EE105 Fall 2006 Microelectronic Devices and Circuits Prof Jan M Rabaey jan eecs Lecture 15 Frequency Domain Second Order Overview Last lecture Frequency Domain Continued Read 10 This lecture Second Order Networks continued 2 1 Second Order Circuits The series resonant circuit is one of the most important elementary circuits 3 Some Key Properties H j 0 1 H j R j 0 0 L R 20 2 j 0 0 L j 0 Q Q 1 20 2 j 0 Q H 0 0 Q 10 Q 100 j 0 2Q 2 0 2 4Q 20 02 1 LC 0 2Q j 0 1 1 4Q 2 L Q 0 R 0 1 0 Q 4 2 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 5 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 6 3 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 7 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 8 4 Underdamped Mag Plot 0 01 0 1 0 2 0 4 0 6 0 8 1 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 10 5 Phase Bode Plot 1 0 01 0 1 0 2 0 4 0 6 0 8 11 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 12 6 Low Noise Amplifier D Shaeffer T Lee ISSCC 97 13 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 14 7 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 15 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 16 8 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 d 2v C LC 2C dt dt dt 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 17 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 0 i L 0 0 V For t 0 the source voltage is Vdd Solve the following non homogeneous equation subject to above IC dv C d 2v C Vdd v C t RC LC dt dt 2 Steady state d 0 dt Vdd v C 18 9 Guess Solution Let s subtract out the steady state solution v C t Vdd v t Vdd Vdd v t RC 0 v t RC dv dt LC 2 d 2v dt 2 d v dv LC dt dt 2 Guess solution is of the following form v t Ae st 0 Ae st 1 RCs LCs 2 0 Ae st RC sAe st LC s 2 Ae st 0 1 RCs LCs 2 19 Again We re Back to Algebra Our guess is valid if we can find values of s that 1 satisfy this equation Q 0 1 RCs LCs 2 1 s 2 s 2 0 2 2 1 0 The solutions are s 1 This is the same equation we solved in the last lecture There we found three interesting cases 1 1 1 Underdamped Critically damped Overdamped 20 10 General Case Solutions are real or complex conj depending on if 1 or 1 s 1 s 2 1 1 s2 vC t Vdd A exp s1t B exp s2t vC 0 Vdd A B 0 i 0 C dvC t 0 As1 exp s1t Bs2 exp s2t t 0 0 dt t 0 As1 Bs2 0 A B Vdd 21 Final Solution General Case s A 1 A Vdd s2 s Vdd 1 1 Vdd V s2 A dd B Vdd A s1 s1 1 1 s2 s2 s1 Vdd Vdd s2 exp s1t exp s2t vC t Vdd s1 s1 1 1 s2 s2 1 s1t s1 s2t e e vC t Vdd 1 s1 s2 1 s 2 Solve for A and B s1 Vdd s2 s 1 1 s2 22 11 Overdamped Case 1 Time constants are real and negative s 1 s 2 1 1 0 s2 1 Vdd 1 2 23 Critically Damped 1 Time constants are real and equal 1 1 s 2 1 lim vC t Vdd 1 e t te t 1 1 Vdd 1 1 24 …
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