EE5 - Fall 5 Microelectronic Device and ircuit Lecture 9 Second-Order ircuit Amplifier Frequency Repone Announcement Homework 8 due tomorrow noon Lab 7 next week Reading: hapter.,.3.
Lecture Material Lat lecture Bode plot Second order function Thi lecture Finih econd-order circuit Frequency repone of amplifier 3 Second Order Tranfer Function So we have: o H R L R To find the pole/zero, let put the H in canonical form: R H L R One zero at D frequency no D current through a capacitor 4
3 5 Pole of nd Order Tranfer Function Denominator i a quadratic polynomial: L R L L R R L R H L R L R H L H R L 6 Finding the pole Let factor the denominator: Pole are complex conugate frequencie The parameter i called the quality-factor or -factor Thi i an important parameter: Re Im ± ± 4 4 R
Reonance without Lo The tranfer function can be parameterized in term of Im lo. Firt, take the lole cae, R: ± 4 ± When the circuit i lole, the pole are at real frequencie, o the tranfer function blow up! At thi reonance frequency, the circuit ha zero imaginary impedance and thu zero total impedance Even if we et the ource equal to zero, the circuit can have a teady-tate repone ocillate Re 7 Magnitude Repone The repone peakine depend on R L H R L H H Δ H 8 4
5 9 How Peaky i it? Let find the point when the tranfer function quared ha dropped in half: H / H /
6 Half Power Frequencie Bandwidth We have the following: / ± m a b b a > ± ± ± ± 4 Δ / Δ Four olution! < > < > b a b a b a b a Take poitive frequencie:
More Notation Often a econd-order tranfer function i characterized by the damping factor a oppoed to the uality factor τ τ τ τ τζ ζ 3 Second Order ircuit Bode Plot uadratic pole or zero have the following form: τ τζ damping ratio The root can be parameterized in term of the damping ratio: ζ τ τ τ ζ > τ ζ ± ζ τ Two equal pole τζ τ τ Two real pole 4 7
Bode Plot: Damped ae The cae of ζ > and ζ i a imple generalization of imple pole zero. In the cae that ζ >, the pole zero are at ditinct frequencie. For ζ, the pole are at the ame real frequency: ζ τ τ τ τ τ log τ 4log τ τ τ τ τ Aymptotic Slope i 4 db/dec Aymptotic Phae Shift i 8 5 Underdamped ae For ζ <, the pole are complex conugate: τ τ ζ ± τζ ζ ζ ± ζ For τ <<, thi quadratic i negligible db For τ >>, we can implify: log τ τζ log τ 4log τ In the tranition region τ ~, thing are tricky! 6 8
Underdamped Mag Plot ζ. ζ ζ. ζ. ζ.4 ζ.6 ζ.8 7 8 9
Underdamped Phae The phae for the quadratic factor i given by: τζ τ τζ tan τ For τ <, the phae hift i le than 9 For τ, the phae hift i exactly 9 For τ >, the argument i negative o the phae hift i above 9 and approache 8 Key point: argument hift ign around reonance 9 Phae Bode Plot ζ ζ....4.6.8
Bode Plot Guideline In the tranition region, note that at the breakpoint: τ τζ ζ ζ From thi you can etimate the peakine in the magnitude repone. Example: For ζ., the Bode magnitude plot peak by log5 ~ 4 db The phae i much more difficult. Note for ζ, the phae repone i a tep function For ζ, the phae i two real pole at a fixed frequency For < ζ <, the plot hould go omewhere in between! Low-Noie Amplifier D. Shaeffer, T. Lee, ISS 97
thin-film Bulk Acoutic Reonator FBAR RF MEMS Agilent Technologie IEEE ISS > Reonate at.9 GHz Pad Thin Piezoelectric Film ell phone duplexer R x R x L x 3 Serie LR Step Repone onider the tranient repone of the following circuit when we apply a tep at input Without inductor, the cap charge with R time contant EES 4 Where doe the inductor come from? Intentional inductor placed in erie Every phyical loop ha inductance! paraitic 4
LR Step Repone: L Small We know the teady-tate repone i a contant voltage of acro capacitor inductor i hort, cap i open For the cae of zero inductance, we know olution i of the following form: v t v t e t / τ t τ 5 LR ircuit ODE Tranient repone olved in next few lide A. Niknead Apply KL to derive governing time-domain equation: v t v t v t v t R Inductor and capacitor current/voltage: dv di i i vl L dt dt d dv d v v L L L dv dt dt v ir R dt R dt We have the following nd order ODE: L v dv d v t v t R L dt dt 6 3
Initial ondition For the olution of a econd order circuit, we need to pecify to initial condition I: v v i i For t >, the ource voltage i. Solve the following non-homogeneou equation ubect to above I: Steady tate: L dv t R dt v d dt v v d L dt 7 Gue Solution! Let ubtract out the teady-tate olution: v t v t dv d v v t R L dt dt dv d v v t R L dt dt Gue olution i of the following form: t Ae R Ae t t L Ae R Ae t L R L v t t Ae 8 4
Again We re Back to Algebra Our gue i valid if we can find value of that atify thi equation: ζ R L τ ζ τ τ The olution are: τ ζ ± ζ Thi i the ame equation we olved in the lat lecture! There we found three intereting cae: < ζ > Underdamped ritically damped Overdamped 9 General ae Solution are real or complex con depending on if ζ > or ζ < ζ ± ζ τ v dv t i dt t B t t Aexp exp t v A B A exp t B exp t t A B A B 3 5
6 3 Final Solution General ae Solve for A and B: A B A A A t t e e t v t t t v exp exp 3 Overdamped ae ζ > : Time contant are real and negative < ζ ζ ± τ ζ τ
ritically Damped ζ > : Time contant are real and equal ζ ± τ lim v ζ t ζ τ t / τ t / τ e te τ ζ 33 Underdamped Now the value are complex conugate a b a b a b t Bexp a b t v t Aexp v at Aexp bt B bt v t e exp * A B * * at * Aexp bt A bt t e exp 34 7
Underdamped cont So we have: v at * Aexp bt A bt at e Re[ Aexp bt ] t e exp v v t at t e A co t φ A φ 35 Underdamped Peaking For ζ <, the tep repone overhoot: τ ζ.5 36 8
Extremely Underdamped τ ζ. 37 ommon Source Amplifer: A i D Bia i problematic: what et GS? 38 9
ommon Source: Dicrete Biaing DD With ideal MOS R R L S v l v R 39 4