EE40Lecture 15Venkat AnantharamImpedanceSome Thoughts on ImpedanceExample: Single Loop CircuitImpedance ExampleWhat happens when w changes?Circuit Analysis Using Complex ImpedancesSteady-State AC AnalysisFind the Equivalent ImpedanceChange the FrequencyFind an Equivalent ImpedanceSeries ImpedanceParallel ImpedanceSlide 1EE40 Spring 2008 Venkat AnantharamEE40Lecture 15Venkat Anantharam2/29/08 Reading: Chap. 5: phasorsSlide 2EE40 Spring 2008 Venkat AnantharamImpedance• AC steady-state analysis using phasorsallows us to express the relationship between current and voltage using a formula that looks likes Ohm’s law:V = IZ• Z is called impedance.Slide 3EE40 Spring 2008 Venkat AnantharamSome Thoughts on Impedance• Impedance depends on the frequency ω.• Impedance is (often) a complex number.• Impedance allows us to use the same solution techniques for AC steady state as we use for DC steady state.Slide 4EE40 Spring 2008 Venkat AnantharamResistor I-V relationshipvR= iRR ………….VR= IRR where R is the resistance,VR= phasor voltage, IR= phasor current,hence ZR = RCapacitor I-V relationshipiC= CdvC/dt ...............Phasor current IC= phasor voltage VC/capacitive impedance ZC: Æ IC= VC/ZC where ZC = 1/jωCInductor I-V relationshipvL= LdiL/dt ...............Phasor voltage VL= phasor current IL/inductive impedance ZL Æ VL= ILZLwhere ZL= jωLSlide 5EE40 Spring 2008 Venkat AnantharamExample: Single Loop Circuit20kΩ+-1µFVC+-10V ∠ 0°f=60 Hz, VC=?How do we find VC?First compute impedances for resistor and capacitor:ZR= R= 20kΩ = 20kΩ∠0°ZC= 1/j (2πf x 1µF) = 2.65kΩ∠-90°Slide 6EE40 Spring 2008 Venkat AnantharamImpedance Example20kΩ ∠ 0°+-VC+-2.65kΩ ∠ -90°10V ∠ 0°Now use the voltage divider to find VC:⎟⎠⎞⎜⎝⎛°∠Ω+°∠Ω°∠Ω°∠=0k2090-k65.290-k65.20 10VCV°∠=4.82- 1.31VCVSlide 7EE40 Spring 2008 Venkat AnantharamWhat happens when ω changes?20kΩ+-1µFVC+-10V ∠ 0°ω = 10Find VCSlide 8EE40 Spring 2008 Venkat AnantharamCircuit Analysis Using Complex Impedances• Suitable for AC steady state.• KVL• Phasor Form KCL• Use complex impedances for inductors and capacitors and follow same analysis as in chap 2.() () ()312312123112 23 3()() ( )12 3()() ( )12 3() () () 0cos cos cos 0Re 000jtjt jtjjjvt vt vtVtVt VtVe Ve VeVe Ve Veωθωθ ωθθθθωθ ωθ ωθ+++++=++ ++ +=⎡⎤++ =⎣⎦++==123V+V+VPhasor Form KVL0=123I+I+ISlide 9EE40 Spring 2008 Venkat AnantharamSteady-State AC AnalysisFind v(t) for ω=2π 30001kΩ0.1µF5mA ∠ 0°+-V1kΩ-j530kΩ5mA ∠ 0°+-VSlide 10EE40 Spring 2008 Venkat AnantharamFind the Equivalent Impedance5mA ∠ 0°+-VZeq()°−∠°−∠×°∠=−−=9.27113290530010530100053010003jjeqZ°−∠Ω=1.622.468eqZ°−∠Ω×°∠==1.622.4680mA5eqIZV°−∠=1.62V34.2V)1.623000t(2cosV34.2)( °−=πtvSlide 11EE40 Spring 2008 Venkat AnantharamChange the FrequencyFind v(t) for ω=2π 4550001kΩ0.1µF5mA ∠ 0°+-V1kΩ-j3.5Ω5mA ∠ 0°+-VSlide 12EE40 Spring 2008 Venkat AnantharamFind an Equivalent Impedance5mA ∠ 0°+-VZeq()°−∠°−∠×°∠=−−=2.01000905.30105.310005.310003jjeqZ°−∠Ω= 8.895.3eqZ°−∠Ω×°∠==8.895.30mA5eqIZV°−∠= 8.89mV5.17V)8.89455000t(2cosmV5.17)( °−=πtvSlide 13EE40 Spring 2008 Venkat AnantharamSeries ImpedanceZeq= Z1+ Z2+ Z3ZeqZ1Z3Z2L2L1Zeq= jω(L1+L2)For example: 1211eqjCjCωω=+ZC1C2Slide 14EE40 Spring 2008 Venkat AnantharamParallel ImpedanceZ3Z1Z2Zeq1/Zeq= 1/Z1+ 1/Z2+ 1/Z3For example:
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