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CU-Boulder ECEN 4517 - Basic CCM SEPIC Example

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Fundamentals of Power Electronics1Appendix B: Computer SimulationBasic CCM SEPIC ExampleFrequency ResponseIdeal SEPIC frequency response.lib switch.libVg 1 0 dc 120VL1 1 2x 800uHRL1 2x 2 1UC1 2 3 100uFL2 3 0 100uHC2 4 0 100uFRL 4 0 40Vc 5 0 dc 0.4 ac 1Rc 5 0 1MXswitch 2 0 4 3 5 CCM1.ac DEC 201 10 100kHz.PROBE.end+–12345CCM1VgL1RL1C1L2C2RL122x 3 450+–RcFundamentals of Power Electronics2Appendix B: Computer SimulationSwitch Library File.subckt CCM1 1 2 3 4 5Et 1 6 value={(1-v(5))*v(3,4)/v(5)}Vdum 6 2 0Gd 4 3 value={(1-v(5))*i(Vdum)/v(5)}.ends12345CCM112345+–+–EtGdVdum6Symbol SubcircuitFundamentals of Power Electronics3Appendix B: Computer SimulationPROBE OutputSEPIC Example: Control-to-output transfer functionMagnitudePhaseFundamentals of Power Electronics4Appendix B: Computer SimulationTransient responseStep change in load resistanceIdeal SEPIC transient response.lib switch.libVg 1 0 dc 120VL1 1 2x 800uH IC=1.5RL1 2x 2 1UC1 2 3 100uF IC=120L2 3 0 100uH IC=2C2 4 0 100uF IC=80SLoad 4 0 6 0 load.MODEL load VSWITCH RON=40ROFF=200VLC 6 0 PULSE(-2 2 0 0 0 100MS200MS)RLC 6 0 1MVc 5 0 dc 0.4 ac 1Rc 5 0 1MXswitch 2 0 4 3 5 CCM1.tran 5US 200MS UIC.PROBE.end+–12345CCM1VgL1RL1C1L2C2122x 3 450+–RcRonRoffSloadFundamentals of Power Electronics5Appendix B: Computer SimulationTransient simulation: PROBE outputSEPIC exampleLoadcurrentOutputvoltageFundamentals of Power Electronics6Appendix B: Computer SimulationInclusion of switch ON-resistancev1(t)Ts=Rond(t)+d′(t)RDd2(t)i1(t)Ts+d′(t)d(t)v2(t)Ts+ VDi2(t)Ts=d′(t)d(t)i1(t)Tsi1(t) i2(t)12345+–idealswitchidealdiodeVDRonRD+v1(t)–+v2(t)–Circuit:Averaged equations:Subcircuit model:i2(t)Tsv2(t)Tsv1(t)Tsi1(t)Tsd+–+–12345CCM2Fundamentals of Power Electronics7Appendix B: Computer SimulationModeling losses in SEPIC example+–+–123450CCM2VgL1RL1RL2L2C1C2RloadVc0.5 Ω0.1 Ω100 µF100 µF50 Ω800 µH100 µH50 V+v–12435RonVD = 0.8 VRD = 0.05 ΩXswitchsee Fig. B.4 fornetlist—DC analysiswith steppedon-resistanceparameterFundamentals of Power Electronics8Appendix B: Computer SimulationResultsConduction losses in SEPICDV/Vg0 0.2 0.4 0.6 0.8 1012345Ron = 0.5 ΩRon = 1 ΩRon = 0ηD0 0.2 0.4 0.6 0.8 120%0%40%60%80%100%Ron = 0.5 ΩRon = 1 ΩRon = 0Fundamentals of Power Electronics9Appendix B: Computer SimulationComparison of simulation approachesTransient response+–Switch network+v1––v2+i1i2R+v(t)–Vgi(t)RLL1C150 µF 20 Ω0.1 Ω15 V15 µHBuck-boost exampleFundamentals of Power Electronics10Appendix B: Computer SimulationTwo approaches+–R+v(t)–VgSq1D1i(t)RLL1C150µF20Ω0.1Ω15V1234+–vc515µH+–R+v(t)–Vgi(t)RLL1C150µF20Ω0.1Ω15V1234+–vc515µHdCCM2RD = 0VD = 0.8 VRon = 0.05 Ω12435XswitchUsing ideal switch Using averaged modelFundamentals of Power Electronics11Appendix B: Computer SimulationResults of simulationsTurn-on transient010 A20 A30 A40 A50 A60 Ai(t)tWaveform obtained by simulationof the switching circuit modelWaveform obtained by simulation of the averaged model0.2 ms0.4 ms 0.6 ms 0.8 ms1 ms1.2 ms0-60 V-50 V-40 V-30 V-20 V-10 V010 Vv(t)Waveform obtained by simulationof the switching circuit modelWaveform obtained by simulation of the averaged modelt0.2 ms0.4 ms 0.6 ms 0.8 ms1 ms1.2


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CU-Boulder ECEN 4517 - Basic CCM SEPIC Example

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