Important Instructions to examiners: 1) The answers should be examined by key words and not as word-to-word as given in the model answer scheme. 2) The model answer and the answer written by candidate may vary but the examiner may try to assess the understanding level of the candidate. 3) The language errors such as grammatical, spelling errors should not be given more Importance (Not applicable for subject English and Communication Skills). 4) While assessing figures, examiner may give credit for principal components indicated in the figure. The figures drawn by candidate and model answer may vary. The examiner may give credit for any equivalent figure drawn. 5) Credits may be given step wise for numerical problems. In some cases, the assumed constant values may vary and there may be some difference in the candidate s answers and model answer. 6) In case of some questions credit may be given by judgement on part of examiner of relevant answer based on candidate s understanding. 7) For programming language papers, credit may be given to any other program based on equivalent concept. Q. 1 Attempt any ten of the following: 20M a) List any two examples of acceptor and donor impurities each. Ans: [Note: Any two of the following] Acceptor Impurities: i) Gallium ii) iii) iv) Indium Aluminum Boron [Note: Any two of the following] Donor Impurities: i) Phosphorus ii) Antimony Page 1 of 26
iii) Arsenic iv) Bismuth b) Sketch V-I characteristics of P-N junction diode. Ans: Characteristics of P-N junction diode: c) Define the following terms with reference to rectifier i) Ripple factor ii) PIV Ans: [Note: Each definition 1 mark] i) Ripple factor: [it is the AC component present in the output is called Ripple] ii) PIV: the maximum value of reverse voltage occurs at the negative alternation of the cycle. This maximum reverse voltage is called PIV. d) Define regulator. State the need of voltage regulator in power supply. Ans: Regulator: A circuit which is used to maintain constant output is called regulator. Necessity: In the operation of electronic devices, phase shift, calibration of instrument, distortion may occurred. So to avoid it, there is a need of voltage regulator. Page 2 of 26
e) Draw the symbol of PNP and NPN transistors. Ans: PNP transistor: NPN transistor: f) State the need of biasing the transistor. Ans: [Note: Each point 1 mark] The transistor biasing is required for, i) proper flow of zero signal collector current ii)maintenance of C-E voltage and B-E voltage so as to get proper amplification. g) State the function of emitter bypass capacitor in small signal amplifiers. Ans: [Note: relevant answer will be considered, any two points 2 marks] Function of emitter bypass capacitor is, To ground the AC current signal To block DC To provide high voltage gain. h) State the maximum efficiency of Class B and Class C push pull amplifier. Ans: Overall maximum efficiency of Class B amplifier is, = 78.5% Overall maximum efficiency of Class C amplifier is, = (100%) more than 95% Page 3 of 26
i) Draw a neat diagram of π filter. Ans: π filter: j) Define I CEO & I CBO. Ans: I CEO : In CE configuration small collector current flows even when base current is zero. This collector current is known as I CEO. I CBO : The minority carriers diffuses across collector base junction and hence produce a certain value of current is called leakage current from collector to base i.e. I CBO. k) Draw the symbol of point contact diode and schottky diode. Ans: Schottky diode: Point contact diode: l) Draw a neat labeled circuit diagram of transistor connected in CE configuration. Ans: [Note: any one diagram ] Page 4 of 26
Q2. Attempt any two of the following: 16M a) Describe the working of forward biasing of PN junction with neat circuit diagram. Ans: Diagram: Explanation: In forward bias P is connected to positive terminal and N is connected to negative terminal of supply. At this time, holes are repelled by positive terminal of battery and moves towards junction. Similarly electrons are repelled by negative terminal of battery and more forwards junction. Some holes and electrons enter the depletion layer and combine themselves. This reduces the barrier. More majority carriers diffuse across junction. Thus large amount of current flows through junction. b) Give the salient features of zener diode. Describe zener breakdown. Ans: Description: Page 5 of 26
Zener breakdown is shown in fig. in which reverse current increases very large with constant voltage. This breakdown occurs when reverse high voltage is applied. This high voltage exerts an electrons in the outer shell and pulled away from their parent nuclei. This ionization, which occurs due to electrons force of attraction, is known as zener effect. This causes increase in free carriers and hence increase in reverse current. Features of zener diode: [Note: Any two pointn1mark each] It operates in reverse bias. It has sharp breakdown characteristics. It generally available in terms of breakdown voltage from 1.8V to 2000V. Power rating 150mw to 50w. The diode is heavily doped as compared to ordinary diode. It can be used as voltage regulator. c) Compare zener diode and P-N junction diode on the basis of i) Symbol ii) Reverse Breakdown iii) Biasing iv)application Ans: [Note: 1 mark for each point] Parameter Zener diode P-N junction diode Symbol Reverse breakdown Zener breakdown occur. Zener breakdown and avalanche Page 6 of 26
breakdown occurred. Biasing It operates in reverse bias. It operates in forward bias. Application It is generally used in voltage regulator circuit, and as clippers / limiters Used in rectifier, power diode, signal diode in communication. d) State the operating principle of LED and photodiode. State one application of each. Ans: LED: Light Emitting Diode LED emits light when forward bias. The amount of light output is proportional to the forward current therefore higher the current higher is the light output. Application: [Note: Any one application] As indicators. In alphanumeric, seven-segment display. In optical switching. Photo diode: Photo diode operates in reverse bias. When light falls on it the reverse current increases. When there is no light, the reverse current is almost negligible. Application: Page 7 of 26
[Note: Any one of it or relevant can do.] Photo detection. Demodulation. Logic circuit. Characteristics recognition. Encoder. e) Compare CB, CE and CC configuration of BJT on the basis of input impedance, output impedance, current gain, application. Ans: [Note: 1 mark for each point] Parameter CB CC CE Input impedance Low (30 to 150 ) High (20 to 500 ) Moderate (1k to 2k ) Output impedance High 500k Low 50 to 1000 Low / Moderate 50k Current gain Unity Highest High 50 to 300 Application Machining circuit with Resistance matching Used practically as low output resistance two way amplifier amplifier. with a high output resistance. f) Why efficiency of Class B amplifier is less than that of class C amplifier? 4M Ans: In class C amplifier The output current flows during a part of the positive half cycle of the input signal. As transistor operate or biased below cut-off region so small amount of current gives less power dissipation and more output power. Whereas class B amplifier conduction cycle for total half cycle i.e. 0 0 to 180 0 during this compositely more current flows through amplifier collector so more power get dissipate and output voltage and percent power reduced. So overall efficiency of class B amplifier is less than that of class C amplifier. Page 8 of 26
Q3. Attempt any four of the following. 16M a) Draw and label full wave bridge rectifier circuit along with capacitive shunt filter. Ans: Diagram 4M b) Compare half wave rectifier, center tapped full wave rectifier and full wave Bridge rectifier on the basis of i. Ripple factor ii. TUF iii. Rectification efficiency iv. PIV. Ans: each Parameters Half wave rectifier Center tapped full wave Full wave Bridge rectifier rectifier Ripple factor 1.21 0.482 0.482 TUF 0.287 0.693 0.812 Page 9 of 26
Rectification efficiency 40.6% 81.2% 81.2% PIV. V m 2V m V m c) Which filter is suitable for heavy load application? Why? Ans: L.C or choke input is best suitable for heavy load applications. Reason: 3M The ripple factor of LC filter does not depend on the load resistance at all. It remains constant for all the values of load. This is the biggest advantages of LC filter so we can use it for the heavy as well as highest loads. d) Draw a block diagram of DC regulated power supply and explain the function of each block. Ans: Diagram: Explanation: The block diagram of DC regulated power supply is shown in figure1. Page 10 of 26
A D.C. power supply which maintains the output voltage constant irrespective of A.C. mains fluctuations or load variation is known as regulated D.C. power supply. The D.C. regulated power supply is a combination of three stages viz, a. Bridge rectifier b. Filter section c. Voltage regulator. The rectifier is nothing but a combination of diodes which converts the input 230V AC, 50Hz mains supply into pulsating DC. This pulsating DC is applied at the input of filter circuit which removes the ripple present in the output of rectifier storage and converts this pulsating DC into pure DC voltage form. Finally the voltage regulated is used to maintain the output voltage nearly constant irrespective whether the load current changes or change in input AC voltage. It also additionally helps in reducing the variations in the filtered output voltage. e) Draw a neat circuit diagram of transformer coupled amplifier and state its application. Ans: Diagram Application: Neat labeled circuit diagram of transformer coupled amplifier is drawn in figure 2. Applications of transformer coupled amplifier are as follows: - i. Transformer coupling is mostly employed for impedance matching. Page 11 of 26
ii. It is used for high output gain requirement applications. f) Explain the operation class AB push-pull amplifier with the help of neat circuit diagram, input and output waveforms. Ans: Diagram: 1 ½ M Explanation: 1 ½ M Above fig shows the diagram of class AB push pull power amplifier. The circuit consist of two center tapped transformers T 1 and T 2 two identical transistor Q 1 and Q 2 resistor R and diode D. Working: When there is no input signal, both the transistor Q 1 and Q 2 are cut-off. During the positive half cycle of input signal the base of transistors Q 1 is positive and Q 2 is negative, so, transistor Q 1 conducts, while the transistor Q 2 is off. Similarly, during negative half cycle Q 1 turns off and Q 2 conducts. Thus at any instant, only one transistor in the circuit is conducting. The DC voltage developed across the diode D is connected to the bases of both the transistor through the secondary winding of the input transformer. This voltage acts as a DC bias for the transistor become it is equal to the cut in voltage and they will conduct for complete half cycle period of the input to eliminate the cross-over distortion. Input output Waveform Page 12 of 26
Q4. Attempt any four of the following. 16M a) Define reverse saturation current with reference to the transistors, how it is generated? State the factors on which it depends. Ans: Definition In common base bipolar junction, emitter base junction of transistor is open circuited and I E = 0. The collector base junction is reverse biased due to holes injected from the emitter but forward biased due to thermally generated minority carriers. The minority carriers diffused across the collector base junction and hence produce a certain value of current know as reverse saturation current (I CO ). Generation of reverse saturation current: For a p-n-p transistor, I co consists of holes moving from base to collector and electrons crossing in the opposite direction. This reverse saturation current flowing from collector to base because it is generated by minority carriers then for a p-n-p transistor, I co is negative. For an n-p-n transistor, I co is positive. Factors on which it depends: It depends on V CB and strongly depends on temperature. b) Draw input and output characteristics of NPN transistor in CB mode. Show active region, cutoff region and saturation region in output characteristics. Ans: Input characteristics Page 13 of 26
Output characteristics c) State any four specifications of BJT. Define them. Ans: The four specifications of BJT are as follows: each 1. Collector current (I C ): - It is defined as the maximum value of collector current that a transistor can handle. 2. Collector to base voltage (V CB ): - It is defined as the maximum voltage that applied across the base and collector of a transistor. 3. Forward current gain: - It is defined as the ratio of output current to input current of a transistor. 4. Derating factor: It is defined as the maximum rate of decrease in power dissipation for each degree centigrade increase in temperature. Page 14 of 26
Other specifications of transistor are listed as under:- I. Collector to emitter voltage. (V CEO ). II. Collector saturation voltage (V CEsat ) III. Collector cutoff current (I CEO ). IV. Maximum collector dissipation (P C ) V. Collector to base breakdown voltage(bv CBO ) VI. Collector to emitter breakdown (BV CEO ) VII. Emitter to base breakdown (BV EBO ). d) What are the advantages and disadvantages of fixed bias method and voltage divider method? Ans: Advantages of fixed bias method: i. It is very simple biasing arrangement. ii. The biasing condition can be easily set. iii. There is no loading of the source by the biasing circuit since no resister is employed across base emitter junction. Disadvantages of fixed bias method: i. The saturation of the operating point is very poor. ii. Thermal runaway causes in fixed bias circuit. Advantages of voltage divider method: i. It provides more stability than any other bias method. ii. Thermal runaway is avoided. iii. Operating point is almost independent of β variation. Disadvantages of voltage divider method: i. Biasing arrangement is comparatively complicated. ii. As β value is fixed for a given transistor the relation can be satisfied by keeping R E large. But if R E is large high V CC is necessary. This increases cost as well as precaution necessary while handling. e) With neat diagram describe the base bias method with emitter feedback. Ans: Diagram Page 15 of 26
Explanation: In the reference to the circuit diagram, resister R E improves the stability of Q point considerably. When the current gain increases somehow the collector current in the circuit also increases. It produces more voltage drop across the emitter resistor R E. as a result voltage across the emitter base emitter junction is reduces with a corresponding reduction in the voltage drop across the base resistor R B This causes the base current I B to decrease and hence the collector current also. Thus the increase in collector current due to increase in value of current gain is compensated. The emitter resistor R E provides compensation for the variation in the value of current gain β by reducing the base current I B The resistor R E is present in the output side as well as in the input side of the circuit. Feedback occurs through R E and the feedback voltage is proportional to the emitter current. Hence, this circuit is also called feedback biasing circuit. f) What is thermal Runaway? How it can be avoided. Ans: Thermal Runaway The collector current for a CE configuration is given by I C = β I B + (β + 1) I CBO \ The collector leakage current (I CBO ) is strongly dependent on temperature. The flow of collector current produces heat within the transistor. This raises the transistor temperature and if no stabilization is done the collector leakage current I CBO also increases. Hence if the (I CBO ) increases, the collector current (I C ) increases by (β + 1) I CBO the increase I C will raise the temperature of the transistor which in turn will cause I CBO to increase. Page 16 of 26
This effect is cumulative and in a matter of seconds, the collector current may become very large, causing the transistor to burn out. Hence, from the above discussion, one can say that, The self-destruction of an un-stabilized transistor is known as thermal Runaway. Avoiding the thermal runaway: - In order to avoided thermal runaway and consequent destruction of transistor, it is very essential that operating point is stabilized i.e. I C is kept constant. In practice this is done by causing I B to decrease automatically with temperature increase by β I B will compensate for the increase in (β + 1); keeping I C nearly constant. Q.5 Attempt any four of the following: 16M a) Define gain and bandwidth with respect to the amplifier. Draw necessary diagram. Ans: Gain: (G) Gain of the amplifier is defined as the ratio of output electrical quantity to input electrical quantity of the amplifier. The electrical quantity can be current, voltage or power. Accordingly it can be current gain or voltage gain or power gain. Gain = Output electrical quantity / Input electrical quantity On frequency response of an amplifier gain is shown as, Bandwidth: (BW) Page 17 of 26
The range of frequency over which the voltage gain is equal to or greater than 70.7% of the maximum gain is known as bandwidth. OR The difference between lower cut-off frequency and higher cut-off frequency is called as Bandwidth. BW = Lower cut-off frequency Higher cut-off frequency. On frequency response of amplifier bandwidth is shown as, b) Draw the circuit diagram of single stage CE amplifier with voltage divider biasing and explain its operation. Ans: Diagram: Explanation: Page 18 of 26
i) Biasing Circuit: the resistances R 1, R 2 and R E form the biasing and stabilization circuit. So that, ii) iii) iv) proper operating point is established. Input capacitor C in : Capacitor Cin is used to couple the signal to the base of the transistor. Capacitor C in allows only AC signal to flow and blocks DC also isolates the signal from R 2. Emitter capacitor C E : An emitter bypass capacitor C E is used to provide low reactance path to the amplified AC signal. If it is not used, then amplified AC signal flowing through R E will come a voltage drop across it, thereby reducing the output voltage. Coupling Capacitor C C : the coupling capacitor C C couples one stage of amplification to the next stage. This capacitor isolates the DC of one stage from the next stage but allows the passage of AC signal. c) Draw circuit diagram of RC coupled amplifier. State its advantages and disadvantages. Ans: Diagram: Advantages: [Note: Any two points each having 1 mark] 1. It has excellent frequency response. 2. It has low cast since it employs resistors and capacitors which are cheap. 3. The circuit is very compact, small and light in weight. Disadvantages: [Note: Any two points each having 1 mark] 1. The RC coupled amplifier have low voltage and power gain. 2. They have tendency to become noisy with moist climates. 3. Impedance matching is poor. Page 19 of 26
d) State the factors to be considered while designing circuits for a good transistor amplifier. Ans: [Note: Considered any four points, each point 1 mark ] Factors to be considered while designing circuits for a good transistor amplifier: 1. Voltage gain of the amplifier should be high. 2. Bandwidth of the transistor amplifier should be high i.e. it should good frequency response. 3. The design circuit should have high input impedance. 4. The design circuit should have high output impedance. 5. Operating point (Q point) of the designed circuit should not be affected by temperature i.e. circuit should have good temperature stability. 6. Circuit should be less noisy, compact and light in weight. e) Compare conductors, semiconductors and insulators based on energy bands dig temperature coefficient for resistancivity. Ans: Points Conductor Semiconductor Insulator Temperature coefficient of resistivity. Conductor are having positive temperature coefficient of resistance i.e. when temperature increases conductivity increases. Semiconductors have negative temperature coefficient of resistance i.e. resistivity of semiconductor decreases with increases in Insulators have negative temperature coefficient of resistivity i.e. the resistivity of insulator decreases with increase in temperature. temperature. Energy band diagram Page 20 of 26
f) Draw a neat sketch showing constructional details of NPN transistor. Which region is heavily doped? Why? Ans: Constructional details of NPN transistor: Heavily doped region: In NPN transistor Emitter region is heavily doped. Reason: When biasing supply is connected to NPN transistor, Emitter-base is forward biased. And Emitter region supplies the charge carrier or emitters the electrons to the other two regions. Therefore Emitter region is heavily doped in NPN transistor. Q6. Attempt any four of the following 16M a) How the power amplifiers are classified? State the maximum efficiency of each. Ans: Classification of power amplifier: Power amplifier Page 21 of 26
Class A Class B Class C Class AB Class B push pull Power amplifier Power Power Amplifier Power Power Amplifier Amplifier Amplifier Maximum efficiency of each amplifier: i) Class A 50% ii) Class B 78.5% iii) Class C 100% iv) Class AB - 100% v) Class B push pull 78.5% b) Compare Class A, Class B and Class C amplifiers on the basis of any four factors. Ans: [Note: 1 mark each point] Points Class A Class B Class C Definition Maximum efficiency Power dissipation It is an amplifier in which the transistor bias and amplitude of the input signal is such that the output current flows for the complete cycle of input signal. It is an amplifier in which the transistor bias and amplitude of the input signal are such that the output current flows for only one cycle of input signal. It is an amplifier in which the transistor bias and amplitude of input signal are such that the output current flows for less that half cycle of input signal. 50% 78.5% 100% Maximum power No power dissipation with Power dissipation is less. dissipation with no zero input signal and input signal and no maximum with full input power dissipation with signal. full input signal. Page 22 of 26
Output voltage c) Draw and explain Class AB push pull amplifier. Ans: Diagram: Explanation: Above fig shows the diagram of class AB push pull power amplifier. The circuit consist of two center tapped transformers T 1 and T 2 two identical transistor Q 1 and Q 2 resistor R and diode D. Working: When there is no input signal, both the transistor Q 1 and Q 2 are cut-off. During the positive half cycle of input signal the base of transistors Q 1 is positive and Q 2 is negative, so, transistor Q 1 conducts, while the transistor Q 2 is off. Similarly, during negative half cycle Q 1 turns off and Q 2 conducts. Thus at any instant, only one transistor in the circuit is conducting. The DC voltage developed across the diode D is connected to the bases of both the transistor through the secondary winding of the input transformer. This voltage acts as a DC bias for the Page 23 of 26
transistor become it is equal to the cut in voltage and they will conduct for complete half cycle period of the input to eliminate the cross-over distortion. d) Explain the concept of cross over distortion. List any two specifications of power amplifier. Ans: Concept of cross-over distortion: Diagram: 1 ½ M Explanation: 1 ½ M We know that transistors in class B push pull amplifier are biased at cut-off. It means that when the DC bias voltage is zero, the input signal voltage must exceed the barrier voltage before a transistor conducts. Before of this, there is a time interval between the positive and negative alternations of the input signal when neither transistor is conducting as shown in above fig. the resulting distortion in the output signal is quite common and is called cross-over distortion. Specifications of power amplifier: [Note: Any two can be considered] Specifications of any power amplifier are listed based on following points: 1. Maximum efficiency 2. Power dissipation capability 3. Distortion e) Draw frequency response curve of two stage transformer coupled amplifier and describe the same. Ans: frequency response of transformer coupled amplifier: Page 24 of 26
Explanation: Frequency response of transformer coupled amplifier is shown in above fig. it is clear that the frequency response is poor. i.e. gain is constant only over a small range of frequency. At low frequencies, the reactance of primary begins to fall, resulting in decreased gain. At high frequency, the capacitance between turns of windings acts as a bypass condenser to reduce the output voltage and hence gain. f) Derive the relation between and. Ans: A simple relation exist between and. This can be derived as: 4M. (1) Now, I E = I B + I C. (2) Or Or I E = I B + I C I B = I E - I C Substituting the value of I B in expression (1) we get. (3) Dividing the numerator and denominator of R.H.S. of expression (3) by I E, we get, Page 25 of 26
since Page 26 of 26