Semiconductor — Page 7 | JEE Physics

Q61

Consider the following statements : A. The junction area of solar cell is made very narrow compared to a photo diode. B. Solar cells are not connected with any external bias. C. LED is made of lightly doped p-n junction. D. Increase of forward current results in continuous increase of LED light intensity. E. LEDs have to be connected in forward bias for emission of light. Choose the correct answer from the options given below :

A A, C, E Only

B B, E Only

C A, C Only

D B, D, E Only

Correct Answer

Option B

Solution

Statement A: The junction area of solar cell is made very narrow compared to a photo diode.

Let's think about the purpose of a solar cell.

Its main job is to collect as much sunlight as possible to generate electricity.

To collect more light, it needs a larger surface area.

Therefore, the $p-n$ junction of a solar cell is made with a very large area (typically a few square centimeters).

This statement says the area is narrow, which is the opposite.

So, statement A is Incorrect.

Statement B: Solar cells are not connected with any external bias.

Remember, a solar cell is like a small battery.

It generates an electromotive force (e.m.f.) or voltage when light shines on it.

It gives power; it doesn't take power from an external battery to work.

We don't apply any external bias to it.

We simply connect a load, like a bulb, across it to use the electricity it produces.

So, statement B is Correct.

Statement C: LED is made of lightly doped p-n junction.

An LED (Light Emitting Diode) works by the recombination of electrons and holes at the junction.

To get bright light, we need a very high rate of recombination.

A high rate of recombination is only possible if there are a large number of electrons on the n-side and holes on the p-side.

To achieve this high concentration of charge carriers, the $p-n$ junction of an LED is heavily doped, not lightly doped.

So, statement C is Incorrect.

Statement D: Increase of forward current results in continuous increase of LED light intensity.

It is true that the intensity of light from an LED increases as we increase the forward current.

More current means more electrons and holes are crossing the junction and recombining per second, which produces more light.

However, we cannot increase the current indefinitely.

Every LED has a maximum current rating mentioned by the manufacturer.

If we pass a current higher than this safe limit, the LED will overheat and get permanently damaged.

So, the increase is not "continuous" without any limit.

Therefore, this statement is Incorrect.

Statement E: LEDs have to be connected in forward bias for emission of light.

This is the basic condition for an LED to work.

When we connect it in forward bias, the potential barrier at the junction is lowered.

This allows the majority carriers to flow across the junction and recombine.

This recombination process releases energy in the form of light (photons).

In reverse bias, the barrier height increases, very little current flows, and no light is emitted.

So, statement E is Correct.

Final Conclusion: Based on our step-by-step analysis, the only correct statements are B and E.

Looking at the options provided: Option A: A, C, E Only Option B: B, E Only Option C: A, C Only Option D: B, D, E Only Thus, the correct choice is Option B.

Q62

A zener diode with 5 V zener voltage is used to regulate an unregulated dc voltage input of 25 V . For a

400 \Omega

resistor connected in series, the zener current is found to be 4 times load current. The load current

\left(I_L\right)

and load resistance

\left(R_L\right)

are :

A

\mathrm{I}_{\mathrm{L}}=0.02 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \Omega

B

\mathrm{I}_{\mathrm{L}}=10 \mathrm{~A} ; \mathrm{R}_{\mathrm{L}}=0.5 \Omega

C

\mathrm{I}_{\mathrm{L}}=10 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=500 \Omega

D

\mathrm{I}_{\mathrm{L}}=20 \mathrm{~mA} ; \mathrm{R}_{\mathrm{L}}=250 \Omega

Correct Answer

Option C

Solution

From the circuit diagram,

\begin{aligned} & 5 \mathrm{i}=\frac{20}{400}=\frac{1}{20} \mathrm{~A} \\ & \therefore \mathrm{i}=\frac{1}{100} \mathrm{~A}=10 \mathrm{~mA}=\text{ Load current } \end{aligned}

Also, $\mathrm{V}_{\mathrm{L}}=5 \mathrm{~V}$

\therefore \mathrm{R}_{\mathrm{L}}=\frac{5}{10 \times 10^{-3}} \Omega=500 \Omega

Q63

Consider a n-type semiconductor in which

\mathrm{n}_{\mathrm{e}}

and

\mathrm{n}_{\mathrm{h}}

are number of electrons and holes, respectively. (A) Holes are minority carriers (B) The dopant is a pentavalent atom (C)

\mathrm{n}_{\mathrm{e}} \mathrm{n}_{\mathrm{h}} \neq \mathrm{n}_i^2

(where

\mathrm{n}_i

is number of electrons or holes in semiconductor when it is intrinsic form) (D)

\mathrm{n}_{\mathrm{e}} \mathrm{n}_{\mathrm{h}} \geqslant \mathrm{n}_i^2

(E) The holes are not generated due to the donors Choose the correct answer from the options given below :

A (A), (B), (E) only

B (A), (C), (E) only

C (A), (B), (C) only

D (A), (C), (D) only

Correct Answer

Option A

Solution

(A) n type semiconductor holes are minority carriers and $\mathrm{e}^{-}$ are majority carriers (B) Dopant are pentavalent atom.

(C) $\mathrm{n}_{\mathrm{e}} \cdot \mathrm{n}_{\mathrm{h}}=\mathrm{n}_{\mathrm{i}}^2$ for intrinsic semiconductor (E) In $n$ type semiconductor primary source of holes generation are thermal excitation.

Q64

The Boolean expression

\mathrm{Y}=A \bar{B} C+\bar{A} \bar{C}

can be realised with which of the following gate configurations. A. One 3-input AND gate, 3 NOT gates and one 2-input OR gate, One 2-input AND gate, B. One 3 -input AND gate, 1 NOT gate, One 2 -input NOR gate and one 2 -input OR gate C. 3 -input OR gate, 3 NOT gates and one 2 -input AND gate Choose the correct answer from the options given below:

A B, C Only

B A, B, C Only

C A, B Only

D A, C Only

Correct Answer

Option C

Solution

$\because \overline{\mathrm{A}} \cdot \overline{\mathrm{C}}+\overline{\mathrm{A}+\mathrm{C}} \equiv$ NOR gate

Q65

Match List I with List II: .tg .tg List I List II A. Intrinsic semiconductor I. Fermi-level near the valence bond B. n-type semiconductor II. Fermi-level in the middle of valence and conduction band. C. p-type semiconductor III. Fermi-level near the conduction band D. Metals IV. Fermi-level inside the conduction band Choose the correct answer from the options given below :

A A-III, B-I, C-II, D-IV

B A-II, B-I, C-III, D-IV

C A-I, B-II, C-III, D-IV

D A-II, B-III, C-I, D-IV

Correct Answer

Option D

Solution

(A) Intrinsic semiconductor $\rightarrow$ II (B) n-type semiconductor $\rightarrow$ III (C) p-type semiconductor $\rightarrow 1$ (D) Metals $\rightarrow$ IV

Q66

Given below are two statements : Statement I: In a typical transistor, all three regions emitter, base and collector have same doping level. Statement II: In a transistor, collector is the thickest and base is the thinnest segment. In the light of the above statements, choose the most appropriate answer from the options given below.

A Statement I is correct but Statement II is incorrect

B Both Statement I and Statement II are incorrect

C Statement I is incorrect but Statement II is correct

D Both Statement I and Statement II are correct

Correct Answer

Option C

Solution

.tg .tg Emitter Base Collector Moderate size Thin Thick Maximum Doping Minimum Doping Moderate Doping

Q67

Mobility of electrons in a semiconductor is defined as the ratio of their drift velocity to the applied electric field. If, for an n-type semiconductor, the density of electrons is 1019 m

-

3 and their mobility is 1.6 m2/(V.s) then the resistivity of the semiconductor (since it is an n-type semiconductor contribution of holes is ignored) is close to :

A

2\,\Omega

m

B 4

\,\Omega

m

C 0.4

\,\Omega

m

D 0.2

\,\Omega

m

Correct Answer

Option C

Solution

For semiconductor, Conductivity, $\sigma$ = ne q $\mu$ e + nh q $\mu$ h given that semiconductor is n-type.

So contribution of holes is ignored. $\therefore$ nh q $\mu$ h = 0 $\therefore$ $\sigma$ = ne q $\mu$ e Resistivity, $\rho$ =

{1 \over \sigma }

=

{1 \over {{n_e}q{\mu _e}}}

=

{1 \over {{{10}^{19}} \times 1.6 \times {{10}^{ - 19}} \times 1.6}}

=

{1 \over {1.6 \times 1.6}}

= 0.4

\Omega

m

Q68

The truth table given in fig. represents : .tg .tg A B Y 0 0 0 0 1 1 1 0 1 1 1 1

A AND - Gate

B OR - Gate

C NAND - Gate

D NOR - Gate

Correct Answer

Option B

Solution

Here Y = 1 when, (1) A = 0 and B = 1

\left( {\overline A B} \right)

(2) A = 1 and B = 0

\left( {A\,\overline B } \right)

(3) A = 1 and B = 1

\left( {A\,B} \right)

$\therefore$ Y =

{\overline A }

B + A

{\overline B }

+ AB =

{\overline A }

B + A (

{\overline B }

+ B) =

{\overline A }

B + A [as

{\overline B }

+ B = 1] = (A +

{\overline A }

) (A + B) = A + B So, this represented OR gate. Note : X +

{\overline X }

Y = (X +

{\overline X }

) (X + Y)

Q69

A strip of copper and another of germanium are cooled from room temperature to

80K.

The resistance of

A each of these decreases

B copper strip increases and that of germanium decreases

C copper strip decreases and that of germanium increases

D each of these increases

Correct Answer

Option C

Solution

The resistance of metal (like

Cu

) decreases with decrease in temperature whereas the resistance of a semi-conductor (like

Ge

) increases with decrease in temperature.

Q70

The positive feedback is required by an amplifier to act an oscillator. The feedback here means :

A External input is necessary to sustain ac signal in output.

B A portion of the output power is returned back to the input.

C Feedback can be achieved by LR network.

D The base-collector junction must be forward biased.

Correct Answer

Option B

Solution

Feedback means a portion of the output power is fed to the inputs.