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 Low Noise Amplifier D Shaeffer T Lee ISSCC 97 19 2 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 20 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 21 3 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 22 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 23 4 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 Vdd v C t RC Steady state dv C d 2v C LC dt dt 2 d 0 dt Vdd v C 24 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 dv d v 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 25 5 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 The solutions are s 1 2 1 0 This is the same equation we solved in the last lecture There we found three interesting cases 1 1 1 Underdamped Critically damped Overdamped 26 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 27 6 Final Solution General Case s A 1 A Vdd s2 s Vdd 1 1 Vdd V s2 A dd B Vdd A s s 1 1 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 28 Overdamped Case 1 Time constants are real and negative s 1 s 2 1 1 0 s2 1 Vdd 1 2 29 7 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 30 Underdamped 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 eat A exp jbt B exp jbt s1 Vdd Vdd s2 Vdd A B s1 s2 1 1 s1 1 s s2 1 s 2 vC t Vdd e at A exp jbt A exp jbt 31 8 Underdamped cont So we have vC t Vdd e at A exp jbt A exp jbt vC t Vdd eat 2Re A exp jbt vC t Vdd eat 2 A cos t A Vdd s 1 1 s2 V dd s 1 1 s2 32 Underdamped Peaking For 1 the step response overshoots 1 Vdd 1 5 33 9 Extremely Underdamped 1 Vdd 1 01 34 Common Source Amplifer Ai j DC Bias is problematic what sets VGS 35 10 Common Source Discrete Biasing With ideal MOS VDD R1 CS RL vl vs R2 36 CS Short Circuit Current Gain Transfer function Ai j g m 1 j C gd g m j C gs C gd 37 11 Magnitude Bode Plot Transition frequency Current gain 1 T 38 MOS Unity Gain Frequency Since the zero occurs at a higher frequency than pole assume it has negligible effect Ai gm 1 j Cgs Cgd T gm Cgs Cgd W VGS VT gm 3 VGS VT L T 2 Cgs L2 2 WLCox 3 Performance improves with L2 for long channel devices For short channel devices the dependence is L1 VGS VT Time to Eeff v 3 VGS VT L cross T L channel L2 L L L 2 39 Cox 12
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