1EE105 - Fall 2005Microelectronic Devices and CircuitsLecture 19Second-Order CircuitsAmplifier Frequency Response2AnnouncementsHomework 8 due tomorrow 12 noonLab 7 next weekReading: Chapter 10 (10.2, 10.3.2)23Lecture MaterialLast lectureBode plotsSecond order functionsThis lectureFinish second-order circuitsFrequency response of amplifiers4Second Order Transfer FunctionSo we have:To find the poles/zeros, let’s put the H in canonical form:One zero at DC frequency Æ no DC current through a capacitorRCjLjRVVjHs+ω+ω==ω1)(0+Vo−RCjLCCRjVVjHsω+ω−ω==ω201)(35Poles of 2ndOrder Transfer FunctionDenominator is a quadratic polynomial:LRjjLCLRjRCjLCCRjVVjHsω+ω+ω=ω+ω−ω==ω220)(11)(LRjjLRjjHω+ω+ωω=ω220)()(LC120≡ωQjjQjjH02200)()(ωω+ω+ωωω=ωRLQ0ω≡6Finding the poles…Let’s factor the denominator:Poles are complex conjugate frequenciesThe Q parameter is called the “quality-factor” or Q-factorThis is an important parameter:ReIm0)(2002=ω+ωω+ωQjj22−ω±ω−=ω−ω±ω−=ωQjQQQ4112420020200∞⎯⎯→⎯→0RQ47Resonance without LossThe transfer function can be parameterized in terms of loss. First, take the lossless case, R=0:When the circuit is lossless, the poles are at realfrequencies, so the transfer function blows up!At this resonance frequency, the circuit has zero imaginary impedance and thus zero total impedanceEven if we set the source equal to zero, the circuit can have a steady-state response (oscillates)ReIm02020042ω±=⎟⎟⎟⎠⎞⎜⎜⎜⎝⎛ω−ω±ω−=ω∞→2jQQQ8Magnitude ResponseThe response peakiness depends on QQjQjLRjLRjjH022000022000)(ωω+ω−ωωω=ωωω+ω−ωωωω=ω1=Q10=Q100=Q0ω1)(002020200=ωω+ω−ωω=ωQjQjjH1)(0=ωjH0)0( =HωΔ5910How Peaky is it?Let’s find the points when the transfer function squared has dropped in half:()21)(202220202=⎟⎠⎞⎜⎝⎛ωω+ω−ω⎟⎠⎞⎜⎝⎛ωω=ωQQjH211/1)(202202=+⎟⎟⎠⎞⎜⎜⎝⎛ωωω−ω=ωQjH1/20220=⎟⎟⎠⎞⎜⎜⎝⎛ωωω−ωQ611Half Power Frequencies (Bandwidth)We have the following:1/0220±=ωωω−ωQ02002=ω−ωωωQmabbaQQ>±±=ω+⎟⎠⎞⎜⎝⎛ω±ω±=ω2020042Q0ω=ω−ω=ωΔ−+1/20220=⎟⎟⎠⎞⎜⎜⎝⎛ωωω−ωQQ10=ωωΔFour solutions!0000<−−>+−<−+>++babababaTake positive frequencies:12713More “Notation”Often a second-order transfer function is characterized by the “damping” factor as opposed to the “Quality”factor0)(0220=ωω+ω+ωQjj0)(12=ωτ+ωτ+Qjj01ω=τ02)()(12=ζωτ+ωτ+ jjζ=21Q14Second Order Circuit Bode PlotQuadratic poles or zeros have the following form:The roots can be parameterized in terms of the damping ratio:012)()(2=+ζωτ+ωτ jj22)1(12)()(1 ωτ+=+ωτ+ωτ⇒=ζ jjjdamping ratioTwo equal poles1)1)(1(12)()(12212−ζ±ζ−=ωτωτ+ωτ+=+ζωτ+ωτ⇒>ζjjjjjTwo real poles815Bode Plot: Damped CaseThe case of ζ >1 and ζ =1 is a simple generalization of simple poles (zeros). In the case that ζ >1, the poles (zeros) are at distinct frequencies. For ζ =1, the poles are at the same real frequency:22)1(12)()(1 ωτ+=+ωτ+ωτ⇒=ζ jjj221)1( ωτ+=ωτ+ jjωτ+=ωτ+ jj 1log401log202()()()ωτ+∠=ωτ+∠+ωτ+∠=ωτ+∠ jjjj 1211)1(2AsymptoticSlope is 40 dB/decAsymptotic Phase Shift is 180°16Underdamped CaseFor ζ <1, the poles are complex conjugates:For ωτ << 1, this quadratic is negligible (0dB)For ωτ >> 1, we can simplify:In the transition region ωτ ~ 1, things are tricky!22211012)()(ζ−±ζ=−ζ±ζ−=ωτ=+ζωτ+ωτjjjjωτ=ωτ≈+ζωτ+ωτ log40)(log2012)()(log2022jjj917Underdamped Mag Plot ζ=1ζ=0.01ζ=0.1ζ=0.2ζ=0.4ζ=0.6ζ=0.8181019Underdamped PhaseThe phase for the quadratic factor is given by:For ωτ < 1, the phase shift is less than 90°For ωτ = 1, the phase shift is exactly 90°For ωτ > 1, the argument is negative so the phase shift is above 90° and approaches 180°Key point: argument shifts sign around resonance()⎟⎟⎠⎞⎜⎜⎝⎛ωτ−ωτζ=+ζωτ+ωτ∠−212)(12tan12)()( jj20Phase Bode Plotζ=0.010.10.20.40.60.8ζ=11121Bode Plot GuidelinesIn the transition region, note that at the breakpoint:From this you can estimate the peakiness in the magnitude response. Example: For ζ = 0.1, the Bode magnitude plot peaks by 20 log(5) ~ 14 dBThe phase is much more difficult. Note for ζ = 0, the phase response is a step functionFor ζ = 1, the phase is two real poles at a fixed frequencyFor 0 < ζ < 1, the plot should go somewhere in between!Qjjjj1212)()(12)()(22=ζ=+ζ+=+ζωτ+ωτ22Low-Noise AmplifierD. Shaeffer, T. Lee, ISSCC’971223thin-Film Bulk Acoustic Resonator (FBAR)RF MEMSAgilent Technologies (IEEE ISSCC 2001)Q > 1000Resonates at 1.9 GHzCell phone duplexerC0CxRxLxC1C2R0PadThin Piezoelectric Film 24Series LCR Step ResponseConsider 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)1325LCR Step Response: L SmallWe know the steady-state response is a constant voltage of Vddacross capacitor (inductor is short, cap is open)For the case of zero inductance, we know solution is of the following form:)1()(/0τ−−=tddeVtvddVtv )(0τt26LCR Circuit ODETransient response solved in next few slides (A. Niknejad)Apply KVL to derive governing time-domain equations:Inductor and capacitor currents/voltages:We have the following 2ndorder ODE:)()()()( tvtvtvtvLRCs++=dtdiLvdtdvCiiLCC===22dtvdLCdtdvCdtdLvCCL=⎟⎠⎞⎜⎝⎛=dtdvRCiRvCR==22)()(dtvdLCdtdvRCtvtvCCCs++=1427Initial ConditionsFor the solution of a second order circuit, we need to specify to initial conditions (IC):For t > 0, the source voltage is Vdd. Solve the following non-homogeneous equation subject to above IC:Steady state:V0)0()0(V0)0()0(0====LCiivv22)(dtvdLCdtdvRCtvVCCCdd++=)(0∞=→CddvVdtd28Guess Solution!Let’s subtract out the steady-state solution:Guess solution is of the following form:stAetv =)(()()stststAesLCsAeRCAe20 ++=)()( tvVtvddC+=22)(dtvdLCdtdvRCtvVVdddd+++=22)(0dtvdLCdtdvRCtv ++=()210 LCsRCsAest++=210 LCsRCs ++=1529Again We’re Back to AlgebraOur guess is valid if we can find values of “s”that satisfy this equation:The solutions are:This is the same equation we solved in the last lecture!There we found three interesting cases:210 LCsRCs ++=12−ζ±ζ−=τs0)(2)(12=τ+ζτ+
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