Practice Start Test

Semiconductor

NEET Physics · 93 questions · Page 3 of 10 · Click an option or "Show Solution" to reveal answer

Q21
For the logic circuit shown, the truth table is :
A <table> <thead> <tr> <th>A</th> <th>B</th> <th>Y</th> </tr> </thead> <tbody> <tr> <td>0</td> <td>0</td> <td>0</td> </tr> <tr> <td>0</td> <td>1</td> <td>1</td> </tr> <tr> <td>1</td> <td>0</td> <td>1</td> </tr> <tr> <td>1</td> <td>1</td> <td>1</td> </tr> </tbody> </table>
B <table> <thead> <tr> <th>A</th> <th>B</th> <th>Y</th> </tr> </thead> <tbody> <tr> <td>0</td> <td>0</td> <td>1</td> </tr> <tr> <td>0</td> <td>1</td> <td>1</td> </tr> <tr> <td>1</td> <td>0</td> <td>1</td> </tr> <tr> <td>1</td> <td>1</td> <td>0</td> </tr> </tbody> </table>
C <table> <thead> <tr> <th>A</th> <th>B</th> <th>Y</th> </tr> </thead> <tbody> <tr> <td>0</td> <td>0</td> <td>1</td> </tr> <tr> <td>0</td> <td>1</td> <td>0</td> </tr> <tr> <td>1</td> <td>0</td> <td>0</td> </tr> <tr> <td>1</td> <td>1</td> <td>0</td> </tr> </tbody> </table>
D <table> <thead> <tr> <th>A</th> <th>B</th> <th>Y</th> </tr> </thead> <tbody> <tr> <td>0</td> <td>0</td> <td>0</td> </tr> <tr> <td>0</td> <td>1</td> <td>0</td> </tr> <tr> <td>1</td> <td>0</td> <td>0</td> </tr> <tr> <td>1</td> <td>1</td> <td>1</td> </tr> </tbody> </table>
Correct Answer
Option D
Solution
Y=A‾+B‾‾=A.B‾‾Y = \overline {\overline A + \overline B } = \overline{\overline {A.B}}
=A.B⇒AND= A.B \Rightarrow AND

Gate

Q22
The correct Boolean operation represented by the circuit diagram drawn is :
A OR
B NAND
C NOR
D AND
Correct Answer
Option B
Solution

From the given logic circuit LED will glow, when voltage across LED in high.

This can be verified by the given truth table.

Boolean expression Y =

A.B‾\overline {A.B}

This is output of NAND gate.

Q23
For a p-type semiconductor, which of the following statements is true ?
A Holes are the majority carriers and trivalent atoms are the dopants.
B Holes are the majority carriers and pentavalent atoms are the dopants.
C Electrons are the majority carriers and pentavalent atoms are the dopants
D Electrons are the majority carriers and trivalent atoms are the dopants.
Correct Answer
Option A
Solution

In p-type semiconductor, an intrinsic semiconductor is doped with trivalent impurities, that creates deficiencies of valence electrons called holes which are majority charge carriers.

Q24
In the circuit shown in the figure, the input voltage V i is 20 V, V BE = 0 and V CE = 0. The values of I B , I C and β\beta are given by
A I B = 40 μ\mu A, I C = 10 mA, β\beta = 250
B I B = 25 μ\mu A, I C = 5 mA, β\beta = 200
C I B = 20 μ\mu A, I C = 5 mA, β\beta = 250
D I B = 40 μ\mu A, I C = 5 mA, β\beta = 125
Correct Answer
Option D
Solution

From question, V BE = 0, V i = 20 V V CE = 0 V b = 0 (earthed) We need to find values of base and collector currents I B and I C and amplification factor β\beta, so V CC = I C R C + V CE I C = V CC – V CE /R C = 20 V – 0 V/4000 = 5 mA Base current I B = V i /R B =

20500×103{{20} \over {500 \times {{10}^3}}}

= 40 μA Amplification factor β\beta =

ICIB{{{I_C}} \over {{I_B}}}

=

25×10−340×10−6{{25 \times {{10}^{ - 3}}} \over {40 \times {{10}^{ - 6}}}}

= 125

Q25
In a p-n junction diode, change in temperature due to heating
A affects only reverse resistance
B affects only forward resistance
C does not affect resistance of p-n junction
D affects the overall V - I characteristics of p-n junction
Correct Answer
Option D
Solution

As a result of heating, temperature increases which generates large number of electron-hole pairs which lead to increase in conductivity.

As current increases I = I 0 (

e−qvKT{e^{ - {{qv} \over {KT}}}}

), overall resistance of diode changes which affects forward and reversed biasing.

Q26
In a common emitter transistor amplifier the audio sgnal voltage across the collector is 3 V. The resistance of collector is 3 kΩ\Omega . If current gain is 100 and the base resistance is 2 kΩ\Omega , the voltage and power gain of the amplifier is
A 15 and 200
B 150 and 15000
C 20 and 2000
D 200 and 1000
Correct Answer
Option B
Solution

Given: V i = 3 V, R C = 3 k

Ω\Omega

, R B = 2 k

Ω\Omega

, β\beta = 100 Voltage gain of the CE amplifier A V =

βRcRb\beta {{{R_c}} \over {{R_b}}}

= 100

(32)\left( {{3 \over 2}} \right)

= 150 Power gain = A V β\beta = 150(100) = 15000

Q27
The given electrical network is equivalent to
A OR gate
B NOR gate
C NOT gate
D AND gate
Correct Answer
Option B
Solution

y 1 =

A+B‾\overline {A + B}

y 2 =

y1+y1‾\overline {{y_1} + {y_1}}

=

A+BA + B

y =

y2‾\overline {{y_2}}

=

A+B‾\overline {A + B}

So it is NOR gate.

Q28
For CE transistor amplifier, the aufio signal voltage across the collector resistance of 2 kΩ\Omega is 4 V. If the current amplification factor of the transistor is 100 and the base resistance is 1 kΩ\Omega , then the input signal voltage is
A 10 mV
B 20 mV
C 30 mV
D 15 mV
Correct Answer
Option B
Solution

Voltage gain, A =

βRCRB\beta {{{R_C}} \over {{R_B}}}

=

100×20001000100 \times {{2000} \over {1000}}

= 200 Also, A =

V0Vi{{{V_0}} \over {{V_i}}}

⇒\Rightarrow

Vi=V0A{V_i} = {{{V_0}} \over A}

=

4200{4 \over {200}}

= 20 mA

Q29
What is the output Y in the following circuit, when all the three inputs A, B, C are first 0 and then 1 ?
A 0, 1
B 0, 0
C 1, 0
D 1, 1
Correct Answer
Option C
Solution

Applying De Morgan’s law: Output Y = [(A ⋅ B) ⋅ C]' = A' + B' + C' When A, B, C are 0 ⇒\Rightarrow Y = 1 When A, B, C are 1 ⇒\Rightarrow Y = 0

Q30
A npn transistor is connected in common emitter configuration in a given amplifier. A load resistance of 800 Ω\Omega is connected in the collector circuit and the voltage frop across it is 0.8 V. If the current amplification factor is 0.96 and the input resistance of the circuit is 192 Ω\Omega , the voltage gain and the power gain of the amplifier will respectively be
A 4, 4
B 4, 3.69
C 4, 3.84
D 3.69, 3.84
Correct Answer
Option C
Solution

Voltage gain = Current gain × Resistance gain = 0.96 ×\times

800192{{800} \over {192}}

= 4 Power gain = [Current gain] × [Voltage gain] = 0.96 × 4 = 3.84

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