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Berkeley ELENG 105 - Lecture Notes 5

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Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5Lecture 5: 2ndOrder Circuits in the Time DomainPhysics of ConductionProf. NiknejadDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadLecture Outlinel Second order circuits:– Time Domain Responsel Physics of ConductionDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadSeries LCR Step Responsel Consider the transient response of the following circuit when we apply a step at input l Without inductor, the cap charges with RC time constant (EECS 40)l Where does the inductor come from? – Intentional inductor placed in series – Every physical loop has inductance! (parasitic)Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadLCR Step Response: L Smalll We know the steady-state response is a constant voltage of Vddacross capacitor (inductor is short, cap is open)l For the case of zero inductance, we know solution is of the following form:)1()(/0τtddeVtv−−=ddVtv )(0τtDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadLCR Circuit ODEl Apply KVL to derive governing dynamic equations:l Inductor and capacitor currents/voltages take the form:l We have the following 2ndorder ODE:)()()()( tvtvtvtvLRCs++=dtdiLvdtdvCiiLCC===22dtvdLCdtdvCdtdLvCCL==dtdvRCiRvCR==22)()(dtvdLCdtdvRCtvtvCCCs++=Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadInitial Conditionsl For the solution of a second order circuit, we need to specify to initial conditions (IC):l For t > 0, the source voltage is Vdd. We can now solve for the following non-homogeneous equation subject to above IC:l Steady state:V0)0()0(V0)0()0(0====LCiivv22)(dtvdLCdtdvRCtvVCCCdd++=)(0∞=→CddvVdtdDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadGuess Solution!l Let’s subtract out the steady-state solution:l Guess solution is of the following form:stAetv =)(()()stststAesLCsAeRCAe20 ++=)()( tvVtvddC+=22)(dtvdLCdtdvRCtvVVdddd+++=22)(0dtvdLCdtdvRCtv ++=()210 LCsRCsAest++=210 LCsRCs ++=Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadAgain We’re Back to Algebral Our guess is valid if we can find values of “s” that satisfy this equation:l The solutions are:l This is the same equation we solved last lecture!l There we found three interesting cases:210 LCsRCs ++=12−±−= ζζτs0)(2)(12=++ τζτ ss01ωτ =ζ21=Q><=111ζUnderdampedCritically DampedOverdampedDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadGeneral Casel Solutions are real or complex conj depending on if ?> 1 or ?< 1 =−±−=212)1(1sss ζζτ()()tsBtsAVtvddC 21expexp)(++=0)0(=++=BAVvddC( ) ( )0expexp0)()0(022110=+⇒====ttCtsBstsAsdttdvCiddVBABsAs−=+=+021Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadFinal Solution (General Case)l Solve for A and B:212121212111)1(1ssVssssVssVAVBssVAdddddddddd−=−+−−=−−=−−=ddVAssA −=−+21−−−=tstsddCessessVtv212121111)(( ) ( )tsssVsstsssVVtvddddddC 22121121exp1exp1)(−+−−+=Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadOverdamped Casel ?> 1: Time constants are real and negative0)1(1212<=−±−=sss ζζτ211===ζτddVDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadCritically Dampedl ?> 1: Time constants are real and equalτζζτ1)1(12−=−±−=s()ττζ//11)(limttddCteeVtv−−→−−=111===ζτddVDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadUnderdampedl Now the s values are complex conjugatejbasjbas−=+=21()()tjbaBtjbaAVtvddC)(exp)(exp)(−+++=()()()jbtBjbtAeVtvatddC−++= expexp)(BssVssssVssVAdddddd=−=−=−−=212112*2*1*111()()()jbtAjbtAeVtvatddC−++= expexp)(*Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadUnderdamped (cont)l So we have:()()()jbtAjbtAeVtvatddC−++= expexp)(*()[]jbtAeVtvatddCexpRe2)( +=()φω ++= tAeVtvatddCcos2)(211ssVAdd+=211ssVdd+∠=φDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadUnderdamped Peakingl For ?< 1, the step response overshoots:5.11===ζτddVDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadExtremely Underdamped01.11===ζτddVDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadOhm’s Lawl One of the first things we learn as EECS majors is:l Is this trivial? Maybe what’s really going on is the following:l In the above Taylor exansion, if the voltage is zero for zero current, then this is generally validl The range of validity (radius of convergence) is the important question. It turns out to be VERY large!RIV×=IfIfIffIfV )0('...2/)0('')0(')0()(2≈+++==Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadOhm’s Law Revisitedl In Physics we learned:l Is this also trivial? Well, it’s the same as Ohm’s law, so the questions are related. For a rectangular solid:l Isn’t it strange that current (velocity) is proportional to Force?l Where does conductivity come from?EJσ=IRIALVLVAIJ ====σσDepartment of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadConductivity of a Gasl Electrical conduction is due to the motion of positive and negative chargesl For water with pH=7, the concentration of hydrogen H+ions (and OH-) is:l Typically, the concentration of charged carriers is much smaller than the concentration of neutral moleculesl The motion of the charged carriers (electrons, ions, molecules) gives rise to electrical conduction 313323-103-107cm106cm1002.610mole/cm10mole/L10−−−×=××==Department of EECS University of California, BerkeleyEECS 105 Fall 2003, Lecture 5 Prof. A. NiknejadCollisions in Gasl At a temperate T, each charged carrier will move in a random direction and velocity until it encounters a neutral molecule or another charged carrierl Since the


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Berkeley ELENG 105 - Lecture Notes 5

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