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 amplifiers19Low-Noise AmplifierD. Shaeffer, T. Lee, ISSCC’97320thin-Film Bulk Acoustic Resonator (FBAR)RF MEMSAgilent Technologies (IEEE ISSCC 2001)Q > 1000Resonates at 1.9 GHzCell phone duplexerC0CxRxLxC1C2R0PadThin Piezoelectric Film 21Series 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)422LCR 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 )(0τt23LCR 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++=524Initial 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∞=→CddvVdtd25Guess 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 ++=626Again 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+ ss01ω=τζ=21Q⎪⎩⎪⎨⎧>=<=ζ111UnderdampedCritically dampedOverdamped27General CaseSolutions 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−=+=+021728Final Solution (General Case)Solve for A and B:212121212111)1(1ssVssssVssVAVBssVAdddddddddd−=−+−−=−−=−−=ddVAssA −=−+21⎟⎟⎟⎟⎠⎞⎜⎜⎜⎜⎝⎛⎟⎟⎠⎞⎜⎜⎝⎛−−−=tstsddCessessVtv221121111)(() ()tsssVsstsssVVtvddddddC 22121121exp1exp1)(−+−−+=29Overdamped Caseζ > 1: Time constants are real and negative0)1(1212<⎩⎨⎧=−ζ±ζ−τ=sss211=ζ==τddV830Critically Dampedζ > 1: Time constants are real and equalτ−=−ζ±ζ−τ=1)1(12s()τ−τ−→ζ−−=//11)(limttddCteeVtv111=ζ==τddV31UnderdampedNow the s values are complex conjugatejbasjbas−=+=21()()tjbaBtjbaAVtvddC)(exp)(exp)(−+++=() ( )()jbtBjbtAeVtvatddC−++= expexp)(BssVssssVssVAdddddd=−=−=−−=212112*2*1*111() ( )()jbtAjbtAeVtvatddC−++= expexp)(*932Underdamped (cont)So we have:() ( )()jbtAjbtAeVtvatddC−++= expexp)(*()[]jbtAeVtvatddCexpRe2)( +=()φ+ω+= tAeVtvatddCcos2)(211ssVAdd+=211ssVdd+∠=φ33Underdamped PeakingFor ζ < 1, the step response overshoots:5.11===ζτddV1034Extremely Underdamped01.11===ζτddV35Common Source Amplifer: Ai(jω)DC Bias is problematic: what sets VGS?1136Common Source: Discrete BiasingVDDR1R2CSRL+vsvlWith ideal MOS37CS Short-Circuit Current GainTransfer function:())(/1)(gdgsmgdmiCCjgCjgjA+−=ωωω1238Magnitude Bode Plot ωT Transition frequency:Current gain = 139MOS Unity Gain FrequencySince the zero occurs at a higher frequency than pole, assume ithas negligible effect:1()migs gdgAjC Cω≈=+()mTgsgdgCCω=+2()()3223ox GS TmGSTTgsoxWCVVgVVLCLWLCμμω−−≈= =Performance improves with L2for long channel devices!For short channel devices the dependence is ~ L12()3~2GS TeffGS TTLVVEVVvLLLLLμμμωτ−−≈===Time to cross
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