Chapter 8 Natural and Step Responses of RLC Circuits 8.1-2 The Natural Response of a Parallel RLC Circuit 8.3 The Step Response of a Parallel RLC Circuit 8.4 The Natural and Step Response of a Series RLC Circuit 1 Key points What do the response curves of over-, under-, and critically-damped circuits look like? How to choose R, L, C values to achieve fast switching or to prevent overshooting damage? What are the initial conditions in an RLC circuit? How to use them to determine the expansion coefficients of the complete solution? Comparisons between: (1) natural & step responses, (2) parallel, series, or general RLC. 2 Section 8.1, 8.2 The Natural Response of a Parallel RLC Circuit 1. ODE, ICs, general solution of parallel voltage 2. Over-damped response 3. Under-damped response 4. Critically-damped response 3 The governing ordinary differential equation (ODE) V0, I0, v(t) must satisfy the passive sign convention. dv 1 t v By KCL: C I0 v() t d t 0 . 0 dt L R Perform time derivative, we got a linear 2nd- E of order ODv(t) with constant coefficients: d2 v 1 dv v 0 . dt2 RCdt LC 4 The two initial conditions (ICs) The capacitor voltage cannot change abruptly, (v 0 ) V0 ( 1 ) The inductor current cannot change abruptly, ( 0i )L I, i 0 (C 0 ) iL ( i 0R ) 0 ( I 0 0 V ) , R dvC I0V 0 i(C 0 ) C, v(C 0 ) v ( 0 ) ( 2 ) dt t0 C RC 5 General solution Assume the solution vis () t Ae st , where A, s are unknown constants to be solved. Substitute into the ODE, we got an algebraic (characteristic) equation of s determined by the circuit parameters: s 1 s2 0 . RC LC Since the ODE is linear, linear combination of solutions remains a solution to the equation. The general solution of v(t) must be of the form: s1 t s2 t v t() A1 e 2 A, e where the expansion constants A1, A2 will be determined by the two initial conditions. 6 Neper and resonance frequencies In general, s has two roots, which can be (1) distinct real, (2) degenerate real, or (3) complex conjugate pair. 2 1 1 1 2 2 1s , 2 0 , 2RC RC 2 LC where 1 , …neper frequency 2RC 1 …resonance (natural) frequency 0 LC 7 Three types of natural response How the circuit reaches its steady state depends on the relative magnitudes of and 0: The Circuit is When Solutions real, distinct Over-damped roots s1, s2 complex roots Under-damped * s1 = (s2) real, equal roots Critically-damped s1 = s2 8 Over-damped response ( >0) The complete solution and its derivative are of the form: s1 t s2 t v t () A1 e 2 A, e s1 t s2 t v t () A1 s 1 e 2 A 2 . s e 2 2 where 1s , 2 0 are distinct real. Substitute the two ICs: v( 0 A ) A ( V 1 ) 1 2 0 solve I0V 0 A1, A2. v( 0 s )1 A 1 2 s 2 A ( 2 ) C RC 9 Example 8.2: Discharging a parallel RLC circuit (1) Q: v(t), iC( t), iL( t), iR( t)=? 12 V 30 mA 1 1 12 . 5 kHz ,> , 22 (RC 200 )( 27 10 ) 0 over- 1 1 10 kHz.damped 0 2 7 (LC 5 10 )( 2 10 ) 10 Example 8.2: Solving the parameters (2) The 2 distinct real roots of s are: 2 2 s 1 50 kHz …|s1|< , (slow) 2 2 …|s |>(fast) s 2 200 kHz 2 . The 2 expansion coefficients are: AAV1 2 0 AA 1 2 12 I0V 0 s1 A 1 s 2 A2 5AA 1 20 2 450 C RC 14A1 V ,A2 26 V. 11 Example 8.2: The parallel voltage evolution (3) s1 t s2 t 5000t 20000t v t() A e1 A2 e 14 26 e e V. |s2|>(fast) Converge dominates to zero “Over” |s1|<(slow) damp dominates 12 Example 8.2: The branch currents evolution (4) The branch current through R is: v() t i() t 70e5000130t e20000t mA. R 200 The branch current through L is: t 1 5000t 20000t ( )iL t 30 mA v( t ) d t 56 e26 e mA. 5 0 mH0 The branch current through C is: dv ( )i ( t 0 . 2μ F)14e5000104t e20000t mA. C dt 13 Example 8.2: The branch currents evolution (5) Converge to zero 14 General solution to under-damped response ( <0) The two roots of s are complex conjugate pair: 2 2 1s , 2 0 j d , 2 2 where d 0 is the damped frequency. The general solution is reformulated as: ()j d t ()j d t v() t A1 e A2 e t e A1cos td jsin d t 2 Acosd sin td j t t e A1 A 2 cosd t j1 A 2 sin Ad t t e B1 cosd t 2sin Bd . t 15 Solving the expansion coefficients B1, B2 by ICs The derivative of v(t) is: t t v t() B1 ecos d t d sin ed t t t B2 esin d t d ecos d t t e B 1 Bd 2 cos d t 2 B d 1 sin Bd . t Substitute the two ICs: v( 0 ) B V( 1 ) 1 0 solve I V 0 0 B , B . (v 0 )BB 1 d 2 ( 2 ) 1 2 C RC 16 Example 8.4: Discharging a parallel RLC circuit (1) Q: v(t), iC( t), iL( t), iR( t)=? 0 V -12.25 mA 1 1 0 . 2 kHz , 4 7 < , 2 (2 RC 10 )( 1 . 25 10 ) 0 1 1 under- 1 kHz. 0 7 damped (LC 8 )( 1 . 25 10 ) 17 Example 8.4: Solving the parameters (2) The damped frequency is: 2 2 2 2 d 1 0 0 . 2 0 . 98 kHz . The 2 expansion coefficients are: BV 10 0 ( 1 ) B1 0 , I0V 0 BB 1 d 2 ( 2 ) B2100 V C RC 18 Example 8.4: The parallel voltage evolution (3) t t 200t v t() B e1 cosd t 2 B100sin ed t sin e 980t V . The voltage oscillates (~d ) and approaches the final value (~), different from the over- damped case (no oscillation, 2 decay constants). 19 Example 8.4: The branch currents evolutions (4) The three branch currents are: 20 Rules for circuit designers If one desires the circuit reaches the final value as fast as possible while the minor oscillation is of less concern, choosing R, L, C values to satisfy under-damped condition. If one concerns that the response not exceed its final value to prevent potential damage, designing the system to be over-damped at the cost of slower response. 21 General solution to critically-damped response (=0) Two identical real roots of s make st st st st vtAe() 1 Ae2 (),1 A 2 Ae0 Ae not possible to satisfy 2 independent ICs (V0, I0) with a single expansion constant A0. The general solution is reformulated as: t v t() e D1 t 2. D You can prove the validity of this form by substituting it into the ODE: v()( t RC ) ()(1 v t 1 LC )()0. v t 22 Solving the expansion coefficients D1, D2 by ICs The derivative of v(t) is: t t t t v t D() e1 te D 2 e 1 D 2 D1 . D t e Substitute the two ICs: v( 0 ) D V( 1 ) 2 0 solve D , D . I0V 0 1 2 v( 0 ) D 1 2 D ( 2 ) C RC 23 Example 8.5: Discharging a parallel RLC circuit (1) Q: What is R such that the circuit is critically- damped? Plot the corresponding v(t). R 0 V -12.25 mA 1 1 1 L 1 8 , , R 4 k . 0 2RC LC 2 C 12 . 25 7 10 Increasing R tends to bring the circuit from over- to critically- nd even under-damped. a 24 Example 8.5: Solving the parameters (2) The neper frequency is: 1 1 1 kHz , 2 (2 4RC 10 )(3 1 . 257 10 ) 1 1 ms. The 2 expansion coefficients are: DV 20 0 ( 1 ) -12.25 mA D981 kV s I0V 0 DD1 2 ( 2 ) D2 0 C RC 0.125 F 25 Example 8.5: The parallel voltage evolution (3) t t 1000t v t() D1 te 298 D , e 000te V . 98 V/ms 26 Procedures of solving nature response of parallel RLC 1 Calculate parameters ( 2 RC ) and 0 1 LC. Write the form of v(t) by comparing and : A es1 t A, s2 e t s - 2 , 2 if , 1 2 1 , 2 0 0 t 2 2 v() te B1 cosd t sin2 B d , td 0 , if 0 , et D t, Dif . 1 2 0 Find the expansion constants (A1, A2), (B1, B2), or (D1, D2) by two ICs:(v 0 ) V ( 1 ), 0 I0V 0 (v 0 ) ( 2 ) C RC 27 Section 8.3 The Step Response of a Parallel RLC Circuit 1. Inhomogeneous ODE, ICs, and general solution 28 The homogeneous ODE + Is V0 I0 dv 1 t v By KCL: C I0 v() t d t Is. 0 dt L R Perform time derivative, we got a homogeneous ODE of v(t) independent of the source current Is: 2 d v 1 dv v 0 . dt2 RCdt LC 29 The inhomogeneous ODE i Is L V0 I0 Change the unknown to the inductor current iL( t): dv v C i I , dt L R s d2 i 1 di i I L L L s di dt2 RCdt LC LC v L L , dt 30 The two initial conditions (ICs) i Is L V0 I0 The inductor current cannot change abruptly, (i 0L ) I0 ( 1 ) The capacitor voltage cannot change abruptly, (v 0C ) V0 L ( v 0 ), diL V0 ( 0v )L L , i (L 0 ) ( 2 ) dt t0 L 31 General solution The solution is the sum of final current If =Is and the nature response iL,nature (t): iL() tf I L i, nature( ), t where the three types of nature responses were elucidated in Section 8.2: s1 t s2 t I f A 1 e 2, A if e 0 , t iL()I t ef B1cosd tsin2 Bd , if t 0 , t I ef D1 t, 2 Dif 0 .