The accepter impurity in following is -
- AArsenic
- BPhosphorus
- ✓Indium
- DAntimony
Answer: C.
View full solution →Question types
Semiconductor Electronic: Material, Devices And Simple Circuits question types
478 questions across 7 question groups — pick any mix to generate a Physics paper with step-by-step answer keys.
478
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7
Question groups
5
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01
M.C.Q [1M]
215 Q→02Assertion (A) & Reason (B) MCQ
15 Q→031 Marks Question
46 Q→04Short Answer Type Questions [2M]
52 Q→053 Marks Question
85 Q→064 Marks Questions
12 Q→07Long Answer Type Questions [5M]
53 Q→Sample Questions
Semiconductor Electronic: Material, Devices And Simple Circuits questions
One sample from each question group in this chapter. Select any group above to see the full set with answer keys.
For germanium diode, the junction voltage is about _____.
- A0.7V
- B0.3
- C0.6V
- D3V
In n-type of semiconductor, majority carries are:
- APositron
- BElectron
- CHoles
- DImpure particles
The energy gap between the valence band and the condition band for a material is 6eV. The material is?
- AAn insulator
- BA metal
- CAn intrinsic semiconductor
- DA superconductor
At absolute zero, Si acts as:
- ANon-metal.
- BMetal.
- CInsulator.
- DNone of these.
Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from thecodes(a), (b), (c) and (d) as given below.
Reason: Diode lasers consume less energy.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false and R is also false.
Reason: Diode lasers consume less energy.
Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from thecodes(a), (b), (c) and (d) as given below.
Reason: There is a large energy gap between valence band and conduction band of diamond.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false and R is also false.
Reason: There is a large energy gap between valence band and conduction band of diamond.
Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from thecodes(a), (b), (c) and (d) as given below.
Reason: The drift of electrons and holes is due to thermal excitation.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false and R is also false.
Reason: The drift of electrons and holes is due to thermal excitation.
Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from thecodes(a), (b), (c) and (d) as given below.
Reason: In half-wave rectifier only one diode is used.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false and R is also false.
Reason: In half-wave rectifier only one diode is used.
Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from thecodes(a), (b), (c) and (d) as given below.
Reason: In reversed biased, the depletion layer is reduced.
- Both A and R are true and R is the correct explanation of A.
- Both A and R are true but R is not the correct explanation of A.
- A is true but R is false.
- A is false and R is also false.
Reason: In reversed biased, the depletion layer is reduced.
In an n-type silicon, which of the following statement is true:
- Electrons are majority carriers and trivalent atoms are the dopants.
- Electrons are minority carriers and pentavalent atoms are the dopants.
- Holes are minority carriers and pentavalent atoms are the dopants.
- Holes are majority carriers and trivalent atoms are the dopants.
In an unbiased p-n junction, holes diffuse from the p-region to n-region because,
- Free electrons in the n-region attract them.
- They move across the junction by the potential difference.
- Hole concentration in p-region is more as compared to n-region.
- All the above.
Carbon, silicon and germanium have four valence electrons each. These are characterised by valence and conduction bands separated by energy band gap respectively equal to (Eg)C, (Eg)si and (Eg)Ge. Which of the following statements is true?
- (Eg)Si < (Eg)Ge < (Eg)C
- (Eg)C < (Eg)Ge < (Eg)Si
- (Eg)C > (Eg)Si > (Eg)Ge
- (Eg)C = (Eg)Si = (Eg)Ge
When a forward bias is applied to a p-n junction, it
- Raises the potential barrier.
- Reduces the majority carrier current to zero.
- Lowers the potential barrier.
- None of the above.
Two amplifiers are connected one after the other in series (cascaded). The first amplifier has a voltage gain of 10 and the second has a voltage gain of 20. If the input signal is 0.01 volt, calculate the output ac signal.
In half-wave rectification, what is the output frequency if the input frequency is 50 Hz. What is the output frequency of a full-wave rectifier for the same input frequency.
The $V-I$ characteristic of a silicon diode is shown in the Fig. 14.17. Calculate the resistance of the diode at (a) $I_D=15 mA$ and (b) $V_D=-10 V$.
View full solution →Explain, with the help of a circuit diagram, the working of a p-n junction diode as a half-wave rectifier.
Draw a circuit diagram of n-p-n transistor amplifier in CE configuration. Under what condition does the transistor act as an amplifier?
Write the two processes that take place in the formation of a p-n junction. Explain with the help of a diagram, the formation of depletion region and barrier potential in a p-n junction.
The number of silicon atoms per m3 is 5 × 1028. This is doped simultaneously with 5 × 1022 atoms per m3 of Arsenci and 5 × 1020 per m3 atoms of Indium. Calculate the number of electrons and holes Given that ni = 1.5 × 1016 m-3. Is the material n-type or p-type?
Write the truth table for a NAND gate connected as given in Fig. 14.45.
Hence identify the exact logic operation carried out by this circuit.
Hence identify the exact logic operation carried out by this circuit.
A p-n photodiode is fabricated from a semiconductor with band gap of 2.8 eV. Can it detect a wavelength of 6000 nm?
Graph showing the variation of current versus voltage for a material GaAs is shown in the figure. Identify the region of.
- Negative resistance.
- Where Ohm's law is obeyed.
- State briefly the processes involved in the formation of p-n junction explaining clearly how the depletion region is formed.
- Using the necessary circuit diagrams, show how the V-I characteristics of a p-n junction are obtained in
- Forward biasing.
- Reverse biasing.
How are these characteristics made use of in rectification?
A silicon p-n junction diode is connected to a resistor Rand a battery of voltage VB through milliammeter (mA) as shown in figure. The knee voltage for this junction diode is VN = 0.7V. The p-n junction diode requires a minimum current of 1mA to attain a value higher than the knee point on the I-V characteristics of this junction diode. Assuming that the voltage Vacross the junction is independent of the current above the knee point. A p-n junction is the basic building block of many semiconductordevices like diodes. Important process occurring during the formation of a p-n junction are diffusion and drift. ln an n-type semiconductor concentration of electrons is more as compared to holes. ln a p-type semiconductor concentration of holes is more as compared to electrons.
View full solution →- If VB = 5V, the maximum value of R so that the voltage V is above the knee point voltage is:
- $40\text{k}\Omega$
- $4.3\text{k}\Omega$
- $5.0\text{k}\Omega$
- $5.7\text{k}\Omega$
- If VB = 5V, the value of R in order to establish a current to 6mA in the circuit is:
- $833\Omega$
- $717\Omega$
- $950\Omega$
- $733\Omega$
- If VB = 6V, the power dissipated in the resistor R, when a current of 6mA flows in the circuit is:
- 30.2mW
- 30.8mW
- 31.2mW
- 31.8mW
- When the diode is reverse biased with a voltage of 6V and Vbi = 0.63V. Calculate the total potential.
- 9.27V
- 6.63V
- 5.27V
- 0.63V
- Which of the below mentioned statement is false regarding a p-n junction diode?
- Diodes are uncontrolled devices.
- Diodes are rectifying devices.
- Diodes are unidirectional devices.
- Diodes have three terminals.
P-n junction is a single crystal of Ge or Si doped in such a manner that one half portion of it acts asp-type semiconductor and other half functions as n-type semiconductor. As soon as a p-n junction is formed, the holes from the p-region diffuse into then-region, and electron from n region diffuse in top-region. This results in the development of V 8 across the junction which opposes the further diffusion of electrons and holes through the junction.
- In an unbiased p-n junction electrons diffuse from n-region top-region because:
- Holes in p-region attract them.
- Electrons travel across the junction due to potential difference.
- Electron concentration inn-region is more as compared to that in p-region.
- Only electrons move from n top region and not the vice-versa.
- Electron hole recombination in p-n junction may lead to emission of:
- Light.
- Ultraviolet rays.
- Sound.
- Radioactive rays.
- In an unbiased p-n junction:
- Potential at pis equal to that at n.
- Potential at pis + ve and that at n is - ve.
- Potential at pis more than that at n.
- Potential at pis less than that at n.
- The potential of depletion layer is due to:
- Electrons.
- Holes.
- Ions.
- Forbidden band.
- In the depletion layer of unbiased p-n junction,
- It is devoid of charge carriers.
- Has only electrons.
- Has only holes.
- P-n junction has a weak electric field.
The electron mobility characterises how quickly an electron can move through a metal of semiconductor when pulled by an electric field. There is an analogous quality for holes, called hole mobility. A block of pure silicon at 300K has a length of 10cm and an area of 1.0cm2. A battery of emf 2V is connected across it. The mobility of electron is 0.14m2 V-1s-1 and their number density is 1.5 × 1016m-3. The mobility of holes is 0.05m2 V-1s-1.
View full solution →- The electron current is:
- 6.72 × 10-4A
- 6.72 × 10-5A
- 6.72 × 10-6A
- 6.72 × 10-7A
- The hole current is:
- 2.0 × 10-7A
- 2.2 × 10-7A
- 2.4 × 10-7A
- 2.6 × 10-7A
- The number density of donor atoms which are to be added up to pure silicon semiconductor to produce an n-type semiconductor of conductivity $6.4\Omega^{-1}\text{cm}^{-1}$ is approximately (neglect the contribution of holes to conductivity).
- 3 × 1022m-3
- 3 × 1023m-3
- 3 × 1024m-3
- 3 × 1021m-3
- When the given silicon semiconductor is doped with indium, the hole concentration increases to 4.5 × 1023m-3. The electron concentration in doped silicon is:
- 3 × 109m-3
- 4 × 109m-3
- 5 × 109m-3
- 6 × 109m-3
- Pick out the statement which is not correct.
- At a low temperature, the resistance of a semiconductor is very high.
- Movement of holes is restricted to the valence band only.
- Width of the depletion region increases as the forward bias voltage increases in case of a p-n junction diode.
- ln a forward bias condition, the diode heavily conducts.
In a p-n junction diode, the current I can be expressed as,
$I=I_0\ \text{exp}\Big(\frac{\text{eV}}{2\text{k}_{\text{B}}\text{T}}-1\Big)$
where $I_0$ is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $I$ is the current through the diode, kg is the Boltzmann constant (8.6×10-5 eV/K) and T is the absolute temperature. If for a given diode $I_0$ = 5×10-12 A and T = 300 K, then,
View full solution →$I=I_0\ \text{exp}\Big(\frac{\text{eV}}{2\text{k}_{\text{B}}\text{T}}-1\Big)$
where $I_0$ is called the reverse saturation current, V is the voltage across the diode and is positive for forward bias and negative for reverse bias, and $I$ is the current through the diode, kg is the Boltzmann constant (8.6×10-5 eV/K) and T is the absolute temperature. If for a given diode $I_0$ = 5×10-12 A and T = 300 K, then,
- What will be the forward current at a forward voltage of 0.6V?
- What will be the increase in the current if the voltage across the diode is increased to 0.7V?
- What is the dynamic resistance?
- What will be the current if reverse bias voltage changes from 1V to 2V?
In an intrinsic semiconductor the energy gap Eg is 1.2eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600K and that at 300K? Assume that the temperature dependence of intrinsic carrier concentration ni is given by,
$\text{n}_{\text{i}}=\text{n}_{\text{0}}\ {\text{exp}}\Big(-\frac{\text{E}_\text{g}}{2{\text{K}_{\text{B}}}{\text{T}}}\Big)$
where n0 is a constant.
View full solution →$\text{n}_{\text{i}}=\text{n}_{\text{0}}\ {\text{exp}}\Big(-\frac{\text{E}_\text{g}}{2{\text{K}_{\text{B}}}{\text{T}}}\Big)$
where n0 is a constant.
Write the truth table for the circuits given in Fig. 14.48 consisting of NOR gates only. Identify the logic operations (OR, AND, NOT) performed by the two circuits.
- Describe briefly, with the help of a diagram, the role of the two important processes involved in the formation of a p-n junction.
- Name the device which is used as a voltage regulator. Draw the necessary circuit diagram and explain its working.
- Draw the circuit diagrams of a p-n junction diode in (i) forward bias, (ii) reverse bias. How are these circuits used to study the V - I characteristics of a silicon diode? Draw the typical V - I characteristics.
- What is a light emitting diode (LED)? Mention two important advantages of LEDs over conventional lamps.
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