOTD. XOR

Answer» Correct Answer - A
`Y=bar((AA*B))`=NAND gate
316.

What is the name of the gate obtained by the combination shown in figure ? A. NANDB. ORC. NOTD. XOR

Answer» Correct Answer - B
`Y=bar(bar((A+B)))=A+B=` OR gate
317.

The truth table given below is for: `|(A,B,X),(0,0,0),(0,1,0),(1,0,0),(1,1,1)|`A. ANDB. ORC. NORD. NAND

Answer» Correct Answer - A
318.

Which of the following is not equal to 1 in Boolean algebra ?A. A+1B. `A + bar A`C. `A.barA`D. none of these

Answer» Correct Answer - C
`A*barA=0`
319.

What is the output Y of the gate circuit shown in figure ? A. `bar(AB)`B. `barA. barB`C. `bar(bar(A)B)`D. `bar(A.bar(B))`

Answer» Correct Answer - B
`Y=barA.barB`
320.

NAND gate is-A. a basic gateB. not a universal gateC. a universal gateD. a universal gate

Answer» Correct Answer - C
321.

The truth table given below is for which gate? `|(A,B,C),(0,0,1),(0,1,1),(1,0,1),(1,1,0)|`A. NANDB. XORC. NORD. OR

Answer» Correct Answer - A
322.

What is Boolean expression for the gate circuit shown in figure ? A. A.0=0B. `A. bar A=0`C. `A-1=A`D. A.A=A

Answer» Correct Answer - B
`Y=barA.A=0`
323.

The arrangement shown in figure performs the logic function of a/an….gate :- A. ORB. ANDC. NANDD. NOT

Answer» Correct Answer - A
324.

In the given circuit as shown the two input waveform `A` and `B` are applied simultaneously. The resultant waveform `Y` is A. B. C. D.

Answer» Correct Answer - A
325.

The boolean equation of NOR gate is-A. `C=A+B`B. `C=overline(A+B)`C. `C=A+B`D. `C=overline(A.B)`

Answer» Correct Answer - B
326.

For the logic circuit shown the boolean relation is-A. `Y=ABC`B. `Y=A+BC`C. `Y=(A+B)C`D. `Y=AB+C`

Answer» Correct Answer - C
327.

The boolean equation of NOR gate is-A. C=A+BB. `C=bar(A+B)`C. C=A.BD. `C=bar(A.B)`

Answer» Correct Answer - B
328.

The Boolean equation of OR gate isA. C=A+BB. `C=bar(A+B)`C. `C=A.B`D. `C=bar(A.B)`

Answer» Correct Answer - A
329.

When the two inoputs of a NAND gate are shorted, the resulting gats isA. an OR gateB. an AND gateC. a NOT gateD. NOR gate

Answer» Correct Answer - C
330.

A logic gate is an electronic circuit whichA. makes logic decisionsB. allows electrons flow only in one directionC. works binary algebraD. alternative between 0 and 1 values

Answer» Correct Answer - A
331.

For half wave rectifier if load resistance `R_(d)` is `2 k Omega` and P-N junction resistance `R_(L)` is `2 k Omega` determine rectification efficiency.

Answer» `eta_(HWA) -40.6 ((R_(L))/(R_(d)+R_(L))) =40.6 xx (2 kOmega)/((2+2)k Omega) = 26.3 %`.
332.

NOR gate is a combination ofA. OR gate and NOT gateB. OR gate and AND gateC. OR gate and OR gateD. none of these

Answer» Correct Answer - A
333.

The only function of a NOT gate is toA. stop a signalB. replacement of a signalC. invert an input signalD. act as a universal gate

Answer» Correct Answer - C
334.

In a semiconductor , acceptor imparity isA. AntimonyB. IndiumC. PhosphorusD. Arsenic

Answer» Correct Answer - B
When a pure Ge crystal is doped with a trivalent impurity like indium, we get a p-type semiconductor. The indium impurity accepts the electrons from Ge, hence it is called acceptor impurity.
335.

The current I through `10Omega` resistor in the circuit given below is A. 50 mAB. 20 mAC. 40 mAD. 80 mA

Answer» Correct Answer - D
Here, diode `D_(2)` is reverse biased while `D_(1)` is forward biase (d) So, no current flows across `D_(2)`, current flows through diode `D_(1)`.
`I=(V)/(R )=(2)/(10+15)=(2)/(25)=0.08A=80mA`
336.

The given graph represents `V-I` characteristic for a semiconductor device. which of the following statement is correct? A. It is for a solar cell and points `A` and `B` represent open circuit voltage and currentB. It is for a photodiode and points `A` and `B` represent open circuit voltage and current, respectivelyC. It is for a `LED` and points `A` and `B` represent open circuit voltage and short circuit current respectivelyD. It is `v-I` characteristic for solar cell where point `A` represents open circuit voltage and point `B` short circuit current

Answer» Correct Answer - D
Solar cell:
Open circuit `I = 0`, potential `V = emf`
Short circuit `I = I`, potential = 0.
337.

In a semiconductor , acceptor impurity isA. antimonyB. indiumC. phosphorousD. arsenic

Answer» Correct Answer - B
338.

A pure semiconductorA. extrinsic semiconductorB. intrinsic semiconductorC. p-type semiconductorD. n-type of semiconductor

Answer» Correct Answer - B
339.

In given circuit, current gain of transistor is `beta=100`, the output of amplifier will be A. 10 VB. 0.1 VC. 1.0 VD. 100 V

Answer» Correct Answer - A
`I_("in")=(V)/(R_("in"))=(1xx10^(-3))/(1xx10^(3))=10^(-6)A`
`I_("out")=betaI_("in")=100xx10^(-6)A=10^(-4)A`
`V_("out")=I_("out")xxR_(L)=10^(-4)xx100xx10^(3)=10V`
340.

What will be the input of `A` and `B` for the Boolean expression `bar((A+B)).bar((A.B))=1`?A. 0,0B. 0,1C. 1,0D. 1,1

Answer» Correct Answer - A
`bar((A+B))*bar((A*B))=1`
Now `bar((0+0))*bar((0+0))=1*1=1`
Thus, both inputs A and B should be zero.
341.

The input resistance of a common emitter transistor amplifer, if the output resistance is `500 K Omega`, the current gain `alpha = 0.98` and power gain is `6.0625 xx 10^(6)`, isA. `198 Omega`B. `300Omega`C. `100Omega`D. `400Omega`

Answer» Correct Answer - A
`alpha=0.98`
`beta=(alpha)/(1-alpha)=(0.98)/(1-0.98)=49`
`A_(p)=beta^(2)(R_(L))/(R_("in"))`
`6.0625xx10^(6)=(49)^(2)xx(500xx10^(3))/(R_(in))`
`R_("in")=198Omega`
342.

If the current gain in CB configuration is 0.96, then the current gain in the CE configuration will beA. 24B. 20C. 16D. 12

Answer» Correct Answer - A
`beta=alpha /(1-alpha) = 0.96/(1-0.96)=96/4=24`
343.

The collector supply voltage is 6 V and the voltage drop across a resistor of 600 `Omega` in the collector circuit is 0.6 V, in a circuit of a transistor connected in common emitter mode. What is the base current if the current gain is 20?A. 0.25 mAB. 0.05 mAC. 0. 12 mAD. 0.02 mA

Answer» Correct Answer - B
Collector current,
`I_C="Collector voltage "/"Resistance"=0.6/600=10^(-3) A = 1 mA` and `because beta=I_C/I_B`=20
`therefore I_B=I_C/beta=(1 mA)/20 `=0.05 mA
344.

A common emitter amplifier gives an output of 3 V for an input of 0.01 V. If `beta` of the resistance is 100 and the input resistance is `1 kOmega`. then the collector resistance isA. 1 K`Omega`B. 3 K`Omega`C. 30 K`Omega`D. 30 K`Omega`

Answer» Correct Answer - B
Voltage gain `=V_o/V_i=3/10^(-2)`=300
Current gain = `beta= 100 = I_C/I_B`
`therefore` Resistance gain = `(V.G.)/(C.G.)=300/100=3`
`therefore R_C/R_i`=3 or `R_C =1xx3=3 K Omega`
345.

What is the current through an ideal p-n junction diode shown in figure below ? A. ZeroB. 10 mAC. 20 mAD. 50mA

Answer» Correct Answer - C
The resistance of an ideal diode is zero and the potential on then side is more negative than that of the p-side. `therefore` D is forward biased with net e.m.f.= [-1-(-3)] = 2 V
`therefore I=2/100=0.2 A =` 20 mA
346.

Assuming that the junction diode is ideal the current through the diode is :- A. 200 mAB. 20 mAC. 2 mAD. zero

Answer» Correct Answer - B
`I=(3-1)/(100)=20mA`
347.

What is the energy level below which all levels are completely occupied at Zero Kelvin called?(a) Boson Energy(b) Fermi Energy(c) Stable Energy(d) Ground Energy

Answer» The correct choice is (b) Fermi Energy

To explain: Fermi energy is said to be the energy of the highest possible occupied energy level at 0 K. Below this level, all the states are completely occupied.
348.

Explain any four application of p-n junction diode.

Answer»

1. Solar cell:

1. It converts light energy into electric energy. 

2. It is useful to produce electricity in remote areas and also for providing electricity for satellites, space probes and space stations.

ii. Photodiode: It conducts when illuminated with light.

iii. LED (Light Emitting Diode):

1. It emits light when current passes through it. 

2. House hold LED lamps use similar technology. 

3. They consume less power, are smaller in size and have a longer life and are cost effective.

iv. Solid State Laser: It is a special type of LED. It emits light of specific frequency. It is smaller in size and consumes less power.

349.

How are thermistors fabricated?

Answer»

Thermistors are made from thermally sensitive metal oxide semiconductors. Thermistors are very sensitive to changes in temperature.

350.

What are different ty pes of thermistor and what are their applications?

Answer»

There are two types of thermistor:

i. NTC (Negative Thermal Coefficient) thermistor: Resistance of a NTC thermistor decreases with increase in its temperature. Its temperature coefficient is negative. They are commonly used as temperature sensors and also in temperature control circuits.

ii. PTC (Positive Thermal Coefficient) thermistor: Resistance of a PTC thermistor increases with increase in its temperature. They are commonly used in series with a circuit. They are generally used as a reusable fuse to limit current passing through a circuit to protect against over current conditions, as resettable fuses.