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Electronics Engineering Programme (ESEP) 'Model Question Paper' with 'Synoptic Answer Key' TES041: Basic Electronics-1 Maximum Allowed Time: 180 Minutes Instructions for the students 1. 2. 3. 4. 5. All questions are compulsory. Long Answer type Question (LAQ) is a supply type question of 20 marks, which requires typical answer of about 60-80 lines in about 32-40 minutes. Short Answer type Question (SAQ) is a supply type question of 05 marks, which requires typical answer of about 15-20 lines
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    Electronics Engineering Programme (ESEP)'Model Question Paper' with 'Synoptic Answer Key'TES041: Basic Electronics-1 Maximum Allowed Time : 180 Minutes Maximum Marks : 100 Instructions for the students 1.   All questions are compulsory.2.   Long Answer type Question (LAQ) is a supply type question of 20 marks, which requires typical answer of about 60-80 lines in about 32-40 minutes.3.   Short Answer type Question (SAQ) is a supply type question of 05 marks, which requires typical answerof about 15-20 lines in about 08-10 minutes.4.   Use of non-programmable type of scientific calculator is allowed.5.   Draw neat diagrams wherever necessary.Q.N. Question (Q) Marks Long Answer type Questions (LAQs)  (A) Define Thevenin voltage and Thevenin resistance. For the following circuit diagram, calculateThevenin’s voltage and Thevenin resistance.(B) Define forward biasing and reverse biasing. What is the surface leakage current?(C) A Silicon diode has a saturation current of 5 nA at 25ºC. What is the saturation current at 100ºC?Synoptic Answer KeySN Key Marks(A)Thevenin voltage (V th ): The voltage across the load terminals when the load resistoris open.2Thevenin resistance (R th ): The resistance that an ohmmeter measures across theload terminals.2V th = 24v (from circuit diagram) 201R th = k k k k  663634 =+×+ (from circuit diagram)4(B)FB: The negative battery terminal connected to the n-type material and positiveconnected to p-type material, this connection is called forward bias.2RB: The positive battery terminal connected to the n-type material and negativeconnected to p-type material, this connection is called forward bias.202Surface leakage current:- reverse current on the surface of the crystal.1(C) ∆ T=100º C – 25º C = 75º C1Is = (2 7 ) (5 nA) = 640 nA 2103Is = (1.07 5 ) (640 nA) = 898 nA 2202 (a) Using second approximation, calculates the load voltage, load current, and diode power for thefollowing circuit diagram.20   (b) Write short note on Surface-Mount Diode (SM).(c) Explain operation of Bridge Rectifier with neat circuit diagram. Draw waveforms across loadresistor R L .Synoptic Answer KeySN Key Marks(a) V V V V   L 3.97.010 =−= Load Voltage1 mAK V  I   L 3.913.9 == Load Current201 mW mAP  D 51.6)3.9)(7.0( == Diode Power2(b)SM: Surface-mount diode can be found anywhere there is a need for diodeapplication.1SM diode are small, efficient, and relatively easy to test, remove on the PCB 1Two SM package: SM & SOT (small outline transistor) 1SM has two L-bend leads and a colored band on one end to indicate cathode 102Larger the surface area, larger the current rating. 11(c)1] The bridge rectifier is similar to a full-wave rectifier because it produces a full-wave output voltage.42] Diode D1 and D2 conducts on the positive half cycle, Diode D2 and D4 conductson the negative half-cycle13] As a result, the rectified load current flows during both half cycles. 14] During the both cycles, the load voltage has the same polarity and the loadcurrent is in the same direction.1035] The circuit has changed the ac input voltage to the pulsating dc output voltage. 16] Neat and labeled waveform across load resistor R L 23 (a) For the following figure, calculate dc load voltage and ripple? 20   (b) Explain operation of clipper with neat diagram and draw its waveforms.Synoptic Answer KeySN Key Marks(a)The rms secondary voltage, V V V  2451202 ==  2The peak secondary voltage, V V V   p 34707.024 ==  2V L = 34 V 2Ripple, mAK  RV  I   L L L 8.6534 ===  201 PPPP R V V uF  HzmAV  1.113.1 )100)(60( 8.6 ≈==  202 (b)Clipper: It is a circuit that removes either positive or negative parts of awaveform.14Positive Clipper: The circuit removes all the positive parts of the input signal. 1During the positive half cycle, the diode turns on and looks like a short across theoutput terminals.1On the negative half cycle, the diode is open. So negative half cycle appears acrossoutput.1The series resistor is much smaller than load resistor. 1This is why the negative output peak is shown as -Vp 1   (a) What would be the collector-emitter voltage for the figure shown below using i) Ideal transistor ii)Second approximation iii) Using V BE : 1 V?(b) What would be the voltage between the collector and ground? Also find voltage between collectorand emitter?Synoptic Answer KeySN Key Marks01 (a)In this example we have to compare the three approximations for the case of lowsupply voltage.11] With ideal transitor:  AK V  I   B µ  6.104705 ==  1 mA A I  C  06.1)6.10(100 == µ   1 V K V V  CE  2.11)6.3)(06.1(15 =−=  12] With the second approximation:  AK V  I   B µ  15.94703.4 ==  1 mA A I  C  915.0)15.9(100 == µ   1 V K V V  CE  7.11)6.3)(915.0(15 =−=  143] With the measured V BE :  AK V  I   B µ  51.84700.4 ==  120 mA A I  C  851.0)51.8(100 == µ   1 V K V V  CE  9.11)6.3)(851.0(15 =−=  102 (b) V V V V   E  3.47.05 =−=  1 mAK V  I   E  3.413.4 ==  1 V K mA R I  C C  6.8)2)(3.4( ==  1
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