Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

1. Engineering 43 RC & RL 1st Order Ckts Bruce Mayer, PE Licensed Electrical & Mechanical EngineerBMayer@ChabotCollege.edu

2. C&L Summary

3. Introduction  Transient Circuits • In Circuits Which Contain Inductors And Capacitors, Currents & Voltages CanNOT Change Instantaneously • Even The Application, Or Removal, Of ConstantSources Creates Transient (Time-Dependent) Behavior

4. 1st & 2nd Order Circuits • FIRST 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

5. 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 Analysis The DC Math Model The Ckt

6. 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 fromENGR25 • See Math-4 for More Info on ODEs

7. 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

8. 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

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

10. 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

11. Including the initial conditions the model equation for the capacitor-voltage or the inductor-current will be shown to be of the form General Form of the Response • 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

12. Given xp and xcthe Total Solution to the ODE 1st Order Response Eqns • 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:

13. Sub Into the General (Particular) Eqnxp and dxp/dt 1st Order Response Eqns cont • Next Separate the Variables & Integrate • Recognize LHS as a Natural Log; so • Next, Divide the Homogeneous (RHS=0) Eqn by xc(t) to yield • Next Take “e” to The Power of the LHS & RHS

14. Then 1st Order Response Eqns cont • For This Solution Examine Extreme Cases • t =0 • t → ∞ • Note that Units of TIME CONSTANT, , are Sec • Thus the Solution for a Constant Forcing Fcn • The Latter Case is Called the Steady-State Response • All Time-Dependent Behavior has dissipated

15. 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

16. 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

17. Charging a Cap Time Constant Example • Now let • vC(t = 0 sec) = 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

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

19. 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

20. 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 Example • Model t>0 using KCL at v(t) after switch is made • Find Time Constant; Put Eqn into Std Form • Multiply ODE by R

21. Compare Std-Form with Model Const Force Fcn Model Example cont • 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

22. At t = 0+ The Model Solution Example cont.2 • The Total/General Soln • Recall The IC: v(0−) = VS/2 = v(0+) for a Cap • Then • Check  

23. Find i(t) Given i(0−) = 0 Recognize Single E-Storage Ckt w/ a Constant Forcing Fcn Assume Solution of Form KVL Inductor Exmpl • 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 x→i

24. Casting ODE in Standard form Inductor cont • Next Using IC (t = 0+) • Recognize Time Const • Also Note FC • Thus the ODE Solution • Thus

25. Solution Process Summary • ReWrite ODE in 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 → 

26. Solutions for f(t) ≠ Constant • Use KVL or KCL to Write the ODE • Perform Math Operations to Obtain a CoEfficient of “1” for the “Zeroth” order Term. This yields an Eqn of the form • Find a Particular Solution, xp(t) to the FULL ODE above • This depends on f(t), and may require “Educated Guessing”

27. Solutions for f(t) ≠ Constant • Find the total solution by adding the COMPLEMENTARY Solution, xc(t) to previously determined xp(t). xc(t) takes the form: • The Total Solutionat this Point → • Use the IC at t=0+ to find K; e.g.; IC = 7 • Where M is just a NUMBER

28. Capacitor Example • For Ckt Below Find vo(t) for t>0 (note f(t) = const; 12V) • Assume a Solution of the Form for vc • At t=0+ Apply KCL

29. Step1: By Inspection of the ReGrouped KCL Eqn Recognize  Cap Exmp cont • 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 xp = K1

30. Now The IC If the Switch is Closed for a Long Time before t =0, a STEADY-STATE Condition Exists for NEGATIVE Times Cap Exmp cont Recall: Cap is OPEN to DC • vo(0−) by V-Divider • Recall Reln Between vo and vC for t ≥ 0

31. Step-3: Apply The IC Cap Exmp cont • Recall vo = (1/3)vC • Note: For f(t)=Const, puttingtheODE in Std Form yields  and K1by Inspection: • Now have All the Parameters needed To Write The Solution

32. Obtain The Thevenin Voltage Across The Capacitor, Or The Norton Current Through The Inductor Thevenin/Norton Techniques Thévenin • 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)

33. Thevenin Models for ODE • KCL at node a • KVL for Single Loop

34. 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 Thévenin (or Norton) Equivalent • ReAttach The C or L Load • Use KCL or KVL to arrive at ODE • Put The ODE in Standard Form

35. Find ODE Soln By Thevenin • Recognize the Solution Parameters • For Capacitor •  = RTHC • K1 (Final Condition) = vTH = vOC= xp • For Inductor •  = 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

36. The Variable Of Interest Is The Inductor Current The Thévenin model Inductor Example • Since this Ckt has a CONSTANT Forcing Function (24V), Then The Solution Is Of The Form • Next Construct the Thévenin Equivalent for the Inductor “Driving Circuit”

37. Inductor Example cont • The f(t)=const ODE in Standard Form • The Solution Substituted into the ODE at t = 0+ • From This Ckt Observe • From Std Form K1 = 0 • So

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

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

41. WhiteBoard Work • Let’s Work This Problem • Well, Maybe NEXT time…

42. WhiteBoard Work None Today

43. 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 FINAL & INITIAL Conditions To Find The Parameters K1 and K2

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

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

46. 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

47. Examine Std-Form Eqn to Find K1 Inductor Example cont • 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

48. Now Use Step-3 To Find K2 from IC Remember Current Thru an Inductor Must be Time-Continuous 0 Inductor Example cont.1 • The Answer • Recall that K1 was zero • Construct From the Parameters The ODE Solution