107_29 - 1 Electrical Engineering Technology EET 107...

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Unformatted text preview: 1 Electrical Engineering Technology EET 107 Introduction to Circuit Analysis # 29 Professor Robert Herrick Purdue University © EET 107 - 29 Introduction to Circuit Analysis 2 Electrical Engineering Technology AC and AC with DC Circuit Analysis Purdue University © EET 107 - 29 Introduction to Circuit Analysis 3 Overview u AC Voltage and Current Supplies u Circuit Analysis with AC Supplies u Circuit u Analysis with AC & DC Supplies Thevenin Model example Purdue University © EET 107 - 29 Introduction to Circuit Analysis 4 AC Supplies + es is AC Voltage Supply Note: AC Current Supply lower case letters for ac signal symbols Purdue University © EET 107 - 29 Introduction to Circuit Analysis 5 AC Voltage Supply - example Assumed Standard + 10V 1kHz 10 V rms voltage Unless otherwise specified as Vp peak Vpp peak-to-peak Note: Vdc = Vaverage = 0 V unless otherwise specified Purdue University © EET 107 - 29 Introduction to Circuit Analysis 6 AC Voltage Supply - example + 10V 1kHz Sine wave f = 1 kHz Vrms = 10 Vrms (assumed ) Vp = √2 × 10 V = 14.14 Vp Vpp = 2 × 14.14 V = 28.28 Vpp Purdue University © EET 107 - 29 Introduction to Circuit Analysis 7 AC Voltage Supply - example + 10V 1kHz Triangle wave f = 1 kHz Vrms = 10 Vrms (assumed) Vp = √3 × 10 V = 17.32 Vp Vpp = 2 × 17.32 V = 34.64 Vpp Purdue University © EET 107 - 29 Introduction to Circuit Analysis 8 AC Voltage Supply - example + 10V 1kHz Square wave f = 1 kHz Vrms = 10 Vrms Vp = 1 × 10V = 10 Vp Vpp = 2 × 10V = 20 Vpp Purdue University © EET 107 - 29 Introduction to Circuit Analysis 9 DC Voltage Supply - example + 10V - DC DC f = 0 Hz Vdc = Vaverage = 10 Vdc Vrms = 10 Vdc = 10 Vrms Purdue University © EET 107 - 29 Introduction to Circuit Analysis 10 AC Voltage Supply - DC offset example Sine wave 10V 2VDC 1kHz + f = 1 kHz Vrms ac = 10 Vrms Vdc = 2 Vdc Vrms total signal = Vrms totalsignal = 2 Vdc 2 + Vrms ac (2 Vdc ) + (10 Vrms ) Purdue University © AC only 2 EET 107 - 29 2 = 10.2 Vrms Introduction to Circuit Analysis 11 AC Voltage Supply - DC offset example 10V 2VDC 1kHz + + model Purdue University © ac 10V 1kHz EET 107 - 29 + 2V - dc Introduction to Circuit Analysis 12 Overview u AC Voltage and Current Supplies u Circuit Analysis with AC Supplies u Circuit Analysis with AC & DC Supplies u Thevenin Model example Purdue University © EET 107 - 29 Introduction to Circuit Analysis 13 Circuit Analysis with AC & DC Supplies - example 3kΩ 30VPP 30V DC offset 500Hz + iL 6kΩ 3kΩ Convert supply to AC & DC supply models Purdue University © EET 107 - 29 Introduction to Circuit Analysis 14 AC Voltage Supply - DC offset example + 30VPP + 30VDC 500Hz 15VP 500Hz model Purdue University © EET 107 - 29 + 30V - Introduction to Circuit Analysis 15 Circuit Analysis with AC & DC Supplies - example 3kΩ 15V P 500Hz models iL + 6kΩ 3k Ω + 30V - ® Zero AC supply and solve for DC IL ® Zero DC supply and solve for AC iL ® Superimpose AC & DC results for total iL Purdue University © EET 107 - 29 Introduction to Circuit Analysis 16 Circuit Analysis with AC & DC Supplies - example 3k Ω 6mA Zero es 4mA IL short 6kΩ 3k Ω + 30V DC - Isupply = 30 V / 5 kΩ = 6 mAdc IL = 6 kΩ / 9 kΩ × 6 mA = 4 mAdc Purdue University © EET 107 - 29 Introduction to Circuit Analysis 17 Circuit Analysis with AC & DC Supplies - example 3kΩ 3mA 15V P 500Hz Zero ES 2mA iL + 6kΩ 3k Ω short AC isupply = 15 V / 5 kΩ = 3 mAp iL = 6 kΩ / 9 kΩ × 3 mA = 2 mAp Purdue University © EET 107 - 29 Introduction to Circuit Analysis 18 Circuit Analysis with AC & DC Supplies - example 3kΩ 15VP 30V offset 500Hz + iL 6kΩ 3kΩ DC IL = 4 mAdc AC iLm = 2 mAp sine wave at 500Hz Purdue University © EET 107 - 29 Introduction to Circuit Analysis 19 Circuit Analysis with AC Supplies - example Sketch 2 mA peak amplitude sine wave 500 Hz 4 mA DC offset T = 1/f = 1/500Hz = 2 ms Ip total signal = 4 mAdc + 2 mAp = 6 mAp Imin total signal = 4 mAdc − 2 mAmin = 2 mAmin Ipp total signal = 2 × 2 mA = 4 mApp Purdue University © EET 107 - 29 Introduction to Circuit Analysis 20 Circuit Analysis with AC Supplies - example i L(t) 6mA 2mA 4mA 4mA 2mA 2mA 0 t 2ms Purdue University © EET 107 - 29 Introduction to Circuit Analysis 21 Power Calculations - DC only - example 3kΩ 30VPP 30V DC offset 500Hz + 4mA 6kΩ 3kΩ + 12V - DC only PL = IL × VL dc dc dc = 4mAdc × 12Vdc = Purdue University © 48 mW EET 107 - 29 Introduction to Circuit Analysis 22 Power Calculations - AC only - example 2mAP 3kΩ 30VPP 30V DC offset 500Hz + 6kΩ PLaverage = IL PL ac = 2 mA p Purdue University © 2 3kΩ rms × × VL 6 Vp 2 + 6VP - AC only rms = 6 mW EET 107 - 29 Introduction to Circuit Analysis 23 Power Calculations - Total Signal - example 4mADC 2mAP 3kΩ 30VPP 30V DC offset 500Hz + 6kΩ 3kΩ Total Signal I L rms total signal = (4 mA ) 2 + (2 mA / 2 ) 2 = 4.24 mA rms PL total signal = ( 4.24 mA ) × 3 kΩ = 54 mW 2 Purdue University © EET 107 - 29 Introduction to Circuit Analysis 24 DC & AC Power Superimpose? - example 3kΩ 30VPP 30V DC offset 500Hz + 6kΩ 3kΩ PL dc = 48 mW PL ac = 6 mW PL total signal = 54 mW Purdue University © EET 107 - 29 Yes Total AC average power adds to total DC power to get total signal power. Introduction to Circuit Analysis 25 Overview u AC Voltage and Current Supplies u Circuit Analysis with AC Supplies u Circuit Analysis with AC & DC Supplies Supplies uThevenin Model example Purdue University © EET 107 - 29 Introduction to Circuit Analysis 26 Circuit Analysis with AC & DC Sources - example 3kΩ 30VPP 30V DC offset 500Hz + iL 6kΩ 3kΩ Another example Find the Thevenin Model of the above circuit ! Purdue University © EET 107 - 29 Introduction to Circuit Analysis 27 Circuit Analysis - Thevenin example 3kΩ + 30V - + 6kΩ ΕΤΗ − DC ETH = 6 kΩ / 9 kΩ × 30 V = 20 Vdc Purdue University © EET 107 - 29 Introduction to Circuit Analysis 28 Circuit Analysis - Thevenin example 3kΩ + 30V - 6kΩ ΙSC Short Circuit Current ISC = 30 V / 3 kΩ = 10 mAdc Rth = Eth/Isc = 20 V / 10 mA = 2 kΩ Purdue University © EET 107 - 29 Introduction to Circuit Analysis 29 Circuit Analysis - Thevenin example 3kΩ RTH 6k Ω Thevenin Resistance RTH = 6 kΩ // 3 kΩ = 2 kΩ Purdue University © EET 107 - 29 Introduction to Circuit Analysis 30 Circuit Analysis - Thevenin example 3kΩ 15VP 500Hz + + 6kΩ eth AC eth = 6 kΩ / 9 kΩ × 15 V = 10 Vp Purdue University © EET 107 - 29 Introduction to Circuit Analysis 31 Circuit Analysis - Thevenin example 2kΩ 10VP 500Hz + RTH eth + 20V - ETH Thevenin Model Purdue University © EET 107 - 29 Introduction to Circuit Analysis 32 Circuit Analysis - Thevenin example 2kΩ 10VP 500Hz IL iL + Loaded Model 3kΩ + 20V - IL dc = 20 V / 5 kΩ = 4 mAdc IL rms ac = 10 V / 5 kΩ = 2 mAp Purdue University © EET 107 - 29 AC only Introduction to Circuit Analysis 33 Overview u AC Voltage and Current Supplies u Circuit Analysis with AC Supplies u Circuit u Analysis with AC & DC Supplies Thevenin Model example Purdue University © EET 107 - 29 Introduction to Circuit Analysis ...
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