Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu Engineering 43 Chp 7 – 1st Order Ckts Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege.edu

Introduction  Transient Circuits In Circuits Which Contain Inductors And Capacitors, Currents And Voltages Cannot Change Instantaneously Even The Application, Or Removal, Of Constant Sources Creates Transient (Time- Dependent) Behavior

1st & 2nd Order Circuits FIRST ORDER CIRCUITS SECOND ORDER CIRCUITS Circuits That Contain ONE Energy Storing Element Either a Capacitor or an Inductor SECOND ORDER CIRCUITS Circuits With TWO Energy Storing Elements in ANY Combination

Circuits with L’s and/or C’s Conventional 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 Analysis The Model The Ckt

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 See Math-4 for More Info on ODEs

First Order Circuit Analysis A Method Based On Thevenin 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

Basic Concept Inductors And Capacitors Can Store Energy Under Certain Conditions This Energy Can Be Released RATE OF ENERGY RELEASE Depends on the parameters Of The Circuit Connected To The Terminals Of The Energy Storing Element

Example: Flash Circuit The Battery, Vs, Charges the Cap To Prepare for a Flash Moving the Switch to the Right “Triggers” The Flash i.e., The Cap Releases its Stored Energy to the Lamp Say “Cheese”

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

General Form of the Response Including the initial conditions the model 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

1st Order Response Eqns Given xp and xc the Total Solution to the ODE 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:

1st Order Response Eqns cont Sub Into the General (Particular) Eqn xp and dxp/dt Next Separate the Variables & Integrate Recognize LHS as a Natural Log; so Next, Divide the Homogeneous Eqn by xc(t) to yield Next Take “e” to The Power of the LHS & RHS

1st Order Response Eqns cont Then For This Solution Examine Extreme Cases t =0 t → ∞ Now Rename 1/a as the TIME CONSTANT,  Thus the Solution for a Constant Forcing Fcn The Latter Case is Called the Steady-State Response All Time-Dependent Behavior has dissipated

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

Large vs Small Time Constants Larger Time Constants Result in Longer Decay Times The Circuit has a Sluggish Response Slow to Steady-State Quick to Steady-State

Time Constant Example Charging a Cap Now let vC(0) = 0 V vS(t)= VS (a const) Rearrange the KCL Eqn For the Homogenous Case where Vs = 0 Use KCL at node-a Thus the Time Constant

Time Constant Example cont Charging a Cap “Fully” Charged Criteria vC >0.99VS OR The Solution Can be shown to be

Differential Eqn Approach Conditions for Using This Technique Circuit Contains ONE Energy Storing Device The Circuit Has Only CONSTANT, INDEPENDENT Sources The Differential Equation For The Variable Of Interest is SIMPLE To Obtain Normally by Using Basic Analysis Tools; e.g., KCL, KVL, Thevenin, Norton, etc. The INITIAL CONDITION For The Differential Equation Is Known, Or Can Be Obtained Using STEADY STATE Analysis prior to Switching

Differential Eqn Approach cont Math Property When all Independent Sources Are CONSTANT, then for ANY variable y(t); i.e., v(t) or i(t), in The Circuit The Solution takes the Form The Solution Strategy Use The DIFFERENTIAL EQUATION And The INITIAL & FINAL Conditions To Find The Parameters K1 and K2

Differential Eqn Approach cont If the ODE for y is Known to Take This Form Then Sub Into ODE Equating the TRANSIENT (exponential) and CONSTANT Terms Find We Can Use This Structure to Find The Unknowns. If:

Differential Eqn Approach cont Up to Now So Finally If we Write the ODE in Proper form We can Determine By Inspection  and K1 Next Use the Initial Condition

Example Given the Ckt At Right with Find v(t) for t>0 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 Find Time Constant; Put Eqn into Std Form Multiply ODE by R

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 In This Case Next Check Steady-State (SS) Condition In SS the Time Derivative goes to ZERO Now Use IC to Find K2

Example cont.2   At t = 0+ The Complete Solution The Model Solution The Complete Solution Recall The IC: v(0-) = VS/2 = v(0+) for a Cap Then Check  

Inductor Exmpl Find i(t) Given KVL Find i(t) Given i(0−) = 0 Recognize Single E-Storage Ckt w/ a Constant Forcing Fcn Assume Solution of Form To Find the ODE Use KVL for Single-Loop ckt Now Consider IC By Physics, The Current Thru an Inductor Can NOT Change instantaneously In This Case y→i

Inductor cont Casting ODE in Standard form Next Using IC Recognize Time Const Also Note FC Thus the ODE Solution Thus

Solution Process Summary Cast Eqn Into Standard Form Yields The Time-Constant,  Analyze The Steady-State Behavior Finds The Final Condition Constant, K1 Use the Initial Condition Gives The Exponential PreFactor, K2 Check: Is The Solution Consistent With the Extreme Cases t = 0+ t → 

Capacitor Example For Ckt Below Find vo(t) for t>0 Assume a Solution of the Form for vc At t=0+ Apply KCL

Cap Exmp cont Step1: By Inspection of the Regrouped KCL Eqn Recognize  Step-2: Consider The Steady-State In This Case After the Switch Opens The Energy Stored in the Cap Will be Dissipated as HEAT by the Resistors Now Examine the Reln Between vo and vC A V-Divider So K1

Recall: Cap is OPEN to DC 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 Recall: Cap is OPEN to DC vo(0−) by V-Divider Recall Reln Between vo and vC

Cap Exmp cont Step-3: Apply The IC Recall vo = (1/3)vC Now have All the Parameters needed To Write The Solution Note: With the ODE in Std Form Can Find  and K1 By Inspection

Inductor Example For The Ckt Shown Find i1(t) for t>0 Assume Solution of the Form The Model for t>0 → KVL on single-loop ckt Recognize Time Const Rewrite In Std Form

Inductor Example cont Examine Std-Form Eqn to Find K1 The SS Ckt Prior to Switching Recall An Inductor is a SHORT to DC For Initial Conditions Need the Inductor Current for t<0 Again Consider DC (Steady-State) Condition for t<0 So

Inductor Example cont.1 The Answer Now Use Step-3 To Find K2 from IC Remember Current Thru an Inductor Must be Time-Continuous Recall that K1 was zero Construct From the Parameters The ODE Solution

Thevenin/Norton Techniques Obtain The Thevenin Voltage Across The Capacitor, Or The Norton Current Through The Inductor Thevenin With This approach can Analyze a SINGLE-LOOP, or SINGLE-NODE Ckt to Find Time Constant using RTH Steady-State Final Condition using vTH

Thevenin Models for ODE KCL at node a KVL for Single Loop

Find ODE Soln By Thevenin Break Out the Energy Storage Device (C or L) as the “Load” for a Driving Circuit Analyze the Driving Ckt to Arrive at it’s Thevenin (or Norton) Equivalent ReAttach The C or L Load Use KCL or KVL to arrive at ODE Put The ODE in Standard Form

Find ODE Soln By Thevenin Recognize the Solution Parameters For Capacitor  = RTHC K1 (Final Condition) = vTH = vOC For Inductor  = L/RTH K1 (Final Condition) = vTH/RTH = iSC = iN In Both Cases Use the Initial Condition to Find Parameter K2

Inductor Example Then The Solution Is Of The Form Next Construct the Thevenin Equivalent for the Inductor “Driving Circuit” The Variable Of Interest Is The Inductor Current The Thevenin model

Inductor Example cont Then The ODE in Standard Form The Solution Substituted into the ODE at t = 0+ From This Ckt Observe From Std Form K1 = 0 So

Inductor Example cont.2 Analyzing Ckt with 3H as Short Reveals The Reader Should Verify the Above Now Find K2 Assume Switch closed for a Long Time Before t =0 Inductor is SHORT to DC Then the Entire Solution

Untangle to find iO(0−) Untangle Analyze Be Faithful to Nodes

WhiteBoard Work Well, Maybe NEXT time… Let’s Work This Problem 7e prob 6.16

WhiteBoard Work None Today