First Order Circuit Analysis: Analyzing Circuits with Energy Storing Elements using ODEs, Slides of Electrical Circuit Analysis

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RC & RL 1st Order Ckts Docsity.com C&L Summary e The important (dual) relationships for capacitors and inductors are as follows: di(t) ae) = dt v(t) = - | i(x) dx i(t) = ; i v(x) dx ” do(t) ~‘di(t) P(t) = Co(t)— P(t) = Litt) W(t) = 1/2Cv°(t) W,(t) = 1/2L?(0) * The passive sign convention is used with capacitors and inductors. ¢ In dc steady state a capacitor looks like an open circuit and an inductor looks like a short circuit. e Leakage resistance is present in practical capacitors and inductors. e When capacitors are interconnected, their equivalent capacitance is determined as follows: Capacitors in series combine like resistors in parallel and capacitors in parallel combine like resistors in series. e¢ When inductors are interconnected, their equivalent inductance is determined as follows: Inductors in series combine like resistors in series and inductors in parallel combine like resistors in parallel. Docsity.com Circuits with L’s and/or C’s • Conventional DC Analysis Using Mathematical Models Requires The Determination of (a Set of) Equations That Represent the Circuit Response • Example; In Node Or Loop Analysis Of Resistive Circuits One Represents The Circuit By A Set Of Algebraic Equations The Ckt The DC Math Model Analysis iv  =G Docsity.com Ckt w/ L’s & C’s cont. • When The Circuit Includes Inductors Or Capacitors The Models Become Linear Ordinary Differential Equations (ODEs) • Thus Need ODE Tools In Order To Analyze Circuits With Energy Storing Elements – Recall ODEs from ENGR25 – See Math-4 for More Info on ODEs Docsity.com First Order Circuit Analysis • A Method Based On Thévenin Will Be Developed To Derive Mathematical Models For Any Arbitrary Linear Circuit With One Energy Storing Element • This General Approach Can Be Simplified In Some Special Cases When The Form Of The Solution Can Be Known BeforeHand – Straight-Forward ParaMetric Solution Docsity.com Flash Ckt Transient Response • The Voltage Across the Flash-Ckt Storage Cap as a Function of TIME  Note That the Discharge Time (the Flash) is Much Less Than the Charge-Time Docsity.com General Form of the Response • Including the initial conditions the model equation for the capacitor-voltage or the inductor-current will be shown to be of the form  xp(t) ≡ ANY Solution to the General ODE • Called the “Particular” Solution  xc(t) ≡ The Solution to the General Eqn with f(t) =0 • Called the “Complementary Solution” or the “Natural” (unforced) Response • i.e., xc is the Soln to the “Homogenous” Eqn  This is the General Eqn  Now By Linear Differential Eqn Theorem (SuperPosition) Let ( ) ( ) ( )tftx dt tdx =+τ ( ) ( ) 0=+ tx dt tdxτ Docsity.com 1st Order Response Eqns • Given xp and xc the Total Solution to the ODE ( ) ( ) ( )txtxtx cp +=  Consider the Case Where the Forcing Function is a Constant • f(t) = A  Now Solve the ODE in Two Parts  For the Particular Soln, Notice that a CONSTANT Fits the Eqn: ( ) ( ) ( ) ( ) 0=+ =+ tx dt tdx Atx dt tdx c c p p τ τ ( ) ( ) 0 so and 1 = = dt tdx Ktx p p Docsity.com Effect of the Time Constant Tangent reaches x-axis in one time constant Decreases 63.2% after One Time Constant time Drops to 1.8% after 4 Time Constants 0183.04 =−e Docsity.com Large vs Small Time Constants • Larger Time Constants Result in Longer Decay Times – The Circuit has a Sluggish Response Quick to Steady-State Docsity.com Time Constant Example • Charging a Cap  Use KCL at node-a  Now let • vC(t = 0 sec) = 0 V • vS(t)= VS (a const)  Rearrange the KCL Eqn For the Homogenous Case where Vs = 0  Thus the Time Constant − +vS RS a b C + vc _ dt dvC C S SC R vv − S S S CC S SCC R v R v dt dvC R vv dt dvC =+ = − + 0 SC C CRv dtdv 1 −= SCR=τ ( ) ( ) ( ) τ τ t c c c eKtx tx dttdx −=⇒−= 2 1 Docsity.com Example • Given the RC Ckt At Right with – Initial Condition (IC): • v(0−) = VS/2 • Find v(t) for t>0 • Looks Like a Single E-Storage Ckt w/ a Constant Forcing Fcn – Assume Solution of Form  Model t>0 using KCL at v(t) after switch is made )0();( 0,)( 211 21 +=+∞= >+= − xKKxK teKKtx t τ 0)()( =+− t dt dvC R Vtv S  Find Time Constant; Put Eqn into Std Form • Multiply ODE by R sVtvtdt dvRC =+ )()( Docsity.com Example cont • Compare Std-Form with Model – Const Force Fcn Model  Note: the SS condition is Often Called the “Final” Condition (FC)  In This Case the FC  Now Use IC to Find K2 1Kxdt dx =+τ • In This Case sVtvtdt dvRC =+ )()(  Next Check Steady- State (SS) Condition • In SS the Time Derivative goes to ZERO s s V K Vtvt = →⇒∞→ 1 :confirms )( Docsity.com Example cont.2 • At t = 0+ – The Model Solution  The Total/General Soln  Recall The IC: v(0−) = VS/2 = v(0+) for a Cap • Then 12 0 21 )0( )0( KvK eKKv −=⇒ += − τ 2 2 )0( 2 2 12 S SS VK VVK KvK −= −= −=       −= − RC t S eVtv 5.01)(  Check ( ) sS VeVv 5.05.01)0( 0 =−=+ − ( ) sS VeVv =−=∞ ∞−5.01)(   ( ) or0,21 >+= − teKKtv t τ Docsity.com Solution Process Summary 1. ReWrite ODE in Standard Form – Yields The Time-Constant, τ 2. Analyze The Steady-State Behavior – Finds The Final Condition Constant, K1 3. Use the Initial Condition – Gives The Exponential PreFactor, K2 4. Check: Is The Solution Consistent With the Extreme Cases – t = 0+ – t → ∞ Docsity.com Solutions for f(t) ≠ Constant 1. Use KVL or KCL to Write the ODE 2. Perform Math Operations to Obtain a CoEfficient of “1” for the “Zeroth” order Term. This yields an Eqn of the form 3. Find a Particular Solution, xp(t) to the FULL ODE above – This depends on f(t), and may require “Educated Guessing” ( ) ) (yieldsττ tfx dt dx =+ Docsity.com Solutions for f(t) ≠ Constant 4. Find the total solution by adding the COMPLEMENTARY Solution, xc(t) to previously determined xp(t). xc(t) takes the form: – The Total Solution at this Point → 5. Use the IC at t=0+ to find K; e.g.; IC = 7 – Where M is just a NUMBER ( ) τtc Ketx −= ( ) ( )txKetx pt += − τ ( ) MKxKe p +=⇒+= − 707 0 τ Docsity.com Cap Exmp cont • Now The IC • If the Switch is Closed for a Long Time before t =0, a STEADY- STATE Condition Exists for NEGATIVE Times  vo(0−) by V-Divider  Recall Reln Between vo and vC for t ≥ 0 Vv VVv O O 3 8)0( 12 9 212 423 2)0( = = ++ = − − VVvv tvtv CC OC 812 9 6)0()0( )(3)( ==−=+ ⇒= Recall: Cap is OPEN to DC Docsity.com Cap Exmp cont • Step-3: Apply The IC  Now have All the Parameters needed To Write The Solution )0,80)( 0,)( 6.0 21 >+= >+= − − tVetv teKKtv s t C t C τ VvK vK vKK C C C 8)0( 0)( and )0( 2 1 21 ==∴ =∞= =+ + +  Note: For f(t)=Const, puttingthe ODE in Std Form yields τ and K1 by Inspection: 0)()( 21 =++ cC vtdt dvCRR 1Kτ  Recall vo = (1/3)vC 0,V 3 8)( 6.0 >     = − tetv s t o Docsity.com Thevenin/Norton Techniques • Obtain The Thevenin Voltage Across The Capacitor, Or The Norton Current Through The Inductor  With This approach can Analyze a SINGLE-LOOP, or SINGLE-NODE Ckt to Find • Time Constant using RTH • Steady-State Final Condition using vTH (if vTH a constant) Circuit with resistances and sources Inductor or Capacitor a b Representation of an arbitrary circuit with one storage element Thévenin − +VTH RTH Inductor or Capacitor a b Docsity.com Find ODE Soln By Thevenin • Recognize the Solution Parameters • For Capacitor – τ = RTHC – K1 (Final Condition) = vTH = vOC = xp • For Inductor – τ = L/RTH – K1 (Final Condition) = vTH/RTH = iSC = iN = xp • In Both Cases Use the vC(0−) or iL(0−) Initial Condition to Find Parameter K2 Docsity.com Inductor Example • The Variable Of Interest Is The Inductor Current • The Thévenin model  Since this Ckt has a CONSTANT Forcing Function (24V), Then The Solution Is Of The Form Ω6 Ω6 Ω6 Ω6 H3 V24 − + )(tiO 0=t 0>t;(t)iFind O TH TH O O TH R vi dt di R L =+ 0;)( 21 >+= − teKKti t O τ  Next Construct the Thévenin Equivalent for the Inductor “Driving Circuit” Docsity.com Inductor Example cont  The f(t)=const ODE in Standard Form [ ]( ) sH R L R v TH TH TH 3.0 10 3 106666 0 = Ω == Ω=++= = τ  The Solution Substituted into the ODE at t = 0+ Ω6 Ω6 Ω6 Ω6V24 − + 0>t Thevenin for t>0 at inductor terminals a b 0;03.0 >=+ ti dt dis OO 0 3.0 3.0 3.0213.02 =      ++      − −− tt eKKeK 0;)( 3.02 >= − teKti t O  From Std Form K1 = 0 • So  From This Ckt Observe Docsity.com First Order Inductor Ciruit Transient Solution 0.0 0.5 1.0 1.5 2.0 2.5 3.0 3.5 4.0 4.5 5.0 5.5 -0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8 Time (s) C ur re nt (A ) i(t) (A) file = Engr44_Lec_06-1_Last_example_Fall03..xls [ ] 0;333.5)( 3.0 >= − teAti s t O ( ) ( ) ( ) s A s A s A dt di s i dt di s i dt di s i dt dii dt dis t O O t OOO OO O O 78.17 33 1016 103 316 103 0 103 3.0 03.0 0 0 ≈ ⋅ ⋅ ==∴ +− =⇒ − =∴ − =⇒=+ += += Docsity.com WhiteBoard Work • Let’s Work This Problem t=0 24V 2Ω 4Ω 4Ω 8Ω6Ω VC(t) (1/4)H + -  Well, Maybe NEXT time… Docsity.com WhiteBoard Work Docsity.com Differential Eqn Approach cont • Up to Now  Next Use the Initial Condition  So Finally  If we Write the ODE in Proper form We can Determine By Inspection τ and K1 ( ) 012 0 aa t eK a Aty − += 102 021 0 0 21 0 or )0( )0( KyK yKK yeKKy yy −= =+ =+=+ =+ ( ) 01 0 0 0 aa t e a Ay a Aty −       −+= 00 1 01 a Ay dt dy a aAya dt dya =+⇒=+ τ 1K Docsity.com Inductor Example  The Model for t>0 → KVL on single-loop ckt  Rewrite In Std Form L )(1 ti − + Lv S 9 1 =τ  Recognize Time Const 0)()( 18 2 1 1 =+ tit dt di )0();( 0,)( 12111 211 +=+∞= >+= − iKKiK teKKti t τ ( ) 0)(126 11 =++ tidt diL  For The Ckt Shown Find i1(t) for t>0  Assume Solution of the Form Docsity.com Inductor Example cont • Examine Std-Form Eqn to Find K1  For Initial Conditions Need the Inductor Current for t<0  Again Consider DC (Steady-State) Condition for t<0  The SS Ckt Prior to Switching • Recall An Inductor is a SHORT to DC 0 0)()( 9 1 1 1 1 =⇒ =+ K tit dt di )0(1 −i  So AVi 1 12 12)0(1 =Ω =− Docsity.com