Semiconductor — Page 5 | JEE Physics

Q41

The correct relation between

\alpha

(ratio of collector current to emitter current) and

\beta

(ratio of collector current to base current) of a transistor is :

A

\beta = {\alpha \over {1 + \alpha }}

B

\alpha = {\beta \over {1 - \alpha }}

C

\alpha = {\beta \over {1 + \beta }}

D

\beta = {1 \over {1 - \alpha }}

Correct Answer

Option C

Solution

\alpha = {{{I_C}} \over {{I_E}}}

&

\beta = {{{I_C}} \over {{I_B}}}

&

{I_E} = {I_B} + {I_C}

$\therefore$

{{{I_E}} \over {{I_C}}} = {{{I_B}} \over {{I_C}}} + {{{I_C}} \over {{I_C}}}

\Rightarrow {1 \over \alpha } = {1 \over \beta } + 1 = {{1 + \beta } \over \beta }

\Rightarrow \alpha = {\beta \over {1 + \beta }}

Q42

Consider a situation in which reserve biased current of a particular P-N junction increases when it is exposed to a light of wavelength

\le

621 nm. During this process, enhancement in carrier concentration takes place due to generation of hole-electron pairs. The value of band gap is nearly.

A 1 eV

B 4 eV

C 0.5 eV

D 2 eV

Correct Answer

Option D

Solution

Band gap =

{{hc} \over \lambda } = 2

eV

Q43

Statement I : By doping silicon semiconductor with pentavalent material, the electrons density increases. Statement II : The n-type semiconductor has net negative charge. In the light of the above statements, choose the most appropriate answer from the options given below :

A Statement - I is true but Statement - II is false.

B Statement - I is false but Statement - II is true.

C Both Statement I and Statement II are true.

D Both Statement I and Statement II are false.

Correct Answer

Option A

Solution

Pentavalent activities have excess free e $-$ , so e $-$ density increases but overall semiconductor is neutral.

Q44

For a transistor in CE mode to be used as an amplifier, it must be operated in :

A Both cut-off and Saturation

B Saturation region only

C Cut-off region only

D The active region only

Correct Answer

Option D

Solution

Active region of the CE transistor is linear region and is best suited for its use as an amplifier.

Q45

For a transistor

\alpha

and

\beta

are given as

\alpha = {{{I_C}} \over {{I_E}}}

and

\beta = {{{I_C}} \over {{I_B}}}

. Then the correct relation between

\alpha

and

\beta

will be :

A

\alpha = {{1 - \beta } \over \beta }

B

\beta = {\alpha \over {1 - \alpha }}

C

\alpha \beta = 1

D

\alpha = {\beta \over {1 - \beta }}

Correct Answer

Option B

Solution

\alpha = {{{I_C}} \over {{I_E}}}

,

\beta = {{{I_C}} \over {{I_B}}}

;

{I_E} = {I_C} + {I_B}

\alpha = {{{I_C}} \over {{I_C} + {I_B}}} = {{{I_C}/{I_B}} \over {{{{I_C}} \over {{I_B}}} + 1}} = {\beta \over {\beta + 1}}

1 + {1 \over \beta } = {1 \over \alpha }

{1 \over \beta } = {{1 - \alpha } \over \alpha }

\beta = {\alpha \over {1 - \alpha }}

Q46

Statement - I : To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect a capacitor across the output parallel to the load RL. Statement - II : To get a steady dc output from the pulsating voltage received from a full wave rectifier we can connect an inductor in series with RL. In the light of the above statements, choose the most appropriate answer from the options given below :

A Statement I is true but Statement II is false

B Statement I is false but Statement II is true

C Both Statement I and Statement II are false

D Both Statement I and Statement II are true

Correct Answer

Option D

Solution

To convert pulsating dc into steady dc both of mentioned method are correct.

Q47

For using a multimeter to identify diode from electrical components, choose the correct statement out of the following about the diode :

A It is two terminal device which conducts current in both directions.

B It is two terminal device which conducts current in one direction only.

C It does not conduct current gives an initial deflection which decays to zero.

D It is three terminal device which conducts current in one direction only between central terminal and either of the remaining two terminals

Correct Answer

Option B

Solution

A diode is a two terminal device which conducts current in forward bias only.

Q48

Given below are two statements : One is labelled as Assertion A and the other is labelled as Reason R. Assertion A : n-p-n transistor permits more current than a p-n-p transistor. Reason R : Electrons have greater mobility as a charge carrier. Choose the correct answer from the options given below :

A Both A and R are true, and R is correct explanation of A.

B Both A and R are true but R is NOT the correct explanation of A.

C A is true but R is false.

D A is false but R is true.

Correct Answer

Option A

Solution

(A) is true as n-p-n transistor permits more current than p-n-p transistor as electrons which are majority charge carriers in n-p-n have higher mobility than holes which are majority carriers in p-n-p transistor.

$\Rightarrow$ Statement R is correct explanation of statement A.

Q49

For a transistor to act as a switch, it must be operated in

A Active region.

B Saturation state only.

C Cut-off state only.

D Saturation and cut-off state.

Correct Answer

Option D

Solution

A transistor acts as a switch when it is operated in saturation and cut-off state.

Q50

For a constant collector-emitter voltage of

8 \mathrm{~V}

, the collector current of a transistor reached to the value of

6 \mathrm{~mA}

from

4 \mathrm{~mA}

, whereas base current changed from

20 \,\mu \mathrm{A}

to

25 \,\mu \mathrm{A}

value. If transistor is in active state, small signal current gain (current amplification factor) will be :

A 240

B 400

C 0.0025

D 200

Correct Answer

Option B

Solution

\beta = {{\Delta {I_C}} \over {\Delta {I_B}}}

= {{(6 - 4) \times {{10}^{ - 3}}} \over {(25 - 20) \times {{10}^{ - 6}}}}

= {2 \over 5} \times {10^3}

= 400