Study Guide

DC Circuits | Second-Order Circuits

Second-Order Circuits

A second-order circuit is characterized by a second-order differential equation. It consists of resistors and the equivalent of two energy storage elements

Finding Initial and Final Values

First, focus on the variables that cannot change abruptly; capacitor voltage and inductor current.

There are two key points to keep in mind in determining the initial conditions.

1. Carefully handle the polarity of voltage across the capacitor and the direction of the current through the inductor; and are defined strictly according to the passive sign convention.
2. The capacitor voltage is always continuous so that and the inductor current is always continuous so that , where denotes the time just before switching and is the time just after, assuming that the switching takes place at .

Example: The switch in Fig. 1 has been closed for a long time. It is open at t=0. We are going to find:

a. The switch closed a long time before t = 0 means that the circuit is at dc steady-state at t = 0. Thus, the inductor acts like a short circuit, while the capacitor acts like an open circuit. At t = 0-,

b. At t = 0+, the switch is open and the same current flows through both the inductor and capacitor. Hence, ic(0+) = i(0+) = 2A

Since ic = C(dv/dt), dv/dt = ic/c and

Similarly, since

We now obtain VL by applying KVL to the loop in Fig. 2(b). The result is

Thus,

c. For t > 0, the circuit undergoes transience. But as t -> ∞, the circuit reaches steady-state again. The inductor acts like a short circuit and the capacitor like an open circuit. Thus, we have

The Source-Free Series RLC Circuit

A series RLC circuit is shown in Fig. 3. The circuit is being excited by the energy initially stored in the capacitor and inductor.

• The energy is represented by the initial capacitor voltage and initial inductor current . Thus, at t=0, .
• Applying KVL around the loop and differentiating with respect to t,
• This is a second-order differential equation. The solution is of the form and substituting this to the DE, the characteristic equation is

where

are the two roots of the characteristic equation of the differential. A more compact way of expressing the roots is

where

• The roots and are called natural frequencies, measured in nepers per second (Np/s).
• is known as the resonant frequency or strictly as the undamped natural frequency, expressed in radians per second (rad/s).
• is the neper frequency expressed in Np/s.
• Since there are two possible solutions from the two values of ,
• A complete or total solution would therefore require a linear combination of and . Thus the natural response of the series RLC circuit is , where the constants and are determined from initial values.

There are three types of solutions:

1. If > , the overdamped case; roots are unequal and real.
2. If = , the critically damped case; roots are equal and real.
3. If < , the underdamped case; roots are complex.
• Overdamped Case (> )
• implies > . When this happens, both roots and are negative and real. The response is , which decays and approaches zero as t increases. Fig. 4 illustrates a typical overdamped response.
• Critically Damped Case ( = )
• When = , = and . The second-order differential equation becomes
• Solving the DE gives the natural response of the critically damped circuit: a sum of a negative exponential and a negative exponential multiplied by a linear term, .
• A typical critically damped response is shown in Fig. 5. It is a sketch of , which reaches a maximum value of at , one time constant, and then decays all the way to zero.
• Underdamped Case ( < )
• When < , < . The roots may be written as , where and which is called the damped frequency.
• Both and are natural frequencies because they help determine the natural response.
• Using Euler's identities, and replacing constants with constants , the natural response is .
• The natural response for this case is exponentially damped and oscillatory in nature. It has a time constant of and a period of . Fig. 6 shows a typical underdamped response.

The Source-Free Parallel RLC Circuit

• Consider the parallel RLC circuit shown in Fig. 7. Assume initial inductor current and initial capacitor voltage , and .

The three elements in parallel have the same voltage across. According to the passive sign convention, the current through each element is leaving the top node. Applying KCL at the top node, taking the derivative with respect to t and dividing by C results in

The characteristic equation is obtained as

The roots of the characteristic equation are

• Again, there are three possible solutions, depending on whether > , = , or <

Overdamped Case

• > when > , the roots of the characteristic equation are real and negative. The response is

Critically Damped Case

• For = , , the roots are real and equal so that the response is

Underdamped Case

• When < , < . In this case the roots are complex and may be expressed as .
• The response is
• The constants in each case can be determined from the initial conditions: and . To find ,

Step Response of a Series RLC Circuit

Consider the series RLC circuit shown in Fig. 8.

Applying KVL around the loop for t>0,

Substituting i = C dv/dt,

The solution has two components: the transient response (vt(t)) and the steady-state response (vss(t)) ; that is,

vt(t) for each case:

• Overdamped:
• Critically damped:
• Underdamped:

In the circuit in Fig. 8, the final value of the capacitor voltage is the same as the source voltage VS. Hence the steady-state response is

Complete solutions for the three cases:

• Overdamped:
• Critically damped:
• Underdamped:
• and are, respectively, the voltage across C and the current through L.
• are obtained from the initial conditions: .
• Once the capacitor voltage is known, can be obtained which is the same current through C, L, and R.
• Hence, the voltage across the resistor is , while the inductor voltage is

Step Response of a Parallel RLC Circuit

Consider the parallel RLC circuit shown in Fig. 9.

Applying KCL at the top node for t>0,

Substituting v = L di/dt and dividing by LC,

Again, the complete solution consists of the transient response (it(t)) and the steady-state response (iss(t)) ; that is,

it(t) for each case

• Overdamped:
• Critically damped:
• Underdamped:
• The steady-state response is the final value of . In the circuit, the final value of the current through L is the same as the source current .

Complete solutions for the three cases

• Overdamped:
• Critically damped:
• Underdamped:
• These equations only apply for finding the inductor current .
• are obtained from the initial conditions for .
• Once the inductor current is known, can be obtained which is the same voltage across C, L, and R.
• Hence, the current through the resistor is , while the capacitor current is .

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