Semiconductor — Page 5 | NEET Physics

Q41

In a common emitter (CE) amplifier having a voltage gain G, the transistor used has transconductance 0.03 mho and current gain 25. If the above transistor is replaced with another one with transconductance 0.02 mho and current gain 20, the voltage gain will be

A

{1 \over 3}G

B

{5 \over 4}G

C

{2 \over 3}G

D 1.5G

Correct Answer

Option C

Solution

Formula for trans conductance

g

m is

g

m =

{\beta \over {{r_i}}}

Formula for voltage gain A, A = $\beta$

{{{R_L}} \over {{r_i}}}

$\Rightarrow$ A =

g

m R L Given A = G $\therefore$ G =

g

m R L $\Rightarrow$ G $\propto$

g

m $\Rightarrow$

{{{G_2}} \over {{G_1}}} = {{{g_{{m_2}}}} \over {{g_{{m_1}}}}}

$\Rightarrow$ G 2 =

{{0.02} \over {0.03}}

$\times$ G =

{2 \over 3}G

Q42

The input resistance of a silicon transistor is 100

\Omega

. Base current is changed by 40

\mu

A which results in a change in collector current by 2 mA. This transistor is used as a common emitter amplifier with a load resistance of 4 k

\Omega

. The voltage gain of the amplifier is

A 2000

B 3000

C 4000

D 1000

Correct Answer

Option A

Solution

Current gain ( $\beta$ ) : $\beta$ =

{{\Delta {I_C}} \over {\Delta {I_B}}}

=

{{2 \times {{10}^{ - 3}}} \over {40 \times {{10}^{ - 6}}}}

= 50 Voltage gain of the amplifier is A V =

\beta {{{R_L}} \over {{R_i}}}

= 50 $\times$

{{4 \times {{10}^3}} \over {100}}

= 2000

Q43

C and Si both have same lattice structure; having 4 bonding electrons in each. However, C is insulator where as Si is intrinsic semiconductor. This is because

A In case of C the valence band is not completely filled at absolute zero temperature.

B In case of C the conduction band is partly filled even at absolute zero temperature.

C The four bonding electrons in the case of C lie in the second orbit, whereas in the case of Si they lie in the third.

D The four bonding electrons in the case of C lie in the third orbit , whereas for Si they lie in the fourth orbit.

Correct Answer

Option C

Solution

Electronic configuration of 6 C 6 C = 1s 2 , 2s 2 2p 2 The electronic configuration of 14 Si 14 Si = 1s 2 , 2s 2 2p 6 , 3s 2 3p 2 As they are away from Nucleus, so effect of nucleus is low for Si even for Sn and Pb are almost mettalic.

Q44

Transfer characteristics [output voltage (V 0 ) vs input voltage (V i )] for a base biased transistor in CE configuration is as shown in the figure. For using transistor as a switch, it is used

A in region III

B both in region (I) and (III)

C in region II

D in region I

Correct Answer

Option B

Solution

In the given graph, Region (I) – Cutoff region Region (II) – Active region Region (III) – Saturation region Using transistor as a switch it is used in cutoff region or saturation region.

Using transistor as a amplifier it is used in active region.

Q45

Two ideal diodes are connected to a battery as shown in the circuit. The current supplied by the battery is

A 0.75 A

B zero

C 0.25 A

D 0.5 A

Correct Answer

Option D

Solution

Here D 1 is in forward bias and D 2 is in reverse bias so, D 1 will conduct and D 2 will not conduct.

So, the current supplied by the battery is I =

{5 \over {10}}

= 0.5 A

Q46

The figure shows a logic circuit with two inputs A and B and the output C. The voltage wave forms across A, B and C are as given. The logic circuit gate is

A OR gate

B NOR gate

C AND gate

D NAND gate

Correct Answer

Option A

Solution

The truth table is The logic circuit is OR gate.

Q47

In a CE transistor amplifier, the audio signal voltage across the collector resistance of 2 k

\Omega

is 2 V. If the base resistance is 1 k

\Omega

and the current amplification of the transistor is 100, the input signal voltage is

A 0.1 V

B 1.0 V

C 1 mV

D 10 mV

Correct Answer

Option D

Solution

Here, R C = 2 k

\Omega

= 2 × 10 3

\Omega

V 0 = 2 V, R B = 1 k

\Omega

= 1 × 10 3

\Omega

, $\beta$ = 100 The output voltage, across the load R C V 0 = I C R C = 2 The collector current (I C ) I C =

{2 \over {2 \times {{10}^3}}}

= 10 -3 A = 1 mA Current gain( $\beta$ ) =

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

= 100 $\Rightarrow$ I B =

{{{I_C}} \over {100}}

=

{{{{10}^{ - 3}}} \over {100}}

= 10 -5 A Input voltage, V i = I B R B = (10 –5 A) (1 × 10 3

\Omega

) = 10 –2 V = 10 mV

Q48

Pure Si at 500 K has equal number of electron (n e ) and hole (n h ) concentrations of 1.5

\times

10 16 m

-

3 . Doping by indium increases n h to 4.5

\times

10 22 m

-

3 . The doped semiconductor is of

A p-type having electron concentration n e = 5

\times

10 9 m

-

3

B n-type with electron concentration n e = 5

\times

10 22 m

-

3

C p-type with electron concentration n e = 2.5

\times

10 10 m

-

3

D n-type with electron concentration n e = 2.5

\times

10 23 m

-

3

Correct Answer

Option A

Solution

(n i ) 2 = n e × n h (1.5 × 10 16 ) 2 = n e (4.5 × 10 22 ) So n e = 5 × 10 9 Now n h = 4.5 × 10 22 $\Rightarrow$ n h

>>

n e Hence, semiconductor is p-type and n e = 5 × 10 9 m –3

Q49

A Zener diode, having breakdown voltage equal to 15 V, is used in a voltage regulator circuit shown in figure. The current through the diode is

A 5 mA

B 10 mA

C 15 mA

D 20 mA

Correct Answer

Option A

Solution

Voltage across 250 Ω resistance = 20 V – 15 V = 5 V Now current through 250

\Omega

resistance: 5/250 = 20 mA If voltage across load resistance 1 k

\Omega

is 15 V, then current through 1 kΩ is 15/1000 = 15 mA The current through the zener diode is = Current through 250

\Omega

resistance – Current through 1 k

\Omega

resistance. = 20 - 15 = 5 mA

Q50

In the following figure, the diodes which are forward biased, are

A (A), (B) and (D)

B (C) only

C (C) and (A)

D (B) and (D)

Correct Answer

Option C

Solution

p-n junction is said to be forward biased when p side is at high potential than n side. It is for circuit (A) and (C).