Digi ana total 2012

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Exercises in Electronics with solutions

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Digi ana total 2012

  1. 1. Courses In Electrical Engineering Volume II ELECTRICAL, DIGITAL AND INDUSTRIAL CIRCUITS EXAM QUESTIONS WITH SOLUTIONS (2012 academic year) By Jean-Paul NGOUNE DIPET I (Electrotechnics), DIPET II (Electrotechnics) M.Sc. (Electrical Engineering) Teacher in the Electrical Department, GTHS KUMBO, Cameroon.Exam questions with solutions_2012_Jean-Paul NGOUNE 1
  2. 2. ForewordThis is a compilation of some exam questions that I gave to my students during thisacademic year. They are accompanied by solutions proposed by me. I will be delightif this book can be of any use for you. I will also be very happy to receive any critic orsuggestion from you. I dedicate this book to my students of Class 6, ElectricalTechnology, GTHS Kumbo, 2012 batch. They are a bit stubborn, but I like to teachthem. May you be blessed as you are using this book.NGOUNE Jean-Paul.17 May 2012.Exam questions with solutions_2012_Jean-Paul NGOUNE 2
  3. 3. AcknowledgementMost of the questions treated in this book are “Probatoire Technique” past questionsproposed by the Cameroon General Certificate of Education Board (GCEB) and the“Office du Baccalaureat du Cameroun” (OBC).Exam questions with solutions_2012_Jean-Paul NGOUNE 3
  4. 4. ContentsItem PageForeword 2Acknowledgment 3Contents 4First sequence exam with solution 5Second sequence exam with solution 13Third sequence exam with solution 30About the author 46Exam questions with solutions_2012_Jean-Paul NGOUNE 4
  5. 5. Courses In Electrical Engineering Volume II DIGITAL ELECTRONICS FIRST SEQUENCE EXAM WITH SOLUTION By Jean-Paul NGOUNE DIPET I (Electrotechnics), DIPET II (Electrotechnics) M.Sc. (Electrical Engineering) Teacher in the Electrical Department, GTHS KUMBO, Cameroon.Exam questions with solutions_2012_Jean-Paul NGOUNE 5
  6. 6. REPUBLIC OF CAMEROON FIRST SEQUENCE EXAMPeace – Work – Fatherland Class: F36 ……………GTHS KUMBO/ ELECT DPT Option: Electrotechnology Duration: 2H30 Coefficient: 4 DIGITAL CIRCUITS No document is allowed except the one given to the candidates by the examiners I TECHNOLOGY 1.1 Give the meaning of the following abbreviations: TTL, CMOS, SSI, LSI,VLSI. 1.2 Give the rated voltage used to supply TTL integrated circuits. 1.3 Give the rated voltage used to supply CMOS integrated circuits. 1.4 Give the difference between digital representation and analogue representation. 1.5 Give two examples of digital apparatus and two examples of analogue apparatus. 1.6 Define the following notions used in the field of integrated circuits: a. Noise immunity; b. Celerity; c. Integration scale. II DIGITAL CIRCUITS Exercise 1: Numeration systems and codes. The information stored in a register of a ROM is given as follows: X = F8DA. 1. What is the meaning of ROM? 2. What is the numeration system used to codify that information? 3. Convert X into binary. 4. Convert X into octal. 5. Convert X into decimal. 6. X is made up of how many bits? 7. Knowing that one byte = 8 bits, Give the length of the memory word X in terms of bytes. Exam questions with solutions_2012_Jean-Paul NGOUNE 6
  7. 7. Exercise 2: Logic gates.The following figure is a digital circuit having four inputs A, B, C, D and one output X.A XBCD 1. Determine the expression of the output X. 2. Draw the truth table of the digital circuit. 3. Draw the logic circuit above using exclusively 2 input AND gates, 2 input OR gates and 2 input NAND gates. 4. Knowing that the IC 4081 is a quad 2-input AND gate, the IC 4011 is a quad 2- input NAND gate and the IC 4071 is a quad 2-input OR gate, determine the number of IC 4081,IC 4011 and IC 4071 that should be used to realise the digital circuit above.Exercise 3: Realisation of gate circuits.Realise the logic circuit corresponding to each of the following expressions: X A.B(C D) Y A B CD E BC D Z A B PQ CD Proposed by Mr. NGOUNE Jean-Paul, PLET Electrotechnics, GTHS KUMBO.Exam questions with solutions_2012_Jean-Paul NGOUNE 7
  8. 8. PROPOSITION OF SOLUTIONI TECHNOLOGY1.1 Meaning of the abbreviations:TTL: Transistor Transistor Logic.CMOS: Complementary Metal Oxide Semiconductor.SSI: Small Scale Integration.LSI: Large Scale Integration.VLSI: Very Large Scale Integration.MSI: Medium Scale Integration.1.2 Rated voltage for the supply of TTL integrated circuit: 5V+/-0.25V.1.3 Rated voltage used for the supply of CMOS IC: 5V, 15V, 18V.1.4 Difference between analogue representation and digital representation: Analogue representation Digital representation Infinitely divisible Discrete (Step by step) Prone to errors of precision Absolute precision1.5 Examples of digital apparatus: electronic watch, computer, mobile phone, digitalcamera… Examples of analogue apparatus: radio, oscilloscope, some model of TV,analogue multimeter.1.6 Definitions:Noise immunity: It is the ability of an integrated circuit not to be disturbed in hisfunctioning by an external signal (electromagnetic signal). A noise can be defined asa signal that disturbs the useful signal of being well treated by an electronic device.Celerity: It is the speed at which electrical information are being treated by anintegrated circuit.Integration scale: it is a range that informs on the amount of transistors used in themanufacture of an integrated circuit. There are many integration scales (SSI, MSI,LSI, VLSI, ULSI).Exam questions with solutions_2012_Jean-Paul NGOUNE 8
  9. 9. II DIGITAL CIRCUITSExercise 1: Numeration systems and codes.The information stored in a ROM is given as follows: X = F8DA. 1. ROM stands for Read Only Memory. It is a type of memory in which data, once they are written, can only be read. 2. The numeration system used to codify that information is the hexadecimal numeration system. 3. Conversion of X into binary: X = 1111100011011010 2. 4. Conversion of X into octal: X = 1743328. 5. Conversion of X into decimal: X = 6370610. 6. X is made up of 16 bits. 7. X = 2Bytes.Exercise 2: Logic gates.Let us consider the following digital circuit:A XBCD 1. Expression of the output X:X AC AC B B C D 2. Truth table of the digital circuit:Exam questions with solutions_2012_Jean-Paul NGOUNE 9
  10. 10. A B C D X 0 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 1 0 1 0 0 1 0 1 0 1 1 0 1 1 0 1 0 1 1 1 1 1 0 0 0 0 1 0 0 1 1 1 0 1 0 1 1 0 1 1 1 1 1 0 0 1 1 1 0 1 1 1 1 1 0 1 1 1 1 1 1 3. Let us draw the circuit using exclusively: 2 input AND gates, 2 input OR gates and 2 input NAND gates:Using gate universality principle, we can convert the logic gates used in the designingof the logic circuit into those required. A /A = A /A A A B = B C A+B+C C A A B = B C ABC CExam questions with solutions_2012_Jean-Paul NGOUNE 10
  11. 11. Then the logic circuit can be redrawn as follows:A XBCD 4. Number of integrated circuit of each type to be used:  Number of AND gates in the circuit: 3; therefore 1 IC 4081is sufficient (one IC contains 4 gates).  Number of NAND gates in the circuit: 2; therefore 1 IC 4011 is sufficient (one IC contains 4 gates).  Number of OR gates in the circuit: 4; therefore 1 IC 4071 is sufficient (one IC contains 4 gates).So, with 1 IC 4081, 1 IC 4011 and 1 IC 4070 the logic circuit can be designed.Exercise 3: Realisation of gate circuits:Let us realise the logic circuits corresponding to each of the following equations: X A.B(C D) Y A B CD E BC D Z A B PQ CDExam questions with solutions_2012_Jean-Paul NGOUNE 11
  12. 12. A B C D XA B C D E YA B C D P Q ZExam questions with solutions_2012_Jean-Paul NGOUNE 12
  13. 13. Courses In Electrical Engineering Volume II ELECTRICAL, DIGITAL AND INDUSTRIAL CIRCUITS SECOND SEQUENCE EXAM WITH SOLUTION By Jean-Paul NGOUNE DIPET I (Electrotechnics), DIPET II (Electrotechnics) M.Sc. (Electrical Engineering) Teacher in the Electrical Department, GTHS KUMBO, Cameroon.Exam questions with solutions_2012_Jean-Paul NGOUNE 13
  14. 14. REPUBLIC OF CAMEROON SECOND SEQUENCE EXAMPeace – Work – Fatherland Class: F36 ……………GTHS KUMBO/ ELECT DPT Option: Electrotechnology Duration: 04H Coefficient: 4 Written paper ELECTRICAL, DIGITAL AND INDUSTRIAL CIRCUITS No document is allowed except the one given to the candidates by the examiners. SECTION ONE : TECHNOLOGY 1.1 Give the value of a resistance having the following colour code: red-orange- yellow-gold. 1.2 Give the meaning of the following abbreviations: TTL, CMOS, JFET, LSI, USB, and EPROM. 1.3 Give two protective means against overheat of semiconductors. 1.4 Name the parameters that characterize the operating point of a bipolar transistor. 1.5 Give two parameters that determine the choice of a Zener diode. 1.6 Define the following terms used in the field of integrated circuits: a) Noise immunity, b) Celerity, c) Integration scale. 1.7 Give the difference between a diode and a thyristor. 1.8 Give the symbol and one application area of each of the following electronic components: a) Transistor, b) Junction diode, c) Zener diode. 1.9 Describe the functioning principle of a thyristor. SECTION TWO: ANALOG CIRCUITS Exercise 1: Alternating current (1). The circuit of figure 1 bellow is supplied by an ac voltage u. i R1 R1 = 220Ω ; R2 = 1kΩ ; C = 4.5µF ; f = 50Hz u(t) u(t) u (t ) 120 2 sin 100 t . R2 C 1. Determine the total impedance of the circuit. Figure1 Exam questions with solutions_2012_Jean-Paul NGOUNE 14
  15. 15. 2. Show that the expression of the current i flowing in the circuit is given by i I 2 sin 100 and determine the value of I and . 3. Show that the expression of the voltage u’(t) is given by u (t ) U 2 sin 100 t and determine the value of U’.Exercise 2: Alternating current (2)The circuit of figure 2 bellow is connected to a voltage v(t ) 220 2 sin 100 t volts. i1 R1 L i i2 C R2 Figure2Given thatR1 = 100Ω; R2 = 150Ω; L = 0.24 H and C = 16µF. 1. Determine the following complex impedances: a. Z1 for the branch (R1 + L) b. Z2 for the branch (C + R2) 2. Calculate the complex values of i1, i2 and i. 3. Draw the phasor diagram of the currenti1, i2 and i.Exercise 3: DC circuitConsider the following circuit in figure 3. A K E1 = 12V, E2 = 6V, Ro= 20Ω, R1 = 10Ω, R2 = 4Ω and R = 5Ω. 1. Determine the characteristics of the Norton R1 E2 equivalent generator seen from terminals A RO R and B when K is opened. R2 E1 2. Deduce the corresponding Thevenin,s equivalent model. 3. Calculate the value of current I in the load R and the voltage drop across it when K is B closed. Figure3Exam questions with solutions_2012_Jean-Paul NGOUNE 15
  16. 16. Exercise 4: Bipolar transistorThe two transistors of figure 4 bellow are in silicon such that V BE1 = VBE2 = 0.7V,β1=100, β2=200. The operating point is such that UEM = 5V for UAM = 20V.R1=1kΩ,R2=1kΩ, R3=10kΩ A 1. Calculate the current i1 flowing through the resistance R1. R3 2. Determine the voltage across the 10k UAM T1 resistance R3 and the current I3 B1 flowing through this resistance. 3. Neglecting IB2 with respect to I1, calculate the base current IB1of the R B2 T2 E transistor T1.UEM 4. Calculate IC2, hence, deduce IB2 R1 1k R2 and verify that IB2 is negligible with 1k respect to I1 5. Calculate the voltage across R and M the current crossing it. Hence, Figure4 determine the resistance R.SECTION THREE: DIGITAL CIRCUITSExercise 5: Multiplicator circuit.The figure 5 below shows the block diagram of an electronic circuit which acceptstwo binary numbers of two bits X1X0 and Y1Y0, and gives at the output the binarynumber Z3Z2Z1Z0 which is equal to the arithmetic product of the two input numbers.For the inputs, X0 and Y0 are the least significant bits (LSB) while for the outputs, Z3is the most significant bit (MSB). X1 Z3 X0 Multiplicator Z2 circuit Z1 Y1 Z0 Y0 Figure5Exam questions with solutions_2012_Jean-Paul NGOUNE 16
  17. 17. 1. What do you understand by the statements “Least significant bit” and “Most significant bit”? 2. Establish the truth table of the system. 3. Write the expression of each output Z3, Z2, Z1 and Z0 as function of X1, X0 Y1 andY0. 4. With the aid of Karnaugh map, simplify the output equations obtained above. 5. Draw the logic diagram of the electronic multiplicator circuit using the simplified output equations.Exercise 6: Parity detectorWe desire to realize a 3-bit parity detector of bits B1, B2 and B3. The operation is asfollows: - If 0 or 2 bits are at high logic level, the output is at the high level. - If 1 or 3 bits are at high logic, the output is at the low level. 1. Draw the corresponding truth table. 2. Give the expression of the output S in terms of B1, B2 and B3. 3. Write the expression of S using the operator exclusive OR only. 4. Draw the logic diagram of S.Exercise 7: Numeration systemLet us consider the following numbers:A = 1101011012 B=6248 C=1A716 1. Convert A into octal and hexadecimal. 2. Convert B and C into binary. 3. Calculate:X(2)=A(2)+ B(2); Y(16)=A(16)+ C(16) W(8)=A(8 ) – B(8) SUBJECT MASTER: NGOUNE Jean-Paul, PLET Electrotechnics, GTHS KUMBO.Exam questions with solutions_2012_Jean-Paul NGOUNE 17
  18. 18. PROPOSITION OF SOLUTIONSECTION ONE: TECHNOLOGY1.1 Value of a resistor having the following colour code: Red – Orange – Yellow – Gold. R = 23 x 104Ω+/- 5% = 230kΩ1.2 Meaning of the abbreviations:  TTL = Transistor Transistor Logic;  CMOS = Complementary Metal Oxide Semiconductor;  JFET = Junction Field Effect Transistor;  LSI = Large Scale Integration;  USB = Universal Serial Bus;  EPROM = Erasable Programmable Read Only Memory.1.3 Two protective means against overheat of semiconductors:  Use of heat sink (radiator)  Use of fan (ventilation)1.4 Parameters characterising the operating point of a bipolar junction transistor:  IBQ = Base current at the quiescent point;  VBEQ = VBE at the quiescent point;  ICQ = Collector current at the quiescent point;  VCEQ = VCE at the quiescent point.1.5 Parameters of choice of a Zener diode:  Zener voltage;  Reverse current; VZ  Coefficient . IZ1.6 Definition of terms:  Noise immunity: Degree of protection of an IC against noise. A noise is an undesirable signal that disturbs the functioning of an IC.  Celerity: Speed of propagation of signal through a circuit. Speed at which an electronic circuit treats information.  Integration scale: Number of components (transistors) integrated per surface unit of a chip. There are many integration scales: SSI, MSI, LSI, VLSI, VLSI.Exam questions with solutions_2012_Jean-Paul NGOUNE 18
  19. 19. 1.7 Difference between diode an thyristor: A diode is a non controlled unidirectional rectifier component, a thyristor is a controlled unidirectional rectifier component.1.8 Symbol an application of electronic components: Component Symbol Application area Bipolar Amplifier, Chopper transistor Junction Rectification diode (Non controlled) Zener diode Stabilisation1.9 Functioning principle of a thyristor: When a positive voltage is applied across a thyristor (VAK > 0), a sufficient gate current (IG > Igm) permits to trigger it on. The current keeps on passing through it even if the gate current is removed (hysteresis effect). To trigger off a thyristor, a negative voltage should be applied across it.SECTION TWO: ANALOGUE CIRCUITSExercise 1: Alternating current (1)Let us consider the following circuit: i R1 u(t) u(t) R1 = 220Ω ; R2 = 1kΩ ; C = 4.5µF R2 C f = 50Hz u (t ) 120 2 sin 100 t1. Total impedance of the circuit:ZT R1 R2 Z C ; 1 1With Z C j j j 707.714 . C 2 f 4.5 10 6 314Hence,Exam questions with solutions_2012_Jean-Paul NGOUNE 19
  20. 20. R2 Z C 1000 j 707.714 707714 90ZT R1 220 220 R2 Z C 1000 j 707.714 1225.095 35.28ZT 220 577.68 54.72 553.65 j 471.58ZT 727.26 40.42 2. Let us show that the expression of the current i flowing in the circuit is given by: i (t ) I 2 sin 100 t . U 120 0I 0.165 40.42 0.165 2 sin 100 t 40.42 ZT 727.26 40.42With I = 0.165 A and 40.42 . 3. Let us show that the voltage u’(t) is given by u (t ) U 2 sin 100 t .The circuit can be redrawn as follows: R1 u(t) u(t) Z0With Z 0 R2 Z C 577.68 54.72 UU Z 0 (Voltage divider). ZT 120 0U 577.68 54.72 95.318 14.30 727.26 40.42u (t ) 95.318 2 sin 100 t 14.30Where U’ = 95.318V and 14.30 .Exercise 2: Alternating current (2)Let us consider the following circuit: i1 R1 L i i2 C R2Exam questions with solutions_2012_Jean-Paul NGOUNE 20
  21. 21. v(t ) 220 2 sin 100 t ; R1 = 100Ω; R2 = 150Ω; L = 0.24 H, C = 16µF 1. Determination of impedances:  Branch (R1 + L)Z1 R1 jL 2 f 100 j 75.36 1.757 37  Branch (C+R2) 1 1Z2 R2 j 150 j 150 j199.044 0.8827 52.99 C 16 10 6 314 2. Complex values of the currents i1, i2 and i. V 220 0I1 1.757 37 1.403 j1.057 A Z1 125.21 37 V 220 0I2 0.8827 52.99 0.531 j 0.704 A Z2 249.235 52.99I I1 I2 1.403 j1.057 0.531 j 0.704 1.934 j 0.353 1.965 10.34 A 4. Phasor diagram.We have already the modulus and the argument of each current:I1 1.757 37 AI2 0.8827 52.99 A Hence, we obtain the following phasor diagram:I 1.965 10.343 A I2 Ref = U O I I1We notice that graphically, I1 + I2 = I.Exercise 3: DC circuits.Let us consider the following figure:E1 = 12V; E2 = 6V; Ro = 20Ω; R1 = 10Ω; R2 = 4ΩExam questions with solutions_2012_Jean-Paul NGOUNE 21
  22. 22. A K R1 E2 RO R R2 E1 B 1. Norton equivalent generator seen from terminals A and B when K is open:The circuit obtained when K is open can be redrawn as follows: A R1 E2 R0 R2 E1 BRN R0 R1 R21 1 1 1 1 1 1 8RN R1 R2 R0 10 4 20 20 20RN 2 .5 8 E1 E2 12 6IN 2 .7 A R1 R2 10 4 2. Thevenin’s equivalent model: RTH A A RN In Eth B BExam questions with solutions_2012_Jean-Paul NGOUNE 22
  23. 23. ETH RN I N 2.5 2.7 6.75V RN RTH 2.5 3. Value of the current flowing in the resistor R when K is closed:Using the Thevenin’s model obtained above, we have: R TH A R E th B ETH 6.75I 0 .9 A RTH R 2 .5 5VR R I 5 0 .9 4.5VExercise 4: Bipolar transistor.Let us consider the following figure: A R3 UAM 10k T1 B1 R B2 T2 E UEM R1 1k R2 1k MVBE1 = VBE2 = 0.7V; β1 = 100; β2 = 200. At the operating point, UEM = 5V; UAM = 20V.R1 = 1kΩ; R2 = 1kΩ; R3 = 10kΩ.1. Current I1 flowing in R1. U EM 5I1 5mA R1 1000Exam questions with solutions_2012_Jean-Paul NGOUNE 23
  24. 24. 2. Voltage across R3 and current through it.U EM VBE1 R3 I 3 U AM 0 R3 I 3 U AM U EM VBE1 20 5 0.7 14.3V 14.3 14.3I3 1.43mA R3 100003. IB2<< I1; calculation of IB1. I1 I E1 1 1 I B1IB2<< I1 I1 5 10 3 I B1 49.504 A 1 1 1014. Determination of IC2. 6 6IC 2 I3 I B1 1.43 10 49.504 10 1.3804mA 3 IC 2 1.3804 10I B2 6.902 A 2 200 I1 5 10 3 6 724.42 , so IB2 is negligible with respect to I1.I B2 6.902 10 5. Voltage across R and value of R.U EM VR VBE 2 R2 I E 2 ; But I E 2 IC 2 3VR U EM VBE 2 R2 I E 2 5 0.7 1000 1.3804 10 2.9196V VR 2.9196R 6 0.403M I B2 6.902 10SECTION THREE: DIGITAL CIRCUITSExercise 5: Multiplicator circuit. 1.  The least significant bit (LSB) of a binary number is the less weighted bit of that number (situated at the extreme right of the number).  The most significant bit of a binary number is the most weighted bit of that binary number (situated at the extreme left of the number). 2. Truth table of the system.Exam questions with solutions_2012_Jean-Paul NGOUNE 24
  25. 25. X1 X0 Y1 Y0 Z3 Z2 Z1 Z0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 1 0 1 1 0 0 0 1 0 0 1 1 1 0 0 1 1 1 0 0 0 0 0 0 0 1 0 0 1 0 0 1 0 1 0 1 0 0 1 0 0 1 0 1 1 0 1 1 0 1 1 0 0 0 0 0 0 1 1 0 1 0 0 1 1 1 1 1 0 0 1 1 0 1 1 1 1 1 0 0 1 3. Expression of each output in function of X1, X0, Y1, Y0.Z3 X 1 X 0Y1Y0Z2 X 1 X 0Y Y0 X 1 X 0Y1Y0 X 1 X 0Y1Y0Z1 X 1 X 0Y1Y0 X 1 X 0Y1Y0 X 1 X 0 Y1Y0 X 1 X 0Y1Y0 X 1 X 0 Y1Y0 X 1 X 0Y1Y0Z0 X 1 X 0 Y1Y0 X 1 X 0Y1Y0 X 1 X 0 Y1Y0 X 1 X 0Y1Y0 4. Simplification of equations using K-maps.Z3 is unchanged:Z3 X 1 X 0Y1Y0Exam questions with solutions_2012_Jean-Paul NGOUNE 25
  26. 26. Z2 Y1Y0X1X0 00 01 11 10 00 01 11 1 10 1 1 Z2 X 1Y1Y0 X 1 X 0Y1 Z1 Y1Y0X1X0 00 01 11 10 00 01 1 1 11 1 1 10 1 1 Z1 X 1Y1Y0 X 1 X 0Y0 X 1 X 0Y1 X 0Y1Y0 Z0 Y1Y0X1X0 00 01 11 10 00 01 1 1 11 1 1 10 Z0 X 0Y0 6. Logic diagram of the multiplicator circuit: Exam questions with solutions_2012_Jean-Paul NGOUNE 26
  27. 27. X1 X0 Y1 Y0 Z3 Z2 Z0 X1 X0 Y1 Y0 Z1Exercise 6: Parity detector. 1. Truth table of the parity detector. Following the description of the functioning of the circuit, the following truth table can be drawn:Exam questions with solutions_2012_Jean-Paul NGOUNE 27
  28. 28. B1 B2 B3 S 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 0 2. Expression of the output S in terms of B1, B2 and B3:S B1.B2 .B3 B1B2 B3 B1 B2 B3 B1B2 B3 3. Expression of S using the operator Exclusive OR only.From the expression above, we have:S B1 B2 .B3 B2 B3 B1 B2 .B3 B2 B3S B1 B2 B3 B1 B2 B3S B1 B2 B3 4. Logic diagram of S. B1 B2 B3 1 3 2 SExam questions with solutions_2012_Jean-Paul NGOUNE 28
  29. 29. Exercise 7: Numeration system:A = 1101011012 B=6248 C=1A7161. Conversion of A into octal and hexadecimal.A = 655(8); A = 1AD(16).2. Conversion of B and C into binary.B = 110010100(2); C = 110100111(2).3. Calculations.X(2) = 1101000001(2); Y(16) = 354(16) ; W (8) = 31(8). END ACKNOWLEDGEMENTAll the exercises solved in this document are past “Probatoire Technique”examination questions proposed by the Cameroon General Certificate of EducationBoard (CGCEB).Exam questions with solutions_2012_Jean-Paul NGOUNE 29
  30. 30. Courses In Electrical Engineering Volume II ELECTRICAL, DIGITAL AND INDUSTRIAL CIRCUITS THIRD SEQUENCE EXAM WITH SOLUTION By Jean-Paul NGOUNE DIPET I (Electrotechnics), DIPET II (Electrotechnics) M.sc. (Electrical Engineering) Teacher in the Electrical Department, GTHS KUMBO, Cameroon.Exam questions with solutions_2012_Jean-Paul NGOUNE 30
  31. 31. REPUBLIC OF CAMEROON THIRD SEQUENCE EXAMPeace – Work – Fatherland Class: F36 ……………GTHS KUMBO/ ELECT DPT Option: Electrotechnology Duration: 04H Coefficient: 4 Written paper ELECTRICAL, DIGITAL AND INDUSTRIAL CIRCUITS No document is allowed except the one given to the candidates by the examiners. SECTION ONE : TECHNOLOGY 1.1 Define: Combinatory logic circuit; sequential logic circuit, decoder, multiplexer, flip-flop. 1.2 Give the meaning of the following abbreviation: PMOS, ECL, USB, ALU, TTL. 1.3 What are the properties of a linear operational amplifier? 1.4 What are the modes of functioning of an OPMP? 1.5 The following symbol is that of the LM741 which is one of the most commonly used OPAMP. Give the name of terminals 1, 2, 3,4,5,6 and 7. 7 1 3 6 2 4 5 1.6 Consider the following table. Indicate by putting a cross in the appropriate cell, the nature of the each component (Active or passive component). Resistor Transistor Inductor Capacitor Diode Active component Passive component 1.7 What is the difference between a multiplexer and a demultiplexer? 1.8 Give two protective means against overheat of semiconductors. Exam questions with solutions_2012_Jean-Paul NGOUNE 31
  32. 32. SECTION TWO: ANALOGUE CIRCUITSExercise 1: Alternating current.Consider the circuit of figure 1 bellow. A e1 = 220V, e2 = j110V, ZL = j103Ω, ZC = -j500Ω, L Z = 103Ω. C 1. Determine the characteristics of the Norton’s Z equivalent generator seen from terminals A and B.e1 e2 2. Determine characteristics of the Thevenin’s B equivalent generator seen from terminals A Figure 1. and B. 3. Using Norton’s equivalent generator, determine the complex value of the current i flowing in the load Z. Deduce its effective value. 4. Using Thevenin’s equivalent generator, determine the complex value of the current i flowing in the load Z. Deduce its complex value.Exercise 2: DC current.The circuit of the figure 2 bellow is a voltage stabilizer. The voltage U1 varies I2 from 10V to 16 V. The Zener diode is idealU1 U2 IB with PZmax = 15mW; RP R Iz Uz = 12V. For the bipolar Uz transistor, take β = 100, VBE = 0.7V. Let R = 300Ω Figure 2. 1. Determine the maximal current IZmax of the Zener diode. 2. For U1 = 16V, determine the values of U2 and RP so that the current in the diode must be maximal. 3. Using the value of RP obtained in question 2 above, determine the maximal value of U1 for which the Zener diode is blocked (Iz = 0). 4. Using the value of RP obtained above, determine the current I2 and the voltage U2 in the following cases: a) U1 = 10V; b) U1 = 14V.Exam questions with solutions_2012_Jean-Paul NGOUNE 32
  33. 33. Exercise 3: Bipolar transistor amplifier.Consider the transistor amplifier circuit presented on the figure 3 below. For the transistor: R2 RC = 99; r = 2kΩ , C2 ICQ = 4.95mA, VBEQ = 0.7V. C1 v2 Take: Vcc = 12V, VCC RG RU R1 = 2kΩ, RC = 2kΩ, v1 R1 RU = 2kΩ, RE CEe RE = 180Ω M A. Static study: Figure 3.Determine: 1. The currents flowing through the base (IB) and the emitter (IE) of the transistor. 2. The voltage VBM between the base and the ground M. 3. The current IP flowing in the resistor R1. 4. The value of the resistance R2. B. Dynamic study: 1. Give the name and the role of capacitors C1, C2 and C3. 2. Draw a.c. equivalent circuit of the amplifier. 3. Determine the input resistance and the output resistance of the amplifier. 4. Calculate the voltage amplification factor.Exercise 4: Operational amplifier.The OPAMPs of figure 4 bellow are ideal. We have R = 10Ω, R1 = 4Ω, R +Vcc R2 = 20Ω, E =100mV and Ve 1 V Vcc = 12V 2 R2 -Vcc Vs 1. Give the operating R1 E modes of the OPAMPs 1 and 2. Figure 4.2. The voltage Ve is a sinusoidal expressed as: Ve 20 cos100 t (mV).Exam questions with solutions_2012_Jean-Paul NGOUNE 33
  34. 34. a. Determine the expression of the of the output voltage V of the OPAMP1. b. Represent in terms of time the voltages Ve and V.3. Draw the waveform of the voltage Vs at the output of OPAMP2 knowing that E is aDC source.SECTION THREE: DIGITAL CIRCUITS. 1. Solve the following operations using 2’s complement: a) 11100002 – 1101112; b) 1001111012 – 110111102; c) 100000002 – 11111112. 2. The figure 5 bellow represents the circuit of a full adder, where A1 and B1 are the in put variables. R1 is the carry while So and Ro are the sum and the reminder respectively. 2.1 Complete the truth table bellow. A1 B1 R1 So Ro 0 0 0 0 0 1 0 1 0 A1 So 0 1 1 B1 Full Adder Ro 1 0 0 R1 1 0 1 1 1 0 Figure 5. 1 1 1 2.2 Simplify the expressions of So and Ro using Boolean algebra method. 2.3 Draw the logigram of this full adder using logic gates. 3. At the input of a decoder, one can place 64 different combinations. Determine: a) The number of ways at the input of this decoder, b) The number of ways at the output of this decoder. SUBJECT MASTER: NGOUNE Jean-Paul, PLET Electrotechnics, GTHS KUMBO.Exam questions with solutions_2012_Jean-Paul NGOUNE 34
  35. 35. ACKNOWLEDGEMENTAll the exercises solved in this document are past “Probatoire Technique”examination questions proposed by the Cameroon General Certificate of EducationBoard (CGCEB) and the “Office du Baccalaureat du Cameroun” (OBC).SECTION ONE: TECHNOLOGY1.1 Definition of terms:  Combinatory logic circuit: It is a logic circuit whose outputs depend only on the combination of its inputs logic states.  Sequential logic circuit: It is a logic circuit whose outputs depend on previous inputs as well as present ones. Thus a sequential logic circuit has a memory.  Decoder: It is a combinatory circuit which functions in such a way that for a given input address, only one of its outputs is activated.  Multiplexer: It is a combinatory logic circuit which permits to direct towards single output information coming from many inputs.  Flip-flop: It is a sequential logic circuit which is able to memorise one bit of information (elementary memory).1.2 Meaning of abbreviations:  PMOS: P-type channel metal oxide semiconductor.  ECL: Emitter coupled logic.  USB: Universal serial bus.  ALU: Arithmetic logic unit.  TTL: Transistor transistor logic.1.3 Properties of a linear operational amplifier:  Infinite voltage gain.  Infinite input impedance.  Zero output resistance.  Zero offsets (voltage and current).  Zero bias current.  Infinite common mode rejection ratio (CMRR).1.4 The modes of functioning of an OPAMP are :  Linear mode,Exam questions with solutions_2012_Jean-Paul NGOUNE 35
  36. 36.  Saturation mode.1.5 Names of the terminals of the OPAMP LM741. 1. Offset null 7 1 2. Inverting input 3 6 3. Non-inverting input 2 4. Negative supply 5. Offset null 4 5 6. Output 7. Positive supply.1.6 Nature of the components Resistor Transistor Inductor Capacitor DiodeActivecomponentPassivecomponent1.7 Difference between multiplexer and demultiplexer.A multiplexer directs towards one output information coming from many inputsmeanwhile a demultiplexer directs towards many outputs (one amongst thoseoutputs) information coming from one input. Thus, the demultiplexer is the reverse ordual circuit of the multiplexer.1.8 Two protective means against overheat of semiconductors:  Use of fan  Use of radiator or heat sinkSECTION TWO: ANALOGUE CIRCUITS.Exercise 1: Alternating current.Let us consider the following network. A L C e1 = 220V, e2 = j110V, ZL = j103Ω, ZC = -j500Ω, Z Z = 103Ω.e1 e2 BExam questions with solutions_2012_Jean-Paul NGOUNE 36
  37. 37. 1. Norton equivalent generator seen from terminals A and B.The circuit above can be transformed as follows: A AI1 ZL Z I2 ZC Ieq ZEQ B B e1 220 I1 j 0.22 A ZL j1000 e2 j110 I2 0.22 AWith ZC j 500 I eq I1 I2 0.22 1 j A IN Z L ZC j1000 j 500 Z eq Z L ZC j1000 ZN Z L ZC j1000 j 500Hence the Norton generator can be represented as follows A IN ZN B2. Thevenin’s equivalent generator seen from terminals A and B.The circuit can be redrawn as follows: A Eth L C e1 e2 B e1.Z C e2.Z L 220 j 500 j110 j1000 j 500 220 j 220ET 220 j 220V Zc Z L j 500 j1000 j 500ZT ZN j1000Exam questions with solutions_2012_Jean-Paul NGOUNE 37
  38. 38. Hence the Thevenin’s equivalent generator can be drawn as follows. A Z th E th B 3. Determination of the current flowing in the impedance Z, using Norton’s equivalent generator. IN A I IN ZN Z BUsing current divider theorem, we can write: I N .Z N 0.22 1 j j1000 j1000 0.22 1 j j 1 I 0.22 0.22 A 0.22 180 ZN Z j1000 1000 1000 1 j 1 jThe effective value of the current can therefore be deduced: I = 0.22A. 4. Determination of the current flowing in the load Z using Thevenin’s equivalent generator. A I Z th Z E th B ET 220 j 220 220 1 jI 0.22 A 0.22 180 A ZT Z j1000 1000 1000 1 jThe effective value of the current can therefore be deduced: I = 0.22A.Exam questions with solutions_2012_Jean-Paul NGOUNE 38
  39. 39. Exercise 2: DC current.Let us consider the following circuit. I2U1 U2 RP IB R Iz Uz 1. Maximal current IZmax of the Zener diode. 3 PZ max 15 10PZùax I Z max U Z I Z max 0.00125 A 1.25mA . U Z max 12 2. U1=16V. Let us determine U2 and RP so that the diode current will be maximal.U2 UZ VBE 12 0.7 11.3V U1 U ZRP I RPBut I RP IB I Z max . On the other hand we have: I2 U2I2 IE 1 IB IB . By replacing in the initial equation, we 1 R 1have: U1 U Z U1 U Z 16 12RP 2460 U2 U2 3 11.3 I Z max I Z max 1.5 10 R 1 R 300 100 3. Maximal value of U1 for which IZ = 0.( Then IRP =IB) U2 11.3U1 RP I B U Z RP UZ 2460 12 12.92V R 100 300 4. For U1 = 10V, we have U1<Uz, then IRP = 0, and UZ = 10. Hence,U2 10 0.7 9.3V U2 9 .3 The I 2 0.031 31mA R 300 For U1 = 14V. Then; Uz = 12V and U2=11.3V U2 11.3I2 37.7 mA R 300Exam questions with solutions_2012_Jean-Paul NGOUNE 39
  40. 40. Exercise 3: Bipolar transistor amplifier.Let us consider the following transistor amplifier circuit. For the transistor: R2 RC = 99; r = 2kΩ , C2 ICQ = 4.95mA, VBEQ = 0.7V. C1 v2 Take: Vcc = 12V, VCC RG RU R1 = 2kΩ, RC = 2kΩ, v1 R1 RU = 2kΩ, RE CEe RE = 180Ω M A. Static study. 1. Determination of base and emitter current. I CQ 4.95 10 3IB 5 10 5 50 A 99IE I CQ I B 50 A 4.95mA 5mA 2. Voltage VBM between the base and the ground. 3VBM VBE RE I E 0.7 180 5 10 1.6V . 3. Determination of the current flowing in the resistor R1. VBM 1 .6IP 0.8mA R1 2000 4. Determination of the value of the resistance R2. VCC VBE RE I EVCC R2 I B IP VBE RE I E R2 IB IP 3 12 0.7 180 5 10 12 1.6 12.235k 5 10 5 0.8 10 3 0.85 10 3 B. Dynamic study. 1. Name of the capacitors C1, C2 and C3.  C1 and C2 are coupling capacitors.  CE is a bypass capacitor.Exam questions with solutions_2012_Jean-Paul NGOUNE 40
  41. 41. 2. ac equivalent circuit of the amplifier. i1 ib iC i2 B C v2 RG v1 RC RU R1 R2 r BiBe E 3. Input and output resistances v1v1 R1 // R2 // r i1 R1 // R2 // r Ri 0.924k i1For the output resistance, the input source e should be rendered inactive (replacedby a short circuit). Then we have,Rout RC 2k 4. Voltage amplification factor. v2 RC // Ru iB RC // RuBy definition, Av v1 riB rThe negative sign shows that the input and the output voltages are in opposition ofphase.Exercise 4: Operational amplifier.The OPAMPs of figure 4 bellow are ideal. R +Vcc Ve 1 V We have R = 10Ω, R1 = 4Ω, 2 R2 -Vcc Vs R2 = 20Ω, E =100mV and R1 E Vcc = 12V 1. The OPAMP1 operates in linear mode (because of its negative feedback); The OPAMP2 operates as a comparator (Saturation mode).Exam questions with solutions_2012_Jean-Paul NGOUNE 41
  42. 42. 2. a. Expression of the output voltage V.Using the voltage divider theorem, we can write (The voltage Ve is directly appliedacross R1since there is no drop across R). V R1 R2 R2Ve R1 V Ve 1 Ve R1 R2 R1 R1 b. Representation of Ve and V in terms of time. 4 20V Ve 6Ve 6 20 cos100 t 120 cos100 t 4 V Ve 120 100 80 60 40 20 0 t T/4 T/2 3T/4 T 5T/4 6T/4 7T/4 2T 9T/4 -120 3. Waveform of the voltage Vs.The OPAMP2 functions as a comparator. So we have:If V<E then Vs = -VccIf V>E then Vs = +Vcc, with E = 100mV.Exam questions with solutions_2012_Jean-Paul NGOUNE 42
  43. 43. 12V Vs V(mV) Ve(mV) 120 100 80 60 40 20 0 t T/4 T/2 3T/4 T 5T/4 6T/4 7T/4 2T 9T/4 -120 -12VSECTION THREE: DIGITAL CIRCUITS 1. Let us solve the following operations using 2’s complement. a) 11100002 - 110111211100002 - 1101112 = 11100002 +2’s compl(1101112)1’scompl(0110111) = 1001000 ( an implied zero has been added in front of thenumber so that the two numbers should have the same number of digits)2’s compl(110111) = 1001000 + 1 = 1001001Hence,11100002 - 1101112 = 1110000 + 1001001 = 10111001The first 1 is rejected. Hence the result of the operation is 111001.Using the same principle, we obtain the following results for the other operations. b) 1001111012 - 110111102 = 10111112 c) 100000002 -11111112 = 12Exam questions with solutions_2012_Jean-Paul NGOUNE 43
  44. 44. 2. Full adder. 2.1 Truth table. A B Ci S Co 0 0 0 0 0 0 0 1 1 0 0 1 0 1 0 0 1 1 0 1 1 0 0 1 0 1 0 1 0 1 1 1 0 0 1 1 1 1 1 1 2.2 Simplification of So and Ro using Boolean algebra method.S A.BCi ABCi AB.Ci ABCi A BCi BCi A B.Ci BCi AB Cin AB CiLet X = B CiS AX AX A X A B Ci S A B CiCo A.BCi ABCi ABCi ABCi The expression will not change if one of the elements of the sum of products isduplicated (After the Boolean additive identity according to which A + A = A, A beinga Boolean variable). So we will duplicate the product ABCi three times in order tosimplify the expression easily.Co A.BCi ABCi ABCi ABCi ABCi ABCi BCi A A ACi B B AB Ci Ci BCi ACi ABCo BCi ACi ABExam questions with solutions_2012_Jean-Paul NGOUNE 44
  45. 45. 2.3 Logigram of the full adder using logic gates.A B Ci Ri S Ro 3. At the input of a decoder, we can place 64 different combinations a) Number of ways at the input of the decoder.We know that, with n ways or inputs, we can have up to 2n different inputcombinations.64 2n n 6 b) Number of ways at the output of the decoder.The number of outputs is equal to the number of inputs combinations, since eachinput combination should permit to select only one output. Hence the number of waysat the output of the decoder is 64.Exam questions with solutions_2012_Jean-Paul NGOUNE 45
  46. 46. ABOUT THE AUTHOR NGOUNE Jean-Paul was born in Foreké-Dschang, Republic of Cameroon in 1984. He is a holder of a Master Degree in electrical engineering, obtained in 2010 in the Doctorate School of the University of Douala (UFD-PSI). He is also a holder of a DIPET II and a DIPET I respectively obtained in 2009 and 2007 in the Advanced Teacher Training College for Technical Education (ENSET de Douala). He is currently a permanent teacher of Electrical Engineering at the Government Technical High School of Kumbo, North-West region, Cameroon. His domain of research concerns the improvement of energy conversion techniques for an efficient generation of electrical energy from renewable sources (especially wind and solar energy, small and medium scale hydropower) and digital designing using FPDs. The author is looking for a Ph.D program in his domain of research (he has not yet found it). Any suggestion for this issue will be warmly welcome. NGOUNE Jean-Paul, M.Sc., PLET. P.O. Box: 102 NSO, Kumbo, Cameroon. Phone: (+237) 7506 2458. Email : jngoune@yahoo.fr Web site: www.scribd.com/jngouneExam questions with solutions_2012_Jean-Paul NGOUNE 46

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