Semiconductor — Page 3 | JEE Physics

Q21

The current voltage relation of diode is given by

{\rm I} = \left( {{e^{100V/T}} - 1} \right)mA,

where the applied voltage

V

is in volts and the temperature

T

is in degree kelvin. If a student makes an error measuring

\pm 0.01\,V

while measuring the current of

5

mA

at

300

K,

what will be the error in the value of current on

mA

?

A

0.2

mA

B

0.02

mA

C

0.5

mA

D

0.05

mA

Correct Answer

Option A

Solution

The current voltage relation of diode is

I = \left( {{e^{1000\,V/T}} - 1} \right)\,\,mA

(given) When,

I = 5mA,{e^{1000\,\,V/T}}\, = 6mA

Also,

dl = \left( {{e^{1000\,\,V/T}}} \right) \times {{1000} \over T}

(By exponential function)

= \left( {6\,mA} \right) \times {{1000} \over {300}} \times \left( {0.01} \right)

=0.2

mA

Q22

A red

LED

emits light at

0.1

watt uniformly around it. The amplitude of the electric field of the light at a distance of

1

m

from the diode is :

A

5.48

V/m

B

7.75

V/m

C

1.73

V/m

D

2.45

V/m

Correct Answer

Option D

Solution

Intensity of light at a distance r,

I = {P \over {4\pi {r^2}}}

[P = power] Again, if the amplitude of the electric field is E0 then

I = {1 \over 2}c{ \in _0}E_0^2

$\therefore$

{P \over {4\pi {r^2}}} = {1 \over 2}c{ \in _0}E_0^2

or,

{E_0} = \sqrt {{P \over {2\pi c{ \in _0}{r^2}}}}

= \sqrt {{{0.1} \over {2 \times 3.14 \times (3 \times {{10}^8}) \times (8.85 \times {{10}^{ - 12}}) \times {1^2}}}}

= 2.45 V/m

Q23

The ratio (R) of output resistance r0, and the input resistance ri in measurements of input and output characteristics of a transistor is typically in the range:

A R ~ 102

-

103

B R ~ 1

-

10

C R ~ 0.1

-

0.01

D R ~ 0.1

-

1.0

Correct Answer

Option B

Solution

R $=$

{{{r_0}} \over {{r_i}}} \equiv 1 - 10

Q24

An experiment is performed to determine the I - V characteristics of a Zener diode, which has a protective resistance of R = 100

\Omega

, and a maximum power of dissipation rating of 1 W. The minimum voltage range of the DC source in the circuit is :

A 0

-

5 V

B 0

-

8 V

C 0

-

12 V

D 0

-

24 V

Correct Answer

Option D

Solution

Potential drop accross zener diode.

Vz = V $-$ 100 I $\therefore$ Power dissiption = Vz I = (V $-$ 100 I) I Given that, (V $-$ 100 I) I = 1 $\Rightarrow$ VI $-$ 100 I2 = 1 $\Rightarrow$ 100 I2 $-$ VI + 1 = 0 As As I is real, So, b2 $-$ 4ac $\ge$ far this quadratic equation.

$\therefore$ V2 $-$ 4(100)1 $\ge$ 0 $\Rightarrow$ V $\ge$ 20 V $\therefore$ Voltage range should be 0 $-$ 24 V

Q25

An unknown transistor needs to be identified as a npn or pnp type. A multimeter, with + ve and − ve terminals, is used to measure resistance between different terminals of transistor. If terminal 2 is the base of the transistor then which of the following is correct for a pnp transistor ?

A + ve termial 1,

-

ve terminal 2, resistance high

B + ve termial 2,

-

ve terminal 1, resistance high

C + ve termial 3,

-

ve terminal 2, resistance high

D + ve termial 2,

-

ve terminal 3, resistance low

Correct Answer

Option A

Solution

+ ve terminal at 1, $-$ ve terminal at 2 and resistance high for pnp transistor.

Q26

What is the conductivity of a semiconductor sale having electron concentration of

5 \times {10^{18}}\,\,{m^{ - 3}},

hole concentration of

5 \times {10^{19}}\,\,{m^{ - 3}},

electron mobility of 2.0 m2 V

-

1 s-1 and hole mobility of 0.01 m2 V

-

1 s

-

1 ? (Take charge of electronas 1.6

\times

10

-

19 c)

A 1.68 (

\Omega

-m)

-

1

B 1.83 (

\Omega

-m)

-

1

C 0.59 (

\Omega

-m)

-

1

D 1.20 (

\Omega

-m)

-

1

Correct Answer

Option A

Solution

Conductivity of semiconductor, $\sigma$ = e

\left( {{\eta _e}{\mu _e} + \eta '{\mu _h}} \right)

= 1.6 $\times$ 10 $-$ 19 (5 $\times$ 1018 $\times$ 2 + 5 $\times$ 1019 $\times$ 0.01) = 1.6 $\times$ 1.05 = 1.68

Q27

The current gain of a common emitter amplifier is 69. If the emitter current is 7.0 mA, collector current is :

A 9.6 mA

B 6.9 mA

C 0.69 mA

D 69 mA

Correct Answer

Option B

Solution

Here, $\beta$ = 69, Ie = 7 mA, Ic = ? $\alpha$ =

{\beta \over {1 + \beta }}

=

{{69} \over {70}}

Also, $\alpha$ =

{{{I_c}} \over {{I_e}}}

$\Rightarrow$

{{69} \over {70}} = {{{I_c}} \over 7}

$\Rightarrow$ Ic =

{{69} \over {70}} \times 7

= 6.9 mA

Q28

In a common emitter amplifier circuit using an n-p-n transistor, the phase difference between the input and the output voltages will be:

A 180°

B 45°

C 90°

D 135°

Correct Answer

Option A

Solution

In common emitter configuration for n-p-n transistor input and output signals are 180° out of phase i.e., phase difference between output and input voltage is 180°.

Q29

In a common emitter configuration with suitable bias, it is given that

{R_L}

is the load resistance and

{R_{BE}}

is small signal dynamic resistance (input side). Then, voltage gain, current gain and power gain are given, respectively, by :

\beta

is curret gain,

{{\rm I}_B},{{\rm I}_C}

and

{{\rm I}_E}

are respectively base, collector and emitter currents.

A

\beta {{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_C}} \over {\Delta {{\rm I}_B}}},{\beta ^2}{{{R_L}} \over {{R_{BE}}}}

B

\beta {{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_E}} \over {\Delta {{\rm I}_B}}},{\beta ^2}{{{R_L}} \over {{R_{BE}}}}

C

{\beta ^2}{{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_C}} \over {\Delta {{\rm I}_E}}},{\beta ^2}{{{R_L}} \over {{R_{BE}}}}

D

{\beta ^2}{{{R_L}} \over {{R_{BE}}}},{{\Delta {{\rm I}_C}} \over {\Delta {{\rm I}_B}}},\beta {{{R_L}} \over {{R_{BE}}}}

Correct Answer

Option A

Solution

Current gain ( $\beta$ ) =

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

Voltage gain =

{{{V_{CE}}} \over {{V_{BE}}}} = \beta {{{R_L}} \over {{R_{BE}}}}

Power gain = voltage gain x current gain =

{\beta ^2}{{{R_L}} \over {{R_{BE}}}}

Q30

An n-p-n transistor operates as a common emitter amplifier, with a power gain of 60 dB. The input circuit resistance is 100

\Omega

and the output load resistance is 10 k

\Omega

. The common emitter current gain

\beta

is :

A 104

B 102

C 6 × 102

D 60

Correct Answer

Option B

Solution

{A_v} \times \beta = {P_{gain}}

60 = 10{\log _{10}}\left( {{P \over {{P_0}}}} \right)

P = {10^6} = {\beta ^2} \times {{{R_{out}}} \over {{R_{in}}}}

= {\beta ^2} \times {{{{10}^4}} \over {100}}

{\beta ^2} = {10^4};\beta = 100