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Engineering 43 Capacitors & Inductors Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] Engineering-43: Engineering Circuit Analysis 1 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Capacitance & Inductance Introduce Two Energy STORING Devices • Capacitors • Inductors Outline • Capacitors – Store energy in their ELECTRIC field (electroStatic energy) – Model as circuit element • Inductors – Store energy in their MAGNETIC field (electroMagnetic energy) – Model as circuit element • Capacitor And Inductor Combinations – Series/Parallel Combinations Of Elements Engineering-43: Engineering Circuit Analysis 2 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt The Capacitor First of the EnergyStorage Devices Basic Physical Model Circuit Representation • Note use of the PASSIVE SIGN Convention Details of Physical Operation Described in PHYS4B & ENGR45 Engineering-43: Engineering Circuit Analysis 3 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Capacitance Defined Consider the Basic Physical Model Where • A The Horizontal Plate-Area, m2 • d The Vertical Plate Separation Distance, m • 0 “Permittivity” of Free Space; i.e., a vacuum – A Physical CONSTANT – Value = 8.85x10-12 Farad/m The Capacitance, C, of the Parallel-Plate 0 A Structure C w/o Dielectric d Engineering-43: Engineering Circuit Analysis 4 Then What are the UNITS of Capacitance, C Typical Cap Values → “micro” or “nano” Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Capacitor Circuit Operation Recall the Circuit Representation LINEAR Caps Follow the Capacitance Law; in DC Q CVc Where The Basic CircuitCapacitance Equation qt Cvc t Engineering-43: Engineering Circuit Analysis 5 • Q The CHARGE STORED in the Cap, Coulombs • C Capacitance, Farad • Vc DC-Voltage Across the Capacitor Discern the Base Units for Capacitance Coulomb Farad Volt Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt “Feel” for Capacitance Pick a Cap, Say 12 µF Recall Capacitor Law Q CVc Solving for Vc Q 15x10 3 Coul Vc C 12x10 6 Coul/Volt Vc 1250 V!!! Now Assume That The Cap is Charged to hold 15 mC • Find V c Engineering-43: Engineering Circuit Analysis 6 Caps can RETAIN Charge for a Long Time after Removing the Charging Voltage • Caps can Be DANGEROUS! Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Forms of the Capacitor Law The time-Invariant Cap Law Q CVc t vC t vC i y dy C dz If vC at − = 0, then the traditional statement of Leads to DIFFERENTIAL the Integral Law Cap Law dvC t dqt i t C dt dt 1 t vC t i y dy C The Differential Suggests If at t0, vC = vC(t0) (a SEPARATING Variables KNOWN value), then the Integral Law becomes i t dt CdvC t 1 t0 1 t vC t i y dy i y dy Leads to The C C t0 INTEGRAL form of the 1 t vC t vC t0 i y dy Capacitance Law C t0 Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis 7 [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Capacitor Integral Law Express the VOLTAGE Across the Cap Using the INTEGRAL Law qt 1 t vC t iC y dy C C If i(t) has NO Gaps in its i(t) curve then 1 t t lim vC t t lim i y dy t 0 t 0 C Even if i(y) has VERTICAL Jumps: lim vC t t vC t t 0 Engineering-43: Engineering Circuit Analysis 8 Thus a Major Mathematical Implication of the Integral law The Voltage Across a Capacitor MUST be Continuous An Alternative View • The Differential Reln dvC t iC t C dt If vC is NOT Continous then dvC/dt → , and So iC → . This is NOT PHYSICALLYBrucepossible Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Capacitor Differential Law Express the CURRENT “Thru” the Cap Using the Differential Law dv t dqt C C dt dt If vC = Constant Then iC t iC 0 This is the DC Steady-State Behavior for a Capacitor Engineering-43: Engineering Circuit Analysis 9 Thus a Major Mathematical Implication of the Differential Law A Cap with CONSTANT Voltage Across it Behaves as an OPEN Circuit Cap Current • Charges do NOT flow THRU a Cap – Charge ENTER or EXITS The Cap in Response to Voltage CHANGES Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Capacitor Current Charges do NOT flow THRU a Cap Charge ENTER or EXITS The Capacitor in Response to the Voltage Across it • That is, the Voltage-Change DISPLACES the Charge Stored in The Cap – This displaced Charge is, to the REST OF THE CKT, Indistinguishable from conduction (Resistor) Current Thus The Capacitor Current is Called the “Displacement” Current Engineering-43: Engineering Circuit Analysis 10 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Capacitor Summary The Circuit Symbol iC vC Note The Passive Sign Convention Compare Ohm’s Law and Capactitance Law Cap Ohm iC (t ) C t dvc (t ) dt 1 vC (t ) iC ( x)dx C 1 vR R vR Ri R iR Engineering-43: Engineering Circuit Analysis 11 From Calculus, Recall an Integral Property t t0 t t0 f x dx f x dx f x dx Now Recall the Long Form of the Integral Relation t t 1 0 1 vC (t ) iC ( x)dx iC ( x)dx C C t0 The DEFINITE Integral is just a no.; call it vC(t0) so t 1 vC (t ) vC (t0 ) iC ( x)dx C t0 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Capacitor Summary cont Consider Finally the Differential Reln dvC t iC t C dt Some Implications • For small Displacement Current dvC/dt is small; i.e, vC changes only a little • Obtaining Large iC requires HUGE Voltage Gradients if C is small Conclusion: A Cap RESISTS CHANGES in VOLTAGE ACROSS It Engineering-43: Engineering Circuit Analysis 12 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt iC Defined by Differential The fact that the Cap Current is defined through a DIFFERENTIAL has important implications... Consider the Example at Left C 5F i (t ) 0 elsewhere i (t ) C dv (t ) dt 60mA i 5 106[ F ] 24 V 20mA 6 103 s Using the 1st Derivative (slopes) to find i(t) • Shows vC(t) – C = 5 µF • Find iC(t) Engineering-43: Engineering Circuit Analysis 13 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Capacitor Energy Storage UNlike an I-src or V-src a Cap Does NOT Produce Energy For a Cap A Cap is a PASSIVE Device that Can STORE Energy Recall from Chp.1 The Relation for POWER Recall also p vi vC iC pC t vC t iC t dvC iC (t ) C (t ) dt Subbing into Pwr Reln dvC pC (t ) CvC (t ) dt By the Derivative CHAIN Then the d d RULE INSTANTANEOUS Power d 1 2 dt dv pC (t ) C vC (t ) p t v t i t dt 2 Engineering-43: Engineering Circuit Analysis 14 C Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt dvC dt Capacitor Energy Storage cont Again From Chp.1 Recall that Energy (or Work) is the time integral of Power • Mathematically t2 wC (t1 , t 2 ) pC ( x)dx t1 Comment on the Bounds • If the Lower Bound is − we talk about “energy stored at time t2” • If the Bounds are − to + then we talk about the “total energy stored” Engineering-43: Engineering Circuit Analysis 15 Integrating the “Chain Rule” Relation 1 2 1 2 wC (t1 , t 2 ) CvC (t 2 ) CvC (t1 ) 2 2 Recall also t qC (t ) 1 vC (t ) iC ( x)dx C C Subbing into Pwr Reln dqC 1 pC (t ) qC (t ) (t ) dt vC t C iC t Again by Chain Rule d d 1 d 1 2 dt dq pC (t ) qc (t ) C dt 2 C Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt dqC dt Capacitor Energy Storage cont.2 Then Energy in Terms of Capacitor Stored-Charge 1 2 1 2 wC (t1 , t 2 ) qC (t 2 ) qC (t1 ) C C Short Example VC(t) C = 5 µF The Total Energy Stored during t = 0-6 ms 1 2 1 2 wC (0,6) CvC (6) CvC (0) 2 2 1 wC (0,6) 5 *10 6 [ F ] * (24) 2 [V 2 ] 2 wC Units? Coul 2 F V V Coul V J V 2 Charge Stored at 3 mS qC (3) CvC (3) qC (3) 5 *10 6 [ F ] *12[V ] 60C Engineering-43: Engineering Circuit Analysis 16 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Some Questions About Example For t > 8 mS, What is the Total Stored CHARGED? vC(t) C = 5 µF qC (t 8mS ) 0 For t > 8 mS, What is the Total Stored ENERGY? wC (t 8mS ) 0 wC (, t 2 ) wC (, t2 ) 1 2 1 CvC t 9 5F 0 2 2 2 DIScharging Current 1 2 1 2 qC t2 11 0 C 5F Engineering-43: Engineering Circuit Analysis 17 CHARGING Current Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Capacitor Summary: Q, V, I, P, W Consider A Cap Driven by A SINUSOIDAL V-Src Charge stored at a Given Time qC (t ) CvC (t ) qC 1 120 8.33mS 2 *10 6 [C ] * sin( )[V ] 0 Current “thru” the Cap v (t ) i(t) C 2 F iC C dvC (t ) dt iC 1 120 8.33mS 2 *106 *130 *120 cos( ) v ( t ) 130 sin (120 t ) iC 1 120 8.33mS 98 mA Energy stored at a given time w (t ) 1 Cv 2 (t ) Find All typical Quantities • Note – 120 = 60∙(2) → 60 Hz Engineering-43: Engineering Circuit Analysis 18 C 2 C 1 1 6 2 2 wC 2 *10 [ F ] *130 sin 240 2 2 wC 16.9 mJ Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Capacitor Summary cont. Consider A Cap Driven by A SINUSOIDAL V-Src v (t ) i(t) C 2 F v ( t ) 130 sin (120 t ) Electric power absorbed by Cap at a given time pC (t ) vC (t )iC (t ) Engineering-43: Engineering Circuit Analysis 19 At 135° = (3/4) iC 1 160 2 *106 *130 *120 cos(0.75 ) iC 1 160 69.3mA vC 1 160 130 sin( 0.75 ) 91.9V pC 1 160 91.9V 69.3mA 6371mW The Cap is SUPPLYING Power at At 135° = (3/4) = 6.25 mS • That is, The Cap is RELEASING (previously) STORED Energy at Rate of 6.371 J/s Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt WhiteBoard Work Let’s Work this Problem • The VOLTAGE across a 0.1-F capacitor is given by the waveform in the Figure Below. Find the WaveForm Eqn for the Capacitor CURRENT vc(t)(V) -12 0 1 2 3 4 5 t(s) -12 Figure P5.14 ANOTHER PROB 0.5 μF iC t 2 Acos50000t + vC(t) - Engineering-43: Engineering Circuit Analysis 20 See ENGR-43_Lec-061_Capacitors_WhtBd.ppt Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt The Inductor Second of the Energy-Storage Devices Basic Physical Model: Ckt Symbol Details of Physical Operation Described in PHYS 4B or ENGR45 Engineering-43: Engineering Circuit Analysis 21 Note the Use of the PASSIVE Sign Convention Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Physical Inductor Inductors are Typically Fabricated by Winding Around a Magnetic (e.g., Iron) Core a LOW Resistance Wire • Applying to the Terminals a TIME VARYING Current Results in a “Back EMF” voltage at the connection terminals Some Real Inductors Engineering-43: Engineering Circuit Analysis 22 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Inductance Defined From Physics, recall that a time varying magnetic flux, , Induces a voltage Thru the Induction Law d vL Where the Constant of Proportionality, L, is called the INDUCTANCE L is Measured in Units of “Henrys”, H • 1H = 1 V•s/Amp Inductors STORE electromagnetic energy For a Linear Inductor The Flux Is Proportional To They May Supply Stored The Current Thru it Energy Back To The diL Circuit, But They LiL vL L CANNOT CREATE dt Energy dt Engineering-43: Engineering Circuit Analysis 23 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Inductance Sign Convention Inductors Cannot Create Energy; They are PASSIVE Devices All Passive Devices Must Obey the Passive Sign Convention Engineering-43: Engineering Circuit Analysis 24 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Inductor Circuit Operation Recall the Circuit Representation Separating the Variables and Integrating Yields the INTEGRAL form t 1 iL (t ) vL ( x)dx L Previously Defined the Differential Form of the Induction Law diL vL L dt Engineering-43: Engineering Circuit Analysis 25 In a development Similar to that used with caps, Integrate − to t0 for an Alternative integral Law t 1 iL (t ) iL (t0 ) vL ( x)dx; t t0 L t0 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Inductor Model Implications From the Differential Law Observe That if iL is not Continuous, diL/dt → , and vL must also → This is NOT physically Possible Thus iL must be continuous diL vL dt Engineering-43: Engineering Circuit Analysis 26 Consider Now the Alternative Integral law t 1 iL (t ) iL (t0 ) vL ( x)dx; t t0 L t0 If iL is constant, say iL(t0), then The Integral MUST be ZERO, and hence vL MUST be ZERO • This is DC Steady-State Inductor Behavior – vL = 0 at DC – i.e; the Inductor looks like a SHORT CIRCUIT to DC Potentials Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Inductor: Power and Energy From the Definition of Instantaneous Power pL (t ) vL (t ) iL (t ) Subbing for the Voltage by the Differential Law diL p L (t ) L (t )iL (t ) dt Again By the Chain Rule for Math Differentiation d d di dt di dt d 1 2 pL t LiL (t ) dt 2 Engineering-43: Engineering Circuit Analysis 27 Time Integrate Power to Find the Energy (Work) t2 d 1 2 wL (t1 , t2 ) LiL ( x) dx dx 2 t1 Units Analysis • J = H x A2 Energy Stored on Time Interval w(t1 , t 2 ) 1 2 1 LiL (t 2 ) LiL2 (t1 ) 2 2 • Energy Stored on an Interval Can be POSITIVE or NEGATIVE Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Inductor: P & W cont. In the Interval Energy Eqn Let at time t1 1 2 LiL (t1 ) 0 2 Then To Arrive At The Stored Energy at a later given time, t 1 2 w(t ) LiL (t ) 2 Thus Observe That the Stored Energy is ALWAYS Positive In ABSOLUTE Terms as iL is SQUARED • ABSOLUTE-POSITIVE-ONLY Energy-Storage is Characteristic of a PASSIVE ELEMENT Engineering-43: Engineering Circuit Analysis 28 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Example Given The iL Current WaveForm, Find vL for L = 10 mH The Derivative of a Line is its SLOPE, m Then the Slopes 10( A / s) 0 t 2ms di 10( A / s) 2 t 4ms dt 0 elsewhere 20 103 A A m 10 s 2 103 s A m 10 s And the vL Voltage di (t ) 10( A / s) 3 vL (t ) 100 10 V 100mV dt L 10 10 3 H The Differential Reln diL vL L dt Engineering-43: Engineering Circuit Analysis 29 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Example Power & Energy The Energy Stored between 2 & 4 mS The Value Is Negative Because The Inductor SUPPLIES Energy PREVIOUSLY STORED with a Magnitude of 2 μJ The Energy Eqn 1 2 1 2 w(2,4) LiL (4) LiL (2) 2 2 Running the No.s w(2,4) 0 0.5 *10 *103 (20 *103 ) 2 w(2,4) 2.0 µJ Engineering-43: Engineering Circuit Analysis 30 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Numerical Example Given The Voltage Wave Form Across L , Find iL if • L = 0.1 H • i(0) = 2A The PieceWise Function t v(t ) 2 v( x)dx 2t; 0 t 2 0 L 0.1H i (t ) 2 20t ; 0 t 2s v(t ) 0; t 2 i (t ) i (2s); t 2s v (V ) A Line Followed by A Constant; Plotting 2 2 t (s) 42 i (A) The Integral Reln t 1 i(t ) i(0) v( x)dx L0 Engineering-43: Engineering Circuit Analysis 31 2 2 t (s) Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Numerical Example - Energy The Current Characteristic 42 The Initial Stored Energy w(0) 0.5 * 0.1[ H ](2 A) 2 0.2 J The “Total Stored Energy” i (A) w() 0.5 * 0.1[ H ] * (42 A) 2 88.2 J Energy Stored between 0-2 2 2 t (s) The Energy Eqn w(t1 , t 2 ) 1 2 1 LiL (t 2 ) LiL2 (t1 ) 2 2 • Energy Stored on Interval Can be POS or NEG Engineering-43: Engineering Circuit Analysis 32 w(2,0) 1 2 1 LiL (2) LiL2 (0) 2 2 w(0,2) 0.5 * 0.1* (42) 2 0.5 * 0.1* (2) 2 w(0,2) 88 J → Consistent with Previous Calculation Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Example w L (t ) Find The Voltage Across And The Energy Stored (As Function Of Time) v (t ) For The Energy Stored Engineering-43: Engineering Circuit Analysis 33 Notice That the ABSOLUTE Energy Stored At Any Given Time Is Non Negative (by sin2) • i.e., The Inductor Is A PASSIVE Element Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt L = 5 mH; Find the Voltage m 20mA 1ms m v 100mV 10 20 ( A / s) 2 1 v 50mV m0 v 0V v (t ) L m di (t ) dt 0 10 ( A / s) 43 v 50mV v 5 103 ( H ) 20( A / s); 0 t 1ms 100mV Engineering-43: Engineering Circuit Analysis 34 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Example Total Energy Stored The Ckt Below is in the DC-SteadyState Shorting-Caps; Opening-Inds Find the Total Energy Stored by the 2-Caps & 2-Inds KCL at node-A Recall that at DC Solving for VA • Cap → Short-Ckt • Ind → Open-Ckt Engineering-43: Engineering Circuit Analysis 35 I out VA VA 6 0 3 A 3 6 6 81 VA V 16.2V 5 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Example Total Energy Stored Continue Analysis of Shorted/Opened ckt I L1 I L 2 3 A 1.8 A 3 A I L1 1.2 A VC2 by Ohm VC 2 I L 2 6 1.8 A 6 Using Ohm and VA = 16.2V I L2 16.2V V 1.8 A 3 6 By KCL at Node-A Engineering-43: Engineering Circuit Analysis 36 VC 2 10.8 V VC1 by Ohm & KVL VC1 9V I L1 6 VC1 9 V 1.2A 6 VC1 9 V 7.2 V Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Example Total Energy Stored Have all I’s & V’s: −1.2 A 1.8 A 10.8 V 16.2 V Next using the E-Storage Eqns 1 wC CVC2 2 wL 1 2 LI L 2 Engineering-43: Engineering Circuit Analysis 37 Then the E- Storage Calculations wtot wk 13.46 mJ Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Caps & Inds Ideal vs. Real A Real CAP has Parasitic Resistances & Inductance: A Real IND has Parasitic Resistances & Capacitance: Generally SMALL Effect Engineering-43: Engineering Circuit Analysis 38 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Ideal vs Real Practical Elements “Leak” Thru Unwanted Resistance Ideal C & L i (t ) i (t ) i (t ) i (t ) v (t ) v (t ) v (t ) v (t ) dv di i (t ) C (t ) v (t ) L (t ) dt dt Engineering-43: Engineering Circuit Analysis 39 v (t ) dv i (t ) C (t ) Rleak dt v (t ) Rleak i (t ) L Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt di (t ) dt Capacitors in Series By KVL for 1-LOOP ckt If the vi(t0) = 0, Then Discern the Equivalent Series Capacitance, CS CAPS in SERIES Combine as Resistors in PARALLEL Engineering-43: Engineering Circuit Analysis 40 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Example Find Spot Caps in Series • Equivalent C • Initial Voltage 1 1 1 1 3 2 1 CS 1 CS 2 3 6 6 1 F Or Can Reduce Two at a Time Use KVL for Initial Voltage vt0 2V 4V 1V 3V 2 F This is the Algebraic Sum of the Initial Voltages • Polarity is Set by the Reference Directions noted in the Diagram Engineering-43: Engineering Circuit Analysis 41 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Numerical Example Two charged Capacitors Are Connected As Shown. Find the Unknown Capacitance 8V + - 12V 30 F 4V C Engineering-43: Engineering Circuit Analysis 42 Recognize SINGLE Loop Ckt → Same Current in Both Caps • Thus Both Caps Accumulate the SAME Charge Q (30 F )(8V ) 240C And Find VC by KVL • VC = 12V-8V = 4V Finally Find C by Charge Eqn QC 240 μC C 60 μC VC 4V Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Capacitors in Parallel By KCL for 1-NodePair ckt Thus The Equivalent Parallel Capacitance CAPS in Parallel Combine as Resistors in SERIES Engineering-43: Engineering Circuit Analysis 43 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Complex Example → Find Ceq 6 F 3 F C eq 3 C eq F 2 2 F Engineering-43: Engineering Circuit Analysis 44 4 F 4 F 12 F Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt 3 F Inductors in Series By KVL For 1-LOOP ckt Use The Inductance Law di vk (t ) Lk (t ) dt Thus di v (t ) LS (t ) dt INDUCTORS in Series add as Resistors in SERIES Engineering-43: Engineering Circuit Analysis 45 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Inductors in Parallel By KCL for 1-NodePair ckt And Thus N i (t0 ) i j (t0 ) j 1 INDUCTORS in Parallel combine as Resistors in PARALLEL Engineering-43: Engineering Circuit Analysis 46 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Example – Find: Leq, i0 4mH 2mH i (t0 ) 3 A 6 A 2 A 1A Series↔Parallel Summary • INDUCTORS Combine as do RESISTORS • CAPACITORS Combine as do CONDUCTORS Engineering-43: Engineering Circuit Analysis 47 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Inductor Ladder Network Find Leq for Li = 4 mH Place Nodes In Chosen Locations a Connect Between Nodes 6mH a 4mH d 4mH 2mH c Leq c d 2mH 2mH b 2mH When in Doubt, ReDraw • Select Nodes Engineering-43: Engineering Circuit Analysis 48 b Leq (6mH || 4mH ) 2mH 4.4mH Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Find Leq for Li = 6mH a ReDraw The Ckt for Enhanced Clarity a 6 || 6 || 6 2mH b Leq • The Electrical Diagram Does NOT have to Follow the Physical Layout Engineering-43: Engineering Circuit Analysis 49 6mH 6mH c Nodes Can have Complex Shapes b 6mH c It’s Simple Now Leq 6 (6 2) || 6 6 48 24 6 mH 14 7 66 Leq mH 7 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt C&L Summary Engineering-43: Engineering Circuit Analysis 50 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt WhiteBoard Work Let’s Work This Problem L 50 H i t 0 it 2te 4t t0 t 0 Find: v(t), tmax for imax, tmin for vmin Engineering-43: Engineering Circuit Analysis 51 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt % Bruce Mayer, PE * 14Mar10 % ENGR43 * Inductor_Lec_WhtBd_Prob_1003.m % t = linspace(0, 1); % iLn = 2*t; iexp = exp(-4*t); plot(t, iLn, t, iexp), grid disp('Showing Plot - Hit ANY KEY to Continue') pause % i = 2*t.*exp(-4*t); plot(t, iLn, t, iexp, t, i),grid disp('Showing Plot - Hit ANY KEY to Continue') pause plot(t, i), grid disp('Showing Plot - Hit ANY KEY to Continue') pause vL_uV =100*exp(-4*t).*(1-4*t); plot(t, vL_uV) i_mA = i*1000; plot(t, vL_uV, t, i_mA), grid disp('Showing Plot - Hit ANY KEY to Continue') pause % % find mins with fminbnd [t_vLmin vLmin] = fminbnd(@(t) 100*exp(-4*t).*(1-4*t), 0,1) % must "turn over" current to create a minimum i_mA_TO = -i*1000; plot(t, i_mA_TO), grid disp('Showing Plot - Hit ANY KEY to Continue') pause [t_iLmin iLmin_TO] = fminbnd(@(t) -1000*(2*t.*exp(-4*t)), 0,1); t_iLmin iLmin = -iLmin_TO plot(t, vL_uV, t, i_mA, t_vLmin, vLmin, 'p', t_iLmin,iLmin,'p'), grid Engineering-43: Engineering Circuit Analysis 52 By MATLAB Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt % Bruce Mayer, PE * 14Mar10 % ENGR43 * Inductor_Lec_WhtBd_Prob_1003.m % t = linspace(0, 1); % iLn = 2*t; iexp = exp(-4*t); plot(t, iLn, t, iexp), grid disp('Showing Plot - Hit ANY KEY to Continue') pause % i = 2*t.*exp(-4*t); plot(t, iLn, t, iexp, t, i),grid disp('Showing Plot - Hit ANY KEY to Continue') pause plot(t, i), grid disp('Showing Plot - Hit ANY KEY to Continue') pause vL_uV =100*exp(-4*t).*(1-4*t); plot(t, vL_uV) i_mA = i*1000; plot(t, vL_uV, t, i_mA), grid disp('Showing Plot - Hit ANY KEY to Continue') pause % % find mins with fminbnd [t_vLmin vLmin] = fminbnd(@(t) 100*exp(-4*t).*(1-4*t), 0,1) % must "turn over" current to create a minimum i_mA_TO = -i*1000; plot(t, i_mA_TO), grid disp('Showing Plot - Hit ANY KEY to Continue') pause [t_iLmin iLmin_TO] = fminbnd(@(t) -1000*(2*t.*exp(-4*t)), 0,1); t_iLmin iLmin = -iLmin_TO plot(t, vL_uV, t, i_mA, t_vLmin, vLmin, 'p', t_iLmin,iLmin,'p'), grid By MATLAB Engineering-43: Engineering Circuit Analysis 53 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt 200 150 100 50 0 -50 0 0.1 0.2 0.3 Engineering-43: Engineering Circuit Analysis 54 0.4 0.5 0.6 0.7 0.8 0.9 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt 1 Irwin Prob 5.26: I(t) & v(t) 200 i(t) (mA) 175 Current (mA) 150 125 100 75 50 25 0 -0.1 0.0 0.1 0.2 file = Engr44_prob_5-26_Fall03..xls Engineering-43: Engineering Circuit Analysis 55 0.3 0.4 0.5 0.6 0.7 0.8 0.9 time (s) Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt 1.0 Irwin Prob 5.26: I(t) & v(t) 200 120 i(t) (mA) v(t) (µV) 100 150 80 125 60 100 40 75 20 50 0 25 -20 0 -0.1 -40 0.0 0.1 0.2 file = Engr44_prob_5-26_Fall03.xls Engineering-43: Engineering Circuit Analysis 56 0.3 0.4 0.5 0.6 0.7 0.8 0.9 time (s) Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt 1.0 Electrical Potential (µV) Current (mA) 175 L = 10 mH; Find the Voltage v 100mV v (t ) L 20 103 A v 10 10 [ H ] 2 103 s 3 Engineering-43: Engineering Circuit Analysis 57 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt di (t ) dt Engineering 43 Appendix Complex Cap Example Bruce Mayer, PE Licensed Electrical & Mechanical Engineer [email protected] Engineering-43: Engineering Circuit Analysis 58 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Numerical Example Given iC, Find vC The Piecewise Fcn for iC C= 4µF vC(0) = 0 > 3 v (t ) 2t 8 10 [V ] 2 t 4ms Integrating & Graphing 1t v (t ) v (0) i ( x )dx; t 0 C0 Linear 0t 2 1t v (t ) v (2) i ( x )dx; t 2 C2 Engineering-43: Engineering Circuit Analysis 59 Parabolic Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Power Example From Before the vC For The Previous Conditions, Find The POWER Characteristic • C = 4 µF • iC by Piecewise curve i (t ) 8 103 t Using the Pwr Reln p(t ) 8t 3 , 0 t 2ms p(t ) 0, elsewhere Engineering-43: Engineering Circuit Analysis 60 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt 2 t 4ms Power Example cont Finally the Power Characteristic Absorbing or Supplying Power? During the CHARGING Period of 0-2 mS, the Cap ABSORBS Power During DIScharge the Cap SUPPLIES power • But only until the stored charge is fully depleted Engineering-43: Engineering Circuit Analysis 61 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Energy Example For The Previous Conditions, Find The ENERGY Characteristic Now The Work (or Energy) is the Time Integral of Power • C = 4 µF • pC by Piecewise curve For 0 t 2 mS p(t ) 8t 3 Engineering-43: Engineering Circuit Analysis 62 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Energy Example cont For 2 < t 4 mS 2t 4 8 t 2 64n t 128 p Taking The Time Integral and adding w(2 mS) Then the Energy Characteristic Engineering-43: Engineering Circuit Analysis 63 Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt If the current is known ... SAMPLE PROBLEM Current through capacitor iC vC C e 0.5t ; t 0 iC (t ) [mA] 0; t 0 C 2F t Voltage at a given time t 1 vC (t ) iC ( x)dx C vC (0) 0[V ] Voltage at a given time t when voltage at time to<t is also known 2 1 1 0.5 x v ( 0 ) e dx vC (2) C 2 *106 C 0 Charge at a given time 2 1 1 1 0.5 x 1 6 1 e 0 . 6321 * 10 e V 6 0.5 2 *10 0.5 0 qC (t ) CvC (t ) qC (2) 2 * 0.6321 t Voltage as a function of time vC (t ) Electric power supplied to capacitor 1 iC ( x)dx C Energy stored in capacitor at a given time w (t ) 64 vC (t ) 0; t 0 pC (t ) vC (t )iC (t ) “Total”Engineering-43: energy stored in the capacitor Engineering Circuit Analysis t 1 vC (t ) vC (t0 ) iC ( x)dx C t0 W 1 2 CvC (t ) J 2 1 wT CvC2 () 2 C t 1 vC (t ) vC (0) e 0.5 x dx C0 106 (1 e 0.5t ); t 0 vC (t ) 0; t 0 1 wT 2 *106 * Bruce (106Mayer, ) 2 PE 106 2 [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt J V SAMPLE PROBLEM Given current and capacitance Compute voltage as a function of time At minus infinity everything is zero. Since current is zero for t<0 we have 0 t 5m sec VC (0) 0 VC (t ) 5 10 VC (t ) 0; t 0 15 A 106 A iC (t ) t 3 3 t 3 *103[ A / s] t 5 ms 10 s 3 t 3 3 *10 xdx [V ] 3 *10 t 2 [V ]; 0 t 5 *103[ s ] 6 4 *10 0 8 In particular 3 *103 * (5 *103 ) 2 75 VC (5ms) [V ] [mV ] 8 8 t (m sec) 5 t 10 ms iC (t ) 10 [A] t 75 75 *103 1 VC (5ms ) [mV ] VC (t ) (10 *106 )[ A / s]dx 6 8 8 4 *10 5*103 Charge stored at 5ms qC (t ) CVC (t ) 75 *103 q(5ms) 4 *10 [ F ] * [V ] 8 6 q(5ms) (75 / 2) [nC ] 75 *103 10 VC (t ) t 5 *103 [V ]; 5 *103 t 10 *103 [ s] 8 4 Total energy stored 1 E CVC2 2 Total means at infinity. Hence 2 Before looking into a formal way to describe the current we will look at additional questions that can be answered. 3 Circuit Analysis 25 * 10 6 Bruce Mayer, PE [ J ] ET 0.5Engineering-43: * 4 *10 Engineering Now, for a formal way to represent piecewise functions.... 65 [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt 8 Formal description of a piecewise analytical signal 0; 3 2 t ; 8 Vc (t ) 75 10 t 5; 8 4 25 ; 8 t0 0 t 5ms 5 t 10 [ms] t 10[ms] Engineering-43: Engineering Circuit Analysis 66 [mV ] Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt Find Ceq for Ci = 4 µF 8 F 8 F 4 F 48 C eq 8 32 8 3 3 Engineering-43: Engineering Circuit Analysis 67 32 F 12 8 F Bruce Mayer, PE [email protected] • ENGR-43_Lec-06-1_Capacitors.ppt 8 F