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FIRST ORDER CIRCUITS

Module by: Dinh Sy Hien

Summary: 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. We carry out the analysis of RC and RL circuits by applying Kirchhoff’s laws. 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.

INTRODUCTION

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.

THE SOURCE-FREE RC 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 Figure 1. (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
v ( 0 ) = V 0 v ( 0 ) = V 0 size 12{v \( 0 \) =V rSub { size 8{0} } } {} (1)
With the corresponding value of the energy stored as
w ( 0 ) = 1 2 CV 0 2 w ( 0 ) = 1 2 CV 0 2 size 12{w \( 0 \) = { {1} over {2} } ital "CV" rSub { size 8{0} } rSup { size 8{2} } } {} (2)
Applying KCL at the top note of the circuit in Figure 1,
i C + i R = 0 i C + i R = 0 size 12{i rSub { size 8{C} } +i rSub { size 8{R} } =0} {} (3)
By definition, iC=Cdv/dtiC=Cdv/dt size 12{i rSub { size 8{C} } =C { ital "dv"} slash { ital "dt"} } {} and iR=v/RiR=v/R size 12{i rSub { size 8{R} } = {v} slash {R} } {}. Thus,
C dv dt + v R = 0 C dv dt + v R = 0 size 12{C { { ital "dv"} over { ital "dt"} } + { {v} over {R} } =0} {} (4)
or
dv dt + v RC = 0 dv dt + v RC = 0 size 12{ { { ital "dv"} over { ital "dt"} } + { {v} over { ital "RC"} } =0} {} (5)
This is a first order differential equation, since only the first derivative of v is involved. To solve it, we arrange the terms as
dv v = − 1 RC dt dv v = − 1 RC dt size 12{ { { ital "dv"} over {v} } = - { {1} over { ital "RC"} } ital "dt"} {} (6)
integrating both sides, we get
ln v = − t RC + ln A ln v = − t RC + ln A size 12{"ln"v= - { {t} over { ital "RC"} } +"ln"A} {}
where ln A is the integration constant. Thus,
ln v A = − t RC ln v A = − t RC size 12{"ln" { {v} over {A} } = - { {t} over { ital "RC"} } } {} (7)
Taking powers of e produces
v ( t ) = Ae − t / RC 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=V0v(0)=A=V0 size 12{v \( 0 \) =A=V rSub { size 8{0} } } {}. Hence,
v ( t ) = V 0 e − t / RC v ( t ) = V 0 e − t / RC size 12{v \( t \) =V rSub { size 8{0} } e rSup { size 8{ - t/ ital "RC"} } } {} (8)
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.
Figure 1: A source free RC circuit.
The natural response is illustrated graphically in Figure 2. Note that at t = 0, we have the correct initial condition as in Equation 1. 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 ττ 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=τt=τ size 12{t=τ} {}, Equation 8 becomes
V 0 e − τ / RC = V 0 e − 1 = 0 . 368 V 0 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
τ = RC τ = RC size 12{τ= ital "RC"} {} (9)
In terms of the time constant, Equation 8 can be written as
v ( t ) = V 0 e − τ / T v ( t ) = V 0 e − τ / T size 12{v \( t \) =V rSub { size 8{0} } e rSup { size 8{ { - τ} slash {T} } } } {} (10)
With a calculator it is easy to show that the value of v(t)/V0v(t)/V0 size 12{ {v \( t \) } slash {V rSub { size 8{0} } } } {} is as shown in Table 1. It is evident from Table 1 that the voltage v(t) is less than 1 percent of V0V0 size 12{V rSub { size 8{0} } } {} after 5 ττ 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 ττ size 12{τ} {} for the circuit to reach its final for every time interval of ττ size 12{τ} {}, the voltage is reduced by 36.8 percent of its previous value, v(t+τ)=v(t)/e=0.368v(t)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/Tv(t)/V0=e−τ/T size 12{ {v \( t \) } slash {V rSub { size 8{0} } =e rSup { size 8{ { - τ} slash {T} } } } } {}
t v ( t ) / V 0 v ( t ) / V 0 size 12{ {v \( t \) } slash {V rSub { size 8{0} } } } {}
τ τ size 12{τ} {} 0.36788
2 ττ size 12{τ} {} 0.13534
3 ττ size 12{τ} {} 0.04979
4 ττ size 12{τ} {} 0.01832
5 ττ size 12{τ} {} 0.00674
Figure 2: The voltage responce of the RC circuit.
Observe from Equation 9 that the smaller the time constant, the more rapidly the voltage decreases, that is, the faster the response. This is illustrated in Figure 3. 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.
Figure 3: Plot of v\Vo = exp(-t/t) for various of the time constant.
With the voltage v(t) in Equation 10, we can find the current iR(t)iR(t) size 12{i rSub { size 8{R} } \( t \) } {},
i R ( t ) = v ( t ) R = V 0 R e − t / τ i R ( t ) = v ( t ) R = V 0 R e − t / τ size 12{i rSub { size 8{R} } \( t \) = { {v \( t \) } over {R} } = { {V rSub { size 8{0} } } over {R} } e rSup { size 8{ - t/τ} } } {} (11)
The power dissipated in the resistor is
p ( t ) = v ( t ) i R ( t ) = V 0 2 R e − 2t / τ p ( t ) = v ( t ) i R ( t ) = V 0 2 R e − 2t / τ size 12{p \( t \) =v \( t \) i rSub { size 8{R} } \( t \) = { {V rSub { size 8{0} } rSup { size 8{2} } } over {R} } e rSup { size 8{ - 2t/τ} } } {} (12)
The energy absorbed by the resistor up to time t is
w R ( t ) = ∫ 0 t pdt = ∫ 0 t V 0 2 R e − 2t / τ dt = − τV 0 2 2R e − 2t / τ ∣ 0 t = 1 2 CV 0 2 ( 1 − e − 2t / τ ) , τ = RC w R ( t ) = ∫ 0 t pdt = ∫ 0 t V 0 2 R e − 2t / τ dt = − τV 0 2 2R e − 2t / τ ∣ 0 t = 1 2 CV 0 2 ( 1 − e − 2t / τ ) , τ = RC 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} } { { {V rSub { size 8{0} } rSup { size 8{2} } } over {R} } e rSup { size 8{ - 2t/τ} } ital "dt"} = - { {τV rSub { size 8{0} } rSup { size 8{2} } } over {2R} } e rSup { size 8{ - 2t/τ} } \rline rSub { size 8{0} } rSup { size 8{t} } = { {1} over {2} } ital "CV" rSub { size 8{0} } rSup { size 8{2} } \( 1 - e rSup { size 8{ { - 2t} slash {τ} } } \) ,τ= ital "RC"} {} (13)
Note that as t→∞t→∞ size 12{t rightarrow infinity } {}, wR(∞)→12CV02wR(∞)→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)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)=V0v(0)=V0 size 12{v \( 0 \) =V rSub { size 8{0} } } {} across the capacitor.
The time constant ττ size 12{τ} {}.
With these two items, we obtain the response as the capacitor voltage vC(t)=v(t)=v(0)e−t/τvC(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 vRvR size 12{v rSub { size 8{R} } } {}, and resistor current iRiR size 12{i rSub { size 8{R} } } {}) can be determined. In finding the time constant τ=RCτ=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=RThR=RTh size 12{R=R rSub { size 8{ ital "Th"} } } {} at its terminals.

THE SOURCE-FREE RL CIRCUIT

Consider the series connection of a resistor and inductor, as shown in Figure 4. 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
i ( 0 ) = I 0 i ( 0 ) = I 0 size 12{i \( 0 \) =I rSub { size 8{0} } } {} (14)
With the corresponding energy stored in the inductor as
w ( 0 ) = 1 2 LI 0 2 w ( 0 ) = 1 2 LI 0 2 size 12{w \( 0 \) = { {1} over {2} } ital "LI" rSub { size 8{0} } rSup { size 8{2} } } {} (15)
Applying KVL around loop in Figure 4,
v L + v R = 0 v L + v R = 0 size 12{v rSub { size 8{L} } +v rSub { size 8{R} } =0} {} (16)
But vL=Ldi/dtvL=Ldi/dt size 12{v rSub { size 8{L} } =L { ital "di"} slash { ital "dt"} } {} and vR=iRvR=iR size 12{v rSub { size 8{R} } = ital "iR"} {}. Thus,
L di dt + Ri = 0 L di dt + Ri = 0 size 12{L { { ital "di"} over { ital "dt"} } + ital "Ri"=0} {}
Or
di dt + R L i = 0 di dt + R L i = 0 size 12{ { { ital "di"} over { ital "dt"} } + { {R} over {L} } i=0} {} (17)
Rearranging terms and integrating gives
∫ I 0 i ( t ) di i = − ∫ 0 t R L dt ∫ 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 ln i / I 0 i ( t ) = − Rt L / 0 t -> ln i ( t ) − ln I 0 = − Rt L + 0 size 12{"ln"i \rline rSub { size 8{I rSub { size 6{0} } } } rSup {i \( t \) } size 12{ {}= - { { ital "Rt"} over {L} } \rline rSub {0} rSup {t} } size 12{ drarrow "ln"i \( t \) - "ln"I rSub {0} } size 12{ {}= - { { ital "Rt"} over {L} } +0}} {}
or
ln i ( t ) I 0 = − Rt L ln i ( t ) I 0 = − Rt L size 12{"ln" { {i \( t \) } over {I rSub { size 8{0} } } } = - { { ital "Rt"} over {L} } } {} (18)
Taking the powers of e, we have
i ( t ) = I 0 e − Rt / L i ( t ) = I 0 e − Rt / L size 12{i \( t \) =I rSub { size 8{0} } e rSup { size 8{ - ital "Rt"/L} } } {} (19)
Figure 4: The current response of the RL circuit.
This shows that the natural response of the RL circuit is an exponential decay of the initial current. The current response is shown in Figure 5. It is evident from Equation 19 that the time constant for the RL circuit is
τ = L R τ = L R size 12{τ= { {L} over {R} } } {} (20)
with