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8.6 Step Response of a Parallel RLC Circuit

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8.6 Step Response of a Parallel RLC Circuit

Consider the parallel RLC circuit sho wn in Fig. 8.22. We want to find i due to a sudden application of a dc current. Applying KCL at the top node for t > 0,

vR+i+Cdvdt=Is(8.46)\frac{v}{R} + i + C\frac{dv}{dt} = I_s \tag{8.46} v=Ldidtv = L \frac{di}{dt}

Substituting for v in Eq. (8.46) and dividing by LC, we get

d2idt2+1RCdidt+iLC=IsLC\frac{d^2i}{dt^2} + \frac{1}{RC}\frac{di}{dt} + \frac{i}{LC} = \frac{I_s}{LC}

(8.47)

which has the same characteristic equation as Eq. (8.29).

The complete solution to Eq. (8.47) consists of the transient response it(t) and the steady-state response iss; that is,

i(t)=it(t)+iss(t)i(t) = it(t) + iss(t)

(8.48)

The transient response is the same as what we had in Section 8.4. The steady-state response is the final value of i. In the circuit in Fig. 8.22, the final value of the current through the inductor is the same as the source current Is. Thus,

i(t)=Is+A1es1t+A2es2ti(t) = I_s + A_1 e^{s_1 t} + A_2 e^{s_2 t}

(Overdamped)
\n

i(t)=Is+(A1+A2t)e−αti(t) = I_s + (A_1 + A_2 t)e^{-\alpha t}

(Critically damped)
\n

i(t)=Is+(A1cos⁡ωdt+A2sin⁡ωdt)e−αti(t) = I_s + (A_1 \cos \omega_d t + A_2 \sin \omega_d t)e^{-\alpha t}

(Underdamped) (8.49)

The constants A1 and A2 in each case can be determined from the initial conditions for i and di∕dt. Again, we should keep in mind that Eq. (8.49) only applies for finding the inductor current i. But once the inductor current iL = i is known, we can find v = L di∕dt, which is the same v oltage across inductor, capacitor, and resistor . Hence, the current through the resistor is iR = v∕R, while the capacitor current is iC = C dv∕dt. Alternatively, the complete response for any variable x(t) may be found directly, using

x(t)=xss(t)+xt(t)x(t) = x_{ss}(t) + x_t(t)

(8.50)

where xss and xt are its final value and transient response, respectively.

Solution:

For t < 0, the switch is open, and the circuit is partitioned into two inde pendent subcircuits. The 4-A current flows through the inductor, so that

i(0)=4 Ai(0) = 4 \text{ A}

Since 30u(− t) = 30 when t < 0 and 0 when t > 0, the voltage source is operative for t < 0. The capacitor acts like an open circuit and the voltage across it is the same as the voltage across the 20-Ω resistor connected in parallel with it. By voltage division, the initial capacitor voltage is

v(0)=2020+20(30)=15 Vv(0) = \frac{20}{20 + 20}(30) = 15 \text{ V}

For t > 0, the switch is closed, and we ha ve a parallel RLC circuit with a current source. The voltage source is zero which means it acts like a short-circuit. The tw o 20-Ω resistors are no w in parallel. They are combined to gi ve R = 20 ‖ 20 = 10 Ω. The characteristic roots are determined as follows:

α=12RC=120×10×8×10−3=6.25\alpha = \frac{1}{2RC} = \frac{1}{20 \times 10 \times 8 \times 10^{-3}} = 6.25

\n

ω0=1LC=120×8×10−3=2.5\omega_0 = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{20 \times 8 \times 10^{-3}}} = 2.5

\n

s1,2=−α±α2−ω02=−6.25±39.0625−6.25s_{1,2} = -\alpha \pm \sqrt{\alpha^2 - \omega_0^2} = -6.25 \pm \sqrt{39.0625 - 6.25}

\n

=−6.25±5.7282= -6.25 \pm 5.7282

or

s1=−11.978,s2=−0.5218s_1 = -11.978, \qquad s_2 = -0.5218

Since α > ω0, we have the overdamped case. Hence,

i(t)=Is+A1e−11.978t+A2e−0.5218ti(t) = I_s + A_1 e^{-11.978t} + A_2 e^{-0.5218t}

(8.8.1)

where Is = 4 is the final value of i(t). We now use the initial conditions to determine A1 and A2. At t = 0,

i(0) = 4 = 4 + A1 + A2 ⇒ A2 = −A1 (8.8.2)

Taking the derivative of i(t) in Eq. (8.8.1),

didt=−11.978A1e−11.978t−0.5218A2e−0.5218t\frac{di}{dt} = -11.978A_1e^{-11.978t} - 0.5218A_2e^{-0.5218t}

so that at t = 0,

di(0)dt=−11.978A1−0.5218A2(8.8.3)\frac{di(0)}{dt} = -11.978A_1 - 0.5218A_2 \tag{8.8.3}

But

Ldi(0)dt=v(0)=15L\frac{di(0)}{dt} = v(0) = 15

⇒\Rightarrow di(0)dt=15L=1520=0.75\frac{di(0)}{dt} = \frac{15}{L} = \frac{15}{20} = 0.75

Substituting this into Eq. (8.8.3) and incorporating Eq. (8.8.2), we get

0.75=(11.978−0.5218)A2⇒A2=0.06550.75 = (11.978 - 0.5218)A_2 \Rightarrow A_2 = 0.0655

Thus, A1 = −0.0655 and A2 = 0.0655. Inserting A1 and A2 in Eq. (8.8.1) gives the complete solution as

i(t)=4+0.0655(e−0.5218t−e−11.978t)i(t) = 4 + 0.0655(e^{-0.5218t} - e^{-11.978t})

A

From i(t), we obtain v(t) = L di∕dt and

iR(t)=v(t)20=L20didt=0.785e−11.978t−0.0342e−0.5218ti_R(t) = \frac{v(t)}{20} = \frac{L}{20} \frac{di}{dt} = 0.785e^{-11.978t} - 0.0342e^{-0.5218t}

A

Find i(t) and v(t) for t > 0 in the circuit of Fig. 8.24.

Answer: 10(1− cos(0.5t)) A, 100 sin(0.5t) V.

Practice Problem 8.8

‒

8.7 General Second-Order Circuits

Now that we have mastered series and parallel RLC circuits, we are prepared to apply the ideas to an y second-order circuit having one or more independent sources with constant values. Although the series and parallel RLC circuits are the second-order circuits of greatest interest, other second-order circuits including op amps are also useful. Given a secondorder circuit, we determine its step response x(t) (which may be voltage or current) by taking the following four steps:

    1. We first determine the initial conditions x(0) and dx(0)∕dt and the final value x(∞), as discussed in Section 8.2.
    1. We turn off the independent sources and find the form of the transient response xt (t) by applying KCL and KVL. Once a second-order differential equation is obtained, we determine its characteristic roots. Depending on whether the response is overdamped, critically damped, or underdamped, we obtain xt(t) with two unknown constants as we did in the previous sections.
    1. We obtain the steady-state response as
xss(t)=x(∞)(8.51)x_{ss}(t) = x(\infty) \tag{8.51}

where x(∞) is the final value of x, obtained in step 1.

  1. The total response is now found as the sum of the transient response and steady-state response
x(t)=xt(t)+xss(t)x(t) = x_t(t) + x_{ss}(t)

(8.52)

We finally determine the constants associated with the transient response by imposing the initial conditions x(0) and dx(0)∕dt, determined in step 1.

We can apply this general procedure to find the step response of any second-order circuit, including those with op amps. The following examples illustrate the four steps.

Find the complete response v and then i for t > 0 in the circuit of Fig. 8.25.

Solution:

We first find the initial and final values. At t = 0−, the circuit is at steady state. The switch is open; the equivalent circuit is shown in Fig. 8.26(a). It is evident from the figure that

v(0−)=12 V,i(0−)=0v(0^-) = 12 \text{ V}, \qquad i(0^-) = 0

At t = 0+, the switch is closed; the equivalent circuit is in Fig. 8.26(b). By the continuity of capacitor voltage and inductor current, we know that

v(0+)=v(0−)=12 V,i(0+)=i(0−)=0v(0^+) = v(0^-) = 12 \text{ V}, \qquad i(0^+) = i(0^-) = 0

(8.9.1)

Figure 8.24 For Practice Prob. 8.8.

A circuit may look complicated at first. But once the sources are turned off in an attempt to find the form of the transient response, it may be reducible to a first-order circuit, when the storage elements can be combined, or to a parallel/series RLC circuit. If it is reducible to a first-order circuit, the solution becomes simply what we had in Chapter 7. If it is reducible to a parallel or series RLC circuit, we apply the techniques of previous sections in this chapter.

Problems in this chapter can also be solved by using Laplace transforms, which are covered in Chapters 15 and 16.

Example 8.9

Equivalent circuit of the circuit in Fig. 8.25 for: (a) t < 0, (b) t > 0.

Obtaining the form of the transient response for Example 8.9.

To get dv(0+)∕dt, we use C dv∕dt = iC or dv∕dt = iC∕C. Applying KCL at node a in Fig. 8.26(b),

i(0+)=iC(0+)+v(0+)2i(0^{+}) = i_{C}(0^{+}) + \frac{v(0^{+})}{2} 0=iC(0+)+122  ⟹  iC(0+)=−6 A0 = i_{C}(0^{+}) + \frac{12}{2} \implies i_{C}(0^{+}) = -6 \text{ A}