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All questions of Semiconductor Electronics: Materials, Devices & Simple Circuits for NEET Exam
Arrange the materials in ascending order of energy band gaps- a)conductor, semiconductor, insulator
- b)insulator, semiconductor, conductor
- c)semiconductor, conductor, insulator
- d)insulator, conductor, semiconductor
Correct answer is option 'A'. Can you explain this answer?
Arrange the materials in ascending order of energy band gaps
a)
conductor, semiconductor, insulator
b)
insulator, semiconductor, conductor
c)
semiconductor, conductor, insulator
d)
insulator, conductor, semiconductor
 | Krishna Iyer answered |
- In the case of insulators, the energy gap is very large while in the case of conductors energy gap is very short.
- This energy gap refers to the energy difference between the valence band and conduction band whereas, in semiconductors, this band gap value is in between that of conductors and insulators.
Can you explain the answer of this question below:When a pure semiconductor is heated, its resistance- A:Goes down.
- B:Goes up.
- C:Remains the same.
- D:None of the above.
The answer is a.
When a pure semiconductor is heated, its resistance
A:
Goes down.
B:
Goes up.
C:
Remains the same.
D:
None of the above.
 | Nikita Singh answered |
As temperature increases electrons from the valence band go to the conduction band, so conductivity increases hence resistance decreases.
The densities of electrons and holes in an extrinsic semiconductor are 8 x 1015 cm-3and 5 x 1012 cm-3 respectively. The mobilities of electrons and holes are respectively 23 x 103 cm3 / Vs and 102 cm3 / Vs. Which type of semiconductor is it?- a)p-type only
- b)n-type only
- c)either n or p type
- d)both n and p type
Correct answer is option 'B'. Can you explain this answer?
The densities of electrons and holes in an extrinsic semiconductor are 8 x 1015 cm-3and 5 x 1012 cm-3 respectively. The mobilities of electrons and holes are respectively 23 x 103 cm3 / Vs and 102 cm3 / Vs. Which type of semiconductor is it?
a)
p-type only
b)
n-type only
c)
either n or p type
d)
both n and p type
 | Maulik Dasgupta answered |
Given data:
- Density of electrons (n) = 8 x 10^15 cm^-3
- Density of holes (p) = 5 x 10^12 cm^-3
- Mobility of electrons (μn) = 23 x 10^3 cm^2/ V s
- Mobility of holes (μp) = 10^2 cm^2/ V s
To determine the type of semiconductor, we need to compare the densities of electrons and holes.
Explanation:
- In an n-type semiconductor, the majority charge carriers are electrons, and the minority charge carriers are holes. The density of electrons is much higher than the density of holes.
- In a p-type semiconductor, the majority charge carriers are holes, and the minority charge carriers are electrons. The density of holes is much higher than the density of electrons.
Now we can compare the given densities of electrons and holes:
- Density of electrons (n) = 8 x 10^15 cm^-3 (very high)
- Density of holes (p) = 5 x 10^12 cm^-3 (very low)
This indicates that the majority charge carriers are electrons, and the semiconductor is n-type.
We can also check the mobility values to confirm this:
- Mobility of electrons (μn) = 23 x 10^3 cm^2/ V s (very high)
- Mobility of holes (μp) = 10^2 cm^2/ V s (very low)
The high mobility of electrons also supports the conclusion that the semiconductor is n-type.
Therefore, the correct answer is option (B) n-type only.
- Density of electrons (n) = 8 x 10^15 cm^-3
- Density of holes (p) = 5 x 10^12 cm^-3
- Mobility of electrons (μn) = 23 x 10^3 cm^2/ V s
- Mobility of holes (μp) = 10^2 cm^2/ V s
To determine the type of semiconductor, we need to compare the densities of electrons and holes.
Explanation:
- In an n-type semiconductor, the majority charge carriers are electrons, and the minority charge carriers are holes. The density of electrons is much higher than the density of holes.
- In a p-type semiconductor, the majority charge carriers are holes, and the minority charge carriers are electrons. The density of holes is much higher than the density of electrons.
Now we can compare the given densities of electrons and holes:
- Density of electrons (n) = 8 x 10^15 cm^-3 (very high)
- Density of holes (p) = 5 x 10^12 cm^-3 (very low)
This indicates that the majority charge carriers are electrons, and the semiconductor is n-type.
We can also check the mobility values to confirm this:
- Mobility of electrons (μn) = 23 x 10^3 cm^2/ V s (very high)
- Mobility of holes (μp) = 10^2 cm^2/ V s (very low)
The high mobility of electrons also supports the conclusion that the semiconductor is n-type.
Therefore, the correct answer is option (B) n-type only.
Can you explain the answer of this question below:What is the order of forbidden gap in the energy bands of silicon?
- A:
3.1 eV
- B:
0.1 eV
- C:
2.1 eV
- D:
1.1 eV
The answer is d.
What is the order of forbidden gap in the energy bands of silicon?
3.1 eV
0.1 eV
2.1 eV
1.1 eV
 | Anjali Iyer answered |
Eg : Silicon & Germanium. b) For silicon forbidden energy gap is 1.1 eV and for germanium 0.72 eV. c) At absolute zero, semiconductors behave as perfect insulators. d) Semiconductors are of two types.
A semiconductor is formed by- a)Covalent bonds
- b)Electrovalent bonds
- c)Coordinate bonds
- d)None of above
Correct answer is option 'A'. Can you explain this answer?
A semiconductor is formed by
a)
Covalent bonds
b)
Electrovalent bonds
c)
Coordinate bonds
d)
None of above
 | Learners Habitat answered |
- The electrons surrounding each atom in a semiconductor are part of a covalent bond. A covalent bond consists of two atoms "sharing" a single electron.
- Each atom forms 4 covalent bonds with the 4 surrounding atoms. Therefore, between each atom and its 4 surrounding atoms, 8 electrons are being shared.
The binary number 10101 is equivalent to decimal number- a)21
- b)13
- c)43
- d)51
Correct answer is option 'A'. Can you explain this answer?
The binary number 10101 is equivalent to decimal number
a)
21
b)
13
c)
43
d)
51
 | Mira Joshi answered |
10101 is the binary number to be converted to a decimal number.
To convert, multiply the numbers individually with 2n where n changes according to place value of the number.
(10101)2= (20× 1) + (0 × 21) + (1×22) + (0×23) + (1×24)
= 1 + 0 + 4 + 0 + 16
= 21
To convert, multiply the numbers individually with 2n where n changes according to place value of the number.
(10101)2= (20× 1) + (0 × 21) + (1×22) + (0×23) + (1×24)
= 1 + 0 + 4 + 0 + 16
= 21
Can you explain the answer of this question below:If A = 1 and B = 0, what is the value of A.A+B?
A: AB: A2 + BC: A.BD: BThe answer is a.
A: A
B: A2 + B
C: A.B
D: B
The answer is a.
 | Ambition Institute answered |
A=1 and B=0
A.A+B
=A2+B [De-Morgan’s theorem)
=1+0
=1
=A
A.A+B
=A2+B [De-Morgan’s theorem)
=1+0
=1
=A
In a semiconductor, the energy gap between valence band and conduction band is abouta)10 eVb)5 eVc)1 eVd)15 eVCorrect answer is option 'C'. Can you explain this answer?
| | Riya Banerjee answered |
In semiconductors, the forbidden gap between the valence band and conduction band is very small. It has a forbidden gap of about 1 electron volt (eV).
Mobilities of electrons and holes in a sample of intrinsic germanium semiconductor at room temperature are 0.36m2/volt-sec and 0.17 m2/volt-sec respectively. If the electron and hole densities are each equal to 2.5 X 1019/m3, the conductivity is ______.- a)1.22 S/m
- b)3.12 S/m
- c)4.12 S/m
- d)2.12 S/m
Correct answer is option 'D'. Can you explain this answer?
Mobilities of electrons and holes in a sample of intrinsic germanium semiconductor at room temperature are 0.36m2/volt-sec and 0.17 m2/volt-sec respectively. If the electron and hole densities are each equal to 2.5 X 1019/m3, the conductivity is ______.
a)
1.22 S/m
b)
3.12 S/m
c)
4.12 S/m
d)
2.12 S/m
| | Krishna Iyer answered |
Given in question,
Mobilities of electrons μe = 0.36×m2V−1s−1
Mobilities of holes μh = 0.17×m2V−1s−1
densities of electron= densities of holes=2.5×1019m−3
As we know, conductivity,
σ=1/p=e(μene+μnnn)
=1.6×10−19[0.36×2.5×1019+0.17×2.5×1019)]
=2.12Sm−1
σ=1/p=e(μene+μnnn)
=1.6×10−19[0.36×2.5×1019+0.17×2.5×1019)]
=2.12Sm−1
So the electrical conductivity of germanium is 2.12Sm−1
Can you explain the answer of this question below:The inputs of a NAND gate are connected together. The resultant circuit is
- A:
NOT gate
- B:
AND gate
- C:
OR gate
- D:
None of the above
The answer is a.
The inputs of a NAND gate are connected together. The resultant circuit is
NOT gate
AND gate
OR gate
None of the above
 | Lavanya Menon answered |
The equation of NAND gate is
Y=Aˉ.Bˉ
When both the terminals are connected together.
A=B=A
Y=Aˉ.Aˉ=Aˉ.Aˉ=Aˉ
Which is nothing but the equation of NOT gate.
Y=Aˉ
Option A is the correct answer.
Y=Aˉ.Bˉ
When both the terminals are connected together.
A=B=A
Y=Aˉ.Aˉ=Aˉ.Aˉ=Aˉ
Which is nothing but the equation of NOT gate.
Y=Aˉ
Option A is the correct answer.
The NOR gate is OR gate followed by- a)AND gate
- b)OR gate
- c)NOT gate
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
The NOR gate is OR gate followed by
a)
AND gate
b)
OR gate
c)
NOT gate
d)
None of the above
| | Rohit Shah answered |
A NOR gate is a logic gate which gives a positive output only when both inputs are negative. Like NAND gates, NOR gates are so-called "universal gates" that can be combined to form any other kind of logic gate.
The name of a diode that can be used to provide a variable capacitance is:- a)varactor diode
- b)varactor capacitor
- c)potential diode
- d)potential capacitor
Correct answer is option 'A'. Can you explain this answer?
The name of a diode that can be used to provide a variable capacitance is:
a)
varactor diode
b)
varactor capacitor
c)
potential diode
d)
potential capacitor
 | Dr Manju Sen answered |
The varactor diode is used in a place where the variable capacitance is required, and that capacitance is controlled with the help of the voltage. The Varactor diode is also known as the Varicap, Voltcap, Voltage variable capacitance or Tunning diode.
The binary number system has a base of- a)8
- b)2
- c)10
- d)16
Correct answer is 'B'. Can you explain this answer?
The binary number system has a base of
a)
8
b)
2
c)
10
d)
16
 | Tanvi Bose answered |
Explanation:
Therefore, the correct answer to the question is option B, which states that the binary number system has a base of 2.
- The binary number system is a positional number system with a base of 2.
- It uses only two digits - 0 and 1 - to represent all numbers and values.
- Each digit in a binary number represents a power of 2, with the rightmost digit representing 2^0 (1), the next digit representing 2^1 (2), the next digit representing 2^2 (4), and so on.
- Binary numbers are used extensively in computer science and digital electronics because they can be easily represented using electronic circuits, and because the two digits can be easily translated into the on/off states of electronic switches.
Therefore, the correct answer to the question is option B, which states that the binary number system has a base of 2.
What is the value of dielectric constant of a air ?- a) Less than 1
- b)Zero
- c)Equal to 1
- d) Not determine
Correct answer is option 'C'. Can you explain this answer?
What is the value of dielectric constant of a air ?
a)
Less than 1
b)
Zero
c)
Equal to 1
d)
Not determine
| | Ambition Institute answered |
The dielectric constant of a medium K = ∈/∈0
where ∈ is the permittivity of the medium.
For air, ∈ = ∈0
⟹ K = 1
where ∈ is the permittivity of the medium.
For air, ∈ = ∈0
⟹ K = 1
The knee voltage of a-n junction diode is 0.8 V and the with of the depletion layer is 2 μm. What is the electric field in the depletion layer?- a)0.4 KV/m
- b)0.4 MV/m
- c)4 MV/m
- d)4 KV/m
Correct answer is option 'B'. Can you explain this answer?
The knee voltage of a-n junction diode is 0.8 V and the with of the depletion layer is 2 μm. What is the electric field in the depletion layer?
a)
0.4 KV/m
b)
0.4 MV/m
c)
4 MV/m
d)
4 KV/m
 | Gaurav Kumar answered |
Knee voltage- it is that forward voltage beyond which current start increasing rapidly, but below knee voltage variation of forward current and applied voltage is linear.
The electric field in a region is given by, E=V/l.
where V is the potential and l is the length or distance of the region in which it has to be measured.
now, E=V/l
E=0.8/2×10-6 m
E= 0.4 MV/m
The electric field in a region is given by, E=V/l.
where V is the potential and l is the length or distance of the region in which it has to be measured.
now, E=V/l
E=0.8/2×10-6 m
E= 0.4 MV/m
In a semiconductor, current conduction is due- a)Only to holes.
- b)Only to electrons
- c)To holes and electrons
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
In a semiconductor, current conduction is due
a)
Only to holes.
b)
Only to electrons
c)
To holes and electrons
d)
None of the above
| | Geetika Shah answered |
There are two types of semiconductors: p-type and n-type. Therefore both holes and electrons conduct current respectively.
Which of the following is the universal gate?- a)OR gate
- b)AND gate
- c)NAND gate
- d)NOT gate
Correct answer is option 'C'. Can you explain this answer?
Which of the following is the universal gate?
a)
OR gate
b)
AND gate
c)
NAND gate
d)
NOT gate
| | Gaurav Kumar answered |
A universal gate is a gate which can implement any Boolean function without need to use any other gate type. The NAND and NOR gates are universal gates. In practice, this is advantageous since NAND and NOR gates are economical and easier to fabricate and are the basic gates used in all IC digital logic families.
Can you explain the answer of this question below:Electrons moving like molecules of a gas exist in:
- A:
all solids
- B:
covalent solids
- C:
metallic solids
- D:
ionic solids
The answer is c.
Electrons moving like molecules of a gas exist in:
all solids
covalent solids
metallic solids
ionic solids
| | Rohan Singh answered |
Metallic solids have unusual properties: in addition to having high thermal and electrical conductivity and being malleable and ductile, they exhibit luster, a shiny surface that reflects light. An alloy is a mixture of metals that has bulk metallic properties different from those of its constituent elements.
How many depletion regions does a transistor have?- a)Three
- b)One
- c)two
- d)Zero
Correct answer is option 'C'. Can you explain this answer?
How many depletion regions does a transistor have?
a)
Three
b)
One
c)
two
d)
Zero
| | Arjun Singhania answered |
There is only one in a P — N junction transistor. In an insulating region or depletion layer within a conductive, doped semiconductor material where the mobile charge carriers have been carried away, or have been forced away by an electric field. The only elements left in the depletion region or layer are ionized donor or acceptor impurities.Therefore, there are two depletion regions. The thickness of the depletion layer will depend on the biasing condition applied to the junction.
In a semiconductor crystal, if current flows due to breakage of crystal bonds, then the semiconductor is called- a)donor
- b)acceptor
- c)extrinsic semiconductor
- d)intrinsic semiconductor
Correct answer is option 'D'. Can you explain this answer?
In a semiconductor crystal, if current flows due to breakage of crystal bonds, then the semiconductor is called
a)
donor
b)
acceptor
c)
extrinsic semiconductor
d)
intrinsic semiconductor
 | Preeti Iyer answered |
In the case of an intrinsic semiconductor (say Si) where each Si is having 4 outermost electrons, its crystal structure consists of making 4 covalent bonds with 4 neighbouring Si atoms. Each bond consists of two electrons.
Now if one of the bonds gets broken due to some reason (collisions or high temperature) then one electron gets free and it will be having sufficient energy to cross the band gap and be ready for conduction- So in intrinsic semiconductors, current flows due to breakage of crystal bonds.
Now if one of the bonds gets broken due to some reason (collisions or high temperature) then one electron gets free and it will be having sufficient energy to cross the band gap and be ready for conduction- So in intrinsic semiconductors, current flows due to breakage of crystal bonds.
The output of a 2-input OR gate is zero only when its:- a)both inputs are 0
- b)either input is 1
- c)both inputs are 1
- d)either input is zero
Correct answer is option 'A'. Can you explain this answer?
The output of a 2-input OR gate is zero only when its:
a)
both inputs are 0
b)
either input is 1
c)
both inputs are 1
d)
either input is zero
| | Ritu Singh answered |
The truth table for 2-input OR gate is as shown in the figure. Thus, the output is zero only when both inputs are zero.
A common emitter (CE) amplifier has a current gain of 49. The voltage drop across the collector terminal is 1.8 V. The collector resistance is 2.2 kilo-ohm. What is the base current in the CE amplifier?- a)1.41 x 10-5 A
- b)1.81 x 10-5 A
- c)1.27 x 10-5 A
- d)1.67 x10-5 A
Correct answer is option 'D'. Can you explain this answer?
A common emitter (CE) amplifier has a current gain of 49. The voltage drop across the collector terminal is 1.8 V. The collector resistance is 2.2 kilo-ohm. What is the base current in the CE amplifier?
a)
1.41 x 10-5 A
b)
1.81 x 10-5 A
c)
1.27 x 10-5 A
d)
1.67 x10-5 A
 | Kiran Sanodiya answered |
The diode current depends on which of the following:- a)the diode voltage, temperature, and forward saturation current
- b)only diode voltage
- c)anode voltage
- d)the diode voltage, temperature, and reverse saturation current
Correct answer is option 'D'. Can you explain this answer?
The diode current depends on which of the following:
a)
the diode voltage, temperature, and forward saturation current
b)
only diode voltage
c)
anode voltage
d)
the diode voltage, temperature, and reverse saturation current
| | Pooja Mehta answered |
Diodes. A diode is a discrete component that allows current to flow in one direction only. ... If the diode is put in the circuit back to front, so that the voltage at the cathode is higher than the voltage at the anode, the diode will not conduct electricity. This is called reverse bias.
A diode as a rectifier converts:- a)a.c. into d.c.
- b)varying d.c. current into constant direct current
- c)high voltage into low voltage and vice-versa
- d)d.c. into a.c.
Correct answer is option 'A'. Can you explain this answer?
A diode as a rectifier converts:
a)
a.c. into d.c.
b)
varying d.c. current into constant direct current
c)
high voltage into low voltage and vice-versa
d)
d.c. into a.c.
| | Sanjeevini Angadi answered |
Rectifier is a device which converts A.C into D.C.....therefore diode as a rectifier converts ac into dc
The power amplification for a common emitter (CE) amplifier is 22110. Its current gain is 67 and internal resistance is 200 ohm. What is the value of the of the output resistance?- a)985 ohm
- b)66000 ohm
- c)41 ohm
- d)9125 ohm
Correct answer is option 'A'. Can you explain this answer?
The power amplification for a common emitter (CE) amplifier is 22110. Its current gain is 67 and internal resistance is 200 ohm. What is the value of the of the output resistance?
a)
985 ohm
b)
66000 ohm
c)
41 ohm
d)
9125 ohm
 | Prakhar Maheshwari answered |
A photodiode is used preferably in- a)to increase voltage
- b)in forward bias condition
- c)to convert electrical energy into light
- d)in reverse bias condition
Correct answer is option 'D'. Can you explain this answer?
A photodiode is used preferably in
a)
to increase voltage
b)
in forward bias condition
c)
to convert electrical energy into light
d)
in reverse bias condition
| | Rohit Shah answered |
In a photo-diode when light is incident, the fractional increase in the majority carriers is much less than the fractional increase in the minority carriers. Consequently, the fractional change due to the photo-effects on the minority carrier dominated reverse bias current is more easily measurable than the fractional change due to the photo-effects on the majority carrier dominated forward bias current. Hence, photo-diodes are preferred to be used in the reverse bias condition to easily observe the variation of current with intensity.
In the binary number 10010, the MSB is- a)100
- b)0
- c)1
- d)10
Correct answer is option 'C'. Can you explain this answer?
In the binary number 10010, the MSB is
a)
100
b)
0
c)
1
d)
10
 | Sushil Kumar answered |
1001=1x23+0x22+0x21+1x20=9
For binary numbers, the digit at the extreme right is referred to as least significant bit (LSB) and the left- most digit is called the most significant bit (MSB). Hence in binary 10010, the most significant bit is 1.
For binary numbers, the digit at the extreme right is referred to as least significant bit (LSB) and the left- most digit is called the most significant bit (MSB). Hence in binary 10010, the most significant bit is 1.
In an n-p-n transistor, the emitter current is- a)equal to base current
- b)slightly less than the collector current
- c)slightly more than the collector current
- d)equal to the collector current
Correct answer is option 'C'. Can you explain this answer?
In an n-p-n transistor, the emitter current is
a)
equal to base current
b)
slightly less than the collector current
c)
slightly more than the collector current
d)
equal to the collector current
| | Sushil Kumar answered |
In the transistor circuit, the current flows from emitter to collector via base.
Ie=Ib+Ic where variables have their usual meanings.
Hence, emitter current is slightly more than the collector current.
Ie=Ib+Ic where variables have their usual meanings.
Hence, emitter current is slightly more than the collector current.
For functioning of a transistor its emitter-base junction has to be:- a)Forward biased
- b)Reverse biased
- c)May be forward biased or reverse biased
- d)May not be biased at all
Correct answer is option 'A'. Can you explain this answer?
For functioning of a transistor its emitter-base junction has to be:
a)
Forward biased
b)
Reverse biased
c)
May be forward biased or reverse biased
d)
May not be biased at all
| | Abhijeet Ingle answered |
When emitter base junction is forward biased and collector base junction is reversed biased ,the electrons move from n to p type region and the holes moves towards the n type .When they reach other they combine enabling a current to flow across a junction.
When a forward bias is applied to a p-n junction, it- a)Raises the potential barrier
- b)Reduces the majority carrier current to zero
- c)Lowers the potential barrier
- d)None of the above
Correct answer is option 'C'. Can you explain this answer?
When a forward bias is applied to a p-n junction, it
a)
Raises the potential barrier
b)
Reduces the majority carrier current to zero
c)
Lowers the potential barrier
d)
None of the above
 | Divey Sethi answered |
When a forward bias is applied to a p-n junction it lowers the value of potential barrier. In the case of a forward bias the potential barrier opposes the applied voltage. Hence the potential barrier across the junction gets reduced.
An electronic device which converts a.c power into d.c power is- a)transistor
- b)amplifier
- c)rectifier
- d)oscillator
Correct answer is option 'C'. Can you explain this answer?
An electronic device which converts a.c power into d.c power is
a)
transistor
b)
amplifier
c)
rectifier
d)
oscillator
| | Meera Dey answered |
Understanding Rectifiers
An electronic device that converts alternating current (a.c.) power into direct current (d.c.) power is known as a rectifier. Here's a deeper look into this essential component:
What is a Rectifier?
- A rectifier is an electronic circuit that allows current to flow in one direction only.
- It plays a crucial role in power supply systems, enabling the use of a.c. sources for d.c. applications.
Types of Rectifiers
- Half-Wave Rectifier: Utilizes only one half of the a.c. signal, resulting in a pulsating d.c. output.
- Full-Wave Rectifier: Uses both halves of the a.c. signal, providing a smoother and more efficient d.c. output.
How Rectifiers Work
- Rectifiers typically use diodes, which are semiconductor devices that permit current flow in one direction.
- When a.c. voltage is applied, the diode conducts during the positive half of the cycle, blocking current during the negative half (in the case of a half-wave rectifier).
- In a full-wave rectifier, multiple diodes are used to harness both halves of the a.c. cycle.
Applications of Rectifiers
- Power supplies for electronic devices: Rectifiers convert mains a.c. to d.c. for devices like computers, TVs, and chargers.
- Battery charging: Rectifiers convert a.c. from the grid to d.c. for charging batteries.
- Signal demodulation: Rectifiers are used in radio and communication systems to retrieve audio signals from radio waves.
In summary, rectifiers are vital for converting a.c. to d.c. power, making them indispensable in various electronic applications.
An electronic device that converts alternating current (a.c.) power into direct current (d.c.) power is known as a rectifier. Here's a deeper look into this essential component:
What is a Rectifier?
- A rectifier is an electronic circuit that allows current to flow in one direction only.
- It plays a crucial role in power supply systems, enabling the use of a.c. sources for d.c. applications.
Types of Rectifiers
- Half-Wave Rectifier: Utilizes only one half of the a.c. signal, resulting in a pulsating d.c. output.
- Full-Wave Rectifier: Uses both halves of the a.c. signal, providing a smoother and more efficient d.c. output.
How Rectifiers Work
- Rectifiers typically use diodes, which are semiconductor devices that permit current flow in one direction.
- When a.c. voltage is applied, the diode conducts during the positive half of the cycle, blocking current during the negative half (in the case of a half-wave rectifier).
- In a full-wave rectifier, multiple diodes are used to harness both halves of the a.c. cycle.
Applications of Rectifiers
- Power supplies for electronic devices: Rectifiers convert mains a.c. to d.c. for devices like computers, TVs, and chargers.
- Battery charging: Rectifiers convert a.c. from the grid to d.c. for charging batteries.
- Signal demodulation: Rectifiers are used in radio and communication systems to retrieve audio signals from radio waves.
In summary, rectifiers are vital for converting a.c. to d.c. power, making them indispensable in various electronic applications.
Which of the following is true for a forward-biased diode?- a)The anode is more positive with respect to the cathode.
- b)The anode is negative with respect to the cathode.
- c)The anode potential is equal to the cathode potential.
- d)None of the above
Correct answer is option 'A'. Can you explain this answer?
Which of the following is true for a forward-biased diode?
a)
The anode is more positive with respect to the cathode.
b)
The anode is negative with respect to the cathode.
c)
The anode potential is equal to the cathode potential.
d)
None of the above
 | Vîshåkhâ Sïñgh answered |
There are two types of ions- anions are the - ve ones & cations are the +ve ones.As you know current flow from +ve to - ve, so During electrolysis or forward biasing anions move to anodes that are +ve in nature whereas cations move to cathodes which are -ve in nature.
you can remember it by a goes to a & c goes to c.
you can remember it by a goes to a & c goes to c.
The conductivity of P – type semiconductor is due to- a)both electrons and holes
- b)holes
- c)electrons
- d)none of the above.
Correct answer is option 'B'. Can you explain this answer?
The conductivity of P – type semiconductor is due to
a)
both electrons and holes
b)
holes
c)
electrons
d)
none of the above.
| | Lavanya Menon answered |
A P-type semiconductor is formed when a trivalent electron deficient impurities such as boron group elements are doped with intrinsic semiconductor. As the impurities are electron deficient, they take electrons from the valence band creating a number of holes. Due this reason conductivity in P-type semiconductor is mainly due to holes rather than electrons.
In an n-p-n transistor, the majority carriers in the emitter and collector are :- a)Free holes
- b)free electrons
- c)Free ions
- d)Valence electrons
Correct answer is option 'B'. Can you explain this answer?
In an n-p-n transistor, the majority carriers in the emitter and collector are :
a)
Free holes
b)
free electrons
c)
Free ions
d)
Valence electrons
| | T.ttttt answered |
Npn transistor is shown:
As we can see from the diagram, the emitter in the npn transistor is of n-type. Therefore majority charge carriers in the emitter of an npn transistor are called electrons
As we can see from the diagram, the emitter in the npn transistor is of n-type. Therefore majority charge carriers in the emitter of an npn transistor are called electrons
An oscillator is nothing but an amplifier with:- a)positive feedback
- b)No feedback
- c)Negative feedback
- d)Reverse feedback
Correct answer is option 'A'. Can you explain this answer?
An oscillator is nothing but an amplifier with:
a)
positive feedback
b)
No feedback
c)
Negative feedback
d)
Reverse feedback
| | Rajat Kapoor answered |
An oscillator is an electronic circuit, which generates alternating voltage. In an oscillator the output power is returned back to the input, in phase with the starting power (i.e., as a positive feedback). Oscillator works on the principle of positive feedback.
What is the voltage gain in a common emitter, where input resistance is 3Ω and the load resistance is 24Ω, β = 0.6- a)2.4
- b)8.4
- c)4.8
- d)480
Correct answer is option 'C'. Can you explain this answer?
What is the voltage gain in a common emitter, where input resistance is 3Ω and the load resistance is 24Ω, β = 0.6
a)
2.4
b)
8.4
c)
4.8
d)
480
| | Lavanya Menon answered |
Voltage gain =β× (Rc/Rb)
Or, Voltage gain =0.6×(24/3)=4.8
Or, Voltage gain =0.6×(24/3)=4.8
In transistors, the base region is thin and lightly doped else- a)collector current will be much larger than the emitter current
- b)the electrons or holes will not be able to reach the collector
- c)the transistor will be very thick
- d)the base current will decrease
Correct answer is option 'B'. Can you explain this answer?
In transistors, the base region is thin and lightly doped else
a)
collector current will be much larger than the emitter current
b)
the electrons or holes will not be able to reach the collector
c)
the transistor will be very thick
d)
the base current will decrease
| | Nikita Singh answered |
In a transistor, base bring interaction between emitter and collector. In a transistor, the base region is made very thin and lightly doped as compared to that of an emitter so that there may be less recombination of electrons and holes in this region. Due to which the base current is quite weak and the collector current is nearly equal to the emitter current.
If the base region of a transistor is doped equally to that of the emitter region, then a large electron hole recombination will take place in base, due to it, the base current would increase but collector current would decrease. Then the very purpose of a transistor would be defeated.
If the base region of a transistor is doped equally to that of the emitter region, then a large electron hole recombination will take place in base, due to it, the base current would increase but collector current would decrease. Then the very purpose of a transistor would be defeated.
Which of the statements is true for p-type semiconductors?- a)Electrons are minority carriers and pentavalent atoms are the dopants
- b)Electrons are majority carriers and trivalent atoms are the dopants
- c)Holes are minority carriers and pentavalent atoms are the dopants.
- d)Holes are majority carriers and trivalent atoms are the dopants
Correct answer is option 'D'. Can you explain this answer?
Which of the statements is true for p-type semiconductors?
a)
Electrons are minority carriers and pentavalent atoms are the dopants
b)
Electrons are majority carriers and trivalent atoms are the dopants
c)
Holes are minority carriers and pentavalent atoms are the dopants.
d)
Holes are majority carriers and trivalent atoms are the dopants
| | Sushil Kumar answered |
In a p-type semiconductor, the holes are the majority carriers, while the electrons are the minority carriers. A p-type semiconductor is obtained when trivalent atoms, such as aluminium, are doped in silicon atoms.
For functioning of a transistor its collector-base junction has to be:- a)May not be biased at all
- b)Reverse biased
- c)May be forward biased or reverse biased
- d)Forward biased
Correct answer is option 'B'. Can you explain this answer?
For functioning of a transistor its collector-base junction has to be:
a)
May not be biased at all
b)
Reverse biased
c)
May be forward biased or reverse biased
d)
Forward biased
| | Vivek answered |
Bcoz only in that case the emitter current will pass through it almost unchanged ( I AM USING *ALMOST* BCOZ THERE IS SOME REDUCTION IN CURRENT IN THE BASE. ) If the collector-base circuit is in forward bias , it's forward current will resist the emitter current as both of them will be in opposite directions.
A solid having uppermost energy – band partially filled with electrons is called- a)none of the above
- b)a conductor
- c)a semi – conductor
- d)an insulator
Correct answer is option 'B'. Can you explain this answer?
A solid having uppermost energy – band partially filled with electrons is called
a)
none of the above
b)
a conductor
c)
a semi – conductor
d)
an insulator
| | Riya Banerjee answered |
Conductor is an object or type of material that allows the flow of an electrical current in one more directions. A metal wire is a common electrical conductor. In metals such as copper or aluminium, the mobile charged particles are welcome.
In an unbiased p-n junction, holes diffuse from the p-region to n-region because- a)Free electrons in the n-region attract them
- b)They move across the junction by the potential difference
- c)Hole concentrate in p-region is more as compared to n-region
- d)All the above
Correct answer is option 'C'. Can you explain this answer?
In an unbiased p-n junction, holes diffuse from the p-region to n-region because
a)
Free electrons in the n-region attract them
b)
They move across the junction by the potential difference
c)
Hole concentrate in p-region is more as compared to n-region
d)
All the above
| | Vîshåkhâ Sïñgh answered |
There are two types of extrinsic semiconductor
p type - when intrinsic(pure form)semiconductor is doped with a trivalent (grp 13)element like B, Al,Ga,In, then its 3 valence electrons form covalent bonds with 3 Ge or Si atoms while the 4th valence electron of Ge or Si is not able to form the bond,thus there remain a hole.
n type - when Ge or Si crystal is doped with pentavalent (grp 15) like P,As,Sb,then it form covalent bond with all 4 electrons of Ge or Si atom and one valence electron is left,thus there is a free electron.
Since p type(+ve) has larger no. of holes (represent +ve)than no. of electrons and anything moves from its higher concentration to its lower concentration if no ext. force is applied, holes move from p to n region.
p type - when intrinsic(pure form)semiconductor is doped with a trivalent (grp 13)element like B, Al,Ga,In, then its 3 valence electrons form covalent bonds with 3 Ge or Si atoms while the 4th valence electron of Ge or Si is not able to form the bond,thus there remain a hole.
n type - when Ge or Si crystal is doped with pentavalent (grp 15) like P,As,Sb,then it form covalent bond with all 4 electrons of Ge or Si atom and one valence electron is left,thus there is a free electron.
Since p type(+ve) has larger no. of holes (represent +ve)than no. of electrons and anything moves from its higher concentration to its lower concentration if no ext. force is applied, holes move from p to n region.
In an n- type silicondiode, which of the following statement is true :- a)Electrons are majority carriers and trivalent atoms are the dopants.
- b)Holes are majority carriers and trivalent atoms are the dopants
- c)Electrons are minority carriers and pentavalent atoms are the dopants.
- d)Holes are minority carriers and pentavalent atoms are the dopants.
Correct answer is option 'D'. Can you explain this answer?
In an n- type silicondiode, which of the following statement is true :
a)
Electrons are majority carriers and trivalent atoms are the dopants.
b)
Holes are majority carriers and trivalent atoms are the dopants
c)
Electrons are minority carriers and pentavalent atoms are the dopants.
d)
Holes are minority carriers and pentavalent atoms are the dopants.
| | Kulsoom Fatima answered |
Ye ese kaise ho skta hai.. pta nhi kyu ye hai ans
An N-type Ge is obtained on doping the Ge- crystal with- a)boron
- b)gold
- c)aluminum
- d)phosphorus
Correct answer is option 'D'. Can you explain this answer?
An N-type Ge is obtained on doping the Ge- crystal with
a)
boron
b)
gold
c)
aluminum
d)
phosphorus
 | Ishani Patel answered |
Explanation:The addition of pentavalent impurities such as antimony, arsenic or phosphorous contributes free electrons, greatly increasing the conductivity of the intrinsic semiconductor.
The efficiency of a full wave rectifier is- a)Double as that of a half wave rectifier
- b)same as that of a half wave rectifier
- c)half as that of a half wave rectifier
- d)One third as that of a half wave rectifier
Correct answer is option 'A'. Can you explain this answer?
The efficiency of a full wave rectifier is
a)
Double as that of a half wave rectifier
b)
same as that of a half wave rectifier
c)
half as that of a half wave rectifier
d)
One third as that of a half wave rectifier
| | Kiran Khanna answered |
Efficiency of a Full Wave Rectifier
The efficiency of a rectifier refers to how effectively it converts alternating current (AC) into direct current (DC). In the case of a full wave rectifier, it is known to have a higher efficiency compared to a half wave rectifier.
1. Half Wave Rectifier
- A half wave rectifier is a simple circuit that uses a single diode to convert AC to DC.
- It works by allowing only one half of the input AC waveform to pass through, while blocking the other half.
- The output waveform produced by a half wave rectifier is characterized by a series of positive half cycles, with the negative half cycles being eliminated.
- The rectified output waveform has a large amount of ripple and contains only half of the input power.
- Therefore, the efficiency of a half wave rectifier is relatively low.
2. Full Wave Rectifier
- A full wave rectifier is a more complex circuit that uses four diodes arranged in a bridge configuration.
- It works by allowing both halves of the input AC waveform to be rectified, resulting in a full wave rectified output waveform.
- The rectified output waveform has a smaller amount of ripple compared to a half wave rectifier, as it includes both positive and negative half cycles.
- The full wave rectifier utilizes the entire input power, resulting in a higher efficiency compared to a half wave rectifier.
3. Comparison of Efficiencies
- The efficiency of a rectifier can be defined as the ratio of the DC power output to the AC power input.
- For a half wave rectifier, the maximum efficiency is around 40.6%.
- In contrast, a full wave rectifier has a maximum efficiency of around 81.2%.
- Therefore, the efficiency of a full wave rectifier is double that of a half wave rectifier.
Conclusion
The efficiency of a full wave rectifier is double that of a half wave rectifier. This is because the full wave rectifier utilizes both halves of the input AC waveform, resulting in a higher power conversion efficiency.
The efficiency of a rectifier refers to how effectively it converts alternating current (AC) into direct current (DC). In the case of a full wave rectifier, it is known to have a higher efficiency compared to a half wave rectifier.
1. Half Wave Rectifier
- A half wave rectifier is a simple circuit that uses a single diode to convert AC to DC.
- It works by allowing only one half of the input AC waveform to pass through, while blocking the other half.
- The output waveform produced by a half wave rectifier is characterized by a series of positive half cycles, with the negative half cycles being eliminated.
- The rectified output waveform has a large amount of ripple and contains only half of the input power.
- Therefore, the efficiency of a half wave rectifier is relatively low.
2. Full Wave Rectifier
- A full wave rectifier is a more complex circuit that uses four diodes arranged in a bridge configuration.
- It works by allowing both halves of the input AC waveform to be rectified, resulting in a full wave rectified output waveform.
- The rectified output waveform has a smaller amount of ripple compared to a half wave rectifier, as it includes both positive and negative half cycles.
- The full wave rectifier utilizes the entire input power, resulting in a higher efficiency compared to a half wave rectifier.
3. Comparison of Efficiencies
- The efficiency of a rectifier can be defined as the ratio of the DC power output to the AC power input.
- For a half wave rectifier, the maximum efficiency is around 40.6%.
- In contrast, a full wave rectifier has a maximum efficiency of around 81.2%.
- Therefore, the efficiency of a full wave rectifier is double that of a half wave rectifier.
Conclusion
The efficiency of a full wave rectifier is double that of a half wave rectifier. This is because the full wave rectifier utilizes both halves of the input AC waveform, resulting in a higher power conversion efficiency.
Transistors are useful in equipments such as hearing aids and small radio receivers because transistors :- a)do not distort sound
- b)Can stand rough use
- c)Consume little power
- d)Are inexpensive
Correct answer is option 'A'. Can you explain this answer?
Transistors are useful in equipments such as hearing aids and small radio receivers because transistors :
a)
do not distort sound
b)
Can stand rough use
c)
Consume little power
d)
Are inexpensive
 | Om Desai answered |
Transistors consume very less energy and are therefore used in these equipment because these equipment don't have space for big power source.
The input voltage applied to a cascaded amplifier (consisting of two amplifiers) is 0.02 V and the output voltage is 6 V. The voltage gain of one of the two amplifiers is 10. What is the voltage gain of the other amplifier?- a)1/30
- b)30
- c)3
- d)1/3
Correct answer is option 'B'. Can you explain this answer?
The input voltage applied to a cascaded amplifier (consisting of two amplifiers) is 0.02 V and the output voltage is 6 V. The voltage gain of one of the two amplifiers is 10. What is the voltage gain of the other amplifier?
a)
1/30
b)
30
c)
3
d)
1/3
| | Roshni Desai answered |
**Given Information:**
- Input voltage (Vin) = 0.02 V
- Output voltage (Vout) = 6 V
- Voltage gain of one amplifier (A1) = 10
**Calculating the Voltage Gain of the Cascaded Amplifier:**
1. The voltage gain of a cascaded amplifier is the product of the voltage gains of the individual amplifiers.
2. Let's assume the voltage gain of the other amplifier (A2) is 'x'.
3. The voltage gain of the cascaded amplifier (A_cascade) can be calculated as follows:
A_cascade = A1 * A2
A_cascade = 10 * x
4. We can calculate the voltage gain of the cascaded amplifier by using the input and output voltages:
A_cascade = Vout / Vin
A_cascade = 6 / 0.02
A_cascade = 300
5. Now we can equate the two expressions for the voltage gain of the cascaded amplifier:
10 * x = 300
6. Solving for 'x', we get:
x = 300 / 10
x = 30
**Conclusion:**
The voltage gain of the other amplifier (A2) is 30. Therefore, the correct answer is option 'B'.
- Input voltage (Vin) = 0.02 V
- Output voltage (Vout) = 6 V
- Voltage gain of one amplifier (A1) = 10
**Calculating the Voltage Gain of the Cascaded Amplifier:**
1. The voltage gain of a cascaded amplifier is the product of the voltage gains of the individual amplifiers.
2. Let's assume the voltage gain of the other amplifier (A2) is 'x'.
3. The voltage gain of the cascaded amplifier (A_cascade) can be calculated as follows:
A_cascade = A1 * A2
A_cascade = 10 * x
4. We can calculate the voltage gain of the cascaded amplifier by using the input and output voltages:
A_cascade = Vout / Vin
A_cascade = 6 / 0.02
A_cascade = 300
5. Now we can equate the two expressions for the voltage gain of the cascaded amplifier:
10 * x = 300
6. Solving for 'x', we get:
x = 300 / 10
x = 30
**Conclusion:**
The voltage gain of the other amplifier (A2) is 30. Therefore, the correct answer is option 'B'.
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