EE105 Fall 2005 Microelectronic Devices and Circuits Lecture 18 Frequency Domain Analysis Second Order Circuits Announcements Homework 8 due next Tuesday Lab 6 this week Lab 7 next week Reading Chapter 10 10 1 2 1 Lecture Material Last lecture Frequency domain analysis Bode plots This lecture More Bode plots Second order functions 3 Get to know your logs dB 20 10 5 3 2 1 ratio 0 100 0 316 0 562 0 708 0 794 0 891 dB 20 10 5 3 2 1 ratio 10 000 3 162 1 778 1 413 1 259 1 122 A margin of 6dB is a factor of 2 power Knowing a few logs can help you calculate logs of different ratios by employing properties of log For instance knowing that the ratio of 2 is 6 dB what s the ratio of 4 4 2 Bode Plot Overview Technique for estimating a complicated transfer function several poles and zeros quickly H G0 j K 1 j z1 1 j z 2 L 1 j zn 1 j p 2 1 j p 2 L 1 j pm Break frequencies i 1 i 5 Summary of Individual Factors Simple Pole 0 dB 1 1 j 1 1 90 Simple Zero 0 dB 90 1 j 0 dB DC Zero 90 j DC Pole 1 j 0 dB 90 6 3 Example Consider the following transfer function H j 10 5 j 1 j 2 1 j 1 1 j 3 1 100 ns 2 10 ns 3 100 ps Break frequencies invert time constants 1 10 Mrad s 2 100 Mrad s j 3 10 Grad s 2 10 H j 1 j 1 j 1 3 5 1 j 7 Breaking Down the Magnitude Recall log of products is sum of logs j 2 10 H j dB 20 log 1 j 1 j 1 3 5 20 log j 105 20 log 1 j 1 j 20 log 1 j 2 20 log 1 j 1 3 Let s plot each factor separately and add them graphically 8 4 Breaking Down the Phase Since a b a b H j 10 5 j 1 j 2 1 j 1 1 j 3 j H j 1 j 2 105 1 j 1 j 1 3 Let s plot each factor separately and add them graphically 9 Magnitude Bode Plot DC Zero 80 60 j 105 40 0 dB 20 104 105 106 107 108 109 1010 1011 20 40 60 80 10 5 Phase Bode Plot DC Zero 180 135 j 105 90 45 104 105 106 107 108 109 1010 1011 45 90 135 180 11 Magnitude Bode Plot Add First Pole 80 1 10 Mrad s j 105 60 dB 40 20 104 105 106 107 108 109 1010 1011 20 40 60 80 1 1 j 107 dB 12 6 Phase Bode Plot Add First Pole 180 135 90 j 105 45 104 105 106 107 108 109 1010 1011 45 90 1 135 1 j 180 10 7 13 Magnitude Bode Plot Add 2nd Zero 80 2 100 Mrad s 1 j 60 108 dB 40 20 104 105 106 107 108 109 1010 1011 20 40 60 80 14 7 Phase Bode Plot Add 2nd Zero 180 135 1 j 90 108 45 104 105 106 107 108 109 1010 1011 45 90 135 180 15 Magnitude Bode Plot Add 2nd Pole 80 60 3 10 Grad s 40 20 104 105 106 107 108 109 1010 1011 20 40 60 1 1 j 1010 dB 80 16 8 Phase Bode Plot Add 2nd Pole 180 135 90 45 104 105 106 107 108 109 1010 1011 45 90 1 j 135 1010 180 17 Comparison to Actual Mag Plot 18 9 Comparison to Actual Phase Plot 19 Power Flow The instantaneous power flow into any element is the product of the voltage and current P t i t v t For a periodic excitation the average power is Pav i v d T In terms of sinusoids we have Pav I cos t i V cos t v d T I V cos t cos i sin t sin i cos t cos v sin t sin v d T I V d cos2 t cos i cos v sin2 sin i sin v c sin t cos t T I V 2 cos i cos v sin i sin v I V 2 cos i v 20 10 Power Flow with Phasors Pav I V cos i v 2 Power Factor Note that if i v then 2 Pav I V From the previous slide P cos 2 0 2 I V 1 1 cos i v Re I V Re I V 2 2 2 21 More Power In terms of the circuit impedance we have 2 V 1 1 V Re Z 1 P Re I V Re V 2 2 2 Z V 2 2 Re Z Z 2 V 2 2Z 2 Re Z V 2 2Z 2 Re Z Check the result for a real impedance resistor Also in terms of current 2 I 1 1 P Re I V Re I I Z Re Z 2 2 2 22 11 Second Order Circuits The series resonant circuit is one of the most important elementary circuits The physics describes not only electrical LCR circuits but also approximates mechanical resonance massspring pendulum molecular resonance microwave cavities transmission lines buildings bridges 23 Series LCR Impedance With phasor analysis this circuit is readily analyzed Z Z j L Z j L 1 R j C 1 1 R R j L 1 j C 2LC 1 Im Z L 1 0 2LC 2 1 LC 24 12 Resonance Resonance occurs when the circuit impedance is purely real Imaginary components of impedance cancel out For a series resonant circuit the current is maximum at resonance V L VC VL VR Vs VC VL VL VR Vs 0 Vs VR Vs VR VC VC 0 0 25 Series Resonance Voltage Gain Note that at resonance the voltage across the inductor and capacitor can be larger than the input voltage VL VL I j 0L VC jQ Vs VR Vs V j 0L s j 0L Z 0 R VC I V 0L V 1 s s j 0L j 0C Z 0 j R jQ Vs L LC 1 L 1 Z0 1 1 Q 0 R C R CR R 0C R 26 13 Second Order Transfer Function So we have V R H j 0 Vs j L 1 R j C Vo To find the poles zeros let s put the H in canonical form V j CR H j 0 Vs 1 2LC j RC One zero at DC frequency no DC current through a capacitor 27 Poles of 2nd Order Transfer Function Denominator is a quadratic polynomial j R L V j CR H j 0 2 1 R Vs 1 LC j RC j 2 j LC L R j L 1 H j 20 R 2 2 0 j j LC L j H j …
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