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Transcript
Principles to Computer Engineering ENG_4_532: Introduction Introduction • Mr Ya Bao (Semester 1); Dr. Zhanfang Zhao (Semester 2) – Room T-815, Tower Block – email: [email protected] – Tel: 020 7815 7588 • Office Hours – Mondays 15:00 – 16:00 – Lecture notes – www.lsbu.ac.uk/vle, Text books (semester 1) Electric Circuits, Nilsson & Riedel, Prentice-Hall, 10E. 2015; ISBN:978-1-292-06054-5 Digital Electronics, Roger L. Tokheim, Mcgraw hill, 8E, 2014; ISBN:978-1-259-06092-2 Overview • • • • • • Module Overview Assessment Methods Log books Good practice in labs Theory for Labs week 3 Summary Learning Outcomes • Develop an understanding of the principles of circuit theory and basic analogue electronics • Introduce the concept of digital systems • To develop your ability to manage and perform the successive steps involved in typical practical engineering experiments • To develop your ability to work as a member of a team • To develop your ability to communicate technical information (report writing) Assessment Methods • Course split 50% Semester 1 & 50% Semester 2 • Phase Test: 20% – Scheduled for 14/Nov/16; Week 8 in LR-313 • Formal report: 20% – Due by Week 13, submit to School Office (T313) by 10th Jan 2017 • Logbooks: 10% – Submit with reports to School Office by 10th Jan 2017 Lab Schedule 1. 2. 3. 4. 5. 6. 7. 8. Introduction to Lab Equipment Voltage, Current, Power & Ohm’s Law Kirchoff’s Laws of Current and Voltage (*) Source Transformations Wheatstone Bridge Transformers and Rectifiers Basic TTL Logic Circuits Timing Circuits and Counters RULES: While you are in the lab • • • • • • No drinking No eating No smoking No dangerous activity Take care with electrical devices Switch off your mobile phone and ipod, distractions waste time Log Books • You MUST keep a log book! Do not use loose paper for intermediate recording before transfer to the log book. • You must use the log book to document your procedure and results AS YOU GO ALONG • Draw graphs only on graph paper, use pencil where needed • Answer all questions from the script in your log • Add ideas/points of interest too • Conclude each experiment with the key points • You MUST get your LOGBOOK stamped at the end of session • You MUST leave your logbook in the lab • Go to stores J205 or good stationers and buy a suitable book with lined paper and add graph paper as needed • Best logbooks are Chartwell Laboratory Books A4-641C • Page number • Date • Names of the group members • Title of the experiment • Brief description of the procedure • Carefully record your results • Do NOT use ‘Tipp-Ex’ • Graphs should have a caption • Make maximum use of the graph paper • Choose an easy scale • Label the axes • Units • Error margins/bars Error Margins • You MUST estimate error margins for any reading you make • You should propagate these errors for any calculations you make • Common measurement devices – Oscilloscope ±3% – Digital Multimeter ±1% – Analogue Meter ±5% The Breadboard Close-Up Vertically Connecte d Brea k Vertically Connecte d Horizontally Connected Break Error Margins • Common components – E12 Resistor ±5% – Electrolytic Capacitor ±20% – Digital Timer ±0.001% • Example logbook notation – 1.00kΩ ±5% – 56.2 s ±0.3s • Avoid quoting too many significant figures – 3 is usually about right Resistor E series tables of values Finally • Please keep your Laboratory Manual safe! • Lab and lecture schedule on VLE • All related materials will be posted on VLE – check often • Please arrive promptly . • Any questions? Electrical Energy • Energy cannot be created or destroyed, however, it may be converted from one form into another. • In the next four lectures we are going to investigate electrical energy and its application, from basic concepts to electric circuits. • A cell is a device that can generate electricity, more precisely, it is a device that converts stored chemical energy into electrical energy • An electrolyte causes EMF or Voltage to appear across the terminals of the cell Conventional Current • In reality the flow of current relates to the movement of charged particles (i.e. electrons) which are in fact negatively charged through conductive material (e.g. metal wires) • However, historically scientists have considered the flow of current from high to low potential (voltage) • This is considered “Conventional Current” and most scientists and engineers use this and not “electron flow”. Charge, Voltage and Energy • Cells have two principle parameters, the Charge stored, Q and the terminal Voltage, V. • Charge is measured in Coulombs [C], • Voltage is measured in Volts [V] • Voltage is the Energy Stored per Coulomb of Charge W E V Q Q • Where W = Energy Stored in Joules [ J ] and Q = Charge • Example: A cell uses 1500 Joules of energy to generate 1000 Coulombs of charge, what is its voltage? Current and Charge • The smallest charge is a single electron which has 1.6x10-19 Coulombs • The rate of flow of electrical charge is termed ‘Current’ Q I t • Where Q = Charge [C] and t = time [s] • Current is measured in Amps [A] • Example: If 1000C of electrons travel through a wire in 100s, what is the current in A and mA? Resistance • Resistance is the property of a material to “resist” the flow of current • Conductors have low resistivity per unit area • Insulators have high resistivity per unit area • The flow of current through a resistive material causes a potential difference (or voltage) to develop across it • Fixed external resistors are very useful circuit components and are made from materials with a known resistivity per unit area Value of resistor can be calculated by the colour ring printed on. Resistor colour coding Voltage Divider Example • Given that V = 10 and the voltage at VB = 3 what are the voltages VAB and VBC? - + I V=10V VBC C VC VAB A B VB V B VA Figure for Question 1.2 A cell and two resistors in series Electrical Power • Electrical power, P, is given by the amount of electrical energy converted (or absorbed) per unit time in Watts [W] E VQ P VI • Hence t t • Where E = Energy [ J] absorbed, t = time [s], Q = Charge [C] and I = Current [A] • Example: A DC motor consumes 2000J of electrical energy per second when it is in use. Find: – a) the power consumed by the motor – b) given that the motor requires 200V to operate deduce the electric current flowing through the motor. Internal Resistance of Cells • All the materials inside cells have some resistance • The resistance inside a cell is called its “internal resistance”, this is denoted by r, and causes a voltage loss when loaded by an external resistance V VL - + + - r - VR + External resistor, R I Internal Resistance and Voltage Drop • Example: A cell V has an internal voltage 1.8 V and the lost voltage VL is 0.3 V (dropped across its internal resistance). What is the terminal voltage V across an external resistor? V =1.8V - VL=0.3 +V - + r - VR + External resistor, R I Connecting Cells in Series • In order for cells to be connected in series the positive terminal of Cell 1 must connect to the negative terminal of Cell 2 Cell 1 VC - Cell 2 + VB - VBC + VA VAB Figure 1.4 Two cells in series • The voltage across Cell 1 VBC = 15 V, and the voltage across Cell 2 is VAB = 15 V. If we set VC = 0, then the voltage across two cells VAC is? Connecting Cells in Parallel • In practice two (or more) cells with the same voltage can be connected in parallel • The voltage across the connected cells is the same as that across any single one of them • The current available from the connected cells is then multiplied by the number of cells (as the internal resistance is effectively reduced) Circuits and Resistance • The simplest circuit is one with an electric source (a cell here) and a resistor, which are connected with metal wires, as shown in Figure 2.1. Cell - + V Metal wire R I Figure 2.1 Electric circuit • As mentioned previously, the flow of electrical current through a resistance causes a voltage drop (as with the internal resistance of a cell) Ohm’s Law • In the early 19th century Georg Ohm proved that the amount of current through a resistance is proportional to the voltage across it and inversely proportional to its resistance. Hence V I R • This can also be written as:V IR • Thus; if there is 1Volt across a resistor and it has 1A flowing through it, the resistor = 1 Ohm [] • Example: 220V is supplied to a DC motor which has a resistance of 22 Ω. Find the current flowing in the motor? • Answer: I = V/R = 220/22 = 10Amps Kirchhoff’s Current Law • Kirchhoff’s current law states that, “The sum of all the currents passing through any node in a circuit equals zero”. • This is the same as saying, “the sum of currents flowing into that node is equal to the sum of V currents flowing out of that See figure - node”. + T R1 I1 R2 IT I2 A R3 I3 • Hence at Node A;I T I1 I 2 I 3 0 or I T I1 I 2 I 3 Kirchhoff’s Current Law: Example 1 • Given that IT = 6 Amps, I1 = 2 Amps, I2 = 1 Amp • What is the value of I3? VT - + R1 I1 R2 IT I2 A R3 I3 • Answer: I3 = IT – I1 – I2 = 6 – 2 – 1 = 3 Amps Kirchhoff’s Current Law: Example 2 • Given that I1 = 1.5 Amps, I2 = 0.5 Amps, I3 = 1 Amp • What is the value of IT? VT - + R1 I1 R2 IT I2 A R3 I3 • Answer: IT = I1 + I2 + I3 = 1.5 + 0.5 + 1 = 3 Amps Kirchhoff’s Voltage Law • Kirchhoff’s voltage law states that, “The sum of all the voltages around a closed loop equals zero”. • This is the same as saying, “the sum of voltage drops in a loop is equal to the sum of voltage rises in a loop” + + V3 V1 – – V2 + – V1 V2 V3 0 equivalent ly V1 V2 V3 0 or simply V1 V2 V3 Kirchhoff’s Voltage Law: Example 1 • Given that V3 = 10 Volts and V2 = 4 Volts • What is the value of V1? + + V3 V1 – – V2 + – • Answer: V1 = V3 – V2 = 10 – 4 = 6 Volts Kirchhoff’s Voltage Law: Example 2 • Given that V1 = 2 Volts and V2 = 7 Volts • What is the value of V3? + + V3 V1 – – V2 + – • Answer: V3 = V1 + V2 = 2 + 7 = 9 Volts Series Resistance • When resistors are connected end to end they are said to be in series • The total resistance can be shown to be the sum of the individual resistances I T VT V1 I1 V2 I2 V3 • IT = I1 = I2 = I3 and VT = V1 + V2 + V3 • VT = I1R1 + I2R2 + I3 R3 = IT(R1 + R2 + R3) • Thus:- RT = VT/IT = R1 + R2 + R3 I3 Series Resistance: Example 1 • Given that, in the circuit below; • R1 = 10, R2 = 20 and R3 = 30 • What is the total series resistance? IT VT R1 I1 R2 I2 R3 I3 • Answer: RT = 10 + 20 + 30 = 60 Series Resistance: Example 2 • Given that, in the circuit below; • What is the total series resistance? • What is the current I? 5V + - 2Ω 3Ω I • Answer: RT = 2 + 3 = 5 • Answer: I = 5/5 = 1 Amp Resistors in Parallel • When components are connected in the same way at both ends, they are said to be in parallel (e.g. resistors in figure below) V T - + IT I1 V1 R1 I2 V2 R2 I3 V3 R3 • Applying Kirchoff’s Laws give us IT = I1 + I2 + I3 and VT = V1 = V2 = V3 Resistors in Parallel (2) • Applying Ohm’s Law to each resistor gives: VT V1 V2 V3 IT I1 I 2 I 3 RT R1 R2 R3 • Since VT = V1 = V2 = V3, they can be cancelled out and hence we have 1 1 1 1 RT R1 R2 R3 • Alternatively this can be re-written specifically for three resistors as: RR R RT 1 2 3 R1 R2 R2 R3 R3 R1 Resistors in Parallel: Example 1 • Given the circuit below, what is the parallel resistance? • Answer: 1 1 1 1 0.1 0.042 0.025 0.167 RT 10 • Alternatively: RT 24 40 RT 1 6 0.167 R1 R2 R3 10 24 40 9600 6 R1 R2 R2 R3 R3 R1 10 24 24 40 10 40 1600 Resistors in Parallel: Example 2 • Given the circuit below, what is the parallel resistance? • Answer: 1 1 1 1 0.05 0.033 0.017 0.1 RT • Alternatively: 20 30 60 RT 1 10 0.1 R1 R2 R3 20 30 60 36000 RT 10 R1 R2 R2 R3 R3 R1 20 30 30 60 20 60 3600 Hybrid connected resistors Series - Parallel Circuits Potentiometers • A three-terminal resistor with a sliding or rotating contact that forms an adjustable voltage divider. • These devices are commonly called “Potentiometers” and are typically used in volume controls on audio equipment 0V Vout Vin Potentiometers continued • It can be shown that the voltage across the bar is a fraction of the input voltage depending on the ratio of the input and V V I output resistances. R R • Hence as I is the same for the entire resistive strip, Ohm’s Law shows us that - + out in out I Vout Vin Rin Rou V t out in Sliding bar Rout Vin Rin Potentiometers: Example • Calculate Rout when we require an output voltage of 10V from a voltage divider, which has the total resistance of 100 Ω and can supply the maximum voltage 50 Rout V. Vout Vin Rin • Answer: Rout • Hence Vout 10 Rin 100 20 Vin 50 Voltage Dividers • Consider the circuit shown in Figure 3.2 below containing two discrete resistors. We can develop an equation that describes the voltage across each resistor R1 and R2 in terms of the input voltage and Vin I a ratio of resistors T R R 1 Vin - V1 IT R1 + IT - V1 + - V2 + R1 R2 Figure 3.2 Potential Divider 2 V2 IT R2 Vin V1 V2 R1 R2 R1 R2 R1 V1 Vin R1 R2 R2 V2 Vin R1 R2 Voltage Dividers: Example • Calculate V1 and V2 when Vin = 24V, R1 = 8Ω and R2 = 40Ω R1 V1 Vin R1 R2 • R2 V2 Vin R1 R2 R1 8 24 8 V1 Vin 24 4V Answer: R1 R2 8 40 48 R2 40 24 40 24 V2 Vin 20V 48 8 40 R1 R2 The Wheatstone Bridge • We use an “Ohmmeter” to measure an unknown resistance • The heart of the simplest Ohmmeter is a socalled “Wheatstone Bridge” circuit • If R1 was a variable resistor, we can adjust it until Vab = 0 The Balanced Wheatstone Bridge • When Vab = 0, a special condition occurs: the bridge is said to be “balanced”, i.e. Va = Vb • This implies that ig = 0, hence from KCL, i4 = i3 and i2 = i1 • Further, from Ohm’s Law; i4R4 = i2R2 and i3R3 = i1R1 The Wheatstone Bridge continued • Hence i1 R1 i3 R3 i2 R2 i4 R4 R3 R1 R2 R4 R1 R4 R3 R2 The Wheatstone Bridge: Example • Calculate R1 in a Wheatstone bridge when it is balanced and when R2 = 300Ω, R3 = 200Ω, R4 = 100Ω . R3 R1 R2 R4 • Answer: R3 200 R1 R2 300 600 R4 100 Capacitors Type of Capacitors Capacitance • Capacitors are devices which store electrical charge • A capacitor consists of two plates separated by an insulator, as shown in Figure 4.1 • The negative plate is connected to a low potential and the positive plate to a high potential Insulator Q - + - + - + V Negative plate Positive plate Figure 4.1 Capacitance continued • The total amount of the charge stored, is denoted by Q and the voltage across the plates by V • The capacitance then is defined as Q • Where C is in Farads C [F ] V • 1 Farad = 1 Coulomb per Volt • 1mF, µF, nF (nano), pF(pico) • 0.1 pF is the smallest available in capacitors for general use in electronic design Energy Stored in a Capacitor • When charged, a capacitor stores electrical energy • Recall the formula for electrical energy in a circuit, which is W = VQ • However, we need to be careful as the voltage between the plates in a capacitor varies from 0 to V • Hence, to be more accurate we should use the average voltage 0 Vab Vab Vm Vab W V Q Q m • So 2 2 and we know Q CVab • Hence W Vab CVab 1 CVab2 2 2 2 Energy Stored in a Capacitor: Example • Question: A capacitor is supplied with 10 V in a circuit. If its capacitance is 150µF, what is the electrical energy stored in the capacitor? • Answer: 1 1 2 W CVab 150 10 6 10 2 75 10 4 J 7.5mJ 2 2 Charging and discharging a capacitor Charging and discharging a capacitor Charging and Discharging a Capacitor • Charging and discharging a capacitor from a DC (direct current) source is shown below V • We assume that the voltage source, V, has no internal resistance • If the switch was held in position 2 for a long time, then the capacitor would be completely discharged, Vc = 0V Charging a Capacitor • If the switch is then moved to position 1, current will start to flow through the resistor R, thereby charging the capacitor, C • The voltage across the plates of the capacitor will rise in time, until after a long time, the capacitor will have the same voltage as the supply, V V VC Discharging a Capacitor • If the switch is then moved back to position 2, current will start to flow through the resistor R, thereby discharging the capacitor, C • The voltage across the plates of the capacitor will fall in time, until after a long time, the capacitor will have no charge at all and again, Vc = 0V. V VC ESSENTIAL FORMULAS Engineering Science Electrics Essential formulas W E V Q Q Where W = Energy Stored in Joules [ J ] and Q = Charge in Coulombs [C] Voltage is measured in Volts [V] Q I t Where Q = Charge [C] and t = time [s] Current I is measured in Amps [A] or mA E VQ P VI t t Where E = Energy [ J] absorbed, t = time [s], Q = Charge [C] and I = Current [A] Power is measured in Watts [W] Ohm’s Law V I R I = Current Amp, [A] V = Voltage, Volts, [V] R = Resistance, Ohms, [Ω] P = Power, Watts, [W] V R I V IR 2 V P VI I R R 2 KC L VT - + R1 I T I1 I 2 I 3 0 or I T I1 I 2 I 3 R2 I1 I2 R3 IT A I3 KVL V1 V2 V3 0 equivalent ly V1 V2 V3 0 or simply V1 V2 V3 + + V3 – V1 – V2 + – IT Series Resistance R =R +R +R T 1 2 3 VT V1 I1 Resistors in Parallel 1 1 1 1 RT R1 R2 R3 V2 I2 - + V3 VT IT I1 V1 R1 I2 V2 R2 I3 V3 R3 I3 - Potentiometer s Vin Vout I Rin Rout I + Vin Rin Rout Vout Vout Sliding bar Rout Vin Rin Vin - Voltage Dividers R1 V2 Vin R2 V1 Vin R R R1 R2 2 1 + IT - V1 + - V2 + R1 R2 Figure 3.2 Potential Divider The Wheatstone Bridge balanced, i.e. Va = Vb R3 R1 R2 R4 R1R4 R2 R3 Capacitance Q C V [F ] 1 Farad = 1 Coulomb per Volt Vab 1 W CVab CVab2 2 2 Charging and Discharging a Capacitor time constant t = RC VC When discharging When charging VC Source Transformation Source transformations are useful method of circuit analysis It is theoretically possible to replace any given arbitrary linear circuit containing any number of sources and resistances with either a Thévenin equivalent or Norton equivalent circuit A simple source transformation allows a voltage source with series resistance to be replaced by a current source with a parallel resistance and vice versa. Both circuits behave in the same way to external loads. a b Thévenin Equivalent Circuit All linear circuits can be modelled by an independent voltage source and a series resistor The voltage of the source is the open circuit voltage of the network across ‘a’ and ‘b’ The resistance is determined from the short circuit current Norton Equivalent Circuit All linear circuits can be modelled by an independent current source and a parallel resistor The current of the source is the short circuit current through ‘a’ and ‘b’ The resistance is determined from the open circuit voltage (same as Thévenin resistance) Example 1 Find both Thévenin and Norton equivalent circuits for the circuit below END