1EE105 - Fall 2006Microelectronic Devices and CircuitsProf. Jan M. Rabaey (jan@eecs)Lecture 15: Frequency Domain + Second Order2Overview Last lecture– Frequency Domain Continued (Read § 10) This lecture– Second Order Networks (continued)23Second Order Circuits The series resonant circuit is one of the most important elementary circuits:4Some Key PropertiesQ10=ωωΔQjQjLRjLRjjH022000022000)(ωω+ω−ωωω=ωωω+ω−ωωωω=ω1=Q10=Q100=Q0ω1)(0=ωjH0)0( =HωΔLC120≡ωRLQ0ω≡2200 000112424jjQQQQωω ωωωω22=− ± − =− ± −35More “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=21Q6Second Order Circuit Bode Plot Quadratic 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 poles47Bode Plot: Damped Case The 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°8Underdamped Case For 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)()(log2022jjj59Underdamped Mag Plot ζ=1ζ=0.01ζ=0.1ζ=0.2ζ=0.4ζ=0.6ζ=0.810Underdamped Phase The 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)()( jj611Phase Bode Plotζ=0.010.10.20.40.60.8ζ=112Bode Plot Guidelines In 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 dB The phase is much more difficult. Note for ζ = 0, the phase response is a step function For ζ = 1, the phase is two real poles at a fixed frequency For 0 < ζ < 1, the plot should go somewhere in between!Qjjjj1212)()(12)()(22=ζ=+ζ+=+ζωτ+ωτ713Low-Noise Amplifier D. Shaeffer, T. Lee, ISSCC’9714thin-Film Bulk Acoustic Resonator (FBAR)RF MEMS Agilent Technologies (IEEE ISSCC 2001) Q > 1000 Resonates at 1.9 GHz Cell phone duplexerC0CxRxLxC1C2R0PadThin Piezoelectric Film815Series 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)16LCR Step Response: L Small We 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)(0τt917LCR Circuit ODE Transient 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++=18Initial Conditions For 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∞=→CddvVdtd1019Guess 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 ++=20Again We’re Back to Algebra Our 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+ ss01ω=τζ=21Q⎪⎩⎪⎨⎧>=<=ζ111UnderdampedCritically dampedOverdamped1121General Case Solutions are real or complex conj depending on if ζ > 1 or ζ < 1 ⎩⎨⎧=−ζ±ζ−τ=212)1(1sss() ( )tsBtsAVtvddC 21expexp)( ++=0)0( =++= BAVvddC() ()0expexp0)()0(022110=+⇒====ttCtsBstsAsdttdvCiddVBABsAs−=+=+ 02122Final Solution (General Case) Solve for A and B:212121212111)1(1ssVssssVssVAVBssVAdddddddddd−=−+−−=−−=−−=ddVAssA −=−+21⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛−−−=tstsddCessessVtv221121111)(() ()tsssVsstsssVVtvddddddC 22121121exp1exp1)(−+−−+=1223Overdamped Case ζ > 1: Time constants are real and negative0)1(1212<⎩⎨⎧=−ζ±ζ−τ=sss211=ζ==τddV24Critically Damped ζ = 1: Time constants are real and equalτ−=−ζ±ζ−τ=1)1(12s()τ−τ−→ζ−−=//11)(limttddCteeVtv111=ζ==τddV1325Underdamped Now the s values are complex conjugatejbasjbas−=+=21()()tjbaBtjbaAVtvddC)(exp)(exp)( −+++=() ( )()jbtBjbtAeVtvatddC−++= expexp)(BssVssssVssVAdddddd=−=−=−−=212112*2*1*111() ( )()jbtAjbtAeVtvatddC−++= expexp)(*26Underdamped (cont) So we have:() ( )()jbtAjbtAeVtvatddC−++= expexp)(*()[]jbtAeVtvatddCexpRe2)( +=()φ+ω+= tAeVtvatddCcos2)(211ssVAdd+=211ssVdd+∠=φ1427Underdamped Peaking For ζ < 1, the step response overshoots:5.11===ζτddV28Extremely Underdamped01.11===ζτddV15Back to AmplifiersFrequency Response ofCommon Source Amplifiers (Read § 10.3.2, 10.4.3-5 )30Common Source: Discrete BiasingVDDR1R2CSRL+vsvlWith ideal MOS1631Where is the Pole?32Common Source Amplifier: Ai(jω)DC Bias
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