ECS 203 (ME2) - Part 4A
Dr.Prapun Suksompong
CHAPTER 7
First-Order Circuits
In Chapter 6, we have studied the relationships between current and
voltage of capacitors/inductors. These elements can be used to store energy
and release energy when needed. In this chapter, we will see how the
voltage or current behaves during the charging/discharging of these storage
elements.
First, we will examine two types of simple circuits with a storage element:
(a) A circuit with a resistor and one capacitor (called an RC circuit);
and
(b) A circuit with a resistor and an inductor (called an RL circuit).
These circuits may look simple but they find continual applications in
electronics, communications, and control systems.
7.0.1. Applying Kirchhoff’s laws to the RC and RL circuits produce
first order differential equations. Hence, the circuits are collectively
known as first-order circuits.
7.0.2. There are two ways to excite the circuits.
(a) By initial conditions of the storage elements in the circuit.
• Also known as source-free circuits
• Assume that energy is initially stored in the capacitive or inductive element.
• This is the discharging process.
(b) By using independent sources
• This is the charging process
• For this chapter, we will consider independent dc sources.
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7. FIRST-ORDER CIRCUITS
Before we start our circuit analysis, it is helpful to consider one mathematical fact which we will use through out this chapter:
7.0.3. The solution of the first-order differential equation
d
x(t) = ax + b
dt
is given by
b
b
(7.3)
x(t) = ea(t−t0 ) x(t0 ) +
− .
a
a
In addition, if a < 0, the solution is given by
(7.4)
x(t) = ea(t−t0 ) (x(t0 ) − x(∞)) + x(∞).
where x(∞) = limt→∞ x(t).
The results above take a bit of effort to proof. However, if you have
MATLAB, you can get 7.3 using one line of code:
dsolve(’Dx = a*x+b’,’x(t0) = x0’, ’t’)
When a < 0, from (7.3), x(∞) = −b/a. We can then get (7.4) by substituting b/a in (7.3) with x(∞).
7.0.4. It turns out that you only have two kinds of plots to worry about
when dealing with (7.4):
7.1. SOURCE-FREE RC CIRCUITS
89
7.1. Source-Free RC Circuits
A source-free RC circuit occurs when its dc source is suddenly disconnected. The energy already stored in the capacitor is released to the
resistor.
Consider a series combination of a resistor an initially charged capacitor.
We assume that at time t = 0, the initial voltage is v(0) = Vo . (Hence, the
initial stored energy is w(0) = 21 CVo2 .)
Applying KCL, we get
i C + iR
dv
v
C +
dt R
dv
v
+
dt CR
dv
dt
= 0
= 0
= 0
1
= −
v
CR
Hence,
t
t
v(t) = Vo e− RC = Vo e− τ ,
where τ = RC.
This shows that the voltage response of 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.
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7. FIRST-ORDER CIRCUITS
7.1.1. Remarks:
(a) τ = RC is the time constant which is the time required for the
response to decay to 36.8 percent of its initial value. (e ≈ 2.718 and
hence 1/e ≈ 0.368.)
(b) The smaller the time constant, the more rapidly the voltage decreases, that is, the faster the response.
7.1. SOURCE-FREE RC CIRCUITS
91
(c) The voltage v(t) is less than 1 percent of V0 after 5τ (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 5τ for the circuit to reach its final state or steady
state when no changes take place with time.
−t
Vo τ
(d) iR (t) = v(t)
R = Re .
2 −2t
(e) pR (t) = viR = VRo e τ
Rt
−2t
(f) wR (t) = 0 p(µ)dµ = 12 CVo2 (1 − e τ ) = wC (0) − wC (t)
• Notice that as t → ∞, wR → 21 CV02 , which is the same as
wC (0), the energy initially stored in the capacitor.
• The energy that was initially stored in the capacitor is eventually dissipated in the resistor.
7.1.2. In summary: The key in working with a source-free RC circuit
is to determine:
1. The initial voltage v(0) = V0 across the capacitor.
2. The time constant τ (= RC)
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7. FIRST-ORDER CIRCUITS
7.1.3. There are basically three ways that we can extend what we have
found.
(a) The initial condition v(t0 ) can be given at non-zero t0 . This has
already been taken care of by (7.4).
(b) There can be more than one resistors. In which case, you first simplify the circuit by finding the equivalent resistance at the capacitor
terminals. See Example 7.1.4.
(c) The initial value of the voltage v(t0 ) might not be given but condition(s) of the circuit before time t0 is given instead. In which case,
you may find v(t−
0 ) by considering the long-term behavior of the circuit before t0 . In such situation, you should review two properties
of capacitors:
(i) For DC, capacitor becomes open circuit.
(ii) Voltage across the capacitor cannot change instantaneously
See Example 7.1.5.
Example 7.1.4. Let vC (0) = 15 V. Determine vC (t), vx (t), and ix (t) for
t > 0.
Remark: The answer is the same if we replace t > 0 in the above
question by t ≥ 0.
7.2. SOURCE-FREE RL CIRCUITS
93
In Example 7.1.4, you may ask how can we have the initial voltage of
15 V across the capacitor in the first place. The next example suggest a
way to get the initial voltage.
Example 7.1.5. The switch in the circuit below has been closed for a
long time, and it is opened at t = 0. Find v(t) for t ≥ 0. Calculate the
initial energy stored in the capacitor.
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7. FIRST-ORDER CIRCUITS
7.2. Source-Free RL Circuits
Consider the series connection of a resistor and an inductor. Assume
that the inductor has an initial current Io or i(0) = Io .(hence, the energy
stored in the inductor is w(0) = 21 LIo2 .)
Applying KVL, we get
vL + vR = 0
di
L + Ri = 0
dt
di R
+ i = 0
dt L
Hence,
i(t) = Io e
−Rt
L
−t
= Io e τ .
Note that:
(a) τ = RL is the time constant which is the time required for the response to decay to 36.8 percent of its initial value.
−t
(b) vR (t) = iR = Io Re τ .
−2t
(c) p(t) = vRRi = Io2 Re τ .
−2t
t
(d) wR (t) = 0 p(µ)dµ = 12 LIo2 (1 − e τ ) = wL (0) − wL (t) .
7.2. SOURCE-FREE RL CIRCUITS
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7.2.1. In summary: The key in working with a source free RL circuit
is to determine:
(a) The initial current i(0) = Io .
(b) The time constant τ (= RL ).
• In general, R is the Thevenin equivalent resistance at the terminal of the inductor.(We take out the inductor L and find
R = RT H at its terminals.)
Example 7.2.2. The switch in the circuit below has been closed for a
long time. At t = 0, the switch is opened. Calculate i(t) for t > 0.
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7. FIRST-ORDER CIRCUITS
Example 7.2.3. Determine i, io and vo for all t in the circuit. Assume
that the switch was closed for a long time.
7.3. UNIT STEP FUNCTION
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7.3. Unit step function
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.
The unit step function u(t) is 0 for negative values of t and 1 for positive
values of t. In mathematical terms,
u(t) =
0, t < 0
1, t > 0
For example, here is a voltage source of V0 u(t)
Similarly, here is a current source of I0 u(t)
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7. FIRST-ORDER CIRCUITS
We can also use the step functions to represent the gate function (pulse)
which may be regarded as a step function that switches on at one value of
t and switches off at another value of t. The gate function below switches
on at t = 2 s and switches off at t = 5 s.
It consists of the sum of two unit step functions:
v(t) = 10u(t − 2) − 10u(t − 5).
When the dc source of an RC circuit is suddenly applied (i.e., this
happens when the capacitor is being charged), the voltage or current source
can be modeled as a step function, and the response is known as a step
response.
7.4. STEP RESPONSE OF AN RC CIRCUIT
99
7.4. Step Response of an RC Circuit
7.4.1. Consider an RC circuit with voltage step input below:
The voltage across the capacitor is v. We assume an initial voltage
V0 on the capacitor. Since the voltage of a capacitor cannot change instantaneously v(0−) = v(0+) = V0 , where v(0−) is the voltage across the
capacitor just before switching and v(0+) is its voltage immediately after
switching.
For t > 0, applying KCL, we have
dv v − Vs
C +
= 0
dt
R
dv
v
Vs
+
=
dt RC
RC
dv
1
Vs
= −
v+
dt
RC
RC
The solution is
−t
v(t) = Vs + (Vo − Vs )e τ , t ≥ 0.
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7. FIRST-ORDER CIRCUITS
7.4.2. In general, the step response of an RC circuit whose switch
changes position at time t = 0 can be written as
−t
v(t) = v(∞) + (v(0) − v(∞))e τ ,
t > 0.
where v(0) is the initial voltage and v(∞) is the final or steady-state value.
We obtain
(a) v(0) from the given circuit for t < 0 and
(b) v(∞) and τ from the circuit for t > 0.
7.4.3. If the switch changes position at time t = to instead of at t = 0,
the response becomes
v(t) = v(∞) + (v(to ) − v(∞))e
−(t−to )
τ
,
t > t0
where v(to ) is the initial value at t = to .
7.4.4. The key in finding the step response of an RC circuit is to determine three values:
1. The initial capacitor voltage v(0) or V (t0 ).
2. The final capacitor voltage v(∞).
3. The time constant τ .
7.4. STEP RESPONSE OF AN RC CIRCUIT
101
Example 7.4.5. The switch in the circuit below has been in position A
for a long time. At t = 0, the switch moves to B. Determine v(t) for t > 0
and calculate its value at t = 1 s and 4 s.
Example 7.4.6. Find v(t) for t > 0 in the circuit below. Assume the
switch has been open for a long time and is closed at t = 0. Numerically
evaluate v(t) at t = 0.5.
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7. FIRST-ORDER CIRCUITS
Example 7.4.7. An electronic flash unit provides a common example
of an RC circuit. This application exploits the ability of the capacitor
to oppose any abrupt change in voltage. A simplified circuit is shown
below. It consists essentially of a high-voltage dc supply, a current-limiting
large resistor R1 , and a capacitor C in parallel with the flashlamp of low
resistance R2 .
• When the switch is in position 1, the capacitor charges slowly due
to the large time constant (τ = R1 C). The capacitor voltage rises
gradually from zero to Vs , while its current decreases gradually from
I1 = Vs /R1 to zero.
• With the switch in position 2, the capacitor voltage is discharged.
The low resistance R2 of the photolamp permits a high discharge
current with peak I2 = Vs /R2 in a short duration.
This simple RC circuit provides a short-duration, high current pulse.
Such a circuit also finds applications in electric spot welding and the radar
transmitter tube.
7.4. STEP RESPONSE OF AN RC CIRCUIT
103
Example 7.4.8. An RC circuit can be used to provide various time
delays. Consider the circuit below. It basically consists of an RC circuit
with the capacitor connected in parallel with a neon lamp. The voltage
source can provide enough voltage to fire the lamp.
• When the switch is closed, the capacitor voltage increases gradually
toward 110 V at a rate determined by the circuit’s time constant,
(R1 + R2 )C. The lamp will act as an open circuit and not emit light
until the voltage across it exceeds a particular level, say 70 V.
• When the voltage level is reached, the lamp fires (goes on), and
the capacitor discharges through it. Due to the low resistance of
the lamp when on, the capacitor voltage drops fast and the lamp
turns off. The lamp acts again as an open circuit and the capacitor
recharges.
• By adjusting R2 , we can introduce either short or long time delays
into the circuit and make the lamp fire, recharge, and fire repeatedly
every time constant τ = (R1 + R2 )C, because it takes a time period
τ to get the capacitor voltage high enough to fire or low enough to
turn off.
• The warning blinkers commonly found on road construction sites
are one example of the usefulness of such an RC delay circuit.
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7. FIRST-ORDER CIRCUITS
7.5. Step Response of an RL Circuit
Here, our goal is to find the inductor current i as the circuit response.
7.5.1. In general, the step response of an RL circuit whose switch
changes position at time t = 0 can be written as
−t
i(t) = i(∞) + (i(0) − i(∞))e τ ,
t > 0.
where i(0) is the initial current and i(∞) is the final or steady-state value.
If the switch changes position at time t = to instead of at t = 0, the
response becomes
i(t) = i(∞) + (i(to ) − i(∞))e
−(t−to )
τ
,
t > t0
where v(to ) is the initial inductor current value at time t = to .
Example 7.5.2. Find i(t) in the circuit below for t > 0. Assume that
the switch has been closed for a long time.
7.5. STEP RESPONSE OF AN RL CIRCUIT
105
Example 7.5.3. The switch in the circuit below has been closed for a
long time. It opens at t = 0. Find i(t) for t > 0.
Example 7.5.4. At t = 0, switch 1 in the circuit below is closed, and
switch 2 is closed 4 s later. Find i(t) for t > 0.