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 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 2 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 3 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 4 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 5 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 0 Q 0 20 j 2 j Q L Q 0 R 28 6 Finding the poles Let s factor the denominator j 2 j 0 20 0 Q 20 0 1 20 0 j 0 1 2 2Q 2Q 4Q 4Q 2 Poles are complex conjugate frequencies Im The Q parameter is called the quality factor or Q factor Re This is an important parameter R 0 Q 29 Resonance without Loss The transfer function can parameterized in terms of Im loss First take the lossless case R 0 20 j 0 0 20 2Q 4Q 2 Q Re When the circuit is lossless the poles are at real frequencies so the transfer function blows up At this resonance frequency the circuit has zero imaginary impedance and thus zero total impedance Even if we set the source equal to zero the circuit can have a steady state response oscillates 30 7 Magnitude Response The response peakiness depends on Q H j H j 0 1 R j 0 0 L 20 R j 0 0 L 2 j 0 Q 20 2 j 0 Q Q 1 H 0 0 H j 0 Q 10 2 j 0 Q 20 20 j 0 0 Q 1 Q 100 31 0 How Peaky is it Let s find the points when the transfer function squared has dropped in half 2 H j 0 Q 2 20 2 2 Q0 2 H j 1 2 2 2 0 1 0 Q 2 1 2 1 2 2 2 2 0 1 0 Q 32 8 Half Power Frequencies Bandwidth We have the following 2 2 2 1 0 0 Q 20 2 0 Q 1 2 m 0 20 0 Q Four solutions a b 0 a b 0 2 0 0 20 a b 2Q 4Q b a a b 0 a b 0 Take positive frequencies 1 0 Q 0 Q 33 More Notation Often a second order transfer function is characterized by the damping factor as opposed to the Quality factor 20 j 2 j 0 0 Q 1 j 2 j 0 Q 1 0 1 j 2 j 2 0 Q 1 2 34 9 Second Order Circuit Bode Plot Quadratic poles or zeros have the following form j 2 j 2 1 0 damping ratio The roots can be parameterized in terms of the damping ratio 1 j 2 j 2 1 1 j 2 Two equal poles 1 2 j j 2 1 1 j 1 1 j 2 j 2 1 35 Two real poles Bode Plot Damped Case The case of 1 and 1 is a simple generalization of simple poles zeros In the case that 1 the poles zeros are at distinct frequencies For 1 the poles are at the same real frequency 1 j 2 j 2 1 1 j 2 1 j 2 1 j 2 Asymptotic Slope is 40 dB dec 2 20 log 1 j 40 log 1 j 1 j 2 1 j 1 j 2 1 j Asymptotic Phase Shift is 180 36 10 Underdamped Case For 1 the poles are complex conjugates j 2 j 2 1 0 j 2 1 j 1 2 For 1 this quadratic is negligible 0dB For 1 we can simplify 20 log j 2 j 2 1 20 log j 2 40 log In the transition region 1 things are tricky 37 Underdamped Mag Plot 0 01 0 1 0 2 0 4 0 6 0 8 1 …
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