Introduction to Analog Electrical CircuitsOutline for Today’s LectureTechnical Subdivisions of EETalk with neighbors and define ...ChargeCurrentVoltagePowerSlide 9Battery and Light BulbOhm’s LawSlide 12More on Battery and Light BulbSlide 14Slide 15Slide 16Slide 17Slide 18Kirchhoff’s Current Law (KCL)Slide 20Kirchhoff’s Voltage Law (KVL)Slide 22More Light Bulb CircuitsSingle BulbSlide 25Voltage DividerSlide 27Application: Light DimmerSlide 29Slide 30Application: Heat and Light SensorsConcluding Remark1ENGR 100Introduction to Analog Electrical CircuitsRichard J. KozickElectrical Engineering Department2ENGR 100Outline for Today’s Lecture•Fundamental quantities,concepts & units:–Charge, current, voltage, power•Battery and light bulb:–Show actual circuit versus “circuit model”•Resistance and Ohm’s Law•Kirchhoff’s Laws; series & parallel circuits•Voltage divider–Light dimmer, volume control, sensors, ...3ENGR 100Technical Subdivisions of EE•Computer Systems•Electronics•Electromagnetics•Electric Power Systems•Signal Processing and Control Systems•Communication Systems4ENGR 100Talk with neighbors and define ...•What is electric charge?•What is electric current?•What is electric voltage?[Note these are things we can’t see or feel directly!]5ENGR 100Charge•Property of matter•Two kinds, + and -•Electrical forces:–Opposite charges attract, like charges repel–Force varies as inverse square of distance between charges (like gravitational force)•Basis for all electrical phenomena•Unit: coulomb (C)6ENGR 100Current•Charges can move•Current = flow rate of charge•Unit: ampere (A) = C/s•Example:–A battery is a supply of charges–Larger current drains the battery faster7ENGR 100Voltage•Potential energy per unit charge–Arises from force between + and - charges•Unit: volt (V) = Joule/coulomb = J/C•Analogy with gravitational potential energy:–P.E. = m • g • h–P.E. per unit mass = g • h•Need a reference to measure voltage:–Analogous to the floor in auditorium–Common voltage reference is ground (earth)8ENGR 100Power•Power = flow rate of energy (W = J/s)•Current = flow rate of charge (A = C/s)•Voltage = P.E. per unit charge (V = J/C)•Say we have a flow of charges (current) that are “giving up” their P.E.:–Power = ??? (W = J/s)9ENGR 100Power•Power = flow rate of energy (W = J/s)•Current = flow rate of charge (A = C/s)•Voltage = P.E. per unit charge (V = J/C)•Say we have a flow of charges (current) that are “giving up” their P.E.:–Power = Voltage × Current (W = J/s)10ENGR 100Battery and Light Bulb•Operation of actual circuit•Circuit model:–Ideal voltage source for battery (9 V always)–“Resistor” to model light bulb (R ohms)–Ideal wires(0 resistance)9 VIrR+-VrGround11ENGR 100Ohm’s Law•Resistance:–Characterizes “ease” of charge flow (current)–Depends on material and geometry of wire•Ohm’s Law: Vr = Ir • R9 VIrR+-VrGround12ENGR 100•Georg Simon Ohm (1826):–First clear definition of voltage and current–Showed voltage and current are related –Then he lost his job and was ridiculed!–Finally, he became a university professor in 184913ENGR 100More on Battery and Light Bulb•Vr = _____•Measurement: Ir = ______•Power dissipated by bulb:P = _____________•Ohm’s Law: Vr = Ir • R•R = _____________9 VIrR+-VrGround14ENGR 100More on Battery and Light Bulb•Vr = 9 V•Measurement: Ir = ______•Power dissipated by bulb:P = _____________•Ohm’s Law: Vr = Ir • R•R = _____________9 VIrR+-VrGround15ENGR 100More on Battery and Light Bulb•Vr = 9 V•Measurement: Ir = 32.5 mA•Power dissipated by bulb:P = _____________ •Ohm’s Law: Vr = Ir • R•R = _____________9 VIrR+-VrGround16ENGR 100More on Battery and Light Bulb•Vr = 9 V•Measurement: Ir = 32.5 mA•Power dissipated by bulb:P = Vr • Ir = 0.29 W•Ohm’s Law: Vr = Ir • R•R = _____________9 VIrR+-VrGround17ENGR 100More on Battery and Light Bulb•Vr = 9 V•Measurement: Ir = 32.5 mA•Power dissipated by bulb:P = Vr • Ir = 0.29 W•Ohm’s Law: Vr = Ir • R•R = Vr / Ir = 277 ohms 9 VIrR+-VrGround18ENGR 100More on Battery and Light Bulb•Vr = 9 V•Measurement: Ir = 32.5 mA•Power dissipated by bulb:P = Vr • Ir = 0.29 W•Ohm’s Law: Vr = Ir • R•R = Vr / Ir = 277 ohms  •What if we use an 18 V battery?9 VIrR+-VrGround19ENGR 100Kirchhoff’s Current Law (KCL)•“The total current entering a node equals the total current leaving a node.”•Why? Because charge is conserved (neither created nor destroyed), and charge is not accumulated at nodes.•Find I1, I2, I3:9 VI1GroundI2I34 A2 A 1 A20ENGR 100Kirchhoff’s Current Law (KCL)•“The total current entering a node equals the total current leaving a node.”•Why? Because charge is conserved (neither created nor destroyed), and charge is not accumulated at nodes.•Find I1, I2, I3:9 V2 AGround1 A1 A4 A2 A 1 A21ENGR 100Kirchhoff’s Voltage Law (KVL)•“Around any closed loop, the sum of voltage rises equals the sum of voltage drops.”•Why? Energy is conserved!•Find Va and Vb 9 V+ -5 V + -1 V++--VaVb22ENGR 100Kirchhoff’s Voltage Law (KVL)•“Around any closed loop, the sum of voltage rises equals the sum of voltage drops.”•Why? Energy is conserved!•Find Va and Vb 9 V+ -5 V + -1 V++--4 V3 V23ENGR 100More Light Bulb Circuits•Bulbs in series•Bulbs in parallelHow does power per bulb compare with single bulb?9 VR9 VR+-GroundR+-R24ENGR 100Single Bulb9 VIrR+-VrGroundVr = 9 VMeasurement: Ir = 32.5 mAPower dissipated by bulb:P = Vr • Ir = 0.29 WR = Vr / Ir = 277 ohms 25ENGR 100More Light Bulb Circuits•Bulbs in series•P = (Vr / 2) • (Ir / 2 ) = 1/4 power•Bulbs in parallel•P = Vr • Ir = same powerFor parallel, battery provides twice as much power.9 VIr / 2R9 VIrR+-VrGroundR+-Vr / 2R26ENGR 100Voltage Divider•Important building block of analog circuits–Behind most “knob” and “slider” controls!–Light dimmer, volume control, treble/bass, …–Used for “filters” (equalizers, crossovers)–Basis for sensors (temperature, light, …)•Easy to derive equations using KCL, KVL, and Ohm’s Law (please try it if interested)27ENGR 100Voltage DividerVs+ -V1+-V2SourceVoltageR1R2Describes the “split” of source voltage across series resistors:ssVRRRVVRRRV --2122211128ENGR 100Application: Light