Case Study Questions Class 12 Physics Semiconductor electronics: Materials, Devices and Simple Circuits

Case study questions class 12 physics chapter 14 semiconductor electronics: materials, devices and simple circuits.

CBSE Class 12 Case Study Questions Physics Semiconductor electronics: Materials, Devices and Simple Circuits. Term 2 Important Case Study Questions for Class 12 Board Exam Students. Here we have arranged some Important Case Base Questions for students who are searching for Paragraph Based Questions Semiconductor electronics: Materials, Devices and Simple Circuits.

CBSE Case Study Questions Class 12 Physics Semiconductor electronics: Materials, Devices and Simple Circuits

Materials are classified on the basis of their conductivity as metals, semiconductors and insulators. Metals are having low resistivity and high conductivity. While semiconductors are having resistivity and conductivity in between metals and insulators. And finally insulators are those which are having high resistivity or very low conductivity. Semiconductors may exist as elemental semiconductors and also compound semiconductors. Si and Ge are elemental semiconductor and CdS, GaAs, CdSe, anthracene, polypyrrole etc. are the compound semiconductors. Each electron in an atom has different energy level and such different energy levels continuing forms the band of energy called as energy bands. Those energy band which has energy levels of Valence electrons is called as Valence band. And the energy band which is present above the Valence band is called as conduction band. On the basis of energy bands materials are also defined as metals, semiconductors and insulators. In case of metals, conduction band and Valence band overlaps with each other due to which electrons are easily available for conduction. In case of insulators, there is some energy gap between conduction band and Valence band due to which no free electrons are easily available for conduction. And in semiconductors, there is a small energy gap between conduction band and Valence band and if we give some external energy then electron from Valence band goes to conduction band  due to which conduction will be possible. These semiconductors are classified as intrinsic semiconductors and extrinsic semiconductors also. Intrinsic semiconductors are those semiconductors which exist in pure form. And intrinsic semiconductors has number of free electron is equal to number of holes. The semiconductors doped with some impurity in order to increase its conductivity are called as extrinsic semiconductors. Two types of dopants are used they are trivalent impurity and pentavalent impurity also. The extrinsic semiconductors doped with pentavalent impurity like Arsenic, Antimony, Phosphorus etc are called as n – type semiconductors. In n type semiconductors electrons are the majority charge carriers and holes are the minority charge carriers. When trivalent impurity is like Indium, Boron, Aluminium etc are added to extrinsic semiconductors then p type semiconductors will be formed. In p type semiconductors holes are majority charge carriers and electrons are the minority charge carriers.

Q 1.) In case of p-type semiconductors___

d) n h = n e = 0

Q 3.) If the energy band gap Eg> 3 eV then such materials are called as

Answer key:

Q 1.) c) n h >> n e

Q 5.) When trivalent impurity like B, Al, In are added to extrinsic semiconductor like Ge or Si then p-type Ge or Si semiconductor is formed. And when pentavalent impurity like As, Sb, P are added to extrinsic semiconductors like Ge or Si then n-type Ge or Si semiconductor is formed.

NOT gate is that logic gate which gives inverted input as output such as if 1 is the input given then output will be 0. The other logic gates OR, AND, NAND, NOR has two inputs and only one output.

Q 2.) c) 1.8 eV

Q 5.) The electric potential required to restrict the movement of electron from n-side to p-side across p-n junction is called as barrier potential.

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case study on rectifier class 12

CBSE 12th Standard Physics Subject Semiconductor Electronics Materials Devices And Simple Circuits Chapter Case Study Questions 2021

By QB365 on 21 May, 2021

QB365 Provides the updated CASE Study Questions for Class 12 , and also provide the detail solution for each and every case study questions . Case study questions are latest updated question pattern from NCERT, QB365 will helps to get  more marks in Exams 

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Cbse 12th standard physics subject semiconductor electronics materials devices and simple circuits case study questions 2021.

12th Standard CBSE

Final Semester - June 2015

When the diode is forward biased, it is found that beyond forward voltage V = V k , called knee voltage, the conductivity is very high. At this value of battery biasing for p-n junction,the potential barrier is overcome and the current increases rapidly with increase in forward voltage. When the diode is reverse biased, the reverse bias voltage produces a very small current about a few microamperes which almost remains constant with bias. This small current is reverse saturation current. (i) In which of the following figures, the p-n diode is forward biased.

(ii) Based on the V-I characteristics of the diode, we can classify diode as

case study on rectifier class 12

(iv) In the case of forward biasing of a p-n junction diode, which one of the following figures correctly depicts the direction of conventional current (indicated by an arrow mark)?

case study on rectifier class 12

case study on rectifier class 12

case study on rectifier class 12

(v) With an ac input from 50 Hz power line, the ripple frequency is

case study on rectifier class 12

= 0 < 3eV > 3eV

(ii) In a semiconductor, separation between conduction and valence band is of the order of

(iii) Based on the band theory of conductors, insulators and semiconductors, the forbidden gap is smallest in

(iv) Carbon, silicon and germanium have four valence electrons each. At room temperature which one of the following statements is most appropriate?

(v) Solids having highest energy level partially filled with electrons are

case study on rectifier class 12

(ii) The schematic symbol of light emitting diode is (LED)

(iii) An LED is constructed from a p-n junction based on a certain Ga-As-P semiconducting material whose energy gap is 1.9 eV. Identify the colour of the emitted light.

(iv) Which one of the following statement is not correct in the case of light emitting diodes?

(v) The energy of radiation emitted by LED is

case study on rectifier class 12

(ii) Three photo diodes D 1  D 2 and D 3 are made of semiconductors having band gap of 2.5 eV, 2 eV and 3 eV, respectively. Which one will be able to detect light of wavelength 6000 \(\dot A\) ?

, and D both

(iii) Photodiode is a device

(iv) To detect light of wavelength 500 nrn, the photodiode must be fabricated from a semiconductor of minimum bandwidth of

(v) Photo diode can be used as a photo detector to detect

*****************************************

Cbse 12th standard physics subject semiconductor electronics materials devices and simple circuits case study questions 2021 answer keys.

(i) (c) :The p-n diode is forward biased when p-side is at a higher potential than n-side. (ii) (c) (iii) (d) : Forward bias resistance,  \(R_{1}=\frac{\Delta V}{\Delta I}=\frac{0.8-0.7}{(20-10) \times 10^{-3}}=\frac{0.1}{10 \times 10^{-3}}=10\) \(\text { Reverse bias resistance, } R_{2}=\frac{10}{1 \times 10^{-6}}=10^{7}\) Then, the ratio of forward to reverse bias resistance \(\frac{R_{1}}{R_{2}}=\frac{10}{10^{7}}=10^{-6}\) (iv) (d) : In p-region the direction of conventional current is same as flow of holes. In n-region the direction of conventional current is opposite to the flow of electrons. (v) (c) : In the given circuit the junction diode is forward biased and offers zero resistance \(\therefore \text { The current, } I=\frac{3 \mathrm{~V}-1 \mathrm{~V}}{200 \Omega}=\frac{2 \mathrm{~V}}{200 \Omega}=0.01 \mathrm{~A}\)

(i) (a) :The rms value of the output voltage at the load resistance  \(V_{\mathrm{rms}}=\frac{V_{0}}{\sqrt{2}}\) (ii) (d) (iii) (a) (iv) (c) : The given circuit works as a half wave rectifier. In this circuit, we will get current through R when p-n junction is forward biased and no current when p-n junction is reverse biased. Thus the current (I) through resistor (R) will be shown in option (c). (v) (c)

(i) (c) :In insulator, energy band gap is > 3 eV (ii) (b) : In conductor, separation between conduction and valence bands is zero and in insulator, it is greater than 1 eV. Hence in semiconductor the separation between conduction and valence band is 1 eV. (iii) (a) : According to band theory the forbidden gap in conductors E g = 0, in insulators E g > 3 eV and in semiconductors E g < 3 eV. (iv) (a) : The four valence electrons of C, Si and Ge lie respectively in the second, third and fourth orbit.Hence energy required to take out an electron from these atoms (i.e. ionisation energy Eg) will be least for Ge, followed by Si and highest for C. Hence, the number of free electrons for conduction in Ge and Si are significant but negligibly small for C. (v) (b)

(i) (b) : The l- V characteristics of an LED is similar to that of a Si junction diode. But the threshold voltages are much higher and slightly different for each colour. (ii) (b) (iii) (b) :  \(\text { As } E_{g}=\frac{h c}{\lambda} \quad \therefore \lambda=\frac{h c}{E_{g}}\) Here, E g = 1.9 eV, he = 1240 eV nm \(\therefore \quad \lambda=\frac{1240 \mathrm{eVnm}}{1.9 \mathrm{eV}}=652.6 \mathrm{nm}\) Hence, the emitted light is of red colour. (iv) (c) : A light emitting diode is a heavily doped p-n junction diode which emits light only when it is forward biased. (v) (d)

(i) (c) :  \(\lambda_{\max }=\frac{h c}{E}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2.5 \times 1.6 \times 10^{-19}}\) \(=5000 \dot A\) \(\therefore \quad \lambda=4000 \dot A<\lambda_{\max }\) (ii) (b) : Energy of incident photon  \(E=\frac{h c}{\lambda}\) \(=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{6 \times 10^{-7} \times 1.6 \times 10^{-19}}=2.06 \mathrm{eV}\) The incident radiation can be detected by a photo diode if energy of incident photon is greater than the band gap. As D 2 = 2 eV, therefore D 2 will detect these radiations. (iii) (a) : Photodiode is a device which is always operated in reverse bias. (iv) (c)  : Let E g be the required bandwidth. Then \(E_{g}=\frac{h c}{\lambda}\) Here, he = 1240 eV nm, \(\lambda\)  = 500 nm \(\therefore \quad E_{g}=\frac{1240 \mathrm{eV} \mathrm{nm}}{500 \mathrm{nm}}=2.48 \mathrm{eV}\) (v) (a) : A photo diode is a device which is used to detect optical signals.

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Case Study Questions for Class 12 Physics Chapter 14 Semiconductor Electronics

  • Last modified on: 2 years ago
  • Reading Time: 8 Minutes

Case Study Questions for Class 12 Physics Chapter 14 Semiconductor Electronics

Question 1:

Light Emitting Diode: It is a heavily doped p-n junction which under forward bias emits spontaneous radiation. The diode is encapsulated with a transparent cover so that emitted light can come out. When the diode is forward biased, electrons are sent from n → p (where they are minority carriers) and holes are sent from p → n (where they are minority carriers). At the junction boundary, the concentration of minority carriers increases as compared to the equilibrium concentration (i.e., when there is no bias). Thus at the junction boundary on either side of the junction, excess minority carriers are there which recombine with majority carriers near the junction. On recombination, the energy is released in the form of photons. Photons with energy equal to or slightly less than the band gap are emitted. When the forward current of the diode is small, the intensity of light emitted is small. As the forward current increases, intensity of light increases and reaches a maximum. Further increase in the forward current results in decrease of light intensity. LED’s are biased such that the light emitting efficiency is maximum. The V-I characteristics of a LED is similar to that of a Si junction diode. But, the threshold voltages are much higher and slightly different for each colour. The reverse breakdown voltages of LED’s are very low, typically around 5 V. So care should be taken that high reverse voltages do not appear across them. LED’s that can emit red, yellow, orange, green and blue light are commercially available.

Read the above case/passage and answer the following questions:

(i) LED is a: (a) lightly doped p-n junction diode. (b) heavily doped p-n junction diode. (c) moderately doped p-n junction diode. (d) two back to back p-n junction diode.

(ii) LED emits light: (a) when reversed biased. (b) when forward biased. (c) when forward or reverse biased. (d) when heated.

(iii) During recombination at the junction, emitted photons have: ( a) energy equal to or slightly less than the band gap. (b) energy greater than the band gap. (c) energy which has no relation with the band gap. (d) very low energy compared to band gap.

(iv) Threshold voltage of LED is: (a) lower compared to other p-n junction diodes and slightly different for each colour. (b) lower compared to other p-n junction diodes and same for all colours. (c) higher compared to other p-n junction diodes and same for all colours. (d) higher compared to other p-n junction diodes and slightly different for each colour.

(v) The reverse breakdown voltages of LED’s are: (a) very low and typically around 0.5 V. (b) very high and typically around 50 V. (c) very low and typically around 5 V. (d) very low and typically around 0.05 V.

Case Study Question for Class 12 Physics Chapter 7 Alternating Current

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Last modified on:2 years agoReading Time:8MinutesCase Study Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Question 1: Light Emitting Diode:It is a heavily doped p-n junction which under forward bias emits spontaneous radiation. The diode is encapsulated with a transparent cover so that emitted light can come out. When the diode is forward biased, electrons…

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Class 12 Physics Chapter 14 Case Study Question Semiconductor Electronics: Materials, Devices, and Simple Circuits PDF Download

In Class 12 Boards there will be Case studies and Passage Based Questions will be asked, So practice these types of questions. Study Rate is always there to help you. Free PDF Download of CBSE Class 12 Physics Chapter 14 Semiconductor Electronics: Materials, Devices, and Simple Circuits Case Study and Passage Based Questions with Answers were Prepared Based on Latest Exam Pattern. Students can solve NCERT Class 12 Physics Semiconductor Electronics : Materials, Devices and Simple Circuits to know their preparation level.

case study on rectifier class 12

In CBSE Class 12 Physics Paper, Students will have to answer some questions based on  Assertion and Reason . There will be a few questions based on case studies and passage-based as well. In that, a paragraph will be given, and then the MCQ questions based on it will be asked.

Semiconductor Electronics : Materials, Devices and Simple Circuits Case Study Questions With answers

Here, we have provided case-based/passage-based questions for Class 12 Physics  Chapter 14 Semiconductor Electronics : Materials, Devices and Simple Circuits

Case Study/Passage Based Questions

When the diode is forward biased, it is found that beyond forward voltage V = Vk, called knee voltage, the conductivity is very high. At this value of battery biasing for p-n junction, the potential barrier is overcome and the current increases rapidly with an increase in forwarding voltage. When the diode is reverse biased, the reverse bias voltage produces a very small current about a few microamperes which almost remains constant with bias. This small current is reverse saturation current.

In which of the following figures, the p-n diode is forward biased

case study on rectifier class 12

Answer:(d) : All options are basic properties of nuclear forces. So, all options are correct.

Based on the V-I characteristics of the diode, we can classify diode as (a) bi-directional device (b) ohmic device (c) non-ohmic device (d) passive element

Answer:(d) passive element

The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is

case study on rectifier class 12

(a) 100 (b) 10 6 (c) 10 (d) 10 –6

Answer:(d) 10–6

In the case of forwarding biasing of a p-n junction diode, which one of the following figures correctly depicts the direction of conventional current (indicated by an arrow mark)?

case study on rectifier class 12

If an ideal junction diode is connected as shown, then the value of the current I is

case study on rectifier class 12

(a) 0.013 A (b) 0.02 A (c) 0.01 A (d) 0.1 A

Answer:(a)0.013 A

A photodiode is an optoelectronic device in which current carriers are generated by photons through photo-excitation i.e., photoconduction by light. It is a p-n junction fabricated from a photosensitive semiconductor and provided with a transparent window so as to allow light to fall on its junction. A photodiode can turn its current ON and OFF in nanoseconds. So, it can be used as the fastest photo-detector

case study on rectifier class 12

The photodiode is a device (a) that is always operated in reverse bias. (b) which is always operated in forwarding bias. (c) in which photocurrent is independent of the intensity of incident radiation. (d) which may be operated in both forward or reverse bias

Answer:(a) that is always operated in reverse bias.

To detect light of wavelength 500 nm, the photodiode must be fabricated from a semiconductor of a minimum bandwidth of (a) 1.24 eV (b) 0.62 eV (c) 2.48 eV (d) 3.2 eV

Answer:(c) 2.48 eV ​

The photodiode can be used as a photodetector to detect (a) optical signals (b) electrical signals (c) both (a) and (b) (d) None of these

Answer:(b) electrical signals ​

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Case Study on Semiconductor Electronics - Materials Devices & Simple Circuits Class 12 Physics PDF

Better preparation of Case Study on Semiconductor Electronics - Materials, Devices & Simple Circuits Class 12 Physics can help students score good marks in the CBSE Class 12 Board examination. Additionally, it helps build confidence and enables students to deepen their knowledge of Semiconductor Electronics - Materials, Devices & Simple Circuits. Because case-based questions are equally important for learning and board exam preparation, our team has prepared Case Study on Semiconductor Electronics - Materials, Devices & Simple Circuits Class 12 Physics in a PDF file for free distribution among students.

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Class 12 Physics (India)

Course: class 12 physics (india)   >   unit 14, half wave rectifiers.

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case study on rectifier class 12

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NCERT Solutions for Class 6, 7, 8, 9, 10, 11 and 12

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions

December 6, 2019 by Sastry CBSE

Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions Very Short Answer Type

Question 1. State the reason, why GaAs is most commonly used in making of a solar cell. (All India 2008) Answer: GaAs is most commonly used in making of a solar cell because : (i) It has high optical absorption (~ 104 cm -1 ) . (ii) It has high electrical conductivity.

Question 2. Why should a photodiode be operated at a reverse bias? (All India 2008) Answer: As fractional change in minority charge carriers is more than the fractional change in majority charge carriers, the variation in reverse saturation current is more prominent.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 204

Question 6. In a transistor, doping level in base is increased slightly. How will it affect (i) collector current and (ii) base current? (Delhi 2011) Answer: Increasing base doping level will decrease base resistance and hence increasing base current, which results in a decrease in collector current.

Question 7. What happens to the width of depletion layer of a p-n junction when it is (i) forward biased, (ii) reverse biased? (Delhi 2011) Answer: (i) In forward biased, the width of depletion layer of a p-n junction decreases. (ii) In reverse biased, the width of depletion layer of a p-n junction increases

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 200

Question 10. How does the depletion region of a p-n junction diode get affected under reverse bias? (Comptt. Delhi 2011) Answer: Depletion region widens under reverse bias.

Question 11. How does the width of depletion region of a p-n junction diode change under forward bias? (Comptt. Delhi 2011) Answer: The width of depletion region of a p-n junction

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 5

Question 14. What is the function of a photodiode? (Comptt. All India 2013) Answer: A photodiode is a special purpose p-n junction diode fabricated with a transparent window to allow light to fall on diode. It is operated under reverse bias.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 15

Question 20. Identify the logic gate whose output equals 1 when both of its inputs are 0 each. (Comptt. Delhi 2015) Answer: NAND gate or NOR gate.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 21

Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions Short Answer Type

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 202

Question 36. How is forward biasing different from reverse biasing in a pn junction diode? (Delhi 2011) Answer: Forward biasing : If the positive terminal of a battery is connected to a p-side and the negative terminal to the 72-side, then the p-n junction is said to be forward biased. Here the applied voltage V opposes the barrier voltage VB. As a result of this

  • the effective resistance across the p-n junction decreases.
  • the diffusion of electrons and holes into the depletion layer which decreases its width.

Reverse biasing : If the positive terminal of a battery is connected to the 72-side and negative terminal to the p-side, then the p-n junction is said to be reverse biased. The applied voltage V and the barrier potential V B are in the same direction. As a result of this

  • the resistance of the p-n junction becomes very large.
  • the majority charge carriers move away from the junction, increasing the width of the depletion layer.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 84

Question 39. The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~µA). What is the reason, then, to operate the photodiode in reverse bias? (Delhi 2012) Answer: The fractional increase in majority carriers is much less than the fractional increase in 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 in the majority carrier dominated forward bias current.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 44

When the photodiode is reverse biased then a certain current exits in the circuit even when no light is incident on the p-n junction of photodiode. This current is called dark current. A photodiode can turn its current ON and OFF in nanoseconds. Hence it can be used to detect the optical signals.

Question 44. Mention the important considerations required while fabricating a p-n junction diode to be used as a Light Emitting Diode (LED). What should be the order of band gap of an LED if it is required to emit light in the visible range? (Delhi 2013) Answer: The important considerations required while fabricating a p-n junction diode to be used as a Light Emitting Diode (LED) are : (i) The Light Emitting efficiency is maximum. (ii) The reverse breakdown voltage of LEDs are very low. Care should be taken that high reverse voltages do not appear across them. (iii) The semiconductor used for fabrication of visible, LEDs must have a band gap of 1.8 eV (spectral range of visible light is from about 0.4 µm to 0.7 µm i.e. from about 3 eV to 1.8 eV).

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 48

Question 66. State briefly the underlying principle of a- transistor oscillator. Draw a circuit diagram showing how the feedback is accomplished by inductive coupling. Explain the oscillator action. (All India 2008) Answer: Principle of transistor oscillator : “Sustained a.c. signals can he obtained from an amplifier circuit without any external input signal by giving a positive feedback to the input circuit through inductive coupling or RC/LC network.”

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 81

Question 76. Draw the transfer characteristic of a base-biased transistor in CE configuration. Mark the regions where the transistor can be used as a switch. Explain briefly its working. (Comptt. Delhi 2011) Answer: Transistor as a switch. The circuit diagram of transistor as a switch is shown in Figure 1. Transfer characteristics. The graph between V 0 and V i is called the transfer characteristics of the base-biased transistor, shown in Figure 2.

When the transistor is used in the cut off or saturation state, it acts as a switch.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 94

Question 81. With what considerations in view, a photodiode is fabricated? State its working with the help of a suitable diagram. Even though the current in the forward bias is known to be more than in the reverse bias, yet the photodiode works in reverse bias. What is the reason? (Delhi 2014) Answer: (a) Why is photodiode fabricated?

  • It is fabricated with a transparent window to allow light to fall on diode.

(b) Working of photodiode : When the photodiode is illuminated with photons of energy (hv > E g ) greater than the energy gap

  • of the semiconductor, electron-holes pairs are generated. These get separated due to the Junction electric field (before they recombine) which produces an emf.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 99

(d) Reason. It is easier to observe the change in the current, with change in light intensity, if a reverse bias is applied.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 100

Role of energy levels in conduction and valence bands : In the energy band diagram of n-type Si semiconductor, the donor energy level E D is slightly below the bottom E C of the conduction band and electrons from this level moves into conduction band with very small supply of energy. At room temperature, most of the donor atoms get ionised, but very few (~ 10 -12 ) atoms of Si atom get ionised. So the conduction band will have most electrons coming from donor impurities, as shown in the figure.

For p-type semiconductor, the acceptance energy level E A is slightly above the top E V of the valence band. With very small supply of energy, an electron from the valence band can jump to the level E A and ionise the acceptor negatively. At room temperature, most of the acceptor atoms get ionised leaving holes in the valence band.

Question 87. Draw a plot of transfer characteristic (V 0 vs V i and show which portion of the characteristic is used in amplification and why? Draw the circuit diagram of base bias transistor amplifier in CE configuration and briefly explain its working. (Comptt. All India 2014) Answer: (i) Transistor as a switch. The circuit diagram of transistor as a switch is shown in Figure 1. Transfer characteristics. The graph between V 0 and V i is called the transfer characteristics of the base-biased transistor, shown in Figure 2.

Question 88. (i) Write the functions of three segments of a transistor. (ii) Draw the circuit diagram for studying the input and output characteristics of n-p-n transistor in common emitter configuration. Using the circuit, explain how input, output characteristics are obtained. (Delhi 2014) Answer: (i) (a) All the three segments of a transistor have different thickness and their doping levels are also different. A brief description of the three segments of a transistor is given below :

  • Emitter: This is the segment on one side of the transistor. It is of moderate size and heavily doped. It supplies a large number of majority carriers for the current flow through the transistor.
  • Base : This is the central segment. It is very thin and lightly doped.
  • Collector : This segment collects a major portion of the majority carriers supplied by the emitter. The collector side is moderately doped and larger in size as compared to the emitter.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 165

Question 89. (i) Explain with the help of a diagram the formation of depletion region and barrier potential in a pn junction. (ii) Draw the circuit diagram of a half wave rectifier and explain its working. (All India 2016) Answer: (a) (i) Depletion layer. The layer containing unneutralized acceptor and donor ion across a p-n junction is called depletion layer. It is called depletion layer because it is depleted of mobile charge carriers. (ii) Barrier potential. The electric field between the acceptor and donor ions is called the barrier. The difference of potential from one side of the barrier to the other side is called barrier potential. (i) The increase of doping concentration will reduce width of depletion layer in semi conductor. (ii) depletion layer widens under reverse bias and vice versa.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 113

Question 91. Give reasons for the following : (i) High reverse voltage do not appear across a LED. (ii) Sunlight is not always required for the working of a solar cell. (ill) The electric field, of the junction of a Zener diode, is very high even for a small reverse bias voltage of about 5V. (Comptt. Delhi 2016) Answer: (i) It is because reverse breakdown voltage of LED is very low, i.e., nearly 5V. (ii) Solar cell can work with any light whose photon energy is more than the band gap energy. (iii) The heavy doping of p and n sides of pn junction makes the depletion region very thin, hence for a small reverse bias voltage, electric field is very high.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 115

Due to the diffusion of electrons and holes across the junction, a region of (immobile) positive charge is created on the n-side and a region of (immobile) negative charge is created on the p-side, near the junction; this is called depletion region.

Barrier potential is formed due to loss of electrons from n-region and gain of electrons by p-region. Its polarity is such that it opposes the movement of charge carriers across the junction.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 119

This is because, in the breakdown region, Zener voltage remains constant even though the current through the Zener diode changes.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 150

Using the circuit arrangements shown in fig. (i) and fig (ii), we study the variation of current with applied voltage to obtain the V-I characteristics. From the V-I characteristics of a junction diode, it is clear that it allows the current to pass only when it is forward biased. So when an alternatively voltage is applied across the diode, current flows only during that part of the cycle when it is forward biased.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 121

Question 105. State the reason, why the photodiode is always operated under reverse bias. Write the working principle of operation of a photodiode. The semiconducting material used to fabricate a photodiode, has an energy gap of 1.2 eV. Using calculations, show whether it can detect light of wavelength of 400 nm incident on it. (Comptt. All India 2017) Answer: (a) Why is photodiode fabricated?

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 133

Question 106. Draw the circuit diagram of a common emitter transistor amplifier. Write the expression for its voltage gain. Explain, how the input and output signals differ in phase by 180°. (Comptt. All India 2017) Answer:

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 135

Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions Long Answer Type

Question 110. How is a zener diode fabricated so as to make it a special purpose diode? Draw I-V characteristics of zener diode and explain the significance of breakdown voltage. Explain briefly, with the help of a circuit diagram, how a p-n junction diode works as a half wave rectifier. Answer: Zener diode is fabricated by heavily doping both p and n-sides. Due to this, depletion region formed is very thin (< 10 -6 n and the electric field of the junction is extremely high (~5 × 10 6 V / m) even for a small reverse bias voltage of 5 volts. It is seen that when the applied reverse bias voltage (V) reaches the breakdown voltage (Vz) of the Zener diode, there is a large change in the current. After the breakdown voltage Vz, a large change in the current can be produced by almost insignificant change in the reverse bias voltage. In other words, Zener voltage remains constant even though current through the Zener diode varies over a wide range.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 144

Question 112. (a) With the help of the circuit diagram explain the working principle of a transistor amplifier as an oscillator. (b) Distinguish between a conductor, a semiconductor and an insulator on the basis of energy band diagrams. (Delhi 2017) Answer: (a) Principle of transistor oscillator : “Sustained a.c. signals can he obtained from an amplifier circuit without any external input signal by giving a positive feedback to the input circuit through inductive coupling or RC/LC network.”

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 148

(b) Light emitting diode (LED) is a heavily doped p-n junction which under forward bias emits spontaneous radiations. Two important advantages of LEDs: (i) Low operational voltage and less power. (ii) Fast on-off switching capacity.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 152

Principle of transistor oscillator : “Sustained a.c. signals can he obtained from an amplifier circuit without any external input signal by giving a positive feedback to the input circuit through inductive coupling or RC/LC network.”

Question 121. (a) Why is the base region of a transistor thin and lightly doped? (b) Draw the circuit diagram for studying the characteristics of an n-p-n transistor in com-mon emitter configuration. Sketch the typical (i) input and (ii) output characteristics in this configuration. (c) Describe briefly how the output characteristics can be used to obtain the current gain in the transistor. (Comptt. Delhi 2012) Answer: (a) The base is made very thin so as to control current flowing between emitter and collector. The base is lightly doped to make a thin depletion layer between emitter and collector.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 169

(c) Using capacitive filter circuits connected in parallel to get a pure d.c. output.

Question 123. (a) Define the terms ‘depletion layer’ and ‘barrier potential’ for a p-n junction. How does (i) an increase in the doping concentration and (ii) biasing across the junction, affect the width of the depletion layer? (b) Draw the circuit diagram of a p-n diode used as a half-wave rectifier. Explain its working. (Comptt. All India 2012) Answer: (a) (i) Depletion layer. The layer containing unneutralized acceptor and donor ion across a p-n junction is called depletion layer. It is called depletion layer because it is depleted of mobile charge carriers. (ii) Barrier potential. The electric field between the acceptor and donor ions is called the barrier. The difference of potential from one side of the barrier to the other side is called barrier potential. (i) The increase of doping concentration will reduce width of depletion layer in semi conductor. (ii) depletion layer widens under reverse bias and vice versa.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 174

“These characteristic make use of p-n junction diode as a rectifier. In the positive half cycle of ac, the voltage is positive and the diode is forward biassed and it conducts current. While in the negative half cycle of ac, the voltage is negative, the diode is reverse biassed and it does not conduct current. Thus we get rectified output during positive half cycles only. The output is unidirectional, but varying.”

Question 126. (a) Differentiate between three segments of a transistor on the basis of their size and level of doping. (b) How is a transistor biased to be in active state? (c) With the help of necessary circuit diagram, describe briefly how tt-p-n transistor in CE configuration amplifies a small sinusoidal input voltage. Write the expression for the ac current gain. (Delhi 2012) Answer: (a) All the three segments of a transistor have different thickness and their doping levels are also different. A brief description of the three segments of a transistor is given below :

(b) When the transistor works as an amplifier, with its emitter-base junction forward biased; and the base-collector junction reverse biased, is said to be in Active state.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 177

In case of n-p-n transistor, the negative terminal of V EB repels the electrons of the emitter towards the base and constitute emitter current IE. About 5% of the electrons combine with the holes of the base to give small base current I B .

The remaining 95% of the electrons enter the collector region under the reverse bias and constitute collector current I C .

According to Kirchhoff’s law, the emitter current is the sum of collector current and base current. I E = I C + I B

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 178

Transistor acts as an amplifier in the active region.

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 180

Question 133. Draw the ‘Energy bands’, diagrams for a (i) pure semiconductor (ii) insulator. How does the energy band, for a pure semiconductor, get affected when this semiconductor is doped with (a) an acceptor impurity (b) donor impurity? Hence discuss why the ‘holes’, and the ‘electrons’ respectively, become the ‘majority charge carriers’ in these two cases? Write the two processes involved in the formation of p-n junction. (Comptt. All India 2015) Answer: ‘Energy Band’ diagrams : Distinguishing features between conductors, semiconductors and insulators : (i) Insulator. In insulator, the valence band is completely filled. The conduction band is empty and forbidden energy gap is quite large. So no electron is able to go from valence band to conduction band even if electric field is applied. Hence electrical conduction is impossible. The solid/ substance is an insulator.

(ii) Conductors (Metals). In metals, either the conduction band is partially filled or the conduction and valence band partly overlap each other. If small electric field is applied across the metal, the free electrons start moving in a direction opposite to the direction of electric field. Hence, metal behaves as a conductor.

In the first case, electrons from the valence band, easily jump over to the acceptor lelvel, leaving ‘holes’ behind. Hence, ‘holes’ becomes the majority charge carriers.

In the second case, electrons from the donor level, easily ‘jump over’ to the conduction band. Hence, electrons become the majority charge carriers. The two processes involved in the formation of the p-n junction are : (i) Diffusion (ii) Drift

Important Questions for Class 12 Physics Chapter 14 Semiconductor Electronics Materials Devices and Simple Circuits Class 12 Important Questions 183

Important Questions for Class 12 Physics

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Physics Project On Rectifier For Class 12

Table of Contents

Acknowledgment

Completing this project on rectifiers has been a rewarding and insightful journey, and I extend my heartfelt appreciation to those who have contributed to its successful realization.

Firstly, I would like to express my gratitude to [Teacher’s Name], whose guidance and expertise have been instrumental throughout this project. Their encouragement, valuable suggestions, and unwavering support have played a pivotal role in shaping the direction of this endeavor.

I am also thankful to my classmates and fellow enthusiasts who have shared their knowledge and insights during discussions and collaborative sessions. The exchange of ideas has added depth to the project and enhanced my overall understanding of rectifiers.

Special acknowledgment is due to [Name], who generously shared resources and references, enriching the theoretical foundation of this project. The insights gained from these references have greatly contributed to the depth and accuracy of our exploration into rectifiers.

Furthermore, I extend my appreciation to [Name] for their assistance in the practical aspects of the project. Their hands-on expertise and willingness to share experiences have significantly enriched the experimental phase of this endeavor.

Lastly, I want to express gratitude to my family for their unwavering support and understanding during the course of this project. Their encouragement and patience have been a constant source of motivation.

Each contribution, whether large or small, has played a vital role in the successful completion of this project. This collaborative effort has not only deepened my understanding of rectifiers but has also underscored the importance of teamwork and shared knowledge in the pursuit of scientific exploration. Once again, thank you to everyone who has been part of this endeavor.

Introduction

Rectifiers play a fundamental role in the field of electronics by facilitating the conversion of alternating current (AC) into direct current (DC). This project delves into the intricate workings of rectifiers, focusing on two main types: the Half-Wave Rectifier and the Full-Wave Rectifier, specifically the Bridge Rectifier configuration. The ability to transform AC to DC is essential for numerous electronic applications, ranging from powering everyday devices to facilitating battery charging and industrial processes like electroplating.

In this project, we aim to unravel the principles underlying rectification and, in turn, demonstrate the practical application of rectifier circuits. By understanding the nuances of both half-wave and full-wave rectifiers, we seek to explore their advantages, disadvantages, and the significance of their implementation in various technological domains. This project not only provides a hands-on experience in constructing and observing rectifier circuits but also offers insights into the broader realm of electrical engineering and its real-world applications. As we embark on this exploration, we aim to gain a comprehensive understanding of rectifiers and their pivotal role in shaping the landscape of modern electronics.

The journey through rectifiers holds a crucial place in the realm of electrical engineering, where the transformation of alternating current into a unidirectional flow is a prerequisite for numerous electronic devices and systems. Rectifiers act as the gatekeepers, regulating the flow of electrical currents in a controlled manner, enabling the steady supply of power for a myriad of applications.

This project takes a focused approach to examine two key types of rectifiers — the Half-Wave Rectifier and the Full-Wave Rectifier with the Bridge Rectifier configuration. The Half-Wave Rectifier, with its simple design using a single diode, allows us to explore the foundational principles of rectification by allowing only one half of the AC waveform. On the other hand, the Full-Wave Rectifier, employing a more complex Bridge configuration with four diodes, extends our exploration to harness both halves of the AC waveform, promising a more efficient conversion of AC to DC.

As we embark on constructing and analyzing these rectifier circuits, we will delve into the practical aspects of their operation. The project aims not only to showcase the theoretical underpinnings of rectifiers but also to provide tangible, hands-on experiences that deepen our comprehension of the subject matter. Through systematic observations, comparisons, and measurements of output waveforms, we intend to unravel the nuances of each rectifier type.

Beyond the confines of our experiment, this project invites reflection on the broader implications of rectifiers in real-world applications. From powering our homes with stable DC voltages to enabling the seamless charging of electronic devices, rectifiers form the backbone of modern electrical systems. By grasping the intricacies of rectification, we not only enhance our knowledge of electrical engineering principles but also gain insights into the indispensable role that rectifiers play in advancing technology.

Understanding Rectifier Types

  • Half-Wave Rectifier: Utilizing a single diode, the half-wave rectifier allows only one half of the AC waveform to pass through. This simple yet effective design forms the foundation of AC to DC conversion.
  • Full-Wave Rectifier: Enter the bridge rectifier, a more sophisticated setup employing four diodes in a bridge configuration. This design permits both halves of the AC waveform to be utilized, enhancing efficiency in the conversion process.

case study on rectifier class 12

  • Half-Wave Rectifier Demonstration: a. Connect a diode to the transformer’s secondary winding. b. Introduce a load resistor in series. c. Connect the setup to an AC power supply. d. Employ a voltmeter to measure the output.
  • Full-Wave Rectifier Demonstration: a. Assemble four diodes in a bridge configuration. b. Integrate a load resistor into the circuit. c. Connect the system to an AC power supply. d. Utilize a voltmeter to measure the output.

Observations

Half-wave rectifier.

  • Observe the shape of the waveform.
  • Measure the DC voltage across the load resistor.

Full-Wave Rectifier

case study on rectifier class 12

Comparative Analysis

  • Examine and compare the waveforms obtained from both the half-wave and full-wave rectifiers.
  • Measure and analyze the DC voltages across the load resistors in each case.

Advantages and Disadvantages

  • Discuss the advantages and disadvantages of each rectifier type based on the observed results.
  • Consider factors such as simplicity, efficiency, and cost-effectiveness.

Efficiency and Application Suitability

  • Evaluate the efficiency of each rectifier in converting AC to DC power.
  • Discuss the suitability of each rectifier type for specific applications, considering the observed waveform characteristics.

In the culmination of this project on rectifiers, we find ourselves at the intersection of theoretical exploration and practical application, having traversed the intricacies of converting alternating current (AC) to direct current (DC). Through the examination of both the Half-Wave Rectifier and the Full-Wave Rectifier with the Bridge Rectifier configuration, we have gained valuable insights into the fundamental principles that govern electrical systems.

Our journey began with a theoretical understanding of rectifiers, recognizing their pivotal role in electronics. The hands-on construction and analysis of the Half-Wave Rectifier allowed us to witness the selective passage of one half of the AC waveform, laying the foundation for comprehending the essentials of rectification. Transitioning to the Full-Wave Rectifier, specifically the Bridge Rectifier, expanded our scope, enabling us to harness the complete AC waveform for a more efficient conversion to DC.

Through meticulous observation, comparison of waveforms, and measurement of DC voltages, we have not only validated theoretical concepts but also grasped the practical implications of choosing one rectifier configuration over another. The Half-Wave Rectifier, though simple, revealed its limitations in terms of efficiency, while the Full-Wave Rectifier exhibited a more stable and continuous DC output.

In a broader context, this project has underscored the ubiquity of rectifiers in our daily lives. From powering electronic devices to providing a reliable source of energy, rectifiers form the backbone of our modern electrical infrastructure. The project has, therefore, not only been an academic exercise but a journey into the heart of technology, where theoretical knowledge transforms into tangible applications.

In conclusion, this project on rectifiers has been a comprehensive exploration, encompassing theoretical foundations, practical experiments, and reflections on real-world applications. As we conclude this endeavor, we carry forward a deeper appreciation for the intricate workings of rectifiers and their indispensable role in shaping the landscape of electrical engineering. This project stands as a testament to the fusion of knowledge, experimentation, and the enduring pursuit of understanding the science that powers our world.

Importance in Various Applications

  • Emphasize the significance of understanding rectifiers in the broader context of electronics.
  • Discuss the impact of rectifiers in applications like power supplies, battery charging, and electroplating.

Bibliography

  • Horowitz, Paul, and Winfield Hill. The Art of Electronics. Cambridge University Press, 2015.
  • Sedra, Adel S., and Smith, Kenneth C. Microelectronic Circuits. Oxford University Press, 2014.
  • Boylestad, Robert L., and Nashelsky, Louis. Electronic Devices and Circuit Theory. Pearson, 2018.
  • Streetman, Ben G., and Banerjee, Sanjay. Solid State Electronic Devices. Pearson, 2015.
  • Website: All About Circuits
  • Reference Article: Rectifier Circuits on Electronics Tutorials.
  • Reference Book: Milliman, Jacob, and Halkias, Christos C. Integrated Electronics: Analog and Digital Circuits and Systems. Tata McGraw-Hill, 2008.

Certificate of Completion

This is to certify that I, [Student’s Name], a [Class/Grade Level] student, have successfully completed the “physics project on rectifier for class 12.” The project explores the fundamental principles and key aspects of the chosen topic, providing a comprehensive understanding of its significance and implications.

In this project, I delved into in-depth research and analysis, investigating various facets and relevant theories related to the chosen topic. I demonstrated dedication, diligence, and a high level of sincerity throughout the project’s completion.

Key Achievements:

Thoroughly researched and analyzed Project on physics project on rectifier for class 12 . Examined the historical background and evolution of the subject matter. Explored the contributions of notable figures in the field. Investigated the key theories and principles associated with the topic. Discussed practical applications and real-world implications. Considered critical viewpoints and alternative theories, fostering a well-rounded understanding. This project has significantly enhanced my knowledge and critical thinking skills in the chosen field of study. It reflects my commitment to academic excellence and the pursuit of knowledge.

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Introduction

An electric device that converts alternating current (AC), which periodically reverses direction, to (DC) direct current, which flows in only one direction is known as a rectifier. The process of doing this is known as rectification as it straightens the direction of current . physically describing rectifiers, take a number of forms including diode and vacuum tubes, wet chemical cells, mercury arc valves, stocks of cotton and selenium oxide plates, semiconductor diode, silicon controlled rectifiers and other silicon based semiconductor switches.  rectifiers have many uses but more often they are seen serving as components of DC supply of power and high voltage direct current power transmission system.

Bridge Rectifier

An important part of electronic power supply is the bridge rectifier. Different types of electronic power circuits also require the DC power supply for powering the various electronic components available in AC mains supply. Rectifiers are also found in various wide variety of electronics, AC power devices like the home appliances, modulation process, motor controllers, welding applications etc.

A bridge rectifier is a converter of alternating current (AC) to direct current (DC) that rectifies main AC input to main DC output.  These rectifiers are used in power supply that provide necessary DC voltage for electronics devices or components. Bridge rectifiers are of different types: 

Single Phase and Three Phase Rectifier 

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The nature of supply is dependent on the single phase or three phase rectifiers. A single phase rectifier consists of four bodies for converting the AC current into DC. on the other hand a three phase rectifier uses six diodes, as presented by the figure. 

Uncontrolled Bridge Rectifier

This rectifier uses diodes for rectifying the input. As the name shows that this diode is a unidirectional device so it allows the current to flow in one direction only. With this diode configuration in the rectifier it does not allow the power to vary depending on the load requirement. 

Controlled Bridge Rectifier

This is called an AC/DC converter or rectifier instead of uncontrolled diode controlled solid state devices like SCR, IGBT are used to vary output power at varying voltage.

Full Wave Rectifier

It is one of the methods to improve the cycle of conversions. Two bodies are used in the full wave rectifiers one for each half of the cycle. Multiple winding transformers are used whose secondary winding is equally split into two halves with a common center taped connection. 

This configuration results in each diode conducting in turn when it’s anode terminal is positive with respect to the center of the transformer which is producing output during both half cycles, twice for the half rectifier so it’s 100% efficient. 

Half Wave Rectifier

A rectifier converts AC current to DC. It is done with the help of a diode or group of diodes, but a half wave rectifier uses only one diode whereas a full wave uses multiple diodes. It is the simplest form of rectifier available. Its working is: when a standard waveform is passed through it only half of the AC waveforms remain. It only allows half cycle of AC voltage through and it blocks the other half cycle on the DC side. 

Only one diode is required to initial a half wave cycle. As we know that the DC system is designed in such a manner that the current can flow in one direction only, putting a waveform from negative and positive cycles through a DC device may result in destructive consequences. So we use the half wave rectifier to convert the AC input to the DC output. 

The complete half circuit consists of three main components that are:

A transformer, resistive load, diode. 

Some Applications of Rectifiers

The primary use of a rectifier is to derive DC power from an AC supply. Inside the power supplies of the virtually electronic equipment. DC\AC power supplies are divided into two parts that are linear power supply and switch mode power supplies. The rectifier in such power supply will be in series following the transformer and be filled by a smoothing filter and a voltage regulator. From one volt converting the DC power into another is much more complicated. DC-to-DC first converts power to AC then to change a voltage used by the transformer. And finally rectifies power back to DC. 

For the detection of amplitude modulated radio signals also rectifiers are used. The signal may get amplified before detection. 

Polarised voltage is supplied by rectifiers for welding purposes; in such circuits the output current is controlled, this is achieved sometimes by replacing the diode in a bridge rectifier with thyristors, diodes whose voltage output can be regulated.

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FAQs on Rectifier

Q1. Define an uncontrolled Rectifier?

Ans: The circuit converter which converts AC current to DC current is known as a rectifier. Using diodes the rectifier circuit only is called the uncontrolled rectifiers circuit. Unlike the diodes, SCR does not become or behave like this after conducting immediately after it’s voltage becomes positive. It gets triggered by means of gate pulse signals.

Q2. Explain the reduction of 12 pulse Rectifiers Harmonics?

Ans: A two 6 plus rectifier is used by the 12 pulse rectifier in parallel to feed a common DC bus. An open primary and two secondary winding which is present in the 12 phase rectifier creates a 30 degree phase shift between the two currents waveforms, which starts to eliminate the 7th and the 5th harmonics and reduces current THD to between 10 and 15th percent.

Q3. Mention the components of a Rectifier.

Ans: A rectifier consists of a transformer, a stack and a cabinet. The transformers function is to safely separate the incoming AC voltage that is the primary side, from the secondary side, which is then adjusted to control the output voltage of the rectifier.

Q4. Define 6 plus Rectifiers.

Ans: The down to earth theory for this is: the 6 plus rectifier is used in both the rectifier and inverter as well and for the first time we are with the frying alpha. It has a plus number of 6 and so it can be thought of as a 6 phase or half wave circuit.

  • Electric Circuit

Full Wave Rectifier

Electric circuits that convert AC to DC are known as rectifiers. Rectifiers are classified into two types as Half Wave Rectifiers and Full Wave Rectifiers. Significant power is lost while using a half-wave rectifier and is not feasible for applications that need a smooth and steady supply. For a more smooth and steady supply, we use the full wave rectifiers. In this article, we will be looking into the working and characteristics of a full wave rectifier.

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Defining Full Wave Rectifiers

A full wave rectifier is defined as a rectifier that converts the complete cycle of alternating current into pulsating DC.

Unlike halfwave rectifiers that utilize only the halfwave of the input AC cycle, full wave rectifiers utilize the full cycle. The lower efficiency of the half wave rectifier can be overcome by the full wave rectifier.

Full Wave Rectifier Circuit

The circuit of the full wave rectifier can be constructed in two ways. The first method uses a centre tapped transformer and two diodes. This arrangement is known as a centre tapped full wave rectifier. The second method uses a standard transformer with four diodes arranged as a bridge. This is known as a bridge rectifier. In the next section, we will restrict the discussion to the centre tapped full wave rectifier only. You can read our article on bridge rectifier to learn the construction and working of bridge rectifier in detail.

Full Wave Rectifier Circuit

The circuit of the full wave rectifier consists of a step-down transformer and two diodes that are connected and centre tapped. The output voltage is obtained across the connected load resistor.

Working of Full Wave Rectifier

The input AC supplied to the full wave rectifier is very high. The step-down transformer in the rectifier circuit converts the high voltage AC into low voltage AC. The anode of the centre tapped diodes is connected to the transformer’s secondary winding and connected to the load resistor. During the positive half cycle of the alternating current, the top half of the secondary winding becomes positive while the second half of the secondary winding becomes negative.

During the positive half cycle, diode D 1 is forward biased as it is connected to the top of the secondary winding while diode D 2 is reverse biased as it is connected to the bottom of the secondary winding. Due to this, diode D 1 will conduct acting as a short circuit and D 2 will not conduct acting as an open circuit

During the negative half cycle, the diode D 1 is reverse biased and the diode D 2 is forward biased because the top half of the secondary circuit becomes negative and the bottom half of the circuit becomes positive. Thus in a full wave rectifiers, DC voltage is obtained for both positive and negative half cycle.

Full Wave Rectifier Formula

Peak inverse voltage.

Peak inverse voltage is the maximum voltage a diode can withstand in the reverse-biased direction before breakdown. The peak inverse voltage of the full-wave rectifier is double that of a half-wave rectifier. The PIV across D 1 and D 2 is 2V max .

DC Output Voltage

The following formula gives the average value of the DC output voltage:

RMS Value of Current

The RMS value of the current can be calculated using the following formula:

Form Factor

The form factor of the full wave rectifier is calculated using the formula:

Peak Factor

The following formula gives the peak factor of the full wave rectifier:

Rectification Efficiency

The rectification efficiency of the full-wave rectifier can be obtained using the following formula:

The efficiency of the full wave rectifiers is 81.2%.

Advantages of Full Wave Rectifier

  • The rectification efficiency of full wave rectifiers is double that of half wave rectifiers. The efficiency of half wave rectifiers is 40.6% while the rectification efficiency of full wave rectifiers is 81.2%.
  • The ripple factor in full wave rectifiers is low hence a simple filter is required. The value of ripple factor in full wave rectifier is 0.482 while in half wave rectifier it is about 1.21.
  • The output voltage and the output power obtained in full wave rectifiers are higher than that obtained using half wave rectifiers.

The only disadvantage of the full wave rectifier is that they need more circuit elements than the half wave rectifier which makes, making it costlier.

Frequently Asked Questions – FAQs

What is a full wave rectifier.

Full wave rectifiers convert both polarities of the input AC waveform to pulsating DC.

Why do we use a capacitor in full wave rectifier circuit?

A capacitor is used in the circuit to reduce the ripple factor.

What is a centre tapped full wave rectifier?

A centre tapped full wave rectifier is a type of rectifier that uses a centre tapped transformer and two diodes to convert the complete AC signal into DC signal.

Where is a full wave rectifier used?

A full wave rectifier is used in signal modulation and in electric welding.

What are the disadvantages of full wave rectifiers?

The full wave rectifiers are not suitable to use when a small voltage is required to be rectified. This is because, in a fullwave circuit, two diodes are connected in series and offer double voltage drop due to internal resistances.

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