EE105 Fall 2005 Microelectronic Devices and Circuits Lecture 19 Second Order Circuits Amplifier Frequency Response Announcements Homework 8 due tomorrow 12 noon Lab 7 next week Reading Chapter 10 10 2 10 3 2 2 1 Lecture Material Last lecture Bode plots Second order functions This lecture Finish second order circuits Frequency response of amplifiers 3 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 4 2 Poles of 2nd Order Transfer Function Denominator is a quadratic polynomial j R L V j CR H j 0 2 R 1 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 5 Finding the poles Let s factor the denominator j 2 j 0 20 0 Q 20 0 1 20 0 j 0 1 2Q 2Q 4Q 2 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 6 3 Resonance without Loss The transfer function can be parameterized in terms of Im loss First take the lossless case R 0 20 0 2 j 0 0 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 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 Q 10 H j 0 2 j 0 Q 20 20 j 0 0 Q 1 Q 100 0 8 4 9 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 10 5 Half Power Frequencies Bandwidth We have the following 2 2 2 1 0 0 Q 20 2 0 Q 2 m 0 20 0 Q Four solutions a b 0 a b 0 2 0 0 20 a b 2Q 4Q 1 b a a b 0 a b 0 Take positive frequencies 0 Q 1 0 Q 11 12 6 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 13 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 Two real poles 14 7 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 15 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 16 8 Underdamped Mag Plot 0 01 0 1 0 2 0 4 0 6 0 8 1 17 18 9 Underdamped Phase The phase for the quadratic factor is given by 2 j 2 j 2 1 tan 1 1 2 For 1 the phase shift is less than 90 For 1 the phase shift is exactly 90 For 1 the argument is negative so the phase shift is above 90 and approaches 180 Key point argument shifts sign around resonance 19 Phase Bode Plot 1 0 01 0 1 0 2 0 4 0 6 0 8 20 10 Bode Plot Guidelines In the transition region note that at the breakpoint j 2 j 2 1 j 2 j 2 1 2 1 Q From this you can estimate the peakiness in the magnitude response Example For 0 1 the Bode magnitude plot peaks by 20 log 5 14 dB The phase is much more difficult Note for 0 the phase response is a step function For 1 the phase is two real poles at a fixed frequency For 0 1 the plot should go somewhere in between 21 Low Noise Amplifier D Shaeffer T Lee ISSCC 97 22 11 thin Film Bulk Acoustic Resonator FBAR RF MEMS Agilent Technologies IEEE ISSCC 2001 Q 1000 Pad Resonates at 1 9 GHz Thin Piezoelectric Film Cell phone duplexer C1 C0 Cx R0 Rx C2 Lx 23 Series LCR Step Response Consider the transient response of the following circuit when we apply a step at input Without inductor the cap charges with RC time constant EECS 40 Where does the inductor come from Intentional inductor placed in series Every physical loop has inductance parasitic 24 12 LCR Step Response L Small We know the steady state response is a constant voltage of Vdd across capacitor inductor is short cap is open For the case of zero inductance we know solution is of the following form v 0 t Vdd v 0 t Vdd 1 e t t 25 LCR Circuit ODE Transient response solved in next few slides A Niknejad Apply KVL to derive governing time domain equations v s t v C t v R t v L t Inductor and capacitor currents voltages i iC C vL L d dv C C dt dt d 2v C LC dt 2 dv C di vL L dt dt v R iR RC dv C dt We have the following 2nd order ODE v s t v C t RC dv C d 2v C LC dt dt 2 26 13 Initial Conditions For the solution of a second order circuit we need to specify to initial conditions IC v 0 0 v C 0 0 V i …
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