Announcements Homework 8 due tomorrow 12 noon EE105 Fall 2005 Lab 7 next week Microelectronic Devices and Circuits Reading Chapter 10 10 2 10 3 2 Lecture 19 Second Order Circuits Amplifier Frequency Response 2 Lecture Material Low Noise Amplifier D Shaeffer T Lee ISSCC 97 Last lecture Bode plots Second order functions This lecture Finish second order circuits Frequency response of amplifiers 3 thin Film Bulk Acoustic Resonator FBAR Series LCR Step Response RF MEMS Agilent Technologies IEEE ISSCC 2001 Q 1000 Pad Resonates at 1 9 GHz 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 Thin Piezoelectric Film Where does the inductor come from Intentional inductor placed in series Every physical loop has inductance parasitic Cell phone duplexer C1 C0 Cx R0 Rx 19 C2 Lx 20 21 1 LCR Step Response L Small LCR Circuit ODE Transient response solved in next few slides A Niknejad 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 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 v 0 t Vdd v 0 t Vdd 1 e t d 2v d dv C C LC 2C dt dt dt v s t v C t RC 22 Initial Conditions 23 v C t Vdd v t Vdd Vdd v t RC i 0 i L 0 0 V dv dt LC dv d 2v 0 v t RC LC dt dt 2 For t 0 the source voltage is Vdd Solve the following non homogeneous equation subject to above IC d 2v dt 2 Guess solution is of the following form v t Ae st 0 Ae st 1 RCs LCs 2 2 dv C d vC LC dt dt 2 0 Ae st RC sAe st LC s 2 Ae st d 0 dt Vdd v C 0 1 RCs LCs 2 24 Again We re Back to Algebra 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 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 Underdamped dvC t 0 As1 exp s1t Bs2 exp s2t t 0 0 dt t 0 As1 Bs2 0 Critically damped Overdamped 25 General Case Our guess is valid if we can find values of s that 1 satisfy this equation Q 1 1 1 dv C d 2v C LC dt dt 2 Let s subtract out the steady state solution v 0 0 v C 0 0 V 0 1 RCs LCs 2 dv C dt Guess Solution For the solution of a second order circuit we need to specify to initial conditions IC Steady state v R iR RC We have the following 2nd order ODE t Vdd v C t RC dv C di vL L dt dt 26 A B Vdd 27 2 Final Solution General Case Overdamped Case s A 1 A Vdd Solve for A and B s2 s Vdd 1 1 Vdd V s2 A dd B Vdd A s s 1 1 1 1 s2 s2 s1 Vdd V s vC t Vdd dd exp s1t 2 exp s2t s s 1 1 1 1 s2 s2 1 s1t s1 s2t e e vC t Vdd 1 s1 s2 1 s 2 1 Time constants are real and negative s1 Vdd s2 s 1 1 s2 s 1 s 2 1 1 0 s2 1 Vdd 1 2 28 Critically Damped Underdamped 1 Time constants are real and equal 1 1 s 2 1 lim vC t Vdd 1 e t te t 1 29 Now the s values are complex conjugate s1 a jb s2 a jb vC t Vdd A exp a jb t B exp a jb t vC t Vdd e at A exp jbt B exp jbt s1 Vdd Vdd s2 Vdd A B s1 s s2 1 1 s 1 1 s 1 2 s 1 Vdd 1 1 2 30 Underdamped cont So we have vC t Vdd e at A exp jbt A exp jbt 31 Underdamped Peaking For 1 the step response overshoots vC t Vdd e at A exp jbt A exp jbt vC t Vdd e 2Re A exp jbt at 1 vC t Vdd e at 2 A cos t A Vdd s 1 1 s2 Vdd 1 5 V dd s 1 1 s2 32 33 3 Common Source Amplifer Ai j Extremely Underdamped 1 Vdd 1 01 DC Bias is problematic what sets VGS 34 35 CS Short Circuit Current Gain Common Source Discrete Biasing With ideal MOS VDD R1 CS RL vl vs Transfer function R2 Ai j g m 1 j C gd g m j C gs C gd 36 Magnitude Bode Plot 37 MOS Unity Gain Frequency Since the zero occurs at a higher frequency than pole assume it has negligible effect Ai Transition frequency Current gain 1 gm 1 j C gs Cgd T gm Cgs Cgd W g m Cox L VGS VT 3 VGS VT 2 2 C gs L2 WLCox 3 Performance improves with L2 for long channel devices For short channel devices the dependence is L1 V V GS T E Time to 3 VGS VT v eff L cross T L channel 2 L2 L L L 39 T T 38 4
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