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 )( tdtv .0 0 dt L R
Perform time derivative, we got a linear 2nd- order ODE of v(t) with constant coefficients:
2vd 1 dv v .0 2 RCdt dt LC 4 The two initial conditions (ICs)
The capacitor voltage cannot change abruptly, Vv 0 )1()0( The inductor current cannot change abruptly, L 0,)0( C L R )0()0()0( 00 RVIiiiIi ,
dvC I V00 C )0( Ci , C vv )0()0( )2( dt t0 C RC 5 General solution
Assume the solution is )( Aetv 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 LCRC 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: 1 2tsts )( 1 2eAeAtv ,
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 s 2,1 0 , 2RC 2 LCRC 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: 1 2tsts )( 1 2eAeAtv , 1ts 2ts )( 11 22 esAesAtv .
2 2 where s 2,1 0 are distinct real.
Substitute the two ICs:
)0( VAAv )1( 021 solve I V00 A1, A2. )0( AsAsv 2211 )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 5.12 ,kHz > , 2RC 7 )102)(200(2 0 over- 1 1 10 kHz. damped 0 2 7 LC )102)(105( 10 Example 8.2: Solving the parameters (2)
The 2 distinct real roots of s are:
2 2 s1 0 5 ,kHz …|s1|<(slow) 2 2 …|s |>(fast) s2 0 20 .kHz 2
The 2 expansion coefficients are:
VAA 021 AA 21 12 I V00 AsAs 2211 1 205 AA 2 450 C RC
A1 14 A2 26,V V. 11 Example 8.2: The parallel voltage evolution (3)
1 2tsts 5000t 20000t )( 1 2 14 26eeeAeAtv V.
|s2|>(fast) Converge dominates to zero
“Over” |s |<(slow) damp 1 dominates
12 Example 8.2: The branch currents evolution (4)
The branch current through R is: tv )( ti )( 70e5000t 130e20000t mA. R 200
The branch current through L is:
t 1 5000t 20000t L ti 30)( mA 56)( 26eetdtv mA. mH 05 mH 0
The branch current through C is: dv ti 2.0()( μF) 14e5000t 104e20000t 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 s 2,1 0 jd ,
22 where d 0 is the damped frequency.
The general solution is reformulated as:
d )( tj d )( tj )( 1eAtv 2eA t 1cos d sind 2 cos d sind tjtAtjtAe t 21 cosd 21 sind tAAjtAAe t 1 cosd 2 d tBtBe .sin
15 Solving the expansion coefficients B1, B2 by ICs
The derivative of v(t) is: t t )( 1 cos dd sind teteBtv t t 2 sin dd cosd teteB t 1 d 2 cos d 2 d 1 d tBBtBBe .sin
Substitute the two ICs:
)0( VBv )1( 01 solve I V 00 B , B . v )0( 1 d BB 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 2.0 ,kHz 4 7 < , 2RC )1025.1)(102(2 0 1 1 under- 1 kHz. 0 7 damped LC )1025.1)(8( 17 Example 8.4: Solving the parameters (2)
The damped frequency is:
2222 d 0 98.02.01 .kHz
The 2 expansion coefficients are:
VB 01 )1(0 B1 ,0 I V00 1 d BB 2 )2( B2 100 V C RC
18 Example 8.4: The parallel voltage evolution (3)
t t 200t )( 1 cosd 2 sind eteBteBtv 980sin100 t .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 )( 1 2 21 0eAeAAeAeAtv ,)(
not possible to satisfy 2 independent ICs (V0, I0)
with a single expansion constant A0.
The general solution is reformulated as:
t )( DtDetv 21 .
You can prove the validity of this form by substituting it into the ODE: 1 1 tvLCtvRCtv .0)()()()()( 22 Solving the expansion coefficients D1, D2 by ICs
The derivative of v(t) is:
t t t t )( 1 2 21 1 etDDDeDteeDtv .
Substitute the two ICs:
)0( VDv )1( 02 solve D , D . I V00 1 2 )0( DDv 21 )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 2 1025.1 7
Increasing R tends to bring the circuit from over- to critically- and even under-damped. 24 Example 8.5: Solving the parameters (2)
The neper frequency is: 1 1 1 ,kHz 2RC 3 7 )1025.1)(104(2 1 1 ms.
The 2 expansion coefficients are:
VD 02 )1(0 -12.25 mA D1 98 skV I V00 DD 21 )2( D2 0 C RC 0.125 F 25 Example 8.5: The parallel voltage evolution (3)
t t 1000t )( 1 2eDteDtv 000,98 te .V 98 V/ms
26 Procedures of solving nature response of parallel RLC
1 Calculate parameters RC )2( and 0 1 LC .
Write the form of v(t) by comparing and :
1 2tsts , seAeA - 2 2 , if , 1 2,12 0 0 t 22 tv )( 1 cosd 2 tBtBe ,sin dd 0 , if 0 , t DtDe , if . 21 0
Find the expansion constants (A1, A2), (B1, B2), or (D1, D2) by two ICs: Vv ),1()0( 0
I V00 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 )( tdtv Is. 0 dt L R Perform time derivative, we got a homogeneous
ODE of v(t) independent of the source current Is: 2 vd 1 dv v .0 2 RCdt dt 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 2id 1 di i I L sLL di 2 RCdt dt LC LC Lv L , dt
30 The two initial conditions (ICs)
i Is L V0 I0
The inductor current cannot change abruptly, L Ii 0 )1()0( The capacitor voltage cannot change abruptly, C 0 vVv L ),0()0(
diL V0 L Lv iL )2()0(,)0( dt t0 L 31 General solution
The solution is the sum of final current If =Is and
the nature response iL,nature (t):
L )( ,natureLf tiIti ),( where the three types of nature responses were elucidated in Section 8.2:
1 2tsts f 1 2eAeAI , if 0, t L ti )( f 1cosd 2 d tBtBeI ,sin if 0, t f DtDeI 21 , if 0.
32 Example 8.7: Charging a parallel RLC circuit (1)
Q: iL( t)= ?
I =0 I = 0 s 24 mA V0 =0 625
1 1 32 ,kHz 8 < , 2RC )105.2)(625(2 0 1 1 under- 40 kHz. 0 2 8 damped LC )105.2)(105.2( 33 Example 8.7: Solving the parameters (2)
The complete solution is of the form: t L )( s 1cosd 2 d tBtBeIti ,sin
22 22 where d 0 243240 .kHz
The 2 expansion coefficients are:
s IBI 01 )1(0 B1 24 ,mA V0 1 d BB 2 )2(0 B2 32 mA L
34 Example 8.7: Inductor current evolution (3)
000,32 t 000,32 t L ti 2424)( e 32)000,24cos( et t)000,24sin( .mA
35 Example 8.9: Charging of parallel RLC circuits (1)
Q: Compare iL( t) when the resistance R = 625 (under-damp), 500 (critical damp), 400 (over-damp), respectively.
I0=0 Is= 24 mA V0 =0
+ + Initial & final conditions remain: iL(0 )=0, i'L(0 )=0,
If =24 mA. Different R’s give different functional forms and expansion constants. 36 Example 8.9: Comparison of rise times (2)
The current of an under-damped circuit rises faster than that of its over-damped counterpart.
37 Section 8.4 The Natural and Step Response of a Series RLC Circuit
1. Modifications of time constant, neper frequency
38 ODE of nature response
V0, I0, i(t) must satisfy the passive sign convention.
di 1 t By KVL: LRi V0 tdti .0)( 0 dt C 2id R di i By derivative: 2 .0 dt L dt LC 1 in parallel RLC RC 39 The two initial conditions (ICs)
The inductor cannot change abruptly, Ii 0 )1()0( The capacitor voltage cannot change abruptly, C 0,)0( L C R )0()0()0( 00 RIVvvvVv ,
diL 00 RIV L Lv ,)0( L ii )0()0( )2( dt t0 L 40 General solution
st Substitute )( Aeti into the ODE, we got a different characteristic equation of s: R 1 s2 s .0 s 2 2 . L LC 2,1 0
The form of s1,2 determines the form of general
solution: 1 2tsts 1 2eAeA , if 0 t ti )( 1 cosd 2 d tBtBe ,sin if 0 t DtDe 21 , if 0
R 1 22 where , 0 , d 0 . 2L LC 1 RC)2( in parallel RLC 41 Example 8.11: Discharging a series RLC circuit (1)
Q: i(t), vC( t)=?
I0 =0 - + vC(0 )=100 V, V =-100 V 0 100 V
R 560 8.2 ,kHz < 2L )1.0(2 0, 1 1 under- 10 kHz. 0 7 damped LC )101)(1.0( 42 Example 8.11: Solving the parameters (2)
The damped frequency is:
22 22 d 0 6.98.210 .kHz
The 2 expansion coefficients are:
IB 01 )1(0 -100 V B1 ,0 RIV 00 B 2.104 mA 1 d BB 2 )2( 2 9.6 L kHz 100 mH
43 Example 8.11: Loop current evolution (3)
t 800,2 t )( 1 cosd 2 sind etBtBeti 600,9sin2.104 t .mA
44 Example 8.11: Capacitor voltage evolution (4)
c tiLtRitv )()()( e 800,2 t t 600,9sin17.29600,9cos100 t .V
When the capacitor energy starts to decrease, the inductor energy starts to increase. Inductor energy starts to decrease before capacitor energy decays to 0. 45 ODEs of step response
I 0 + Vs V 0
di 1 t By KVL: LRi V0 Vtdti s.)( 0 dt C The homogeneous and inhomogeneous ODEs
of i(t) and vC (t) are: 2id R di i 2vd R dv v V ,0 and C sCC . dt2 L dt LC dt2 L dt LC LC 46 The two initial conditions (ICs)
I 0 + Vs vC (t) V0
The capacitor voltage cannot change abruptly, C Vv 0 )1()0( The inductor current cannot change abruptly, L 0 iIi C ),0()0(
dvC I0 C )0( Ci vC )2()0(, dt t0 C 47 General solution
The solution is the sum of final voltage Vf =Vs
and the nature response vC,nature (t):
C )( ,natureCf tvVtv ),( where the three types of nature responses were elucidated in Section 8.4.
1 2tsts f 1 2eAeAV , if 0 , t C tv )( f 1cosd 2 d tBtBeV ,sin if 0, t f DtDeV 21 , if 0.
48 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.
49