ELECTRICAL AND ELECTRONIC CIRCUITS E030. Professor Dr Ir Mostafa Afifi

Transcription

1 ELECTRICAL AND ELECTRONIC CIRCUITS E030 Professor Dr Ir Mostafa Afifi

2 INTRODUCTION Lectures are 3 Hrs / week (one week two consecutive periods 2 & 3 of Tuesday, and the other week only in the first period S7) Two Hours of exercises every week. 3M3 & 3M4 on Wednesday 4 th period in S6 3M & 3M2 on Wednesday 5 th period in S6 One Hour of Laboratories every week All on Thursday 3 rd period for (3M / 3M2) and the 4 th period for (3M3 / 3M4)

3 Grading Final exam is 60% Exercise works and Quizzes 0% Laboratory Works 20% Attendance 5% Midterm Exam 5%

4 INTRODUCTION The handling and operating electric circuits by mechanical engineers is proper for the advance of civil prosperity. In all aspects of our life electromechanical integration is now very evident. Mechanical engineers are the first to use electric motors to operate and manipulate engine operations. Also they are the first to apply digital handling of information using rotating shafts and timing activities.

5 Many applications of Electrical Engineering in Automobiles For Safety: Antiskid brakes, inflatable restraints, Collision warning and avoidance, Blind-zone detections, infrared night vision systems, Heads-up displays and accident notifications. For Communications and entertainment: AM/FM Radios, Digital audio Broadcastings, CD / Tape players, Cellular phones, Computer / . For convenience: Electronic navigation, Personalized seat/mirror/radio settings, electronic door locks. For Emission, performance & Fuel economy: Vehicle instrumentation, electronic ignition, Tire inflation sensors, computerized performance evaluation and maintenance scheduling and adaptable suspension systems. For alternative propulsion systems: Electric vehicles, advanced batteries and Hybrid vehicles.

6 Output of a transducer for a Knocking Engine

7 Other important applications are in Household Appliances (Mechatronics) with intimate integration and harmonious blending of many different technologies Keypads for operator control Sensors Electronic displays Microcomputer Chips Electronic digital switches Heating Elements Motors

8 Major application subdivisions Communication systems: with major modern Information technology using Internet broadcastings in ground and Satellite systems (including roadside sensors and GPS global positioning Systems ) Computer Systems: Process and store of information in digital formats, Examples of wide scale applications is the control systems of modern cars, including more than 50 microcomputers in one car Control Systems: gathering information with sensors and processing for physical control, examples are the heating and air cooling in buildings, temperature and flow rates in chemical processes and control of motion in tall buildings Electromagnetism: using microwaves in ovens, manufacturing of plywood, cell phones and satellite applications. Electronics: applied in every field of engineering and science, many useful circuits are planned for this course to introduce and build major engineering capabilities. Power systems: for interchanging conversion of mechanical and electrical energies with applications in transportation and increase of efficiency. Signal Processing: where transducers are used to convert physical phenomenon to electrical signals that can be processed for desired performance by computers. Major example of this is the control of ignition in cars to optimise performance and avoid damaging knock of engines.

9 The Need to Study Electrical Systems applications Professional Engineers (PE): Need to pass electrical Engineering for Qualification Fundamentals of Engineering (FE): exams are needed to modernize productivity in different disciplines, especially in concern of mechatronics. Electrical Engineering is interwoven with nearly all design projects in other fields of engineering Those who strict their knowledge to their own fields are destined to be directed by others.

10 An Overview of Electric Circuits Fluid Flow analogy The Battery is analogous to a pump Charges are analogous to fluid Wire conductors correspond to pipes Voltage correspond to pressure Electrical Resistance is analogous to constriction in a fluid system resulting in turbulence and conversion of energy to heat Current is a measure of flow of charge through the cross section of the element Voltage is measured between the ends of the circuit element or between any other two points in the circuit

11 Basic Definitions of Electric Quantities Electric Charge "Q" )the mechanical force "for Q and Q2 at distance R" F = QQ2/εR 2 (. Electric Current "I" )I=dQ/dt(. Electric Potential "V"

12 Head Light Electric circuit in the CAR

13 An Electric circuit Consisting of a source, resistance, capacity and Inductance

14 Current is the time rate of charge flow in a wire i( t) dq( t) dt q( t) t i( t) dt q( t0) t0

15 Mathematical example for currents and charges q(t) = 2 2 e -00 t i(t) = 200 e -00 t

16 Direct Currents and Alternating Currents The directions of currents in the circuit elements are very important. Directions are assumed and determined by analysis. The direction of current (relative to the assumed direction) is determined by the analytic function generated by analysis, using the zero sum for continuity of the currents at junctions and the voltage of the loop circuits (Kirchhoff s KCL for the current, and KVL for the voltage). Continuity of currents and conservation of energy around circuit loops are the only logic rules applied to analyze electric circuits.

17 Basics of Units and Dimensions The force between two charges is given by F = Q Q 2 / ( έ d 2 ). The mechanical dimensions for the force, related to the mechanical dimensions of the other variables is given by: [F] = [Q] 2 / έ [d] 2. or [F] = M L T -2 = [Q] 2 / έ L 2. Or [Q] = έ /2 M /2 L 3/2 T - And the current [I]=[dQ/dt]=έ /2 M /2 L 3/2 T -2

18 The Resistance of Wire conductors Aluminum 2.73 x 0-8 Carbon (amorphous) 3.5 x 0-5 Copper.72 x 0-8 Gold 2.27 x 0-8 Nichrome.2 x 0-6 Silver.63 x 0-8 Tungsten 5.44 x 0-8 Semiconductors Silicon (device grade) 0-5 to Dependent on the impurity concentration Insulators Fused Quartz > 02 Glass (typical) x 02 Teflon x 09 R = ρl / A, The conductivity ρ is as shown to the right in Ohm M

19 Strain Gauge Relations ΔR = Expanding the denominator yields (ρl/a)( + έ)/(- ΔA/A) - R And using the relation ΔA/A = Δl/l, proved later yields ΔR = R ( + έ)2 R = R(+2 έ) R ΔR = 2 έ R = G έ R The Gauge Factor G = R / (R έ) = 2

20 Series and Parallel Resistors i = v v = v + v2 + v3 = R i + R2 i + R3 i = ( R + R2 + R3 ) i = Req i i = i + i 2 + i 3 = [ R R v R v R v 2 R3 2 R3 ] = v R eq

21 Kirchhoff s Current Law (KCL) Three Node Partial Circuits Illustrating Kirchhoff s Current Law (KCL). At node a i + i2 i3 = 0 At node b : At node c i5 + i6 + i7 = 0 i3 = i4

22 Kirchhoff s Voltage Law (KVL) Loop : - va + vb + vc = 0 Loop 2: - vc vd + ve = 0 Loop 3: va vb + vd ve = 0

23 Independent Voltage Sources Independent Voltage Sources

24 Independent Current Sources Independent Current Sources.

25 Dependent Voltage and Dependent Current Voltage Sources Dependent Voltage and Dependent Current Voltage Sources

26 Dependent Current Source Dependent Current Source.

27 Ohm s Law v = i R Voltage is proportional to current in an ideal resistor

28 Examples for Problems of The circuit shown contains a voltage controlled current source. Solve for vs. The circuit shown contains a current controlled voltage source. Solve for i s. Controlled Sources

29 Node Voltages v x = v 2 - v 3, v y = v 2 v and v z = v 3 v Reference Node in the bottom

30 Superposition Application Voltage Adders Superposition can be applied to generate the transfer function to v from v and v 2. v o =v RR 2 /(RR +RR 2 +R R 2 ), v o2 =v 2 RR /(RR +RR 2 +R R 2 ) Then v = v RR 2 / (RR +RR 2 +R R 2 ) + v 2 RR / (RR +RR 2 +R R 2 ) =(v R 2 +v 2 R )R/(RR +RR 2 +R R 2 ) When R =R 2 =R o v=(v +v 2 )R /(2R+R o )

31 the Wheatstone bridge At Balance Condition i = i 3 and i 2 = i 4 i R = i 2 R 2 and i 3 R 3 = i 4 R x R 3 / R = R x / R 2

32 Thevenin and Norton Equivalent Sources R n = v oc / i sc = R th V oc = Thevenin Source voltage i sc = Norton Source Current R th = R L for Max Power Transfer

33 Performance of the Differential Amplifier differential input voltage v d = V cc / A = approximately (5 / = 0.5 mv)

34 Non-Inverting OP-Amp V (-) = V in V in = V out R /(R +R f ) (V out / V in ) = (R +R f ) / R The Unity Gain Op-Amp Circuit

35 Negative Feed Back in Car Power Steering

36 Inverting Op-Amp i = V / R = - i f V out / V = - R f / R

37 Difference Op-Amp Circuit V + = V 2 [ R f / (R + R f ) ] i = [V -V + ] / R i f = - [ V out - V + ] / R f i f = - i V out = (R f / R )(V 2 - V )

38 Capacitor Current and Fluid Flow Analogy

39 Capacitor Concept with Current, Charge and Voltage Relations i = dq / dt = C dv / dt The current is the rate of change of the charge The Current also is the capacity multiplied by the rate of the change of the voltage

40 Voltage, Charge and Current Relations

41 Current Integration Relations for the Voltage and the Charge t q( to) v( t) i( t) dt C to C t q( t) i( t) d q( to) to t

42 Power and Energy Relations for the Capacitor p(t) = v(t) i(t) = C v dv / dt w( t) t p( t) dt 0 W ( t ) = ½ C v 2 ( t ) = ½ v ( t ) q ( t ) = q 2 ( t ) / 2 C

43 Capacitors in Parallel and Capacitors in Series i = i +i 2 +i 3 = C dv/dt + C 2 dv/dt + C 3 dv/dt = (C +C 2 +C 3 ) dv/dt v = v + v 2 + v 3 = (q / C ) + (q / C 2 ) + (q / C 3 ) = q / C

44 C = ε A / d Capacitor Construction and Calculations the Dielectric Constant ε = ε r ε o Fm ε o = Material ε r Air Diamond 5.5 Mica 7 Polyester 3.4 Quartz 4.3

45 Discharge of the Capacitor C dv c /dt + v c /R = 0 v c = K e s t K = V i RCK s e s t + K e s t = 0 RC s + = 0 s = - /RC v c = V i e t / RC

46 Charging of a Capacitor RC d v c / d t + v c = V s v c = K + K 2 e s t ( + RC s) K 2 e s t + K = V s ( + RC s)= 0 s = - / RC K = V s V c = V s V s e t / RC = V s ( e t / RC )

47 Integrator and differentiator Using the Capacitors Vo t 0 Vin R C dt i in = v in / R i in =C d v in / dt v o = - RC d v in / dt

48 Inductors with Voltage and Current Relations i( t) L t to v( t) dt i( to) v(t) = L di / dt

49 Inductor Power, Energy and Its Current with Applied Constant Voltage p(t) = v(t)i(t) = L i(t) di(t) / dt i( t) t L to v( t) dt i( to) w(t) = L i di = ½ L i 2 (t) i(t) = 0 t / L = 5 t Ampere

50 Reactance of Capacitor and Inductor with Alternating Voltage For the Capacitor v(t) = V m sin (ωt) and for the Inductor i(t) = C dv / dt = ωc V m cos (ωt) = V m cos (ωt) / ( / ωc) = - j Vm sin (ωt) / ( / ωc). v(t) = Vm sin (ωt) i(t) = V m sin (ωt) / ( /jωc) Capacitor reactance = /jωc = X C i(t) = - V m cos (ωt) / ωl = - V m sin (ωt+π/2) / ωl = - j V m sin (ωt) / ωl i(t) i(t) = = Vm Vm sin sin (ωt) (ωt)/ / X i(t) = V m sin (ωt) / j ωl Inductor reactance = jωl = X L

51 RLC Circuits and its Mechanical Analogy m d 2 x / d t 2 + K d x / d t + Cm x = F X = D + A e s t + B e s2 t the mass (m) equivalent to the inductance (L), spring action (C m ) equivalent to the inverted capacity (/C) viscous damping (K) equivalent to the Resistance (R) L d 2 Q / d t 2 + R dq / d t + (/C) Q = V s Q c = D + A e s t + B e s2 t Displacement x is equivalent the Charge Q

52 Ohm s Law Experiment Ohm's Law Experiment 25.0 R (KΩ) (ma) I (Volts) V I (analysis) R meas Voltage V Current in ma Ohm's Law Measured values

53 Series and Parallel Resistors Total Current in ma Total Currents of parallel Resistors Current in ma Performance of Series Connected Resistors Total value of the resistors Total Series Resistance Calculated Currents Measured total Currents Calculated currents Measured Currents V 0 Ω R Ω R2 Rt=R+R2 ma I I mesured Series Resistors

54 Maximum Power Vo 2 ma A m W m mw Power Transfered in mw Maximum Power Transfer KΩ RL K Ω R K Ω R V 0 m V I m I P Pm Load Resistance in K Ohm Power Transfered Measured Values