EE40 Lec 11 Phasors and Power Prof Nathan Cheung 10 06 2009 Reading Hambley Chapter 5 Power Triangle and Sectional 5 6 are optional EE40 Fall 2009 Slide 1 Prof Cheung Example Single Loop Circuit 10V 0 20k 1 F VC Vout Vin 377 How do we find VC First compute impedances for resistor and capacitor ZR 20k 20k 0 ZC 1 j j 377 377 1 F 2 65k 65 90 90 Then use Voltage Divider Formula EE40 Fall 2009 Slide 2 Prof Cheung What happens when changes 10V 0 20k 1 F VC 10 Find VC 20k 0 10V 0 EE40 Fall 2009 Complex impedance of C and L will change Slide 3 VC ZC 1 j 10 1 F 100k 90 90 Prof Cheung Phasor Domain Analysis Step by step EE40 Fall 2009 Slide 4 Prof Cheung Phasor Diagram Visualization Convert this circuit to the phasor domain v s t 20 cos 500t 30 V o KVL Vs V KVL V VR V VL V VC EE40 Fall 2009 Slide 5 Prof Cheung Absolute Phasor Diagram Vs VR VL VC EE40 Fall 2009 Slide 6 Prof Cheung ac Power With ac signals we need to think about the instantaneous and average power being delivered When Wh it was d dc th these ttwo were th the same thi thing We will also define a root mean square value EE40 Fall 2009 Slide 7 Prof Cheung Instantaneous 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 is 2 p t i t R Average power is the integral of the waveform over one period T 1 T p t p t dt T 0 EE40 Fall 2009 Slide 8 Prof Cheung Root mean Square rms Value Definition X RMS 1 T 2 2 X t dt X t T 0 B Bar iindicating di ti titime average here h Example Mean Square value of a sinusoidal voltage v t V 0 cos t V0 v RMS 2 EE40 Fall 2009 Slide 9 Prof Cheung Average Power of Impedance Z p t v t i t Re V0e e j j t V0e j j t Re e Z 1 Re V0e j e j t 2 1 V02 Re 2 Z 1 Re V I 2 V I applied here Z V0e j j t e Z j 2 V e V j 0 0 Note V I V0e Z Z 1 p t Re V I 2 EE40 Fall 2009 Slide 10 Prof Cheung Maximum Average Power Transfer Zs vs t ZL vL t What s the load impedance should be to get the maximum power transferred from the source t the to th load l d Power consumption on the load source l d load ZL VL Vs ZL Zs p L t 1 Re VL I L 2 and 1 ZL Therefore p L t Re 2 ZL Zs IL 1 Re Z L Vs Vs 2 2 Z L Z s 2 2 Vs Z L Zs 2 However impedance is complex both real and imaginary parts need to be considered to optimize the power EE40 Fall 2009 Slide 11 Prof Cheung Max Power Transfer Cont Z L RL jX L Z s Rs jX s and 1 RL p L t Vs 2 2 2 R L R s X L X s 2 Since the numerator doesn t have the X components to maximize the average power the term contains the X components in the denominator should be 0 XL Xs Th f Therefore If take t k d p L t 0 dR L and d ZL Zs EE40 Fall 2009 1 RL p L t Vs 2 2 R L R s 2 RL Rs Conjugate of Zs gives max max power transfer Slide 12 Prof Cheung Other Power Relationships For sinusoid v t and i t v t Vm cos t i t I m cos t 1 Vm I m Vm I m cos i p t Re V I Re e 2 2 2 or EE40 Fall 2009 p t Vrms I rms cos Slide 13 Prof Cheung Other Power quantities P p t Vrms I rms cos Power Factor PF cos voltage current Reactive Power A Apparent Power EE40 Fall 2009 Q Vrms I rms sin AP Vrms I rms P 2 Q 2 Slide 14 Prof Cheung Appendix Some useful complex number identities for phasors 1 Z Z Z is a complex number then Re Z 2 2 2 A B are complex numbers A numbers then 3 If j t A A0 e e EE40 Fall 2009 j and A B B B0 e j t AB e then Re A Re B Slide 15 j 1 Re A B 2 Prof Cheung Appendix Proof of identity 3 A A B B AB A B A B AB Re A Re B 2 2 4 Now examine each term j T A B e 1 T 2 j t j 2 j t 0 0 AB A0 B0 e e dt e 0 0 0 T T 1 T A B A0 B 0 e j e 2 j t dt 0 T 0 1 T A B A0 B0 e j dt A0 B0 e j A B T 0 AB AB A B AB AB AB Re A Re B 4 4 1 Re AB 2 2 T 2 Re AB 4 EE40 Fall 2009 Slide 16 Prof Cheung
View Full Document
Unlocking...