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 &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 &Capacitors Can Store Energy • Under Certain Conditions This Energy Can Be Released • RATE of Energy Storage/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 (K1) 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. 1st Order Ckt Solution Plan • Use some form of KVL/KCL, and to develop an ODE • Isolate the “0th” order term which reveals the • Time Constant • Forcing Function • “EyeBall” the ODE to “Guesstimate the Particular Solution, or

18. 1st Order Ckt Solution Plan • CHECK that the particular Solution Satisfies the ODE • Separate the Variables in the Homogeneous Equation and Integrate to obtain the Complementary Solution or • A the particular and complementary solutions to obtain the total Solution;

19. 1st Order Ckt Solution Plan • The Solution Should include a Negative Exponential with an UnKnownPreFactor • Use the initial condition in the total solution to determine the Prefactor which completes the total solution • Check the total Solution for extreme cases: • t = 0+ • t → ∞

20. Example  1st Order CktSoln • For the Ckt Below Find for

29. Charging a Cap Time Constant Example • Now let • vC(t = 0 sec) = 0 V • vS(t)= VS (a const) • ReArrangethe KCL Eqn For the Homogenous Case where Vs = 0 Reserve Remainder for HW Study • Use KCL at node-a • Thus the Time Constant

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

31. 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 • Based On: Cap is OPEN, Ind is SHORT

32. 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

33. 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

34. 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  

35. 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

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

37. 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 → 

38. 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”

39. 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

40. 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

41. 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 • With xp = K1 = SSsoln

42. 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

43. 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

44. 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)

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

46. 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

47. 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

48. The Variable Of Interest Is The Inductor Current The Thévenin model Inductor Example • Since this Ckt has a CONSTANT Forcing Function of 0V (the 24V source is switched OUT at t = 0), Then The Solution Is Of The Form • Next Construct the ThéveninEquivalent for the Inductor “Driving Circuit”

49. Inductor Example cont • The f(t)=const ODE in Standard Form • The Solution Substituted into the ODE at t > 0 • From This Ckt Observe • Combining the K2 terms shows shows K1= 0, thus

50. 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 Indas Short Reveals • The Reader Should Verify the Above • Then the Entire Solution