First order circuits

Now that we have considered the three passive elements (resistors, capacitors, and inductors) and one active element (the op amp) individually, we are prepared to consider circuits that contain various combinations of two or three of passive elements. In this chapter, we shall examine two types of simple circuits: a circuit comprising a resistor and capacitor and a circuit comprising a resistor and an inductor. These are called RC and RL circuits, respectively. As simple as these circuits are, they find continual applications in electronics, communications, and control system, as we shall see.

We carry out the analysis of RC and RL circuits by applying Kirchhoff’s laws, as we did for resistive circuits. The only difference is that applying Kirchhoff’s laws to purely resistive circuits, results in algebraic differential equations, which are more difficult to solve than algebraic equations. The differential equations resulting from analyzing RC and RL circuits are of the first order. Hence, the circuits are collectively known as first-order circuits.

A first-order circuit is characterized by a first-order diferential equation.

In addition to there being two types of first-order circuits (RC and RL), there are two ways to excite the circuits. The first way is by called source-free circuits, we assume that energy is initially stored in the capacitive or inductive element. The energy circuit and is gradually dissipated in the resistors. Although source-free circuits are by definition free of independent sources, they may have dependent sources. The second way of exiting first-order circuits is by independent sources. In this chapter, the independent sources we will consider are dc sources. The two types of first-order circuits and the two ways of exiting them and add up to the four possible situations we will study in this chapter.

Finally, we consider four typical applications of RC and RL circuits delay and relay circuits, a photoflash unit, and an automobile ignition circuit.

A source-free RC circuit occurs when its dc source is suddenly disconnected. The energy already stored in the capacitor is released to the resistors.

Consider a series combination of a resistor and an initially charged capacitor, as shown in [link]. (The resistor and capacitor may be the equivalent resistance and equivalent capacitance of combination of resistors and capacitors.) Our objective is to determine the circuit response, which, for pedagogic reasons, we assume to be the voltage v(t) across the capacitor. Since the capacitor is initially charged, we can assume that at time t = 0, the initial voltage is

With the corresponding value of the energy stored as

Applying KCL at the top note of the circuit in [link],

By definition, iC=Cdv/dt size 12{i rSub { size 8{C} } =C { ital "dv"} slash { ital "dt"} } {} and iR=v/R size 12{i rSub { size 8{R} } = {v} slash {R} } {}. Thus,

or

This is a first order differential equation, since only the first derivative of v is involved. To solve it, we arrange the terms as

integrating both sides, we get

ln v = − t RC + ln A size 12{"ln"v= - { {t} over { ital "RC"} } +"ln"A} {}

where ln A is the integration constant. Thus,

Taking powers of e produces

v ( t ) = Ae − t / RC size 12{v ( t ) = ital "Ae" rSup { size 8{ - t/ ital "RC"} } } {}

But from the initial conditions, v(0)=A=V0 size 12{v ( 0 ) =A=V rSub { size 8{0} } } {}. Hence,

This shows that the voltage response of the RC circuit is an exponential decay of the initial voltage. Since the response is due to the initial energy stored and the physical characteristics of the circuit and not due to some external voltage or current source, it is called the natural response of the circuit.

The natural response of a circuit refers to the behavior (in terms of voltages and currents) of the circuit itself, with no external sources of excitation.

The natural response is illustrated graphically in [link]. Note that at t = 0, we have the correct initial condition as in [link]. as t increases, the voltage decreases toward zero. The rapidity with which the voltage decreases is expressed in terms of the time constant, denoted by τ size 12{τ} {}, the lower case greek letter tau.

The time constant of a circuit is the time required for the response to decay by a factor of 1/e or 36.6 percent of its initial value.

This implies that at t=τ size 12{t=τ} {}, [link] becomes

V 0 e − τ / RC = V 0 e − 1 = 0 . 368 V 0 size 12{V rSub { size 8{0} } e rSup { size 8{ { - τ} slash { ital "RC"} } } =V rSub { size 8{0} } e rSup { size 8{ - 1} } =0 "." "368"V rSub { size 8{0} } } {}

or

In terms of the time constant, [link] can be written as

With a calculator it is easy to show that the value of v(t)/V0 size 12{ {v ( t ) } slash {V rSub { size 8{0} } } } {} is as shown in [link]. It is evident from [link] that the voltage v(t) is less than 1 percent of V0 size 12{V rSub { size 8{0} } } {} after 5 τ size 12{τ} {} (five time constants). Thus, it is customary to assume that the capacitor is fully discharged (or charged) after five time constants. In other words, it takes 5 τ size 12{τ} {} for the circuit to reach its final for every time interval of τ size 12{τ} {}, the voltage is reduced by 36.8 percent of its previous value, v(t+τ)=v(t)/e=0.368v(t) size 12{v ( t+τ ) = {v ( t ) } slash {e=0 "." "368"v ( t ) } } {}, regardless of the value of t.

Values of v(t)/V0=e−τ/T size 12{ {v ( t ) } slash {V rSub { size 8{0} } =e rSup { size 8{ { - τ} slash {T} } } } } {}

Observe from [link] that the smaller the time constant, the more rapidly the voltage decreases, that is, the faster the response. This is illustrated in [link]. A circuit with a small time constant gives a fast response in that it reaches the steady state (or final state) quickly due to quick dissipation of energy stored, whereas a circuit with a large time constant gives a slow response because it takes longer to reach steady state. At any rate, whether the time constant is small or large, the circuit reaches steady state in five time constants.

With the voltage v(t) in [link], we can find the current iR(t) size 12{i rSub { size 8{R} } ( t ) } {},

The power dissipated in the resistor is

The energy absorbed by the resistor up to time t is

Note that as t→∞ size 12{t rightarrow infinity } {}, wR(∞)→12CV02 size 12{w rSub { size 8{R} } ( infinity ) rightarrow { {1} over {2} } ital "CV" rSub { size 8{0} } rSup { size 8{2} } } {}, which is the same as wC(0) size 12{w rSub { size 8{C} } ( 0 ) } {}, the energy initially stored in the capacitor. The energy that was initially stored in the capacitor is eventually dissipated in the resistor.

In summary:

The key to working with a source-free RC circuit is finding:

1. The initial voltage v(0)=V0 size 12{v ( 0 ) =V rSub { size 8{0} } } {} across the capacitor.

The time constant τ size 12{τ} {}.

With these two items, we obtain the response as the capacitor voltage vC(t)=v(t)=v(0)e−t/τ size 12{v rSub { size 8{C} } ( t ) =v ( t ) =v ( 0 ) e rSup { size 8{ { - t} slash {τ} } } } {}. Once the capacitor voltage is first obtained, other variables (capacitor current iC, resistor voltage vR size 12{v rSub { size 8{R} } } {}, and resistor current iR size 12{i rSub { size 8{R} } } {}) can be determined. In finding the time constant τ=RC size 12{τ= ital "RC"} {}, R is often the Thevenin equivalent resistance at the terminals of the capacitor; that is, we take out the capacitor C and find R=RTh size 12{R=R rSub { size 8{ ital "Th"} } } {} at its terminals.

Consider the series connection of a resistor and inductor, as shown in [link]. Our goal is to determine the circuit response, which we will assume to be the current i(t) through the inductor. We select the inductor current as the response in order to take advantage of the idea that the inductor current can not change instantaneously. At t = 0, we assume that the inductor has an initial current I0, or

With the corresponding energy stored in the inductor as

Applying KVL around loop in [link],

But vL=Ldi/dt size 12{v rSub { size 8{L} } =L { ital "di"} slash { ital "dt"} } {} and vR=iR size 12{v rSub { size 8{R} } = ital "iR"} {}. Thus,

L di dt + Ri = 0 size 12{L { { ital "di"} over { ital "dt"} } + ital "Ri"=0} {}

Or

Rearranging terms and integrating gives

∫ I 0 i ( t ) di i = − ∫ 0 t R L dt size 12{ Int cSub { size 8{I rSub { size 6{0} } } } cSup {i ( t ) } { { { ital "di"} over {i} } } size 12{ {}= - Int cSub {0} cSup {t} { { {R} over {L} } ital "dt"} }} {}

ln i / I 0 i ( t ) = − Rt L / 0 t -> ln i ( t ) − ln I 0 = − Rt L + 0 size 12{"ln"i line rSub { size 8{I rSub { size 6{0} } } } rSup {i ( t ) } size 12{ {}= - { { ital "Rt"} over {L} } line rSub {0} rSup {t} } size 12{ drarrow "ln"i ( t ) - "ln"I rSub {0} } size 12{ {}= - { { ital "Rt"} over {L} } +0}} {}

or

Taking the powers of e, we have

This shows that the natural response of the RL circuit is an exponential decay of the initial current. The current response is shown in [link]. It is evident from [link] that the time constant for the RL circuit is

with τ size 12{τ} {} again having the unit of seconds. Thus, [link] may be written as

With the current in [link], we can find the voltage across the resistor as

The power dissipated in the resistor is

The energy absorbed by the resistor is

w R ( t ) = ∫ 0 t pdt = ∫ 0 t I 0 2 Re − 2t / τ dt = − 1 2 τI 0 2 Re − 2t / τ / 0 t , τ = L R size 12{w rSub { size 8{R} } ( t ) = Int cSub { size 8{0} } cSup { size 8{t} } { ital "pdt"} = Int cSub { size 8{0} } cSup { size 8{t} } {I rSub { size 8{0} } rSup { size 8{2} } "Re" rSup { size 8{ - 2t/τ} } ital "dt"= - { {1} over {2} } } τI rSub { size 8{0} } rSup { size 8{2} } "Re" rSup { size 8{ - 2t/τ} } line rSub { size 8{0} } rSup { size 8{t} } ,τ= { {L} over {R} } } {}

or

Note that as t→∞ size 12{t rightarrow infinity } {}, wR(∞)→12LI02 size 12{w rSub { size 8{R} } ( infinity ) rightarrow { {1} over {2} } ital "LI" rSub { size 8{0} } rSup { size 8{2} } } {}, which is the same as wL(0) size 12{w rSub { size 8{L} } ( 0 ) } {}, the initial energy stored in the inductor as in [link]. Again, the energy initially stored in the inductor is eventually dissipated in the resistor.

In summary:

The key to working with a source-free RL circuit is to find:

1. The initial current i ( 0 ) = I 0 size 12{i ( 0 ) =I rSub { size 8{0} } } {} through the inductor.

The time constant τ size 12{τ} {} of the circuit.

With the two terms, we obtain the response as the inductor current iL(t)=i(t)=i(0)e−τ/T size 12{i rSub { size 8{L} } ( t ) =i ( t ) =i ( 0 ) e rSup { size 8{ { - τ} slash {T} } } } {}. Once we determine the inductor current iL size 12{i rSub { size 8{L} } } {}, other variables (inductor voltage vL size 12{v rSub { size 8{L} } } {}, resistor voltage vR size 12{v rSub { size 8{R} } } {}, and resistor current iR size 12{i rSub { size 8{R} } } {}) can be obtained. Note that in general, R in [link] is the Thevenin resistance at the terminals of the inductor.

Before going on with the second half of this chapter, we need to digress and consider some mathematical concepts that will aid our understanding of transient analysis. A basic understanding of singularity functions will help us make sense of the response of first-order circuits to a sudden application of an independent dc voltage or current source.

Singularity functions (also called switching functions) are very useful in circuit analysis. They serve as good approximations to the switching signals that arise in circuits with switching operations. They are helpful in the neat, compact description of some circuit phenomena, especially the step response of RC or RL, circuits to be discussed in the next sections. By definition,

Singularity functions are functions that either are discontinuous or have discontinuous derivatives.

The three most widely used singularity functions in circuit analysis are the unit step, the unit impulse, and the unit ramp functions.

The unit step function u(t) is 0 for negative values of t and 1 for positive values of t.

The unit step function is undefined at t = 0, where it changes abruptly from 0 to 1. It is dimensionless, like other mathematical functions, such as sine and cosine. [link] depicts the unit step function.

It is the same as saying that u(t) is delayed by t0 size 12{t rSub { size 8{0} } } {} seconds, as shown in [link], we simply replace every t by t−t0 size 12{t - t rSub { size 8{0} } } {}.

u(t) is advanced by t0 size 12{t rSub { size 8{0} } } {} seconds, as shown in [link]b.

We use the step function to represent an abrupt change in voltage or current, like the changes that occur in the circuits of control systems and digital computers.

We may express in terms of the unit step function as

If we let t0=0 size 12{t rSub { size 8{0} } =0} {}, then v(t) is simply the step voltage V0u(t) size 12{V rSub { size 8{0} } u ( t ) } {}. A voltage source of V0u(t) size 12{V rSub { size 8{0} } u ( t ) } {} is shown in [link](a); its equivalent circuit is shown in [link](b). It is evident in [link](b) that terminals a-b are short circuited (v = 0) for t < 0 and that v=V0 size 12{v=V rSub { size 8{0} } } {} appears at the terminals for t > 0. Similarly, a current source of I0u(t) size 12{I rSub { size 8{0} } u ( t ) } {} is shown in [link](a), while its equivalent circuit is in [link](b). Notice that for t < 0, there is an open circuit (i = 0), and that i=I0 size 12{i=I rSub { size 8{0} } } {} flows for t > 0.

The derivative of the unit step function u(t) is the unit impulse function δ(t) size 12{δ ( t ) } {}.

The unit impulse function - also known as the delta function - is shown in [link].

The unit impulse function δ(t) size 12{δ ( t ) } {} is zero everywhere except at t = 0, where it is undefined.

Impulsive currents and voltages occur in electric circuits as a result of switching operations or impulsive sources. Although the unit impulsive function is not physically realizable (just like ideal sources, ideal resistors etc.), it is a very useful mathematical tool.

The unit impulse may be regarded as an applied or resulting shock. It may be visualized as a very short duration pulse of unit area. This may be expressed mathematically as

Where t=0− size 12{t=0 rSup { size 8{ - {}} } } {} denotes the time just before t = 0 and t=0+ size 12{t=0 rSup { size 8{+{}} } } {} is the time just after t = 0. For this reason, it is customary to write 1 (denoting unit area) beside the arrow that is used to symbolize the unit impulse function. When an impulse function has a strength other than unity, the area of the impulse is equal to its strength. For example, an impulse function 10 δ(t) size 12{δ ( t ) } {} has an area of 10. [link] shows the impulse functions 5δ(t+2) size 12{5δ ( t+2 ) } {}, 10 δ(t) size 12{δ ( t ) } {}, and −4δ(t−3) size 12{ - 4δ ( t - 3 ) } {}.

To illustrate how the impulse function effects to other functions, let us evaluate the integral

where a<t0<b size 12{a<t rSub { size 8{0} } <b} {}. Since δ(t−t0)=0 size 12{δ ( t - t rSub { size 8{0} } ) =0} {} except t=t0 size 12{t=t rSub { size 8{0} } } {}, the integrand is zero except at t0 size 12{t rSub { size 8{0} } } {}. Thus,

∫ a b f ( t ) δ ( t − t 0 ) dt = ∫ a b f ( t 0 ) δ ( t − t 0 ) dt f ( t 0 ) ∫ a b f ( t ) δ ( t − t 0 ) dt = f ( t 0 ) alignl { stack { size 12{ Int cSub { size 8{a} } cSup { size 8{b} } {f ( t ) } δ ( t - t rSub { size 8{0} } ) ital "dt"= Int cSub { size 8{a} } cSup { size 8{b} } {f ( t rSub { size 8{0} } ) } δ ( t - t rSub { size 8{0} } ) ital "dt"} {} # =f ( t rSub { size 8{0} } ) Int cSub { size 8{a} } cSup { size 8{b} } {f ( t ) δ ( t - t rSub { size 8{0} } ) ital "dt"} =f ( t rSub { size 8{0} } ) {} } } {}

Or

This shows that when a function is integrated with the impulse function, we obtain the value of the function at the point where the impulse occurs. This is a highly useful property of the impulse function known as the sampling or sifting property. The special case of [link] is for t0=0 size 12{t rSub { size 8{0} } =0} {}. Then [link] becomes

Integrating the unit step function u(t) results in the unit ramp function r(t); we write

The unit ramp function is zero for negative values of t and has a unit slope for positive values of t.

[link] shows the unit ramp function. In general, a ramp is a function that changes at a constant rate.

The unit ramp function may be delayed or advanced as shown in [link].

We should keep in mind that the three singularity functions (impulse, step, and ramp) are related by differentiation as

or by integration as

although there are many more singularity functions, we are only interested in these three (the impulse function, the unit step function, and the ramp function) at this point.

When the dc source of an RC circuit is suddenly applied, the voltage or current source can be modeled as a step function, and the response is known as a step response.

The step response of a circuit is its behavior when the excitation is the step function, which may be a voltage or a current source.

The step response is the response of the circuit due to a sudden application of a dc voltage or current source.

Consider RC circuit in [link](a) which can be replaced by the circuit in [link](b), where VS size 12{V rSub { size 8{S} } } {}is a constant, dc voltage source. Again, we select the capacitor voltage as the circuit response to be determined. We assume an initial voltage V0 size 12{V rSub { size 8{0} } } {} on the capacitor, although this is not necessary for the step response. Since the voltage of a capacitor cannot change instantaneously,

where v(0−) size 12{v ( 0 rSup { size 8{ - {}} } ) } {} is voltage across the capacitor just before switching and v(0+) size 12{v ( 0 rSup { size 8{+{}} } ) } {} is its voltage immediately after switching. Applying KCL, we have

C dv dt + v − V s u ( t ) R = 0 size 12{C { { ital "dv"} over { ital "dt"} } + { {v - V rSub { size 8{s} } u ( t ) } over {R} } =0} {}

or

where v is the voltage across the capacitor. For t > 0, [link] becomes

Rearranging terms gives

dv dt = v − V s RC size 12{ { { ital "dv"} over { ital "dt"} } = { {v - V rSub { size 8{