EE40 Lec 11Phasors and PowerProf. Nathan Cheung10/06/2009Reading: Hambley Chapter 5 (Power Triangle and Sectional 5.6 are optional)Slide 1EE40 Fall 2009 Prof. CheungExample: Single Loop Circuit20kΩ+20kΩ+-1µF10V ∠ 0°VC+-Vout+-ω= 377VinHow do we find VC?First compute impedances for resistor and capacitor:First compute impedances for resistor and capacitor:ZR= 20kΩ= 20kΩ∠0°ZC= 1/j (377•1µF) = 2.65kΩ∠ -90°Slide 2EE40 Fall 2009 Prof. CheungC/j(377µ).6590Then use Voltage Divider FormulaWhat happens when ω changes?20kΩ+V+Complex impedance of C-1µF10V ∠ 0°VC-impedance of C and L will changeω=10ω= 10Find VC20kΩ ∠0°++ZC= 1/j (10• 1µF) 100kΩ∠90°+-10V ∠ 0°VC+-= 100kΩ∠-90°Slide 3EE40 Fall 2009 Prof. CheungPhasor Domain Analysis: Step-by-stepSlide 4EE40 Fall 2009 Prof. CheungPhasor Diagram: Visualization• Convert this circuit to the phasor domain()V30t50020)t(o()V30t500cos20)t(vso+=KVL V VVVSlide 5EE40 Fall 2009 Prof. CheungKVL: Vs=VR+VL+VCAbsolute Phasor DiagramVs=VR+VL+VCSlide 6EE40 Fall 2009 Prof. Cheungac Power– With ac signals we need to think about the instantaneous and average power being delivered(Wh it d th t th thi )•(When it was dc, these two were the same thing)– We will also define a root-mean-square valueSlide 7EE40 Fall 2009 Prof. CheungInstantaneous and Average Values• Instantaneous power is how you thought about power for dc, now it is a function of time– For example, the instantaneous power in a resistor isR)t(i)t(p2=• Average power is the integral of the waveform over one period, T.waveform over one period, T.∫=Tdt)t(p1)t(pSlide 8EE40 Fall 2009 Prof. Cheung∫0)(pT)(pRoot-mean Square (rms) ValueDefinition()()tXdttX1X2T2==∫()()tXdttXTX0RMS==∫B“” i di ti tihBar “--” indicating time-average hereExample: Mean-Square value of a sinusoidal voltage: ()()θ+ωtcosVtv()()θ+ω=tcosVtv0Vv0=Slide 9EE40 Fall 2009 Prof. Cheung2vRMS=Average Power of Impedance Z()()()[]tjj0tjjeVθθV()()()[]()*tjj0tjjtj0tjj0eV1eZeVReeeVRetitvtp=⋅=θθωωθhere)applied ZVI(=()20tj0tjj0V1eZeVeeVRe21 ⋅=ωωθθ−2jVeV[]**0IVRe1ZVRe21 ⋅===⋅=⋅θ*0*j0j0*ZVZeVeVIV:Note[]IVRe2 ⋅=()[]*IVRe1tp⋅=∴Slide 10EE40 Fall 2009 Prof. Cheung()[]IVRe2tp ⋅=∴Maximum Average Power TransferZs()+What’s the load impedance should be to getthe maximum power transferred from the source tthl d?()ZL()tvs-ldto the load?()tvLPower consumption on the loadsourceload()[]*LLLIVRe21tp ⋅=Z*VssLLLVZZZV+=()*sLs*LZZVI+=andTherefore,()[]2s2sLL2s2sLLLVZZZRe21VZZZRe21tp+=+=Slide 11EE40 Fall 2009 Prof. CheungHowever, impedance is complex, both real and imaginary parts need to beconsidered to optimize the power.Max. Power Transfer Cont.LLLjXRZ +=andsssjXRZ+=()()()2s2sL2sLLLVXXRRR21tp+++=Since the numerator doesn’t have the X components, to maximize the averagepower, the term contains the X components in the denominator should be 0.Th fXX()2LVR1tpdTherefore,sLXX−=()()s2sLLLVRR2tp+=and()LtpdIf t kRR*ZZConjugate ofZgives max power transfer()0=LLdRtpdIf takesLRR=⇒Slide 12EE40 Fall 2009 Prof. Cheung sLZZ=∴Conjugate of Zsgives max. power transferOther Power RelationshipsFor sinusoid v(t) and i(t) :()()()()φ+ω=θ+ω=tcosItitcosVtvm()()φ+ω=tcosItim()[][])cos(IVIV1mm)(imm*φ−θφθ()[][]2)cos(IVeRe2IVIVRe21tp mm)(imm*φθ==⋅=∴φ−θ())cos(IVtp rmsrmsφ−θ••=orSlide 13EE40 Fall 2009 Prof. CheungOther Power quantities())cos(IVtp P rmsrmsφ−θ••==Power Factor PF= cos (θvoltage- θcurrent)Reactive Power A)sin(IVQ rmsrmsφ−θ••=Apparent Power 22rmsrmsQPIV AP+=•≡Slide 14EE40 Fall 2009 Prof. CheungAppendixSome useful complex number identities for phasors:1. []2Re*ZZZ+=2ABare complex numbers thenZ is a complex number, then()***ABBA=2.A, Bare complex numbers, then()ABBA=3.[] [][]*Re21ReRe BABA ⋅=⋅If θωjtjeeAA0= and φωjtjeeBB0=, thenSlide 15EE40 Fall 2009 Prof. CheungAppendixProof of identity 3.[][]422ReRe******ABBABAABBBAABA+++=+⋅+=⋅[][]422Now examine each term.()()∫∫===++TtjTjtjjeeBAdteeBAAB20020001ωφθωφθ=Tπω2∫∫eTdteeBATAB00000T()∫==−+−TtjjdteeBATBA0200**01ωφθ∫T0() ()BAeBAdteBATBAjTj *00000*1===−−∫θφθφ**ABAB =[] []()[][]******14Re244ReRe ABABABABBABA =+=+=∴Slide 16EE40 Fall 2009 Prof. Cheung[]*Re21
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