Chapter 8 Engr228 Circuit Analysis Dr Curtis Nelson Chapter 8 Objectives • Be able to determine the natural and the step response of parallel RLC circuits; • Be able to determine the natural and the step response of series RLC circuits. Engr228 - Chapter 8, Nilsson 10e 1 RL and RC Circuit Review • Transient, natural, or homogeneous response – Fades over time; – Resists change. • Forced, steady-state, particular response – Follows the input; – Independent of time passed. • Assume that the total response is of the form k1 + k 2 e − t τ The RLC Circuit • RLC circuits contain both an inductor and a capacitor. • These circuits have a wide range of applications, including oscillators and frequency filters. • They can also model automobile suspension systems, temperature controllers, airplane responses, etc. • The response of RLC circuits with DC sources and switches will also consist of a natural response and a forced response: v(t) = vf(t)+vn(t) The complete response must satisfy both the initial conditions and the “final conditions” of the forced response. Engr228 - Chapter 8, Nilsson 10e 2 Source-Free Parallel RLC Circuits We will first study the natural response of second-order circuit by studying a source-free parallel RLC circuit. iC(t) Parallel RLC Circuit C iR(t) + R v(t) L - iL(t) iR + iL + iC = 0 v(t ) 1 dv (t ) + ∫ v(t )dt + C =0 R L dt d 2 v(t ) 1 dv (t ) 1 C + + v(t ) = 0 dt 2 R dt L Second-order Differential equation Source-Free Parallel RLC Circuits This second-order differential equation can be solved by assuming the form of a solution. st Assume a solution is of the form: v(t ) = Ae C d 2v(t ) 1 dv (t ) 1 + + v(t ) = 0 dt 2 R dt L 1 1 Ase st + Ae st = 0 R L 1 1 Ae st (Cs 2 + s + ) = 0 R L 1 1 Cs 2 + s + = 0 R L CAs 2 e st + which means • This is known as the characteristic equation. Engr228 - Chapter 8, Nilsson 10e 3 Source-Free Parallel RLC Circuits Cs 2 + 1 1 s+ =0 R L Using the quadratic formula, we get 2 s1 = − 1 1 ⎛ 1 ⎞ + ⎜ ⎟ − 2 RC ⎝ 2 RC ⎠ LC 2 1 1 ⎛ 1 ⎞ s2 = − − ⎜ ⎟ − 2 RC ⎝ 2 RC ⎠ LC Both v(t ) = A1e s t and v(t ) = A2e s t are solutions to the equation. 1 2 st Therefore, the complete solution is v(t ) = A1e 1 + A2e s 2t A1 and A2 are determined by the initial conditions of the circuit. Source-Free Parallel RLC Circuits 2 1 1 ⎛ 1 ⎞ s1, 2 = − ± ⎜ ⎟ − 2 RC ⎝ 2 RC ⎠ LC 1 Define resonant frequency: ω0 = LC 1 Define damping factor: α= 2 RC Then: s1, 2 = −α ± α 2 − ω02 We will now divide the circuit response into three cases according to the sign of the term under the radical. Engr228 - Chapter 8, Nilsson 10e 4 Second-Order Differential Equation Solution 2 1 1 ⎛ 1 ⎞ s1, 2 = − ± ⎜ ⎟ − 2 RC ⎝ 2 RC ⎠ LC s1, 2 = −α ± α 2 − ω02 • Case 1 – overdamped response α > ω0 (inside the square root is a positive value) • Case 2 – critically damped response α = ω0 (inside the square root is zero) • Case 3 – underdamped response α < ω0 (inside the square root is a negative value) Circuit Responses Engr228 - Chapter 8, Nilsson 10e 5 Overdamped Case (α > ω0 ) Find v(t) in the circuit at the right. Given initial conditions: vc (0) = 0, iL(0) = -10A α= 1 = 3.5 2 RC ω0 = 1 = 6 LC α > ω0 therefore this is an overdamped case s1, 2 = −α ± α 2 − ω02 s1 = -1, s2 = -6 Overdamped Case - continued −t The solution is in form of v(t ) = A1e + A2e −6t Use initial conditions to find A1 and A2 From vc (0) = 0 at t = 0: v(0) = 0 = A1e0 + A2e0 = A1 + A2 From KCL taken at t = 0: iR + iL + iC = 0 v ( 0) dv (t ) + (−10) + C =0 R dt t =0 0 1 + (−10) + − A1e −t − 6 A2 e −6t R 42 (− A1 − 6 A2 ) = 420 ( Engr228 - Chapter 8, Nilsson 10e ) t =0 =0 6 Overdamped Case - continued Solving the two equations, we get A1 = 84 and A2 = -84 The solution is v(t ) = 84e −t − 84e −6t = 84(e −t − e −6t )V v(t) v(t ) = 84(e −t − e −6t ) t Example: Overdamped RLC Circuit Find vC(t) for t > 0 vC(t) = 80e−50,000t − 20e−200,000t V for t > 0 Engr228 - Chapter 8, Nilsson 10e 7 Critically Damped Case (α = ω0 ) Find v(t) in the circuit at the right. iC(t) 1/42f Given initial conditions: vc (0) = 0, iL(0) = -10A α= 1 = ω0 = 2 RC + v(t) - iR(t) iL(t) 8.573Ω 7H 1 = 2.45 LC Critically damped when α = ω0 s1 = s2 = -2.45 The complete solution in this case is of the form v(t ) = A1test + A2e st Critically Damped Case - continued Use initial conditions to find A1 and A2 From vc (0) = 0 at t = 0: v(0) = 0 = A1 (0)e0 + A2e0 = A2 −2.45t Therefore A2 = 0 and the solution is reduced to v(t ) = A1te Find A1 from KCL at t = 0: iR + iL + iC = 0 v(0) dv (t ) + (−10) + C =0 R dt t =0 0 1 + (−10) + A1t (−2.45)e − 2.45t + A1e − 2.45t R 42 1 − 10 + ( A1 ) = 0 42 ( Engr228 - Chapter 8, Nilsson 10e ) t =0 =0 8 Critically Damped Case - continued Solving the equation, A1 = 420 The solution is v(t ) = 420te−2.45tV v(t) v(t ) = 420te−2.45t t Critically Damped Example Find R1 such that the circuit is critically damped for t > 0 and R2 so that v(0)=2V. Answer: R1 = 31.63 kΩ, R2=0.4Ω Engr228 - Chapter 8, Nilsson 10e 9 Underdamped Case (α < ω0 ) s1, 2 = −α ± α 2 − ω02 For the underdamped case, the term inside the bracket will be negative and s will be a complex number. Define ωd = ω02 − α 2 s1, 2 = −α ± jωd Then v(t ) = A1e( −α + jωd )t + A2e( −α − jωd )t v(t ) = e −αt ( A1e jωd t + A2e − jωd t ) Underdamped Case - continued v(t ) = e −αt ( A1e jωd t + A2e − jωd t ) Using Euler’s Identity e jθ = cosθ + j sin θ v(t ) = e −αt ( A1 cos ωd t + jA1 sin ωd t + A2 cos ωd t − jA2 sin ωd t ) v(t ) = e −αt (( A1 + A2 ) cos ωd t + j ( A1 − A2 ) sin ωd t v(t ) = e −αt ( B1 cos ωd t + B2 sin ωd t ) v(t ) = e −αt ( B1 cos ωd t + B2 sin ωd t ) Engr228 - Chapter 8, Nilsson 10e 10 Mechanical Analogue A pendulum is an example of an underdamped second-order mechanical system. displacement(t) t Underdamped Case - Example iC(t) Find v(t) in the circuit at the right. + v(t) - 1/42f Given initial conditions: vc (0) = 0, iL(0) = -10A α= 1 =2 2 RC ω0 = iR(t) 10.5Ω iL(t) 7H 1 = 6 LC α < ω0 therefore, this is an underdamped case ωd = ω02 − α 2 = 2 v(t) is of the form Engr228 - Chapter 8, Nilsson 10e v(t ) = e −2t ( B1 cos 2t + B2 sin 2t ) 11 Underdamped Case - continued Use initial conditions to find B1 and B2 From vc (0) = 0 at t = 0: v(0) = e0 ( B1 cos 0 + B2 sin 0) = B1 Therefore B1 = 0 and the solution is reduced to v(t ) = e −2t ( B2 sin 2t ) Find B2 from KCL at t = 0: iR + iL + iC = 0 v(0) dv (t ) + (−10) + C =0 R dt t =0 0 1 + (−10) + 2 B2 e − 2t cos 2t − 2 B2 e − 2t sin 2t R 42 1 − 10 + ( 2 B2 ) = 0 42 ( ) t =0 =0 Underdamped Case - continued Solving: B2 = 210 2 = 297 The solution is: v(t ) = 297 e −2t sin 2tV v(t) v(t ) = 297 e −2t sin 2tV t Engr228 - Chapter 8, Nilsson 10e 12 Underdamped Example Find iL for t > 0 iL = e−1.2t (2.027 cos 4.75t + 2.561 sin 4.75t) A Summary of Transient Responses Engr228 - Chapter 8, Nilsson 10e 13 Source - Free Series RLC Circuit i(t) + vR(t) - R + vC(t) vL(t) L + - C vR + vL + vC = 0 di (t ) 1 + ∫ i (t )dt = 0 dt C d 2i (t ) di (t ) 1 L + R + i (t ) = 0 dt 2 dt C i (t ) R + L Series and Parallel RLC Circuits Parallel RLC Series RLC 2 d v(t ) 1 dv (t ) 1 C + + v(t ) = 0 dt 2 R dt L v(t ) = A1e s1t + A2e s2t 1 1 ⎛ 1 ⎞ =− ± ⎜ ⎟ − 2 RC ⎝ 2 RC ⎠ LC s1, 2 = −α ± α 2 − ω02 α= 1 2 RC Engr228 - Chapter 8, Nilsson 10e ω0 = L d i(t ) di (t ) 1 +R + i(t ) = 0 2 dt dt C i(t ) = A1e s1t + A2e s2t 2 s1, 2 2 1 LC 2 R 1 ⎛ R⎞ s1, 2 = − ± ⎜ ⎟ − 2L ⎝ 2 L ⎠ LC s1, 2 = −α ± α 2 − ω02 α= R 2L ω0 = 1 LC 14 Series RLC Circuit Solution ω0 = 1 Define LC Then if: α > ω0 (overdamped): € € α = ω0 (critically damped): α < ω0 (underdamped): s1 = −α + α 2 − ω 02 R α= 2L s2 = −α − α 2 − ω 02 €i (t ) = A1e s1t + A2e s2t i(t ) = e −αt ( A1t + A2 ) i (t ) = e −αt (B1 cos(ωd t ) + B2 sin (ωd t )) Textbook Problem 8.49 Nilsson 10th The circuit contains no initial energy. Find vo(t) for t ≥ 0 vo(t) = 16 - 16e−400tcos300t − 21.33e−400tsin300t V Engr228 - Chapter 8, Nilsson 10e 15 Example: Initial Conditions Find the labeled voltages and currents at t = 0- and t = 0+ . Answer: iR(0−) = −5 A iL (0−) = 5 A iC (0−) = 0 A vR(0−) = −150 V vL (0−) = 0 V vC (0−) = 150 V iR(0+) = −1 A iL (0+) = 5 A iC (0+) = 4 A vR(0+) = −30 V vL (0+) = 120 V vC (0+) = 150 V Summary: Solving RLC Circuits 1. Identify the series or parallel RLC circuit; 2. Find α and ω0; 3. Determine whether the circuit is overdamped, critically damped, or underdamped; 4. Assume a solution (natural response + forced response); A1e s1t + A2e s2t + V f A1test + A2e st + V f e−αt ( B1 cos ωd t + B2 sin ωd t ) + V f Overdamped Critically damped Underdamped 5. Find A, B, and Vf using initial and final conditions. Engr228 - Chapter 8, Nilsson 10e 16 Forced Response Example t = 0 sec Find Vc (t) 30 S1 4A iL(t) 3H + Vc 1/27f 5A - α= R =5 2L ω0 = 1 =3 LC s1, 2 = −α ± α 2 − ω02 = −1,−9 This is an overdamped case, so the solution is of the form A1e −t + A2e −9t + V f Forced Response Example - continued vC (t ) = A1e−t + A2e−9t + V f The initial conditions are vC(0) = 150 V and iL(0) = -5 A As time goes to infinity, vC(∞) = 150 V and iL(∞) = -9 A Using vC(0) = 150 V , we get 150 = A1 + A2 + V f Using vC(∞) = 150 V , we get 150 = 0 + 0 + V f Therefore, Vf = 150 and A1+A2 = 0 Engr228 - Chapter 8, Nilsson 10e 17 Forced Response Example - continued Also, initial conditions determine that iC(0) = 4A dv (t ) 1 = (− A1e −t − 9 A2 e −9t ) dt 27 1 iC (0) = 4 = (− A1e 0 − 9 A2 e 0 ) 27 108 = − A1 − 9 A2 iC (t ) = C A1 = 13.5, A2 = -13.5 Therefore, vC (t ) = 13.5e −t − 13.5e −9t + 150 V Chapter 8 Summary • Showed how to determine the natural and the step response of parallel RLC circuits; • Showed how to determine the natural and the step response of series RLC circuits. Engr228 - Chapter 8, Nilsson 10e 18