Vidyalankar F.Y. Diploma : Sem. II [CO/CD/CM/CW/IF] Basic Electronics V F V K. V K = 0.7 for Si = 0.3 for Ge

F.Y. Diploma : Sem. II [CO/CD/CM/CW/IF] Basic Electronics Time : 3 Hrs. Prelim Question Paper Solution Marks : 100 Q.1 Attempt any TEN of the following: [20] Q.1(a) Draw characteristics of PN junction diode. State rated value for Si and Ge. [2] (A) Characteristics of PN junction diode: V Z I O I F V K Q.1(b) State two limitations of integrated circuits. [2] (A) Two limitations of Integrated circuits The integrated circuits suffer from some of the following limitations: i) Inductors cannot be fabricated directly. ii) It is neither convenient nor economical to fabricate capacitors and resistors exceeding 30 pf and 100 k respectively. Thus, capacitors and resistors are limited in maximum value. iii) Low tolerances on capacitors and resistors and also small temperature coefficient are difficult to obtain. iv) It is not possible to fabricate high grade PNP unit very easily. Q.1(c) List the different types of fixed capacitors. [2] (A) Fixed Capacitors : The capacitance of fixed capacitors is constant. Different types of fixed capacitors are as follows: i) Impregnated paper capacitor or paper capacitors. ii) Mica capacitors. iii) Ceramic capacitors iv) Plastic film capacitors v) Glass capacitors. vi) Electrolytic capacitors. Q.1(d) List types of Resistors. [2] (A) Types of Resistors Classification of V F V K = 0.7 for Si = 0.3 for Ge Fig. 1

: F.Y. Diploma BE Q.1(e) What is need for coupling? [2] (A) Function of coupling If, it does, the biasing conditions of the next stage are disturbed, then this type of coupling is called interstage coupling. It serves the following two purposes : i) It transfers the AC output of one stage to the input of the next stage. ii) It isolates the DC conditions of one stage to the next stage. It is necessary to prevent the shifting of Q-points. Q.1(f) Describe the zener breakdown mechanism. [2] (A) Zener Breakdown This type of breakdown occurs in P-N junctions which are heavily doped. The heavily doped junctions have a narrow depletion region. When the reverse voltage is increased, the electric field across the junction also increases. A high value of electric field causes a covalent bond to break from the crystal structure. As a result of this, a large number of minority carriers are generated and a large current flows through the junction. This mechanism of breakdown is called Zener breakdown. In certain cases, especially in Silicon P-N junctions, the Zener breakdown can be represented by a straight line which corresponds to a constant reverse voltage as shown in Figure. Q.1(g) Give two advantages of IC's. [2] (A) Advantages of IC's are: (i) The physical size of an IC is extremely small (generally thousand times smaller) than that of discrete circuits. (ii) The weight of an IC is very less as compared to that of equivalent discrete circuits. (iii) The reduction in power consumption is achieved due to extremely small size of IC. (iv) Interconnection errors are non-existent in practice. (v) Temperature differences between components of a circuit are small. (vi) Close matching of components and temperature coefficients is possible. (vii) In case of circuit failure, it is very easy to replace an IC by a new one. (viii) Active devices can be generously used as they are cheaper than passive components. Q.1(h) What is SSI & MSI? [2] (A) Small Scale Integration (SSI) : In this case, the number of circuits contained in one package is less than 30 or number of components less than 50 e.g. NAND/NOR and other logic basic gates. Medium Scale Integration (MSI) : In this case, the number of circuits contained in one package is in between 30 to 100 or number of components in between 50 to 500 i.e. counters, decoders, shift-registers, voltage regulators, amplifiers, etc. Q.1(i) What is the significance of Reverse Saturation current? [2] (A) Significance of Reverse Saturation current When P-N junction is reverse biased, practically no current flows due to majority carriers. However, a small amount of current does flow due to the diffusion of minority carriers across the junction. The minority carriers are the hole electron pairs generated throughout the semiconductor material as a result of thermal energy. The current, so produced is known as reverse saturation current and is designed by I O (or I S ). It has been found experimentally that the value of I C is about double in magnitude for every 10 C rise in temperature. I O is used to compare the temperature sensitivity of semiconducting materials. 2

Prelim Question Paper Solution Q.1(j) Compare NPN and PNP transistor. [2] (A) Comparison between NPN and PNP transistor NPN Transistor PNP Transistor i) P-type material is sandwitched between two N-type materials. N-type material is sandwitched between two P-type materials. ii) N P N P N P E C E C iii) Symbol B B C E Symbol iv) Majority charge carriers are electrons. Majority charge carriers are holes. v) NPN transisters are more preffered. Less preffered as compare to NPN transistor. Q.1(k) Give the classification of IC s. [2] (A) Classification of Integrated Circuits There are several ways of classifying integrated circuits. But the integrated circuits can be broadly divided into three important classes based on the (i) active devices used for fabrication, (ii) function performed (iii)structure/technology employed. 1. Active devices : According to this method, the integrated circuits may be classified into two major types, namely : (a) Bipolar ICs (using bipolar transistors and diodes), (b) Unipolar ICs (generally employing MOSFET) 2. Applications or functions : From the point of view of circuit function performed, integrated circuits may be classified into two major types, namely: (a) Linear (or analog) ICs, and (b) Non-linear (or digital) ICs. 3. Fabrication structure/technology : Integrated circuits may be classified into three important types from the structural/technological considerations, namely: (a) Monolithic (in a single Silicon chip) ICs, (b) Thin film (on an insulating substrate) ICs, and (c) Hybrid (multichip) ICs. Q.1(l) List the different types of MOSFET and draw their symbols. [2] (A) Different types of MOSFET and its Symbol MOSFET is the short form of metal oxide semiconductor field effect transistor. MOSFET are different from JFETS in construction and they are of two types: i) Depletion type MOSFET ii) Enhancement type MOSFET The symbols of both n and p channel depletion type MOSFETs are as shown in figure 1(a) and 1(b). The substrate is connected internally to the source terminal. B B C E 3

: F.Y. Diploma BE Gate (G) Source (S) (a) Symbol of n channel depletion Fig. 1 (b) Symbol of p channel depletion The circuit symbols of p and n channel enhancement MOSFETs are as shown in figure 2. The substrate is internally connected to the source terminal making it a three terminal device. Fig. 2 : Circuit symbols of enhancement type MOSFETs Q.2 Attempt any FOUR of the following : [16] Q.2(a) Comparison between conductors, insulators and semiconductors. [4] (A) Comparison between Conductors, Insulators and Semiconductors Conductors Insulators Semiconductors 1. They have very large They have very low They have moderate concentration of free electrons concentration of free electrons concentration of free electrons. about 10 28 electrons/m 3. about 10 7 electrons/m 3. 2. They have low resistivity of the They have very high They have low resistivity of the order of 10 8 ohm m ( m). resistivity of the order of 10 9 to 10 16 ohm m ( m). order of 10 1 to 10 2 ohm m ( m) 3. They have very high conductivity of the order of 10 7 s/m. They have very low conductivity of the order of 10 10 to 10 16 s/m. They have low conductivity of the order of 10 7 s/m. 4. There is no forbidden energy There is high forbidden There is low forbidden gap (E G = 0) energy gap (E G = 5 ev or energy gap (E G = 1 ev) more) 5. They have positive temperature coefficient of resistance. They have negative temperature coefficient of resistance. They have negative temperature coefficient of resistance. 6. Resistance of metals increases with increase in temperature. Resistance of insulating materials decreases with increase in temperature. Resistance of semiconductor materials decreases with increase in temperature. 7. At room temperature, they act as a good conductor of electricity. At room temperature, they act as a bad conductor of electricity i.e. perfect insulator. At room temperature, they act as a moderate conductor of electricity. 4

Prelim Question Paper Solution Q.2(b) Distinguish between varactor diode and LED. [4] (A) Light Emitting Diodes (LEDs) Just like a P-N junction that can absorb light and produce an electrical current, the opposite is also possible that a P-N junction can also emit light or exhibit electroluminescence. Each recombination of electron-hole pair radiates energy. In an ordinary diode, the radiated energy is in the form of heat because Germanium and Silicon diodes have less probabilities of radiating light. A P-N junction, which emits light when forward biased, is known as a light emitting diode (abbreviated as LED). The emitted light may be visible or invisible, i.e. infrared (IR) by using semiconductor materials such as Gallium Arsenide Phosphide (GaAsP), Gallium Phosphide (GaP), Gallium Arsenide (GaAs), Gallium Nitride (GaN) etc. Construction : Varactor Diode : The choice of semiconductor material for the fabrication of a diode depends on following characteristics : 1. Characteristics : i) It should have high mobility for atleast one of the carriers, ii) It should have low dielectric constant, iii) It should have large band gaps, iv) It should have very small impurity ionization energies, v) It should have large thermal conductivity. 2. Properties : These qualities will provide the following properties : i) It has minimum electrical resistance, ii) It has minimum capacitance, iii) Low saturation currents as also potential operation at high temperatures (useful in harmonic generators), iv) Ability to operate at cryogenic temperatures (useful in obtaining low noise parametric amplification), and v) Good thermal dissipation (for maximum power handling capacity in harmonic generators and for effective cooling in parametric varactors). Fig. : 5

: F.Y. Diploma BE Q.2(c) Draw the net diagram of NPN transistor and describe its working. [4] (A) npn Transistor : Forward-Active Mode Operation If the transistor is used as an amplifying device, the base-emitter (B E) junction is forward biased and the base-collector (B-C) junction is reverse biased, in a configuration called forward-active operating mode or simply the active mode. E R E Base-emitter (B-E) junction i E An npn bipolar transistor biased in the forward-active mode. Emitter of the transistor is n-type mean majority carrier in the emitter region are electrons since the B-E junction is forward biased, electrons from the emitter are injected across the B- E junction into the base as shown by an arrow. Since the B-C junction is reverse biased the electron centralization at the edge of that junction (B-C junction) is approximately zero. So electrons from n-type emitter are injected into Base region and at the same there are less no. of electrons in B-C junction this creates large concentration gradient in the base region. Because of above mentioned concentration gradient, electrons injected from the emitter diffuse across the base into the B-C space charge region, where the electric field sweeps them into the collector region. Q.2(d) State the significance of DC load line and Q-point. [4] (A) Significance of DC load line and Q-point Significance of DC load line The DC load line is actually a set of infinite Q points. Any point on the dc load line can be used as the Q-point. The dc load line is used to set up maximum dc voltage and current of the amplifier. Significance of Q-Point The Q-point gives operating values of voltage and current. Application of transistor depends on the position of the Q-point on the load line as shown below. Application Open switch closed switch amplifier. n P n Electron injection V BE + + + V BB B Base-collector (B-C) junction V CC Position of Q-point In the cut-off region. In the saturation region in the active region. The shape of amplifier output signal depends on the position of Q-point for example if it is selected at the centre of the load line, in the middle of the active region such that positive and negative half cycle at the amplified output is identical than it is called as faithful amplification or linear output. i C C R C 6

Prelim Question Paper Solution Q.2(e) Explain the operating principle of schottky diode with net diagram. [4] (A) Working Principle : Fig. 1 : Fig. 2 Consider the structure of a schottky diode as shown in Fig. 1 (a). The metal region of a Schottky diode is heavily occupied with the conduction-band electrons and the N-type region is lightly doped. There are no minority carriers as in other types of diodes, but there are only majority carriers as electrons. It operates only with majority carriers. When it is forward biased, higher energy electrons in the N regions are injected into the metal region where they give up their excess energy very rapidly. Since there are no minority carriers as in a conventional diodes, there is no charge storage and hence there is no reverse recovery diode when it is switched from the forward-biased condition (i.e. ON state) to the reverse biased condition. (i.e. OFF state). Therefore, it has negligible storage time and hence there is a very rapid response a change to in bias. Because of this property, it acts as a very fast switching diode. The V-I characteristic of a Schottky diode is essentially the same as that of P-N junction diode. It should be noted that the voltage drop across it is much less than that of a P-N junction diode for the same forward current. Q.2(f) Draw the construction of N-channel FET and describe it. [4] (A) Construction of JFET : The figure shows structure and symbol of n channel JFET. (a) Structure and symbol for n channel JFET (b) Structure and symbol for P channel JFET A small bar of extrinsic semiconductor material, N type (it can be of P type also) is taken and at its two ends, two ohmic contacts are made which are the drain and source terminals of FET. Heavily 7

: F.Y. Diploma BE doped electrodes of P type material form p n junctions on each side or the bar. The thin region between the two p gates is called the channel. Since this channel is in the n type bar, the FET is known as N channel JFET. The electrons enter the channel through the terminal called source and leave through the terminal called drain. The terminals taken out from heavily doped electrodes of p type material are called gates. Usually, these electrodes are connected together and only one terminal is taken out, which is called gate, as shown in the figure. The device could be made of p-type bar with two n type gates as shown in the figure (b). Then this will be p channel JFET. The principle of working of these two types of is similar; the only difference being that in n channel JFET the current is carried by electrons while in P channel JFET, it is carried by holes. In the absence of any applied voltage, JFET has gate channel junctions under no bias conditions. The result is a depletion region at each junction, as shown in figure (c). This represents same depletion region of a diode under no bias conditions. Recall also that depletion region is that region which does not have any free carriers and therefore is unable to support conduction through the region. In JFET, the p n junction between gate and source is always kept in reverse biased conditions. Since the current in a reverse biased p n junction is extremely small, practically zero; the gate current in JFET is often neglected and assumed to be zero. Q.3 Attempt any FOUR of the following : [16] Q.3(a) In an NPN transistor, I CEO = 200 A, = 100, I B = 10 A. Find I C and I E. [4] (A) I CEO = 200 A = 100 I B = 10 A I C = (10 A) (100) + 200 A = 1200 A = 1.2 ma I E = I B + I C = 1200 A + 10 A = 1.21 ma I C = I E + I CBO I CEO = (1 + ) I CBO I C = I B + I CEO (c) Junction field effect transistor Q.3(b) Draw the circuit diagram of crystal oscillator and give its advantages, disadvantages and applications. (A) Circuit diagram of Crystal Oscillator +V C C C2 R 1 RF C C C1 [4] Crysta C 2 R 2 RE C E C 1 8

Prelim Question Paper Solution Advantages : As the frequency stability is very high at high frequency. It is used to generate very high stable frequencies. Disadvantages : The frequency of oscillation is inversely proportional to thickness hence to obtain very high frequencies a very low value of thickness is required which makes crystal fragile. Applications : There are various type of crystal oscillators available like frequency stability range of crystal oscillators, voltage controlled crystal oscillators, crystal temperature compensated crystal etc. Above mentioned oscillators are used in base stations for mobile phones, optical transmission systems, measuring equipment etc. Q.3(c) Give the complete classification of oscillators. [4] (A) Classification of Oscillators Sinusoidal (Harmonic) Positive Feedback High Frequency (LC oscillators more than 20 khz) 1) Hartley 2) Colpitts 3) Tuned collector 4) Crystal controlled Negative Resistance Pulse generator 1) UJT 2) Tunnel diode Oscillators Low Frequency (RC oscillators up to 20 khz) 1) RC phase shift 2) Wien bridge Multi Vibrators (Square wave generator) 1) Astable 2) Bistable 3) Monostable 4) Schmitt Trigger Relaxation (Non-sinusoidal waveform generators) Saw tooth (Time base signal) 1) Voltage & current generator Q.3(d) On what factors does the choice of semiconductors for diode depend? [4] (A) The choice of semiconductor material for the fabrication of a diode depends on following characteristics : Characteristics : i) It should have high mobility for atleast one of the carriers, ii) It should have low dielectric constant, iii) It should have large band gaps, iv) It should have very small impurity ionization energies, v) It should have large thermal conductivity. 9

: F.Y. Diploma BE Q.3(e) Draw the neat circuit diagram of two stage Transformer coupled amplifier. [4] (A) Two Stage Transformer Coupled (TC) CE Amplifier Two stage transformer coupled CE amplifier Applications : The applications of a transformer amplifier are as follows : i) It is mostly used for impedance matching between the individual stages. ii) It is widely used as a voltage amplifier in the final stage of multistage amplifier. iii) It is widely used for amplification of radio-frequency (RF) signal. iv) It is used to transfer power to the low impedance load such as loudspeaker. Two Stage Direct Coupled (DC) CE Amplifier : Two stage direct coupled CE amplifier Applications : The applications of direct coupled amplifier are as follows : i) It is used is analog computation. ii) It is used in power supply voltage regulators. iii) It is used for bioelectric measurements. iv) It is used in linear integrated circuits. Q.3(f) Give two applications of LED. [4] (A) Application of Light Emitting Diodes (LEDs) 1. It is used in 7-segment, 16-segment and dot matrix displays which are used to indicate alphanumeric characteristics and symbols in various systems such as digital clocks, calculators, stereo tuners, microwaves ovens etc. 2. It is used for indicating power ON/OFF conditions, power level indicators or stereo amplifiers. 3. It is used in optical switching applications. 4. It is used in optical communication systems. 5. It can be used as solid state video displays in place of CRTs. 6. It is used for checking the linearity, speed etc. of opto-electronic detection circuits. 7. It is used to indicate digital logic state. 8. It is used in image sensing circuits in videophones. 9. It is used to monitor/indicator voltage levels or polarities. 10. It is used in burglar alarm systems. 10

Prelim Question Paper Solution Q.4 Attempt any FOUR of the following : [16] Q.4(a) Distinguish between BJT and FET. (Four points) (A) Comparison of BJT and FET : Parameter BJT FET 1 Control element Current controlled device. Input current I B controls output current I c. Voltage controlled device. Input voltage V GS controls drain current I D. 2 Derive type Current flows due to both, majority and minority carriers and hence Current flows only due to majority carriers and hence unipolar device. bipolar device. 3 Types npn and pnp n-channel and p-channel. 4 Symbols 5 Configurations CE, CB, CC CS, CG, CD 6 Input resistance Less compare to JFET. High compare to BJT. 7 Size Bigger than JFET. Small in construction than BJT, thus making them useful in integratedcircuit (IC) chips. Q.4(b) Draw the circuit diagram to age Rectifier with input and output waveforms. [4] (A) Bridge Rectifier A. C. Suppl V 1 V 2 D 4 D 2 Bridge rectifier with transformer When the input voltage is positive as shown above, the decodes D 1 and D 2 are forward biased and conducts. A voltage is developed across the resistance R L due to the current flow through it. The voltage looks the positive half of the input cycle. At this time the diodes D 3 and D 4 are reverse biased. When the input voltage is negative, diodes D 3 and D 4 are forward biased and conduct same current through R 2. During this time the diodes D 1 and D 2 are reverse biased. As a result of this action, a full-wave rectified output is developed across the resistance R L. V m R L D 1 D3 t V Out Input Waveform D 1 D 2 D 3 D 4 D 1 D 2 D 3 D 4 Output Waveform t 11

: F.Y. Diploma BE Q.4(c) Draw the circuit diagram of single stage CE amplifier. Give function of each components. (A) Common emitter (CE) amplifier: [4] +V CC V i C 1 R 1 R C C 2 V o R 2 R E C E Single stage RC coupled RE amplifier. The capacitance C 1 and C 2 are called as the coupling capacitors as the load resistor R 2 is coupled to the amplifier through the coupling capacitor this amplifier is called as RC coupled amplifier. Resistors R 1, R 2 and R E are used for biasing the transistor in the active region, because for operating the transistor as an amplifier it is necessary to bias it in the active region. The type of biasing used here is voltage divider bias or self bias. R C is the collector resistor used for controlling the collector current. Role of coupling Capacitor C 1 : The input coupling capacitors C 1 is used for coupling the ac input voltage V i to the base of the transistor. As capacitors block dc, this capacitor help to block any dc component present in Vi and couples only the ac component of the input signal. This capacitor also ensures that the dc biasing conductions of the transistor remain unchanged ever after applications of the input signal. Role of C E : This capacitor is connected in parallel with the emitter resistor R E is called as emitter bypass capacitor. This capacitor offers a low reactance to the amplified ac signal. Therefore the emitter resistor R E gets bypassed through C E for only the ac signals. This will increase the voltage gain of the amplifier moreover as C E act as an open circuit for d voltages it does not bypass R E for dc conditions. Role of C 2 : This capacitor couples the amplifier output to the load resistance or to the next stage of the amplifier. It is used for blocking the dc part and passing only the ac part of the amplified signal to the load. Q.4(d) Define rectifier. Give the classification of rectifier. Which rectifier is mostly used? [4] (A) A rectifier may be defined as an electronic device, such as a P-N junction diode, used for converting alternating (AC) voltage or current into unidirectional (DC) voltage or current. Types of Rectifier : Rectifiers may be classified into the following two categories depending upon the period of conduction : 1. Half-wave rectifier : A half-wave rectifier is one which converts an AC voltage into a pulsating voltage using only one half cycle of the applied AC voltage. The rectifying diode conducts during one-half cycle only. 2. Full-wave rectifier : A full-wave rectifier is one which converts an AC voltage into a pulsating voltage using both cycles of the applied AC voltage. The rectifying diodes conduct during positive as well as negative half cycle of AC voltage. The following two circuits are commonly used for full-wave rectification. 12

Prelim Question Paper Solution (a) Centre-tap full wave rectifier. (b) Full-wave bridge rectifier. Vacuum diode rectifiers are no longer used and have been suspended by semiconductor diode rectifiers. We shall discuss the study of semiconductor diode as rectifiers. Full wave rectifier is mostly used as it offers more DC output. Q.4(e) Draw the symbol of following diodes. (i) Zener diode (ii) Schottky diode (iii)varactor diode (iv) Tunnel diode (A) (i) Zener diode (ii) Schottky diode (iii)varactor diode (iv) Tunnel diode : Fig. : Q.4(f) Differentiate between Half wave rectifier and Bridge full wave rectifier (any Four points) (A) Difference between PN-junction diode and Zener diode PN-junction diode Zener diode 1. Operated as fb or rb. always operated in rb region. 2. [4] [4] 3. Used in amplifiers, BJT Used in voltage regulator circuit. 4. Generally current flow due to majority current present due to minority carriers. carriers 13

: F.Y. Diploma BE Q.5 Attempt any FOUR of the following : [16] Q.5(a) Draw the circuit diagram of Astable Multivibrator using transistor and given tis two [4] applications. (A) Astable multivibrators are used in amateur radio equipment to receive and transmit radio signals. Astable multivibrators are also used in morse code generators, timers, and systems that require a square wave, including television broadcasts and analog circuits. Q.5(b) State advantages and disadvantages of bridge rectifier. [4] (A) Advantages of Bridge Rectifier i) It can be used in application allowing floating output terminals, i.e. no output terminal is grounded. ii) The need for centre-tapped transformer is eliminated. iii) If stepping up or stepping down of AC voltage is not needed, then it does not even require any transformer. iv) The transformer is less costly as it is required to provide only half the voltage of an equivalent centre-tapped transformer used in a full wave rectifier. v) The PIV is one-half that of the centre-tap circuit. vi) The output is twice that of the centre-tap circuit for the same secondary voltage. vii) The transformer utilization factor is very large. Disadvantages of Bridge Rectifier i) It requires four semiconducting diodes. ii) Two diodes in series conduct at a time on alternate half cycles. This creates a problem when low DC voltages are required. This leads to poor voltage regulation. Q.5(c) Draw the neat circuit diagram of direct coupled amplifier. Give its two applications. [4] (A) Two stage direct coupled CE amplifier 14

Prelim Question Paper Solution Applications : The applications of direct coupled amplifier are as follows : i) It is used is analog computation. ii) It is used in power supply voltage regulators. iii) It is used for bioelectric measurements. iv) It is used in linear integrated circuits. Q.5(d) Differentiate between zener diode and PN junction diode. [4] (A) PN-junction diode Zener diode 1. Operated as fb or rb. Always operated in rb region. 2. 3. Used in amplifiers, BJT Used in voltage regulator circuit. 4. Generally current flow due to majority carriers current present due to minority carriers. Q.5(e) Define (in words) : (i) Bandwidth (ii) Current gain [4] (iii)power gain (iv) Voltage gain (A) (i) Bandwidth : The range of frequency over which the transistor offers constant peak gain to input without causing distortion is called B.W. of transistor. (ii) Current gain : It is defined as ratio of output current to input current at constant output voltage. (iii)power gain : It is defined as ration of output power to input power. (iv) Voltage gain : It is defined as ratio of output voltage to input voltage with input current constant. Q.5(f) What is the need of filter. List the types of filters. [4] (A) Filter Circuits : The output of a rectifier contains DC component as well as an AC component. This nature of output is known as pulsating DC which is not pure DC. Thus, the rectifiers provide a pulsating DC output. The presence of AC component is most undesirable in electronic circuits and therefore must be kept away from the load, i.e. removed from the rectifier output. We can filter or smooth out the AC variations from the rectified voltage. To do so, we can use a filter or smoothing circuit for removing AC component and allowing only the DC component to reach the load. Some of the passive filters used in the field of electronics are as given below : 1. Series inductor (or choke) filters. 2. Shunt capacitor filter. 3. Choke input (LC or L type) filter. 4. Capacitor input (CLC or n type) filter. Q.6 Attempt any FOUR of the following : [16] Q.6(a) Explain the formation of depletion layer in PN junction with neat sketch. [4] (A) As soon as the P-N junction is formed, some of the holes, which are near the junction, from the P-region diffuse into the N-region. They then combine with the free electrons in the N-region, Similarly, some of the free electrons, which are near the junction, from the N-region diffuse into 15

: F.Y. Diploma BE the P-region. These electrons also combine with the holes. The diffusion of holes from P-region to N-region and free electrons from N-region to P-region take place because they move randomly due to thermal energy and also because there is a difference in their concentrations in the two regions. Thus, some of the holes and free electrons diffuse towards each other across the junction and recombine. Fig. 1 But, in practice, this does not occur. The diffusion of holes and free electrons across the junction occurs for a very short time. In this process, the negative acceptor ions in the P-region and the positive donor ions in the N-region in the immediate neighbourhood of the junction are left uncompensated. This situation is shown in Fig. 1(b). Additional holes trying to diffuse into the N-region are now repelled by the uncompensated positive charge of the donor ions. The free electrons trying to diffuse into the P-region are repelled by the uncompensated negative charge of the acceptor ions. As a result, total recombination of holes and free electrons cannot occur and also the further diffusion of free electrons and holes across the junction is stopped. The region containing the uncompensated acceptor and donor ions in the vicinity of the junction is called depletion region (or layer). It is called so because the mobile charge. Q.6(b) What is biasing? State the requirements of biasing. Which is the most useful biasing [4] method? (A) Need of Biasing i) We know that collector current, I C is the function of I CO, V BE and. i.e. I C = F [I CO, V BE, ] I CO is temperature dependent, I CO increases with temperature, it doubles for every 10 C/14 C rise in temperature for Si/Ge device. V BE decreases with temperature, it decreases by 2.5 mv/ C with rise in temperature. varies from device to device, also increases by 1 for every 1 C rise in temperature. Using biasing techniques we make I C independent of variations in I CO, V BE, due to variation in temperature. Hence making I C independent of variations in temperature biasing is required. ii) We know that transistor can operate in any of the three regions of operations, namely cutoff, active region and saturation. To operate the transistor in these regions, the junctions of a transistor should be forward or reverse biased. The required bias voltage can be obtained using a single power supply by means of biasing. Hence Biasing may be defined as the process by which I C is made independent of and temperature by allowing I B to vary within its tolerable limits. The most useful biasing technique : (1) For protecting device against variation in parameter: Potential divider biasing. (2) For protecting against thermal runaway : Self bias. 16

Prelim Question Paper Solution Q.6(c) Draw the transfer characteristics of JFET. Give the meaning of IdSS and VgsOff. [4] (A) Fig. : Drain characteristics of an n channel JFET Here, the characteristic has been divided into three regions vit. cutoff, saturation and ohmic region. i) I DSS (Source Saturation current) : The value of drain current corresponding to V ES = 0V, is called as the source saturation current and it is denoted by I DSS. I DSS corresponds to the maximum drain current because the channel is widest for V GS = 0V. ii) V P (Pinch off voltage) : The pinch off voltage is the value of V DS, at which the drain current reaches its constant saturation value. Any further increase in V DS does not have any effect on the value of I D. iii) Ohmic Region : The drain current I D varies with variation in the drain to source v/g V DS in the ohmic region as shown in the diagram. The JFET is said to be operating as a voltage variable resistance in the ohmic region. The resistance offered by the JFET decrease with decrease in the value of negative gate to source bias voltage i.e. negative V GS. The FET resistance in the ohmic region is given by, R DS = Fig. : Transfer characteristics of n channel JFET V I P DSS Q.6(d) State the need of regulators. Define load and line regulation. [4] (A) Regulators Voltage regulator is the last block in the dc regulated power supply. The voltage regulator is a circuit which will try to maintain the output voltage constant under all the operating circumstances. The output voltage changes if there is a change in input voltage or change in load current or if the ambient temperature is changed. Regulation: It is applicable to the voltage regulator circuits as well. 17

: F.Y. Diploma BE Load Regulation i) The load regulation (L.R.) is defined as the change in output voltage when the load current is changed from zero (no load) to maximum (full load) value. L.R = V NL V FL where, V NL = Output voltage on no load (zero load current) and V FL = Output voltage on full load (maximum load current). ii) The percent load regulation is defined as: V V V NL FL % L.R. = 100 FL The % L.R. of a power supply should ideally be equal to zero and practically it should be as small as possible. The load voltage can be plotted graphically against the load current as shown in figure 1(a). This graph is called as the "Regulation characteristics" of a power supply. Line Regulation or Source Regulation As shown in figure 2 in power supplies, the 230 V ac voltage is applied at the input of a rectifier and the rectified and filtered output is given as input to the regulator. So if the ac mains voltage fluctuates about 230 V, then the input voltage V in to the regulator will also change and there will be some effect on the output voltage of the regulator. In order to account for this change in the load voltage due to change in supply voltage, we define a performance parameter called "Line Regulation" or "Source Regulation (SR)". The source regulation (SR) is defined as the change in regulated load voltage due to change in line voltage in a specified range of 230 V ± 10% at a constant load current. The source regulation can be mathematically expressed as : S.R. = V LH V LL where, V LH = Load voltage with high line voltage and V u = Load voltage with low line voltage. (a) Regulation characteristics of a power supply The percent line regulation is defined as : S.R. % S.R.= 100 V Nom Fig. 1 (b) Equivalent circuit of a power supply Fig. 2 : Block diagram of a power supply. Where V Nom = The nominal load voltage under typical operating condition. Ideally the line regulation should be zero and practically it should be as small as possible. 18

Prelim Question Paper Solution Q.6(e) The turns ratio of the transformer used in a bridge rectifier is n1 : n2 = 12:1. The primary is connected to 220V, 50Hz power mains. Assume diode voltage drop to be zero. Find dc voltage across the load. What is PIV of the diodes? (A) Given : n 1 : n 2 = 12 : 1 V 1 = 220 V, Voltage connected to primary f = 50 Hz V 0 =? Voltage across load PIV =? Bridge = rectifier For Transformer V 1 2 N 2 V = 2 1 N1 N2 V 2 = N1 N N = 12 1 1 2 N N = 1 12 1 12 V V L = 220V 1 ( N 1 : N 2 = 12 : 1) = 18.33 V RMS value of secondary voltage is 18.33 V. The maximum value of the secondary voltage is V m = 2 V 2 = 2 18.33 = 25.92 V DC output voltage V dc = 0.636 V m = 0.636 25.92 = 16.48 V PIV = V m = 25.92 V Q.6(f) Draw the frequency response of RC coupled amplifier. State the reasons for decreasing gain on both sides of frequency Range. (A) Frequency response of RC coupled amplifier (1) Low frequency region : In low frequency region, the voltage gain (or output voltage) decreases with the decrease in frequency of an input AC signal due to the increased reactance of the coupling and bypass capacitors. (2) High frequency region : In high frequency region, the voltage gain (or output voltage) decreases with the increase in frequency of an input AC signal due to the BJT internal capacitances and stray capacitance. [4] [4] 19