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Lecture Electric circuit theory: First-order crcuits include all of the following: Introduction to transient analysis, initial conditions, the source-free RC circuit, the source-free RL circuit, step response of an RC circuit, step response of an RL circuit, the classical method, first-order Op Amp circuits
Nguyễn Công Phương Electric Circuit Theory First-Order Circuits Contents I Basic Elements Of Electrical Circuits II Basic Laws III Electrical Circuit Analysis IV Circuit Theorems V Active Circuits VI Capacitor And Inductor VII First-Order Circuits VIII.Second Order Circuits IX Sinusoidal Steady State Analysis X AC Power Analysis XI Three-phase Circuits XII Magnetically Coupled Circuits XIII.Frequency Response XIV.The Laplace Transform XV Two-port Networks First-Order Circuits - sites.google.com/site/ncpdhbkhn First-Order Circuits Introduction to Transient Analysis Initial Conditions The Source-free RC Circuit The Source-free RL Circuit Step Response of an RC Circuit Step Response of an RL Circuit The Classical Method First-order Op Amp Circuits First-Order Circuits - sites.google.com/site/ncpdhbkhn Introduction to Transient Analysis (1) – + – + 10 V 5Ω 10 V t=0 0.1 H 0.1 H i 2A i Any change in an electrical circuit, which brings about a change in energy distribution, will result in a transient-state 5Ω Steady-state Transient-state Steady-state t 12 V v Steady-state Transient-state Steady-state t – + – + 10 V 5Ω 12 V t=0 0.1 mF First-Order Circuits - sites.google.com/site/ncpdhbkhn 6Ω 0.1 mF + v – Introduction to Transient Analysis (2) vL(t) Inductors in DC circuits Transient -state Old steady-state Short-circuit Capacitors in DC circuits New steady-state t Not short-circuit Short-circuit iC(t) Transient -state Old steady-state New steady-state Open-circuit t Not open-circuit Open-circuit First-Order Circuits - sites.google.com/site/ncpdhbkhn Introduction to Transient Analysis (3) First-Order Circuits - sites.google.com/site/ncpdhbkhn First-Order Circuits Introduction to Transient Analysis Initial Conditions The Source-free RC Circuit The Source-free RL Circuit Step Response of an RC Circuit Step Response of an RL Circuit The Classical Method First-order Op Amp Circuits First-Order Circuits - sites.google.com/site/ncpdhbkhn 20 V Initial Conditions (1) –+ –+ 10 V 5Ω t=0 i 0.1 H i A Steady-state/Initial condition 2A Steady-state/Initial condition 5Ω –+ 10 V t=0 i 0– 0.1 H t Prior to switching 0+ After switching First-Order Circuits - sites.google.com/site/ncpdhbkhn Initial Conditions (2) • 1st switching rule/law: the current (magnetic flux) in an inductor just after switching is equal to the current (flux) in the same inductor just prior to switching iL(0+) = iL(0–) λ(0+) = λ(0–) • 2nd switching rule/law: the voltage (electric charge) in a capacitor just after switching is equal to the voltage (electric charge) in the same capacitor just prior to switching vC(0+) = vC(0–) q(0+) = q(0–) First-Order Circuits - sites.google.com/site/ncpdhbkhn Initial Conditions (3) Ex The switch has been at A for a long time, and it moves to B at t = 0; find I0? − i (0 ) = A i (0+ ) = i (0− ) → i(0+ ) = A → I = A –+ 10 V A 5Ω t=0 B i 0.1 H Ex The switch has been at A for a long time, and it moves to B at t = 0; find I0? 20 i (0 ) = = 4A –+ –+ − i (0+ ) = i (0− ) 20 V → i(0+ ) = A → I = A 10 V A 5Ω t=0 B First-Order Circuits - sites.google.com/site/ncpdhbkhn i 0.1 H 10 E1 E2 The Classical Method (2) –+ t=0 –+ R L E2 i E2 E1 E2 − RL t i(t ) = + − e R R R E1 E2 A= − R R R − E2 E2 E1 L i (0) = + Ae = +A= R R R E1 –+ R L i(0) R –+ R if Ri f − E2 = E → if = R in L Rin + Lin' = in = Aest → RAe st + LAse st = → ( R + Ls ) Ae st = R − t E2 i= + Ae L R E I = i(0) = R → R + Ls = R →s=− L → in = Ae First-Order Circuits - sites.google.com/site/ncpdhbkhn R − t L 27 E1 E2 The Classical Method (3) –+ –+ t=0 R v + C – v(t ) = E2 + ( E1 − E2 )e − E2 R –+ v(0) = E2 + Ae E1 RC R C + – + R ( Cvn' ) = = Aest t RC v f = E2 A = E1 − E2 − + vf – → Ae st + RCAse st = → (1 + RCs ) Ae st = = E2 + A = E1 –+ v = E2 + Ae R + C v(0) – − t RC V0 = v (0) = E1 → + RCs = →s=− RC → = Ae First-Order Circuits - sites.google.com/site/ncpdhbkhn − t RC 28 The Classical Method (4) E1 E2 E1 –+ t=0 –+ R L i i = i f + in I = E1 / R i f = E2 / R R − t L in = Ae E I = + A → A R E2 E1 E2 − RL t i = + − e R R R Write the general form Find the initial condition Find the forced response Deactivate source(s), find the natural response (with the unknow integration constant) Find the integration constant Write the complete response –+ t=0 R –+ v E2 + C – v = v f + V0 = E1 v f = E2 = Ae − t RC V0 = E2 + A → A v = E2 + ( E1 − E2 )e First-Order Circuits - sites.google.com/site/ncpdhbkhn − t RC 29 The Classical Method (5) = Ae −t RC −t = Ae 100×0.05×10−3 – v f = 36 V = Ae−200 t V0 = 36 + Ae0 = 36 + A = 12 → A = −24 → v = 36 − 24e −200 t V + 100 Ω v 12 V V0 = 12 V t=0 + v = v f + 50 Ω + The switch has been at A for a long time, and it moves to B at t = 0; find v for t ≥ 0? B 0.05 mF – – Ex A 36 V Write the general form Find the initial condition Find the forced response Deactivate source(s), find the natural response (with the unknow integration constant) Find the integration constant Write the complete response First-Order Circuits - sites.google.com/site/ncpdhbkhn 30 The Classical Method (6) t=0 + 100 Ω v – 12 V – V B + 50 Ω + v = 36 − 24e −200 t A – 0.05 mF 36 V v 36V 12V t + + + V0 – 12 V – V0 = 12 V B 100 Ω 0.05 mF – = Ae −200t 100 Ω + vf + 50 Ω B – A – 36 V v f = 36 V First-Order Circuits - sites.google.com/site/ncpdhbkhn 31 The Classical Method (7) 50 Ω v = v f + 12 V −12 10 = −2 V 50 + 10 80 vf = 36 = 16 V 100 + 80 V0 = = Ae −t RC −t = Ae 80×100 ×0.05×10 −3 80+100 10 Ω = Ae −450 t V0 = 16 + Ae = 16 + A = −2 → A = −18 → v = 16 − 18e −450 t V B t=0 + v 0.05 mF – 80 Ω + – + The switch has been at A for a long time, and it moves to B at t = 0; find v for t ≥ 0? 100 Ω A – Ex 36 V Write the general form Find the initial condition Find the forced response Deactivate source(s), find the natural response (with the unknow integration constant) Find the integration constant Write the complete response First-Order Circuits - sites.google.com/site/ncpdhbkhn 32 Ex i = i f + in if = 120 = 6A 20 R − t L − in = Ae = Ae t=0 0.5 H 5A I0 = A 20 t 0.5 50 Ω = Ae−40 t I = + Ae0 = + A = → A = −1 → i = − e−40t A B 20 Ω + The switch has been at A for a long time, and it moves to B at t = 0; find i for t ≥ 0? A – The Classical Method (8) i 120 V Write the general form Find the initial condition Find the forced response Deactivate source(s), find the natural response (with the unknow integration constant) Find the integration constant Write the complete response First-Order Circuits - sites.google.com/site/ncpdhbkhn 33 i = − e −40t A t=0 0.5 H B 20 Ω i 5A i + 50 Ω A – The Classical Method (9) 120 V 6A 5A t B B 20 Ω I0 0.5 H in 20 Ω if 5A I0 = A + 50 Ω – A 120 V in = Ae−40t if = A First-Order Circuits - sites.google.com/site/ncpdhbkhn 34 The Classical Method (10) Ex The switch has been at A for a long time, and it is opened at t = 0; find i for t ≥ 0? i in = Ae R − t L = Ae − 50 + 30 t 0.25 = Ae −320t I = 1.88 + Ae = 1.28 → A = −0.60 t=0 0.25 H 40 Ω i = i f + in 1/ 50 I0 = = 1.28 A 1/ 50 + 1/ 40 + 1/ 30 1/ 50 if = = 1.88 A 1/ 50 + 1/ 30 50 Ω if in 50 Ω 100 Ω 5A 30 Ω 0.25 H 100 Ω 5A 30 Ω 50 Ω 100 Ω 0.25 H → i = 1.88 − 0.6e−320 t A First-Order Circuits - sites.google.com/site/ncpdhbkhn 30 Ω 35 First-Order Circuits Introduction to Transient Analysis Initial Conditions The Source-free RC Circuit The Source-free RL Circuit Step Response of an RC Circuit Step Response of an RL Circuit The Classical Method First-order Op Amp Circuits First-Order Circuits - sites.google.com/site/ncpdhbkhn 36 Ex First-Order Op Amp Circuits (1) t = 50 kΩ 100 kΩ C A B The switch has been at A for a long time, and it moves to B at t = 0; find vo for t ≥ 0? – + – v + V µF + vo + B – 50 kΩ C 100 kΩ – µF – + v = v f + v (0) = V vf = vB − vC vB − −6 dvn 1× 10 = = dt 50 × 10 50 × 103 vB = −vn dvn → + =0 → = Ae −20t dt 0.05 v (0) = + Ae = → A = → v = 5e−20 t V – + von + First-Order Circuits - sites.google.com/site/ncpdhbkhn – 37 Ex First-Order Op Amp Circuits (2) t = 50 kΩ 100 kΩ C A B The switch has been at A for a long time, and it moves to B at t = 0; find vo for t ≥ 0? −20 t V vB = − v R1 + – R2 vi ii + vo – R2 vo = − vi R1 vo + → vo = 10e−20t V v V àF 100 ì 103 vo = vB = −2vB 50 × 10 – + – + v = 5e – First-Order Circuits - sites.google.com/site/ncpdhbkhn io 38 First-Order Op Amp Circuits (3) v = v f + t=0 A + Find vo(t)? 10 kΩ – Ex v(0) = 20 kΩ B 60 kΩ + 20 vAf = = 4V 10 + 20 60 vof = 1 + v Af = 12 V 30 + vo 30 kΩ µF – 6V + v – – v f = vBf − vof = vAf − vof = − 12 = −8 V −6 io – dvn → + 60 × 10 × 10 =0 dt → = Ae−16.67t vi + + 60 × 10 i = ii R1 vo R2 vo = 1 + vi R1 R2 First-Order Circuits - sites.google.com/site/ncpdhbkhn 39 v = v f + – v f = −8V 20 kΩ + v(0) = t=0 A + Find vo(t)? 10 kΩ – Ex First-Order Op Amp Circuits (4) + B 60 kΩ vo 30 kΩ µF 6V + v – – = Ae −16.67t v(0) = −8 + Ae = → A = → v = −8 + 8e −16.67 t V vo = vB − v = − ( −8 + 8e −16.67 t ) = 12 − 8e −16.67 t V First-Order Circuits - sites.google.com/site/ncpdhbkhn 40 First-Order Op Amp Circuits (5) Ex 10 kΩ v = v f + Find v(t)? A V0 = 5V = Ae −41.67 t B 60 kΩ vi 40RkΩ v – R2 vo = − vi R1 + ii 60 kΩ – + V0 = v (0) = −6 + Ae = → A = + µF R1 t 1×10−6 – 40×10 + 60×10 + 40×103 ×60×103 + = Ae 40 kΩ – t=0 20 20 vBf = − v Af = − = −10 V 10 10 60 60 vf = vBf = ( −10) = −6 V 40 + 60 40 + 60 − 20 kΩ → v = −6 + 6e −41.67 t V First-Order Circuits - sites.google.com/site/ncpdhbkhn µFvo – io 41 ... Transient -state Old steady-state Short -circuit Capacitors in DC circuits New steady-state t Not short -circuit Short -circuit iC(t) Transient -state Old steady-state New steady-state Open -circuit. .. open -circuit Open -circuit First-Order Circuits - sites.google.com/site/ncpdhbkhn Introduction to Transient Analysis (3) First-Order Circuits - sites.google.com/site/ncpdhbkhn First-Order Circuits... The Source-free RC Circuit The Source-free RL Circuit Step Response of an RC Circuit Step Response of an RL Circuit The Classical Method First-order Op Amp Circuits First-Order Circuits - sites.google.com/site/ncpdhbkhn