{"id":88452,"date":"2021-06-24T16:54:42","date_gmt":"2021-06-24T11:24:42","guid":{"rendered":"https:\/\/kseebsolutions.guru\/?p=88452"},"modified":"2021-07-01T11:40:00","modified_gmt":"2021-07-01T06:10:00","slug":"1st-puc-electronics-question-bank-chapter-6","status":"publish","type":"post","link":"https:\/\/kseebsolutions.guru\/1st-puc-electronics-question-bank-chapter-6\/","title":{"rendered":"1st PUC Electronics Question Bank Chapter 6 Semi-Conductors, Diodes and Applications of Diodes"},"content":{"rendered":"

Karnataka 1st PUC Electronics Question Bank Chapter 6 Semi-Conductors, Diodes and Applications of Diodes<\/h2>\n

One Mark Questions and Answers<\/span><\/p>\n

Question 1.
\nWhat are semiconductors?
\nAnswer:
\nSemiconductors are substances whose electrical conductivity lies between that conductors and insulators.<\/p>\n

\"1st<\/p>\n

Question 2.
\nWhat are conductors?
\nAnswer:
\nConductors are those substances that allow electric charges to flow through them easily.<\/p>\n

Question 3.
\nDefine valence band.
\nAnswer:
\nThe valence band is the range of energies possessed by valence electrons.<\/p>\n

Question 4.
\nDefine conduction band.
\nAnswer:
\nThe range of energies occupied by free electrons is called the conduction band.<\/p>\n

Question 5.
\nWhat is the forbidden energy gap?
\nAnswer:
\nThe gap between the conduction band and valence band in the energy band diagram is known as the forbidden energy gap.<\/p>\n

Question 6.
\nWhat is doping?
\nAnswer:
\nThe process of adding suitable impurity to a semiconductor to increase its conductivity is called doping.<\/p>\n

\"1st<\/p>\n

Question 7.
\nName any one donor impurity.
\nAnswer:
\nArsenic.<\/p>\n

Question 8.
\nName anyone acceptor impurity.
\nAnswer:
\nGallium.<\/p>\n

Question 9.
\nName majority charge carriers in the n-type semiconductor.
\nAnswer:
\nElectrons.<\/p>\n

Question 10.
\nName majority charge carriers in a p-type semiconductor.
\nAnswer:
\nHoles.<\/p>\n

\"1st<\/p>\n

Question 11.
\nDraw the symbol of the p-n junction diode.
\nAnswer:
\n\"1st<\/p>\n

Question 12.
\nHow p-n junction is formed?
\nAnswer:
\nthe p-n junction is formed by doping a donor impurity into one side and an acceptor impurity into another side of a single semiconductor crystal-like silicon or germanium.<\/p>\n

Question 13.
\nWhat is meant by biasing a p-n junction?
\nAnswer:
\nBiasing a p-n junction is connecting an external source of emf to the p-n junction.<\/p>\n

Question 14.
\nWhat is the depletion region?
\nAnswer:
\nThe depletion region is the region near the p-n junction where mobile charge carriers namely electrons and holes are depleted or emptied.<\/p>\n

Question 15.
\nWhat is barrier potential?
\nAnswer:
\nBarrier potential is the small voltage developed across the p-n junction due to positive and negative ions.<\/p>\n

\"1st<\/p>\n

Question 16.
\nWhat is meant by forward biasing?
\nAnswer:
\nA diode is said to be forward if its anode (or p-type material) is connected to the positive terminal of the battery and cathode (or n-type material) is connected to the negative terminal of the battery.<\/p>\n

Question 17.
\nWhat is meant by reverse biasing?
\nAnswer:
\nA diode is said to be reverse-biased if its anode (or p-type material) is connected to the negative terminal of the battery and its cathode (or n-type material) is connected to the positive terminal of the battery.<\/p>\n

Question 18.
\nIn the figure shown, is the diode D forward or reverse biased?<\/p>\n

\"1st
\nAnswer:
\nThe diode is forward biased as the anode is at 10V which is greater than -10V at the cathode of the diode.<\/p>\n

\"1st<\/p>\n

Question 19.
\nIn which type of biasing is the p-n junction diode resistance high?
\nAnswer:
\nIn the reverse biasing of the p-n junction.<\/p>\n

Question 20.
\nWhat is the effect of forwarding bias on the width of the depletion region?
\nAnswer:
\nThe width of the depletion region decreases.<\/p>\n

Question 21.
\nWhat is the effect of reverse bias on the width of the depletion region?
\nAnswer:
\nWidth of the depletion region increases<\/p>\n

Question 22.
\nWhat is reverse saturation current?
\nAnswer:
\nReverse saturation current is the current flowing in a reverse-biased diode due to leakage of minority charge carriers through the depletion region.<\/p>\n

\"1st<\/p>\n

Question 23.
\nIs reverse saturation current dependent on temperature?
\nAnswer:
\nYes. The reverse saturation current doubles for every 10\u00b0C rise in temperature.<\/p>\n

Question 24.
\nWhat do you mean by the breakdown of the junction?
\nAnswer:
\nThe p-n junction is said to be under breakdown when the reverse current rises sharply even though reverse voltage is varied by a small value.<\/p>\n

Question 25.
\nWhat is the static resistance of a diode?
\nAnswer:
\nStatic or DC resistance is the resistance offered by the p-n junction diode when it is used in a DC circuit.<\/p>\n

Question 26.
\nDefine the dynamic resistance of a junction diode.
\nAnswer:
\nAC or dynamic resistance is defined as the ratio of a small change in the forward AC voltage to the corresponding change in AC current in the linear portion of the curve.<\/p>\n

\"1st<\/p>\n

Question 27.
\nName the capacitive effect exhibited by a p-n junction when it is reverse biased.
\nAnswer:
\nTransition or space charge capacitance.<\/p>\n

Question 28.
\nWrite the expression for transition capacitance.
\nAnswer:
\nC = \\(\\frac{\\in \\mathrm{A}}{\\mathrm{W}}\\)
\nWhere \u2208 = permittivity of the material
\nA = Surface area of the junction
\nW = Width of depletion region.<\/p>\n

Question 29.
\nWhat is the value of the potential barrier of a silicon diode?
\nAnswer:
\n0.7 volt.<\/p>\n

Question 30.
\nMention the diode equation.
\nAnswer:
\nShockley’s equation or diode equation is
\nI = Is[\\(e^{\\frac{V_{\\mathrm{D}}}{\\eta V_{T}}}\\) – 1]
\nWhere I = Diode current
\nIs<\/sub> = Revese saturation current
\nVD<\/sub> = Voltage across the diode
\nVT<\/sub> – Thermal voltage
\n\u03b7 = 1 for Ge diode, called ideality or quality factor
\n\u03b7 = 1 for si diode above knee voltage
\n\u03b7 = 2 for silicon diode below knee voltage.<\/p>\n

\"1st<\/p>\n

Question 31.
\nWhat is an ideal diode?
\nAnswer:
\nAn ideal diode is a device that offers zero resistance (perfect conductor) when forward biased and offers infinite resistance when it is reverse biased.<\/p>\n

Question 32.
\nDraw the equivalent circuit of a forward-biased ideal diode.
\nAnswer:
\n\"1st<\/p>\n

Question 33.
\nWhat is the power rating of a diode?
\nAnswer:
\nThe power rating of a diode is the maximum power a diode can withstand without getting damaged.<\/p>\n

\"1st<\/p>\n

Question 34.
\nSketch the shape of output waveform for the circuit shown below assuming the diode to be ideal.?
\nAnswer:
\n\"1st<\/p>\n

Question 35.
\nSketch the shape of the output voltage waveform for the circuit shown below assuming the diode ideal.
\nAnswer:
\n\"1st<\/p>\n

Question 36.
\nWhat is meant by rectification?
\nAnswer:
\nRectification is the process of converting AC voltage into a pulsating DC voltage.<\/p>\n

Question 37.
\nWhat is peak inverse voltage?
\nAnswer:
\nPeak inverse voltage is the maximum reverse voltage a diode can withstand.<\/p>\n

Question 38.
\nMention the value of the ripple factor for the half-wave rectifier.
\nAnswer:
\n1.21.<\/p>\n

\"1st<\/p>\n

Question 39.
\nWhat is meant by filter?
\nAnswer:
\nA filter is a circuit that removes the AC component from the rectifier output and allows pure DC to reach the load.<\/p>\n

Question 40.
\nMention the property of the p-n junction, which is used for rectification.
\nAnswer:
\nUnidirectional property or allowing the current to flow only in one direction.<\/p>\n

Question 41.
\nName the component used to construct an adjustable voltage regulator?
\nAnswer:
\nICLM317.<\/p>\n

Question 42.
\nUnder what bias is LED operated?
\nAnswer:
\nForward bias.<\/p>\n

Question 43.
\nWhat happens to light emission in LED as the forward current is increased? ‘
\nAnswer:
\nIncreases.<\/p>\n

Question 44.
\nWhat is a Schottky diode?
\nAnswer:
\nSchottky diode is a semiconductor diode with a low forward voltage drop and a very fast switching action.<\/p>\n

\"1st<\/p>\n

Question 45.
\nDraw the symbol of the Schottky diode.
\nAnswer:
\n\"1st<\/p>\n

Question 46.
\nWhat type of junction is formed in a Schottky diode?
\nAnswer:
\nThe Schottky diode has a metal to semiconductor junction instead of the p-n junction.<\/p>\n

Question 47.
\nMention applications of Schottky diode.
\nAnswer:
\nThese diodes are used in<\/p>\n

    \n
  1. high-frequency applications<\/li>\n
  2. digital circuits to increase switching speed,<\/li>\n
  3. detector circuits in communication systems.<\/li>\n<\/ol>\n

    Two Mark Questions and Answers<\/span><\/p>\n

    Question 1.
    \nClassify extrinsic semiconductors.
    \nAnswer:<\/p>\n

      \n
    1. n-type semiconductors.<\/li>\n
    2. p-type semiconductors.<\/li>\n<\/ol>\n

      Question 2.
      \nMention majority and minority charge carriers in the n-type semiconductor.
      \nAnswer:
      \nIn an n-type semiconductor, free electrons are majority charge carriers and holes are minority charge carriers.<\/p>\n

      Question 3.
      \nDraw the circuit diagrams for (i) forward biased diode and (ii) reverse-biased diode.
      \nAnswer:
      \n\"1st<\/p>\n

      Question 4.
      \nWhat do you mean by the transition capacitance of a diode?
      \nAnswer:
      \nTransition capacitance or junction capacitance is the capacitance associated with the reverse-biased p-n junction.<\/p>\n

      \"1st<\/p>\n

      Question 5.
      \nWrite any two examples of wave shaping circuits.
      \nAnswer:<\/p>\n

        \n
      1. clipping circuits<\/li>\n
      2. clamping circuits.<\/li>\n<\/ol>\n

        Question 6.
        \nMention any four applications of a diode.
        \nAnswer:
        \nA diode is used<\/p>\n

          \n
        1. as a rectifier in DC power supplies.<\/li>\n
        2. in AM and FM detectors in communication systems.<\/li>\n
        3. in clipping and clamping circuits.<\/li>\n
        4. in tuned diode oscillator.<\/li>\n<\/ol>\n

          Question 7.
          \nMention the primary conditions of the clamping circuit.
          \nAnswer:
          \nThe discharging time constant of the capacitor should be at least ten times the time period of the input signal.<\/p>\n

          Question 8.
          \nWhat is the difference between positive and negative clamper?
          \nAnswer:
          \nA positive clamper clamps the positive peak of a signal to the desired DC level. A negative clamper clamps the negative peak of a signal to the desired DC level.<\/p>\n

          Question 9.
          \nDefine the ripple factor and give its significance.
          \nAnswer:
          \nRipple factor is the ratio of RMS value of AC component of the load voltage to the average DC value of load voltage.
          \nThe smaller the value of the ripple factor, the greater is the PC component in the output and hence better is the rectifier.<\/p>\n

          Question 10.
          \nHow many diodes are used in an (i) centre tapped full wave rectifier and (ii) bridge rectifier?
          \nAnswer:<\/p>\n

            \n
          1. Centre tapped full wave rectifier requires two diodes.<\/li>\n
          2. The bridge rectifier requires four diodes.<\/li>\n<\/ol>\n

            \"1st<\/p>\n

            Question 11.
            \nWhat is the maximum efficiency of a full-wave rectifier and halfwave rectifier?
            \nAnswer:
            \nThe efficiency of the full-wave rectifier is 81.2% and the efficiency of the halfwave rectifier is 40.6%.<\/p>\n

            Question 12.
            \nDraw the circuit diagram of the +12V voltage regulator.
            \nAnswer:
            \n\"1st<\/p>\n

            Question 13.
            \nDraw the circuit diagram of an adjustable voltage regulator.
            \nAnswer:
            \n\"1st<\/p>\n

            Question 14.
            \nWhat is voltage regulation? Mention types of voltage regulation.
            \nAnswer:
            \nVoltage regulation is a measure of the ability of the DC source to maintain a constant output voltage even when there is variation in the load current and input voltage.<\/p>\n

            Two types of voltage regulation are<\/p>\n

              \n
            1. line regulation.<\/li>\n
            2. load regulation.<\/li>\n<\/ol>\n

              Question 15.
              \nWrite any four difference between germanium and silicon diodes.
              \nAnswer:
              \n\"\"<\/p>\n

              Three Mark Questions and Answers<\/span><\/p>\n

              Question 1.
              \nDefine ideal diode. Draw its V-I characteristics.
              \nAnswer:
              \nAn ideal diode is a device that offers zero resistance when forward biased and offers infinite resistance when reverse biased.
              \n\"1st<\/p>\n

              Question 2.
              \nState any two applications of LED. Draw the diagram of the seven-segment display.
              \nAnswer:
              \nLEOs are used<\/p>\n

                \n
              1. in indicator lamps and displays<\/li>\n
              2. in an optical communication system. A Diagram of seven segment display is<\/li>\n<\/ol>\n

                \"1st<\/p>\n

                Question 3.
                \nDistinguish between centre tapped full-wave and half-wave rectifiers.
                \nAnswer:
                \n\"1st<\/p>\n

                \"1st<\/p>\n

                Question 4.
                \nDraw the circuit diagram of a centre tapped full wave rectifier indicating input and output waveforms.
                \nAnswer:
                \n\"1st<\/p>\n

                Question 5.
                \nDraw the circuit of the negative clamper and show input and output waveforms.
                \nAnswer:
                \nA negative clamper is a circuit that clamps the negative peak of a signal to a desired de level.<\/p>\n

                Question 6.
                \nMention any two comparisons of three rectifiers.
                \nAnswer:
                \n\"1st
                \n\"1st<\/p>\n

                Question 7.
                \nWhat is Zener breakdown? Draw V-l characteristics of Zener diode.
                \nAnswer:
                \nWhen a heavily doped diode that has a narrow depletion region is subjected to a reverse voltage, the high electric field at the junction breaks covalent bonds. Hence a large number of electron holes are produced and the diode is said to be under Zener breakdown.
                \n\"1st
                \nV – I characteristics of Zener diode
                \n\"1st<\/p>\n

                Question 8.
                \nExplain the second approximation of the diode.
                \nAnswer:
                \nIn second approximation treats the diode as a switch in series with a battery of potential VB.
                \n\"1st
                \nID = \\(\\frac{\\mathrm{E}-\\mathrm{V}_{\\mathrm{B}}}{\\mathrm{R}}\\)<\/p>\n

                Question 9.
                \nWhat is meant by clipping? Mention one application of clipping circuit.
                \nAnswer:
                \nClipping is removing an unwanted portion of input signals to protect devices from large amplitude signals.<\/p>\n

                Clipping circuits are used in<\/p>\n

                  \n
                1. radio receivers to remove noise.<\/li>\n
                2. to generate square waves or rectangular waves.<\/li>\n<\/ol>\n

                  \"1st<\/p>\n

                  Question 10.
                  \nDraw the circuit of series positive clipper and show input and output waveforms.
                  \nAnswer:
                  \n\"1st<\/p>\n

                  Question 11.
                  \nExplain the third approximation of the semiconductor diode.
                  \nAnswer:
                  \nThe equivalent circuit for the third approximation is a switch with a battery and a resistor. The voltage across diode is VD<\/sub> = VB<\/sub> + ID<\/sub>RB<\/sub>.
                  \nID<\/sub> = \\(\\frac{\\mathrm{V}_{\\mathrm{E}}-\\mathrm{V}_{\\mathrm{D}}}{\\mathrm{R}_{\\mathrm{B}}}\\)
                  \n\"1st<\/p>\n

                  Question 12.
                  \nMention the properties of semiconductors.
                  \nAnswer:<\/p>\n

                    \n
                  1. Semiconductors are tetravalent and hence covalent bonding takes place between their atoms.<\/li>\n
                  2. At absolute zero, semiconductors behave like a perfect insulator.<\/li>\n
                  3. Resistance of semiconductors decrease when their temperature is increased (a temperature coefficient of resistance is negative)<\/li>\n
                  4. The conductivity of semiconductors can be increased by adding suitable impurities to them.<\/li>\n<\/ol>\n

                    Question 13.
                    \nWhat is a bleeder resistor? Mention its uses.
                    \nAnswer:
                    \nA bleeder resistor is a resistor connected across filter output so as to maintain minimum current through the choke at all times.<\/p>\n

                    It is also used to<\/p>\n

                      \n
                    1. improve voltage regulation of the power supply.<\/li>\n
                    2. provide a path for the filter capacitor to discharge so as to avoid the possibility of electric shock to technicians.<\/li>\n<\/ol>\n

                      Question 14.
                      \nWhat is an LCD? Does LCD radiate its own light? Mention any two important applications of LCD.
                      \nAnswer:
                      \nLCD is a liquid crystal display, is a device that operates by applying a varying electric voltage to the layer of liquid crystal. It does not radiate its own light but reflects light.<\/p>\n

                      LCDs are used in laptop computers, LCD monitors and LCD TVs, calculators, and electronic watches.<\/p>\n

                      \"1st<\/p>\n

                      Question 15.
                      \nWhat is a tunnel diode? Draw its symbol and name any two of its applications
                      \nAnswer:
                      \nA tunnel diode is a heavily doped diode with a narrow depletion layer. Its symbol is
                      \n\"1st
                      \nIt is used in tuned circuits and oscillator circuits.<\/p>\n

                      Question 16.
                      \nWhat is a varactor diode? Draw its symbol and mention any two of its applications.
                      \nAnswer:
                      \nVaractor diode is a diode which acts as a variable capacitor under reverse biased condition.
                      \n\"1st
                      \nThey are used in LC tuned circuits, TV receivers, FM receivers.<\/p>\n

                      Question 17.
                      \nDraw the equivalent circuits for (i) reverse biased ideal diode (ii) second approximation diode and (iii) third approximation diode.
                      \nAnswer:
                      \n\"1st<\/p>\n

                      Five Mark Questions and Answers<\/span><\/p>\n

                      Question 1.
                      \nClassify solids based on the energy band diagram.
                      \nAnswer:
                      \n\"1st
                      \nConductors: These are materials that allow the current to flow through them easily. In conductors, the valence band and conduction band overlap completely due to which the valence electrons drift to the conduction band easily. Even at room temperature, there is a large number of free electrons. Hence due to a small voltage applied a large current flows through them.<\/p>\n

                      Semiconductors: These are the materials with electrical conductivity lying between those of conductors and insulators. The forbidden energy gap between the valence band and conduction band is 0.7 eV for germanium and 1. 1 eV for silicon. At absolute zero, a semiconductor acts as an insulator. As the temperature increases, more valence electrons jump to the conduction band. Hence the conductivity of semiconductors increases with an increase in their temperature.<\/p>\n

                      Insulators: Insulators are materials that do not allow the electric current to flow through them easily. The valence band is completely filled and the conduction band is completely empty in insulators. The forbidden energy band between the conduction band and valence band is nearly 5eV. In the conduction band, there are no free electrons present at room temperature. Hence, the insulators do not conduct at all at room temperature.<\/p>\n

                      \"1st<\/p>\n

                      Question 2.
                      \nBriefly explain the n-type semiconductor.
                      \nAnswer:
                      \nn-type semiconductors are the semiconductors that are doped with pentavalent impurities like phosphorous, antimony and arsenic. Each pentavalent atom donates one electron and hence is called donor impurity.<\/p>\n

                      When a pentavalent atom is added, the four valence electrons form covalent bonds with neighbouring four silicon atom and its fifth electron becomes a free electron. The free-electron produced by the pentavalent atom does not create a hole.<\/p>\n

                      When few covalent bonds break due to thermal energy, electrons and holes are created. Thus there will be a large number of free electrons. Hence in n-type material, there are electrons as majority charge carriers and holes as minority charge carriers. The pentavalent impurity atom becomes an immobile positive ion after donating an electron.<\/p>\n

                      The n-type material is electrically neutral because the number of free electron equals the sum of positive holes and positive ions.
                      \n\"1st<\/p>\n

                      Question 3.
                      \nBriefly explain p-type semiconductor.
                      \nAnswer:
                      \nA p-type semiconductor is obtained, by adding (doping) trivalent impurity atoms like boron, gallium, indium to a semiconductor. The impurity doping atoms are known as acceptor impurity because they can accept one valence electron from the semiconductor atom.<\/p>\n

                      When trivalent atom-like gallium is added to a silicon crystal, three valence electrons of the gallium atom will form covalent bonds with three valence electrons of three neighbouring silicon atoms. The fourth covalent bond remains incomplete and the vacancy in the covalent bond is called a hole. Thus, each gallium atom creates a hole without generating a free electron.<\/p>\n

                      Due to the breaking up of few covalent bonds, few electrons and holes are produced. Thus in p-type semiconductor, holes are majority charge carriers and electrons are the minority charge carriers. Due to breaking up few covalent bonds, few electrons and holes are produced.<\/p>\n

                      Thus in p-type semiconductor, holes are majority charge carriers and electrons are the minority charge carriers. Each trivalent atom accepts an electron and hence is called acceptor impurity. On accepting an electron, the trivalent atom becomes a negative ion. The p-type material is electrically neutral because the number of holes are equal to the sum of a number of electrons and negative ions.
                      \n\"1st<\/p>\n

                      Question 4.
                      \nHow is the depletion region formed in a pn-junction?
                      \nAnswer:
                      \nA pn-junction is formed when one side of a pure semiconductor crystal is doped with a trivalent impurity and the other side with a pentavalent impurity.<\/p>\n

                      The n-type material has electrons as majority charge carriers and holes as minority charge carriers. The p-type material has holes as majority charge carriers and electrons as minority charge carriers. Due to this, there is a concentration gradient across the junction and electrons move from n-type to p-type whereas holes move from p-type to n-type region.<\/p>\n

                      \"1st<\/p>\n

                      The transfer of electrons and holes across the junction is known as diffusion. The recombination of electrons and holes makes the region near the p-n junction depleted of electrons and holes.<\/p>\n

                      The depletion region is the region near the p-n junction which is depleted of mobile charge carriers namely electrons and holes.<\/p>\n

                      The small potential difference developed across the p-n junction due to immobile positive and negative ions is called barrier potential. The barrier voltage is 0.3V for Germanium and 0.7V for silicon. The barrier potential stops the diffusion of charge carriers across the pn-junction.
                      \ni\"1st<\/p>\n

                      Question 5.
                      \nExplain the working of the p-n junction when it is forward biased.
                      \nAnswer:
                      \n\"1st
                      \nThe diode is said to be forward-biased if its p-type material is connected to the positive terminal of the battery and the n-type material is connected to the negative terminal of the battery. The electrons in the type material are repelled towards p-n junction by negative terminal of the battery and holes in the p-type material are repelled by the positive terminal of the battery. This decreases the width of the depletion layer and the height of the potential barrier.<\/p>\n

                      Hence the more number of majority charge carriers diffuse across the junction. When the forward bias voltage exceeds the barrier potential, free electrons move across the junction and fall into the p region. Similarly, holes move across the junction and recombine with free electrons in the n region.<\/p>\n

                      For each recombination that takes place near the junction, a covalent bond in the p region near the positive terminal breaks. The released free-electron enters the positive terminal of the battery and the hole moves towards the junction. For every free electron that enters the positive terminal of the battery, an electron moves from the negative terminal to the n region and move towards the junction. Hence a large current flows across the junction. Therefore, a forward-biased p-n junction offers very low resistance (few ohms).<\/p>\n

                      \"1st<\/p>\n

                      Question 6.
                      \nExplain the working of p-n junction under reverse bias.
                      \nAnswer:
                      \nA p-n junction is said to be reverse-biased when the positive terminal of the battery is connected to the n region (cathode) and the negative terminal of the battery is connected to the p region (anode) of the diode.
                      \n\"1st
                      \nUnder reverse bias, the majority of electrons in the n region are attracted by the positive terminal of the battery and the majority of holes in the p region are attracted by the negative terminal of the battery. Thus the majority of carriers are made to move away from the junction. The number of positive ions in the n region and the number of negative ions in the p region increase. This makes the width of the depletion region increase and the barrier potential also increases.<\/p>\n

                      The majority of charge carrier is prevented from diffusing across the junction by the large potential barrier. Hence the reverse-biased pn junction offers a very large resistance. The minority charge carriers, namely electrons in the p region and holes in n region, move across the junction easily. Hence a very small negligible reverse current flows through the diode under reverse bias.<\/p>\n

                      As the reverse current is due to leakage of minority charge carriers through the depletion region, this current is called the leakage current. The number of minority charge carriers depends on temperature and it is independent of reverse voltage. At constant temperature, the leakage current remains almost constant even if the reverse bias voltage is varied. Due to this, reverse current is also called reverse saturation current. The reverse saturation current doubles for every I 0\u00b0C rise in temperature.<\/p>\n

                      Question 7.
                      \nDraw and explain the V-I characteristics of the diode.
                      \nAnswer:
                      \nThe V-I characteristics of a diode is the graph between the voltage across the diode and the current flowing through it.<\/p>\n

                      The V-I characteristics can be divided into two parts namely<\/p>\n

                        \n
                      1. Forward characteristics<\/li>\n
                      2. Reverse characteristics.<\/li>\n<\/ol>\n

                        1. Forward characteristics of semiconductor diode:
                        \n\"1st
                        \nThe circuit for plotting V-I characteristics of the diode is as shown. The resistance R limits the current to a: permissible value. A voltmeter measures the voltage across the diode and the milliammeter reads the current flowing through the diode.<\/p>\n

                        The forward voltage VF<\/sub> is increased in steps of 0.1 V and the corresponding forward current IF<\/sub> are noted. A graph of VF<\/sub> along the x-axis and corresponding IF<\/sub> along the y-axis gives the forward characteristics curve.<\/p>\n

                        Diode current IF<\/sub> is zero, when VF<\/sub> = 0. When VF is more than barrier potential, current IF<\/sub> increases. The forward voltage VF<\/sub> at which the forward current starts to increase rapidly is called the knee voltage or cut-in voltage or threshold voltage (VK<\/sub>). For germanium diode, VK<\/sub> = 0.3 V and for silicon diode, VK<\/sub> = 0.7V.<\/p>\n

                        \"1st<\/p>\n

                        Reverse characteristics of a pn junction :
                        \nThe circuit for plotting reverse characteristics of a diode is shown below
                        \n\"1st
                        \nThe reverse voltage is increased gradually till the diode starts conducting and the corresponding reverse currents are noted. On platting a graph between reverse voltage and reverse current, the characteristic curve is as shown.<\/p>\n

                        Due to reverse bias applied on pn junction, an extremely small current IR (few pA) flows through the diode. The reverse current remains almost constant up to the breakdown voltage. After this point, for a small change in reverse voltage, the reverse current increases very rapidly. This sudden flow of large current may destroy the diode due to overheating. Hence, the diode should be operated in the reverse bias with reverse voltage less than break down voltage.<\/p>\n

                        Question 8.
                        \nWrite a note on diode approximations.
                        \nAnswer:
                        \nDiode approximation is the conventional method used to represent a diode by a combination of ideal diode and linear circuit elements.<\/p>\n

                        Depending on the circuit conditions, three approximations of diodes are used.<\/p>\n

                        First approximation of ideal diode:
                        \nAn ideal diode is a device that conducts with zero resistance (perfect conductor or closed switch) when forward biased and offers infinite resistance (perfect insulator and open switch) when reversing biased.
                        \n\"1st
                        \nSecond approximation (practical diode)
                        \nA diode conducts only when a forward bias voltage is more than the barrier potential. In the second approximation, the diode is replaced by an ideal diode in series with a barrier potential VB<\/sub> (0.3V for Germanium and 0.7V for Silicon).
                        \n\"1st
                        \nIdeal diode under forward bias. {ID<\/sub> = \\(\\frac{\\mathrm{E}}{\\mathrm{R}}\\)}
                        \n\"1st
                        \nIdeal diode under reverse bias. {VD<\/sub> = E}
                        \n\"1st
                        \nThird approximation:
                        \nIn the third approximation, a diode is represented as an ideal diode in series with barrier potential VB<\/sub> and bulk resistance rB<\/sub>.
                        \n\"1st
                        \n\"1st<\/p>\n

                        Question 9.
                        \nExplain the action of a diode as a half-wave rectifier.
                        \nAnswer:
                        \nHalf wave rectifier is a circuit that removes one-half cycle of the AC input and produces a pulsating DC output voltage.
                        \n\"1st
                        \nIt has diode D in series with a load resistor RL. The AC mains voltage to be rectified is applied to the primary of a transformer and voltage induced across secondary is applied to the rectifier. During the positive half cycle, A is positive and B is negative. The diode is forward biased and conducts and the diode acts as a closed switch. Hence, the positive half cycle of the voltage is developed across the load resistor RL.<\/p>\n

                        During a negative half cycle, A is negative and B is positive. The diode is reverse biased and does not conduct as it acts as an open switch. Hence there is no output. The output waveform across RL has only positive half cycles. Ripple factor is the ratio of the RMS value of (r) the AC component of the load voltage to the average value of load voltage.
                        \nr = \\(\\frac{\\mathrm{rms} \\text { value of AC component }}{\\text { value of dc component }}\\) = 1.21<\/p>\n

                        Rectifier efficiency is the ratio of de output power to ac input power.
                        \n\u03b7 = \\(\\frac{\\text { DC output power }}{\\text { AC input power }}\\) = 0.406 or 40.6%<\/p>\n

                        Peak inverse voltage is the maximum voltage the diode can withstand when it is reverse biased and is equal to the peak value of the secondary voltage Vm<\/sub><\/p>\n

                        \"1st<\/p>\n

                        Question 10.
                        \nDescribe the working of a full wave centre tapped rectifier.
                        \nAnswer:
                        \nFull-wave rectifier is a circuit that converts the entire AC input cycle into varying DC output.
                        \n\"1st
                        \n\"1st
                        \nDuring the positive half cycle of AC input, A is positive and B is negative. The diode D1<\/sub> is forward biased and conducts. The diode D2<\/sub> is reverse biased and does not conduct. The current flows along AD1<\/sub>MNC.<\/p>\n

                        During the negative half cycle of AC input, A is negative and B is positive. The diode D1<\/sub> is reverse biased and does not conduct. Hence, the current flows along BD2<\/sub>MNC.<\/p>\n

                        Thus the load current flows in the same direction M toN during both the half cycles of AC input. The peak inverse voltage (PIV) of this rectifier is 2Vm<\/sub> where Vm<\/sub> is the peak or maximum value of the secondary voltage.<\/p>\n

                        RMS value of voltage, Vrms = \\(\\frac{\\mathrm{V}_{\\mathrm{m}}}{\\sqrt{2}}\\)
                        \nRipple factor (r) is the ratio of the RMS value of the AC component of load voltage to the average value of load voltage.<\/p>\n

                        r = \\(\\frac{\\mathrm{V}_{\\mathrm{rms}}}{\\mathrm{V}_{\\mathrm{avg}}}\\) = 0 .4 8<\/p>\n

                        Efficiency is the ratio of DC output power to AC input power.
                        \n\u03b7 = \\(\\frac{\\text { DC power to load }}{\\text { AC input power }}\\) = 0.812 or 81.2%<\/p>\n

                        Question 11.
                        \nExplain the working of the bridge rectifier.
                        \nAnswer:
                        \nThe bridge rectifier does not require a centre tap transformer. It uses four diodes D1,<\/sub> D2<\/sub>, D3<\/sub> and D4<\/sub> connected in the form of a bridge across the secondary of the transformer.
                        \n\"1st
                        \nDuring the positive half cycle of AC input, A is at positive potential and B is not at negative potential. Diodes D1<\/sub> and D3<\/sub> are forward biased and conduct. The current flows through AD1<\/sub> MND3<\/sub>B. D2<\/sub> and D4<\/sub> do not conduct as they are reverse biased. Thus the positive half cycle of voltage is obtained across the load resistor RL.<\/p>\n

                        During the negative half cycle of AC input, A is at negative potential and B is at positive potential. Diodes D2<\/sub> and D4<\/sub> are forward biased and conduct. Diodes D1<\/sub> and D3<\/sub> are reverse biased and do not conduct. Hence current flows along BD2MND4A. Thus a positive half cycle of voltage is developed across the load resistor RL. In both the half cycles, current flows through RL in the same direction from M to N.<\/p>\n

                        Thus varying DC output is obtained across RL.<\/p>\n

                        RMS value of voltage is Vrms<\/sub> = \\(\\frac{\\mathrm{V}_{\\mathrm{m}}}{\\sqrt{2}}\\)<\/p>\n

                        Ripple factor is the ratio of the RMS value of the AC component of the load voltage to the average value of load voltage.
                        \nr = Ripple factor = \\(\\frac{\\mathrm{V}_{\\text {rms }}}{\\mathrm{V}_{\\mathrm{avg}}}=\\sqrt{\\left(\\frac{\\mathrm{I}_{\\mathrm{ms}}}{\\mathrm{I}_{\\mathrm{d}}}\\right)^{2}}\\) – 1 = 0.48<\/p>\n

                        Efficiency (\u03b7) is the ratio of output DC power to the input AC power.
                        \n\u03b7 = \\(\\frac{\\text { DC output power }}{\\text { AC input power }}\\) r = 0.812or81.2%<\/p>\n

                        \"1st<\/p>\n

                        Question 12.
                        \nMention any five comparisons of three rectifiers.
                        \nAnswer:
                        \nComparison of rectifiers
                        \n\"1st<\/p>\n

                        Question 13.
                        \nExplain the working of the Zener diode as a voltage regulator.
                        \nAnswer:
                        \nIn a DC power supply, the output voltage should remain almost constant even as the input and load vary.<\/p>\n

                        Voltages regulation is a measure of the ability of a DC source to maintain a constant output voltage, even when there is variation in the load resistance or input voltage.<\/p>\n

                        Zener diode as a voltage regulator<\/p>\n

                        Zener diode can be used as a voltage regulator to provide a constant output voltage from a source whose voltage may vary.<\/p>\n

                        The two types of voltage regulator are<\/p>\n

                          \n
                        1. Line regulation<\/li>\n
                        2. Load regulation.<\/li>\n<\/ol>\n

                          Load regulation:
                          \n\"1st
                          \nThe Zener diode starts conducting only when the source voltage Vs<\/sub> is greater than the Zener breakdown voltage Vz<\/sub>.<\/p>\n

                          When Vs is increased from OV, load current IL increases with applied voltage. The current through Zener diode will be zero.<\/p>\n

                          If Vs<\/sub> is more than Vz<\/sub>, the Zener diode breaks down and conducts. Increasing the input voltage further increases the voltage across the resistor Rs<\/sub> and voltage across load VL<\/sub> remains constant at Vz<\/sub>.<\/p>\n

                          Line regulation :
                          \n\"1st
                          \nFor load regulation, Vs<\/sub> is made greater than Vz<\/sub>. Vs<\/sub> is kept constant and load resistance RL is varied. If RL is decreased, current IL<\/sub> increases and Iz<\/sub> decreases by the same value, but output voltage VL<\/sub> = Vz<\/sub> remains constant.<\/p>\n

                          If RL is increased, IL decreases and Iz increases, but output voltage VL<\/sub> = Vz<\/sub> remains constant.<\/p>\n

                          Question 14.
                          \nCompare LED and LCD.
                          \nAnswer:<\/p>\n\n\n\n\n\n\n\n\n\n\n
                          LED<\/strong><\/td>\nLCD<\/strong><\/td>\n<\/tr>\n
                          1. It emits light when forward biased<\/td>\n1. It reflects light.<\/td>\n<\/tr>\n
                          2. Operating voltage 2 to 5V DC.<\/td>\n2. Operating voltage V to 20V DC.<\/td>\n<\/tr>\n
                          3. It requires high power<\/td>\n3. It requires low power<\/td>\n<\/tr>\n
                          4. It has a wide viewing range<\/td>\n4. Less viewing range<\/td>\n<\/tr>\n
                          5. These are sturdy<\/td>\n5. These are fragile<\/td>\n<\/tr>\n
                          6. Operate at very high frequency<\/td>\n6. Operate at low frequency<\/td>\n<\/tr>\n
                          7. It has more brightness<\/td>\n7. It has less brightness.<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n

                          \"1st<\/p>\n

                          Filters:
                          \nA filter is a circuit that removes the AC component from the rectifier output and allows pure DC to reach the load.<\/p>\n

                          Series inductor filter:
                          \n\"1st
                          \nIt has an inductor L in series with rectifiers output and loads RL<\/sub>.<\/p>\n

                          The inductor opposes any change in current flowing through it. When the current through the inductor tries to change, a back emf is induced in the inductor which prevents the current from changing. Hence inductor offers a large opposition to the AC component and blocks AC and allows only DC to pass through. Thus DC voltage is developed across RL<\/sub>. ‘<\/p>\n

                          Shunt capacitor filter:
                          \n\"1st
                          \nIt has a large value capacitor connected in parallel with load RL<\/sub>.<\/p>\n

                          The capacitor blocks DC and allows AC to pass through. Hence it bypasses the AC component allowing DC to reach the load. Thus shunt capacitor filter removes most of the AC component.<\/p>\n

                          When the output voltage of the rectifier is increasing, the capacitor charges to the peak value of the input. Beyond peak value, the rectifier diode becomes reverse biased and stops conducting.<\/p>\n

                          Now capacitor discharges through RL. The capacitor discharge until the output of the rectifier again increases to a value greater than capacitor voltage. Now diode becomes forward biased and conducts. Thus the almost constant output is obtained across RL<\/sub>.<\/p>\n

                          LC filter or L type filter:
                          \n\"1st
                          \nThis filter combines the current smoothing action of a series inductor with the voltage stabilising action of the shunt capacitor filter. The inductor allows DC current to pass through easily offering high reactance to AC.<\/p>\n

                          Any variations or ripple that remains after passing through the inductor is bypassed to the ground by the capacitor. Hence DC voltage reaches the load.<\/p>\n

                          1st PUC Electronics Question Bank with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"

                          Karnataka 1st PUC Electronics Question Bank Chapter 6 Semi-Conductors, Diodes and Applications of Diodes One Mark Questions and Answers Question 1. What are semiconductors? Answer: Semiconductors are substances whose electrical conductivity lies between that conductors and insulators. Question 2. What are conductors? Answer: Conductors are those substances that allow electric charges to flow through them …<\/p>\n","protected":false},"author":5,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[81],"tags":[],"jetpack_sharing_enabled":true,"jetpack_featured_media_url":"","_links":{"self":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/88452"}],"collection":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/users\/5"}],"replies":[{"embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/comments?post=88452"}],"version-history":[{"count":0,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/posts\/88452\/revisions"}],"wp:attachment":[{"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/media?parent=88452"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/categories?post=88452"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/kseebsolutions.guru\/wp-json\/wp\/v2\/tags?post=88452"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}