(a) RLC Circuits
Lecture 7 RLC Circuits 1. Topics covered (a) RLC Circuits (b) The basic series and parallel RLC circuits (c) Deriving the RLC differential equation (d) Solving the RLC differential equation (e) Underdamped, overdamped, and critically damped responses (f) An application: A high voltage pulse forming circuit 2. RLC circuits (a) As its name implies, an RLC circuit contains resistors, inductors, and capacitors (b) RLC circuits have many applications: pulse forming networks, oscillator circuits, medical equipment, magnetic launchers, ignition systems, etc. (c) In general RLC circuits are more complicated to analyze than RL or RC circuits (d) The reason is that the differential equation describing the circuit is a second order ODE as opposed to the first order ODEs characterizing an RC or RL circuit (e) Even so, the general mathematical method of analysis is the same • Find a particular solution • Find a homogeneous solution (with 2 free constants C1 and C2 resulting from the fact that we are solving a 2nd order ODE) • Specify 2 initial conditions to find C1 and C2 (f) In this lecture we consider RLC circuits driven by DC sources (g) Sometimes the source will appear explicitly (h) Other times a source can be present which drives the circuit to steady state DC conditions after which a switch is thrown eliminating the source from the circuit. The decay of the stored electric and magnetic energy is then the issue of primary concern. 1
(i) Qualitatively an RLC circuit can respond to DC sources in one of three ways • Underdamped • Overdamped • Critically damped (j) We discuss these cases shortly (k) The basic reason for the three types of responses is associated with the fact that an RLC circuit has two independent characteristic time scales as opposed to a single time scale in an RC or RL circuit (l) The relative sizes of these time scales determines the qualitative behavior of the circuit 3. The basic series and parallel RLC circuits (a) The simplest RLC circuit that we shall consider is the voltage driven series circuit as shown below I R VS
L
C
+ -
Figure 7.1 (b) Typically we want to calculate I(t). (c) The dual circuit is the current driven parallel RLC circuit as shown below
I
R
L
Figure 7.2 (d) Here typically we want to calculate V (t) 2
C
+ V -
(e) To obtain an intuitive picture of the circuit response, consider the series circuit (f) For short times, right after the switch is closed the inductor L acts like an open circuit so that the current I ≈ 0. (g) For long times after the switch is closed the capacitor acts like an open circuit so that the current I ≈ 0. (h) There are two qualitatively different ways this behavior can occur: underdamped and overdamped
I
I
t
OVERDAMPED
UNDERDAMPED Figure 7.3
(i) The transition between these two regimes is defined as the critically damped regime
I
t CRITICALLY DAMPED Figure 7.4 4. Deriving the RLC differential equation (a) Let us begin by deriving the ordinary differential equation describing the behavior of the series RLC circuit 3
+VR -
I
+VL -
R VS
L
+VC -
C
+ -
Figure 7.5 (b) Kirchoff’s voltage law yields the simple relation VS = VR + VL + VC
(1)
(c) For the series circuit it is convenient to eliminate all variables in terms of the capacitor voltage VC . (d) The reason for the convenience, as shown below, is that the final ODE is obtained without the need of differentiating Kirchoff’s voltage equation. (e) If one were to differentiate KVL, this differentiation would introduce another unknown integration constant into the solution. (f) While this new constant can be readily dealt with through a derived initial condition, it is convenient not to have to do so. (g) Returning to the series RLC circuit the relation between the various unknown quantities and VC is given as follows I = C
dVC dt
dVC dt d2 VC dI = L = LC 2 dt dt
VR = RI = RC VL
(2)
(h) Substituting into KVL yields VS = LC
d2 VC dVC + RC + VC dt2 dt
4
(3)
(i) The equation can be slightly simplified by introducing the characteristic natural frequencies of the circuit which are defined as follows ωo2 =
1 LC
,
ν = 2ω0 ξ =
R L
,
ξ=
R C 2 L
1/2
(4)
(j) The equation now reduces to d2 VC dVC + 2ω0 ξ + ω02 VC = ω02 VS 2 dt dt
(5)
(k) We shall see that the quantity ω0 is the natural oscillation frequency of the circuit. (l) Similarly, ν is the natural damping coefficient of the circuit. The quantity ξ is a dimensionless quantity representing the damping coefficient, normalized to the oscillation frequency. (m) A similar analysis holds for the parallel RLC circuit.
IS
IL
IR
IC
+ V -
Figure 7.6 (n) In this case we use KCL: IS = IR + IL + IC and eliminate all quantities in terms of IL . V
= L
dIL dt
V L dIL = R R dt dV d2 IL = C = LC 2 dt dt
IR = IC
5
(6)
(o) The equations for the parallel RLC circuit in standard and simplified forms are given by d2 IL L dIL + + IL = IS dt2 R dt dIL d2 IL + 2ω0 ξ + ω02 IL = ω02 IS 2 dt dt LC
(p) Here, ω02
1 = LC
,
1 L ξ= 2R C
(7)
1/2
(8)
(q) Note the similarities in form for the series and parallel circuit equations. The main difference is that the normalized damping coefficients ξ are inversely defined for each circuit. 5. Solving the RLC differential equation (a) Consider now the solution to the series RLC circuit. (b) This is a linear, second order ODE with constant coefficients and an inhomogeneous driving term. For a DC source the driving term is a constant (as a function of time) (c) The procedure to solve the equation is to find a particular solution, the two homogenous solutions, and then to apply two initial conditions to determine the two unknown constants associated with the homogeneous solutions. (d) For the series RLC equation with a constant driver VS , the particular solution is easily found by inspection Vcp = VS
(9)
(e) This solution obviously satisfies the differential equation (f) The homogeneous solutions are found by setting the driving term to zero. The homogeneous solutions thus satisfy d2 Vch dVch + 2ω0 ξ + ω02 Vch = 0 2 dt dt
6
(10)
(g) A linear, homogenous ODE with constant coefficients has solutions of the form Vch = cest (11) (h) Here, s is an unknown parameter to be determined by substituting the proposed solution into the differential equation. The result is
c s2 + 2ω0 ξs + ω02 est = 0
(12)
(i) For non-trivial solutions to exist (i.e. C = 0) the unknown quantity s must satisfy the quadratic, algebraic equation given by s2 + 2ω0 ξs + ω02 = 0
(13)
(j) Note than any value of c corresponds to a solution as long as the algebraic equation is satisfied. (k) The algebraic equation has two solutions for s which can be written as s1 = ω0 [−ξ + (ξ 2 − 1)1/2 ] s2 = ω0 [−ξ − (ξ 2 − 1)1/2 ]
(14)
(l) Note that the roots for s have the following properties • s1 , s2 both real and negative for ξ > 1 • s1 , s2 complex conjugates with a negative real part for ξ < 1 • s1 , s2 have a double root for ξ = 1 (m) Note also that s1 , s2 are known quantities that can be expressed in terms of the circuit elements R, L, and C . (n) The significance of these three possibilities is discussed shortly. (o) In the meantime, we note that since either s1 or s2 satisfies the algebraic equation there are, as expected, two homogeneous solutions with two arbitrary multiplicative constants (p) The total solution is thus given by VC = VCp + Vch = VS + C1 es,t + C2 es2 t 7
(15)
(q) The constants C1 and C2 are evaluated by applying initial conditions. As a specific example consider the situation where just before the switch is closed there is zero current flowing in the circuit and there is no charge (i.e. voltage) across the capacitor. (r) In terms of VC the initial conditions become VC (0) = 0 I(0) dVC (0) = = 0 dt C
(16)
(s) If we apply these two conditions we obtain two, linear algebraic equations for C1 and C2 Vc (0) = 0 : dVc (0) = 0: dt
C1 + C2 = −Vs s1 C1 + s2 C2 = 0
(17)
(t) Solving for C1 and C2 we find s2 Vs s1 − s2 s1 = − Vs s 1 − s2
C1 = C2
(18)
(u) Substituting back yields the complete solution for VC and I
VC = VS I = C
s2 s1 1+ es1 t − es2 t s 1 − s2 s 1 − s2
dVC s1 s2 es1 t − es2 t = CVS dt s1 − s2
(19)
(v) We next proceed to analyze this solution for the three cases of interest as defined by the value of ξ 6. Underdamped, overdamped, and critically damped responses (a) In this section we examine the details of the solution for the series RLC circuit for the three cases of interest ξ > 1, ξ < 1, and ξ = 1. 8
(b) Consider first ξ > 1. In this regime the two roots for s are both real (c) Intuition about the behavior can be obtained by assuming the inequality is strongly satisfied: ξ >> 1. In this limit (which corresponds to R >> (L/C)1/2 the resistance in the circuit is “large” and we would therefore expect a large amount of damping in the circuit response. (d) For ξ >> 1 we can simplify the expression for the roots by using the Taylor expansion for the square root as follows (ξ 2 − 1)1/2 = ξ(1 −
1 1/2 1 1 ) = ξ(1 − 2 + . . .) ≈ ξ − 2 ξ 2ξ 2ξ
(20)
(e) The two roots for s become s1 s2
1 ω0 = ω0 −ξ + (ξ − 1) ≈ ω0 −ξ + ξ − ≈− 2ξ 2ξ 1 = ω0 −ξ + (ξ 2 − 1)1/2 ≈ ω0 −ξ − ξ − ≈ −2ω0 ξ 2ξ (21) 2
1/2
(f) We see that one root s1 is much smaller than the characteristic frequency ω0 . This corresponds to a slowly decaying exponential. (g) The other root s2 is much larger than ω0 corresponding to a rapidly decaying exponential. (h) The total solution for the current can be approximated as follows
I = CVS ≈
s 1 s2 (es1 t − es2 t ) s1 − s 2
VS −t /2ξ − e−2ξt ) (e R
t = ω0 t
(22)
(i) The solution is sketched below. Note that the current rises rapidly (as compared to the corresponding ω0 time scale) and then decays to zero on a very long time scale. Also the peak current and the time at which the current peaks can be easily found
9
I
VS /R
1
ω0 t
Imax ≈ VS /R ω0 tmax ≈ (1/ξ)n(2ξ) 1, this regime is known as the “overdamped regime” (k) We next consider the regime in which ξ < 1. In this regime the two roots for s are complex conjugates. As in the overdamped case we consider the limit where the inequality is strongly satisfied: ξ
(i) Qualitatively an RLC circuit can respond to DC sources in one of three ways • Underdamped • Overdamped • Critically damped (j) We discuss these cases shortly (k) The basic reason for the three types of responses is associated with the fact that an RLC circuit has two independent characteristic time scales as opposed to a single time scale in an RC or RL circuit (l) The relative sizes of these time scales determines the qualitative behavior of the circuit 3. The basic series and parallel RLC circuits (a) The simplest RLC circuit that we shall consider is the voltage driven series circuit as shown below I R VS
L
C
+ -
Figure 7.1 (b) Typically we want to calculate I(t). (c) The dual circuit is the current driven parallel RLC circuit as shown below
I
R
L
Figure 7.2 (d) Here typically we want to calculate V (t) 2
C
+ V -
(e) To obtain an intuitive picture of the circuit response, consider the series circuit (f) For short times, right after the switch is closed the inductor L acts like an open circuit so that the current I ≈ 0. (g) For long times after the switch is closed the capacitor acts like an open circuit so that the current I ≈ 0. (h) There are two qualitatively different ways this behavior can occur: underdamped and overdamped
I
I
t
OVERDAMPED
UNDERDAMPED Figure 7.3
(i) The transition between these two regimes is defined as the critically damped regime
I
t CRITICALLY DAMPED Figure 7.4 4. Deriving the RLC differential equation (a) Let us begin by deriving the ordinary differential equation describing the behavior of the series RLC circuit 3
+VR -
I
+VL -
R VS
L
+VC -
C
+ -
Figure 7.5 (b) Kirchoff’s voltage law yields the simple relation VS = VR + VL + VC
(1)
(c) For the series circuit it is convenient to eliminate all variables in terms of the capacitor voltage VC . (d) The reason for the convenience, as shown below, is that the final ODE is obtained without the need of differentiating Kirchoff’s voltage equation. (e) If one were to differentiate KVL, this differentiation would introduce another unknown integration constant into the solution. (f) While this new constant can be readily dealt with through a derived initial condition, it is convenient not to have to do so. (g) Returning to the series RLC circuit the relation between the various unknown quantities and VC is given as follows I = C
dVC dt
dVC dt d2 VC dI = L = LC 2 dt dt
VR = RI = RC VL
(2)
(h) Substituting into KVL yields VS = LC
d2 VC dVC + RC + VC dt2 dt
4
(3)
(i) The equation can be slightly simplified by introducing the characteristic natural frequencies of the circuit which are defined as follows ωo2 =
1 LC
,
ν = 2ω0 ξ =
R L
,
ξ=
R C 2 L
1/2
(4)
(j) The equation now reduces to d2 VC dVC + 2ω0 ξ + ω02 VC = ω02 VS 2 dt dt
(5)
(k) We shall see that the quantity ω0 is the natural oscillation frequency of the circuit. (l) Similarly, ν is the natural damping coefficient of the circuit. The quantity ξ is a dimensionless quantity representing the damping coefficient, normalized to the oscillation frequency. (m) A similar analysis holds for the parallel RLC circuit.
IS
IL
IR
IC
+ V -
Figure 7.6 (n) In this case we use KCL: IS = IR + IL + IC and eliminate all quantities in terms of IL . V
= L
dIL dt
V L dIL = R R dt dV d2 IL = C = LC 2 dt dt
IR = IC
5
(6)
(o) The equations for the parallel RLC circuit in standard and simplified forms are given by d2 IL L dIL + + IL = IS dt2 R dt dIL d2 IL + 2ω0 ξ + ω02 IL = ω02 IS 2 dt dt LC
(p) Here, ω02
1 = LC
,
1 L ξ= 2R C
(7)
1/2
(8)
(q) Note the similarities in form for the series and parallel circuit equations. The main difference is that the normalized damping coefficients ξ are inversely defined for each circuit. 5. Solving the RLC differential equation (a) Consider now the solution to the series RLC circuit. (b) This is a linear, second order ODE with constant coefficients and an inhomogeneous driving term. For a DC source the driving term is a constant (as a function of time) (c) The procedure to solve the equation is to find a particular solution, the two homogenous solutions, and then to apply two initial conditions to determine the two unknown constants associated with the homogeneous solutions. (d) For the series RLC equation with a constant driver VS , the particular solution is easily found by inspection Vcp = VS
(9)
(e) This solution obviously satisfies the differential equation (f) The homogeneous solutions are found by setting the driving term to zero. The homogeneous solutions thus satisfy d2 Vch dVch + 2ω0 ξ + ω02 Vch = 0 2 dt dt
6
(10)
(g) A linear, homogenous ODE with constant coefficients has solutions of the form Vch = cest (11) (h) Here, s is an unknown parameter to be determined by substituting the proposed solution into the differential equation. The result is
c s2 + 2ω0 ξs + ω02 est = 0
(12)
(i) For non-trivial solutions to exist (i.e. C = 0) the unknown quantity s must satisfy the quadratic, algebraic equation given by s2 + 2ω0 ξs + ω02 = 0
(13)
(j) Note than any value of c corresponds to a solution as long as the algebraic equation is satisfied. (k) The algebraic equation has two solutions for s which can be written as s1 = ω0 [−ξ + (ξ 2 − 1)1/2 ] s2 = ω0 [−ξ − (ξ 2 − 1)1/2 ]
(14)
(l) Note that the roots for s have the following properties • s1 , s2 both real and negative for ξ > 1 • s1 , s2 complex conjugates with a negative real part for ξ < 1 • s1 , s2 have a double root for ξ = 1 (m) Note also that s1 , s2 are known quantities that can be expressed in terms of the circuit elements R, L, and C . (n) The significance of these three possibilities is discussed shortly. (o) In the meantime, we note that since either s1 or s2 satisfies the algebraic equation there are, as expected, two homogeneous solutions with two arbitrary multiplicative constants (p) The total solution is thus given by VC = VCp + Vch = VS + C1 es,t + C2 es2 t 7
(15)
(q) The constants C1 and C2 are evaluated by applying initial conditions. As a specific example consider the situation where just before the switch is closed there is zero current flowing in the circuit and there is no charge (i.e. voltage) across the capacitor. (r) In terms of VC the initial conditions become VC (0) = 0 I(0) dVC (0) = = 0 dt C
(16)
(s) If we apply these two conditions we obtain two, linear algebraic equations for C1 and C2 Vc (0) = 0 : dVc (0) = 0: dt
C1 + C2 = −Vs s1 C1 + s2 C2 = 0
(17)
(t) Solving for C1 and C2 we find s2 Vs s1 − s2 s1 = − Vs s 1 − s2
C1 = C2
(18)
(u) Substituting back yields the complete solution for VC and I
VC = VS I = C
s2 s1 1+ es1 t − es2 t s 1 − s2 s 1 − s2
dVC s1 s2 es1 t − es2 t = CVS dt s1 − s2
(19)
(v) We next proceed to analyze this solution for the three cases of interest as defined by the value of ξ 6. Underdamped, overdamped, and critically damped responses (a) In this section we examine the details of the solution for the series RLC circuit for the three cases of interest ξ > 1, ξ < 1, and ξ = 1. 8
(b) Consider first ξ > 1. In this regime the two roots for s are both real (c) Intuition about the behavior can be obtained by assuming the inequality is strongly satisfied: ξ >> 1. In this limit (which corresponds to R >> (L/C)1/2 the resistance in the circuit is “large” and we would therefore expect a large amount of damping in the circuit response. (d) For ξ >> 1 we can simplify the expression for the roots by using the Taylor expansion for the square root as follows (ξ 2 − 1)1/2 = ξ(1 −
1 1/2 1 1 ) = ξ(1 − 2 + . . .) ≈ ξ − 2 ξ 2ξ 2ξ
(20)
(e) The two roots for s become s1 s2
1 ω0 = ω0 −ξ + (ξ − 1) ≈ ω0 −ξ + ξ − ≈− 2ξ 2ξ 1 = ω0 −ξ + (ξ 2 − 1)1/2 ≈ ω0 −ξ − ξ − ≈ −2ω0 ξ 2ξ (21) 2
1/2
(f) We see that one root s1 is much smaller than the characteristic frequency ω0 . This corresponds to a slowly decaying exponential. (g) The other root s2 is much larger than ω0 corresponding to a rapidly decaying exponential. (h) The total solution for the current can be approximated as follows
I = CVS ≈
s 1 s2 (es1 t − es2 t ) s1 − s 2
VS −t /2ξ − e−2ξt ) (e R
t = ω0 t
(22)
(i) The solution is sketched below. Note that the current rises rapidly (as compared to the corresponding ω0 time scale) and then decays to zero on a very long time scale. Also the peak current and the time at which the current peaks can be easily found
9
I
VS /R
1
ω0 t
Imax ≈ VS /R ω0 tmax ≈ (1/ξ)n(2ξ) 1, this regime is known as the “overdamped regime” (k) We next consider the regime in which ξ < 1. In this regime the two roots for s are complex conjugates. As in the overdamped case we consider the limit where the inequality is strongly satisfied: ξ
