EE105 Fall 2006 Microelectronic Devices and Circuits Prof Jan M Rabaey jan eecs Lecture 15 Frequency Domain Second Order Overview Last lecture Common Drain Read 8 9 2 Frequency Domain Read 10 1 2 This lecture Frequency Domain Continued Read 9 2 1 Bode Plots 1 Plot magnitude H in dB vs log scale 2 Plot phase H in degrees vs log scale 3 Get to know your logs dB 20 10 6 3 2 1 ratio 0 10 0 32 0 50 0 71 0 79 0 89 dB 20 10 6 3 2 1 ratio 10 00 3 16 2 00 1 41 1 26 1 12 A margin of 6dB is a factor of 2 power Knowing a few logs by memory 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 Low Pass Filter H dB 40 20 1 10 10 1 100 20 40 Phase H 90 1 10 10 1 100 90 5 The High Pass Filter C vs t R vr t Vs 1 j C R Vr V H r Vs 6 3 Magnitude Bode Plot H dB j 1 dB j dB dB 1 j 1 j 7 Graphical Addition of Magnitudes H dB 40 20 1 10 1 10 100 20 40 8 4 Phase Bode Plot for HPF Phase H 90 1 10 10 1 100 90 9 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 10 5 Summary of Individual Factors Simple Pole 0 dB 1 1 j 1 1 90 90 Simple Zero 0 dB 1 j DC Zero 0 dB 90 0 dB 90 j DC Pole 1 j 11 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 H j 2 100 Mrad s 3 10 Grad s j 1 j 5 10 2 1 j 1 j 1 3 12 6 Breaking Down the Magnitude Recall log of products is sum of logs H j dB 20 log 20 log j 1 j 5 10 2 1 j 1 j 1 3 j 20 log 1 j 5 10 2 20 log 1 j 20 log 1 j 1 3 Let s plot each factor separately and add them graphically 13 Breaking Down the Phase Since a b a b H j 10 5 j 1 j 2 1 j 1 1 j 3 H j 1 j j 1 j 5 2 10 1 j 1 3 Let s plot each factor separately and add them graphically 14 7 Magnitude Bode Plot DC Zero 80 60 j 105 40 0 dB 20 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 11 20 40 60 80 15 Phase Bode Plot DC Zero 180 135 j 105 90 45 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 11 45 90 135 180 16 8 Magnitude Bode Plot Add First Pole 80 1 10 Mrad s j 105 60 dB 40 20 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 11 20 40 60 1 80 1 j 107 17 dB Phase Bode Plot Add First Pole 180 135 90 j 105 45 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 11 45 90 135 180 1 1 j 107 18 9 Magnitude Bode Plot Add 2nd Zero 80 2 100 Mrad s 1 j 60 108 dB 40 20 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 11 20 40 60 80 19 Phase Bode Plot Add 2nd Zero 180 135 1 j 90 108 45 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 11 45 90 135 180 20 10 Magnitude Bode Plot Add 2nd Pole 80 60 3 10 Grad s 40 20 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 11 20 1 40 1 j 60 1010 dB 80 21 Phase Bode Plot Add 2nd Pole 180 135 90 45 10 4 10 5 10 6 10 7 10 8 10 9 10 10 10 11 45 90 135 1 j 1010 180 22 11 Comparison to Actual Mag Plot 23 Comparison to Actual Phase Plot 24 12 Average Power and Phasors Integrate P t over one period T i t Z T P i t v t dt I cos t I V cos t V dt 0 v t 0 IV IV 2 Result cos t cos 2 t sin t sin 2 t cos I cos V sin I sin V P IV 2 cos I V Re I V 25 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 26 13 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 27 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 VL VC VL Vs VR VC VL VL VR Vs 0 Vs VR Vs VR VC VC 0 0 28 14 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 Vs V j 0L s j 0L R Z 0 jQ Vs VR VC I V 0L V 1 s s j 0L j 0C Z 0 j R jQ Vs L 1 1 LC 1 L 1 Z0 Q 0 0C R R C R CR R 29 Second Order Transfer Function So we have Vo V R H j 0 Vs j L 1 R j C To find the poles zeros let s put the H in canonical form V j CR H j 0 2 Vs 1 LC j RC One zero at DC frequency no DC current through a capacitor 30 15 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 20 …
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