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Electric Circuits (Fall 2015) Pingqiang Zhou
Lecture 8
- First-Order Circuits
10/29/2015
Reading: Chapter 7
1Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Overview
• This chapter examines RC and LC circuits’ reaction to
switched sources.
The circuits are referred to as first-order circuits.
• Three special functions, the unit step, unit impulse, and
unit ramp function are also introduced.
• Both source free and switched sources are examined.
2Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
First-Order Circuits
• A circuit that contains only sources,
resistors and an inductor is called an RL
circuit.
• A circuit that contains only sources,
resistors and a capacitor is called an RC
circuit.
• RL and RC circuits are called first-order
circuits because their V-Is are described
by first-order differential equations.
• There are also two ways to excite the
circuits:
initial conditions
independent sources
–+
vs L
R
i–+
vs C
R
i
3Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Response of a Circuit
• Transient response of an RL or RC circuit is
Behavior when voltage or current source are suddenly applied to
or removed from the circuit due to switching.
Temporary behavior
• Steady-state response (aka. forced response)
Response that persists long after transient has decayed
• Natural response of an RL or RC circuit is
Behavior (i.e., current and voltage) when stored energy in the
inductor or capacitor is released to the resistive part of the
network (containing no independent sources).
4Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Natural Response
RL Circuit
• Inductor current cannot change instantaneously
• In steady state, an inductor behaves like a short circuit.
RC Circuit
• Capacitor voltage cannot change instantaneously
• In steady state, a capacitor behaves like an open circuit
R
i
L
+
v
–
RC
Electric Circuits (Fall 2015) Pingqiang Zhou
Source Free RC Circuit
• A source free RC circuit occurs when its dc source is
suddenly disconnected.
The energy stored in the capacitor is released to the resistors.
• Consider a series combination of a resistor and a initially
charged capacitor as shown:
6
00v Vat t=0
0C Ri i 0dv v
dt RC
a first-order differential equation
/t RCv t Ae /
0
t RCv t V eln lnt
v ARC
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Natural Response of RC
• The result shows that the voltage
response of the RC circuit is an
exponential decay of the initial
voltage.
• The speed at which the voltage
decays can be characterized by
how long it takes the voltage to
drop to 1/e of the initial voltage.
called the time constant and is
represented by .
𝜏 = 𝑅𝐶.
• The voltage thus be expressed as
7
/
0
tv t V e
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Time Constant 𝜏 (= 𝑅𝐶)
• After five time constants the voltage on the capacitor is
less than one percent.
After five time constants, a capacitor is considered to be either fully
discharged or charged
• A circuit with a small time constant has a fast response
and vice versa.
8Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
RC Discharge
• With the voltage known, we can
find the current:
• The power dissipated in the
resistor is:
• The energy absorbed by the
resistor is:
9
/0 t
R
Vi t e
R
2
2 /0 tVp t e
R
2 2 /
0
11
2
t
Rw t CV e
/
0
t RCv t V e
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Example
• In the circuit below, let 𝑣𝐶 0 = 15V. Find 𝑣𝐶 , 𝑣𝑥, and 𝑖𝑥 for
𝑡 > 0.
10
𝑅𝑒𝑞 = 4Ω 𝜏 = 𝑅𝑒𝑞𝐶 = 0.4s 𝑣𝐶 𝑡 = 𝑣𝐶 0 𝑒−𝑡/𝜏 = ⋯
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Source Free RL Circuit
• Now lets consider the series connection
of a resistor and inductor.
In this case, the value of interest is the
current through the inductor.
Since the current cannot change
instantaneously, we can determine its value
as a function of time.
11
00i I
0L Rv v 0di
L Ridt /
0
Rt Li t I e
The time constant in this case is:L
R
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Example
• 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.
12
When 𝑡 < 0
𝑖 0 = 6A
When 𝑡 > 0
𝑖 𝑡 = 6𝑒−4𝑡A
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Natural Response Summary
RL Circuit
• Inductor current cannot change instantaneously
• time constant
RC Circuit
• Capacitor voltage cannot change instantaneously
• time constant
R
L
/)0()(
)0()0(
teiti
ii
R
i
L
+
v
–
RC
/)0()(
)0()0(
tevtv
vv
RC
Electric Circuits (Fall 2015) Pingqiang Zhou
Singularity Functions
• Before we consider the response of a circuit to an external
voltage, we need to cover some important mathematical
functions.
• Singularity functions (also known as switching functions)
serve as good approximations to the switching signals
that arise in circuits with switching operations.
• The three most common singularity functions are
unit step
unit impulse
unit ramp
14Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
The Unit Step u(t)
• A step function is one that maintains a constant value
before a certain time and then changes to another constant
afterwards.
• The prototypical form is 0 before 𝑡 = 0 and 1 afterwards.
See the graph for an illustration.
15
0, 0
1, 0
tu t
t
switching time may be shifted to 𝑡 = 𝑡0 by
0
0
0
0,
1,
t tu t t
t t
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Equivalent Circuit of Unit Step
• The unit step function has
an equivalent circuit to
represent when it is used
to switch on a source.
• The equivalent circuits for
a voltage and current
source are shown.
16Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Example
• Express the voltage pulse below in terms of the unit step.
17Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
The Unit Impulse Function 𝜹(𝒕)
• The derivative of the unit step
function is the unit impulse function.
• This is expressed as:
• Voltages of this form can occur during
switching operations.
18
0 0
Undefined 0
0 0
t
t t
t
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
The Unit Ramp Function 𝒓(𝒕)
• Integration of the unit step function
results in the unit ramp function:
• Much like the other functions, the
onset of the ramp may be adjusted.
19
0, 0
, 0
tr t
t t
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Example
• Express the sawtooth function in
terms of singularity functions
20Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Step Response of RC Circuit
• When a DC source is suddenly
applied to a RC circuit, the source
can be modeled as a step function.
The circuit response is known as the step
response.
• Let’s consider the circuit shown here.
• We can find the voltage on the
capacitor as a function of time.
22Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Step Response of RC Circuit
• We assume an initial voltage of V0 on the capacitor.
• Applying KCL:
• For t>0 this becomes:
Integrating both sides and introducing initial conditions finally yields:
23
sVdv vu t
dt RC RC
sVdv v
dt RC RC
0
/
0
, 0
0t
s s
V tv t
V V V e t
𝑣 0−
= 𝑣 0+
= 𝑣0
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Step Response of RC Circuit
• This is known as the complete response,
or total response.
• We can consider the response to be
broken into two separate responses:
The natural response of the capacitor or
inductor due to the energy stored in it.
The second part is the forced response.
24
0
/
0
, 0
0t
s s
V tv t
V V V e t
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Forced Response
• The complete response
can be written as:
where the nature response is:
and the forced response is:
• Note that the eventual response of the circuit is to reach
Vs after the natural response decays to zero.
25
n fv v v
/
0
t
nv V e
/1 t
f sv V e
0
/
0
, 0
0t
s s
V tv t
V V V e t
𝑣 0−
= 𝑣 0+
= 𝑣0
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Another Perspective
• Another way to look at the response is to break it up into
the transient response and the steady state response:
𝑣 𝑡 = 𝑣 ∞
steady 𝑣𝑠𝑠
+ 𝑣 0 − 𝑣(∞) 𝑒−𝑡/𝜏
transient 𝑣𝑡
where the transient is:
and the steady state is:
26
/
0
t
t sv V V e
ss sv V
0
/
0
, 0
0t
s s
V tv t
V V V e t
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Example
• The switch has been in position A for a long time. At 𝑡 = 0,
the switch moves to B. Find 𝑣(𝑡).
27
𝑣 0 = 15V 𝑣 ∞ = 30V 𝜏 = 2s
𝑣 𝑡 = 𝑣 ∞ + 𝑣 0 − 𝑣 ∞ 𝑒−𝑡/𝜏
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Step Response of RL Circuit
• Now we can look at the step response of
a RL circuit.
• We will use the transient and steady
state response approach.
• We know that the transient response will
be an exponential:
28
/t
ti Ae
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Step Response of RL Circuit
• After a sufficiently long time, the current
will reach the steady state:
• This yields an overall response of:
• To determine the value of A we need to keep in mind that
the current cannot change instantaneously.
29
sss
Vi
R
/t sVi Ae
R
00 0i i I
/t
ti Ae
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Step Response of RL Circuit
• Thus we can use the t=0 time to establish A
• The complete response of the circuit is thus:
• Without an initial current, the circuit response is shown
here.
30
0sV
A IR
/
0
ts sV Vi t I e
R R
/t sVi Ae
R
00 0i i I
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Example
• Find 𝑖(𝑡) in the circuit for 𝑡 > 0. Assume that the switch
has been closed for a long time.
31
𝑖 0 = 5A 𝑖 ∞ = 2A 𝜏 =1
15s
𝑖 𝑡 = 𝑖 ∞ + 𝑖 0 − 𝑖 ∞ 𝑒−𝑡/𝜏
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Procedure for Finding RC/RL Response
1. Identify the variable of interest
• For RL circuits, it is usually the inductor current iL(t).
• For RC circuits, it is usually the capacitor voltage vc(t).
2. Determine the initial value (at t = t0- and t0
+) of the
variable
• Recall that iL(t) and vc(t) are continuous variables:
iL(t0+) = iL(t0
) and vc(t0+) = vc(t0
)
• Assuming that the circuit reached steady state before t0 , use the
fact that an inductor behaves like a short circuit in steady state or
that a capacitor behaves like an open circuit in steady state.
32Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Procedure (cont’d)
3. Calculate the final value of the variable (its value as
t ∞)
• Again, make use of the fact that an inductor behaves like a
short circuit in steady state (t ∞) or that a capacitor behaves
like an open circuit in steady state (t ∞).
4. Calculate the time constant for the circuit
t = L/R for an RL circuit, where R is the Thévenin equivalent
resistance “seen” by the inductor.
t = RC for an RC circuit where R is the Thévenin equivalent
resistance “seen” by the capacitor.
33Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Application: Delay Circuit
• The RC circuit can be used to delay the turn on of a
connected device.
• For example, a neon lamp which only triggers when a
voltage exceeds a specific value can be delayed using
such a circuit.
34Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Delay Circuit II
• When the switch is closed, the capacitor charges.
The voltage will rise at a rate determined by:
Once the voltage reaches 70 volts, the lamp triggers.
• Once on, the lamp has low resistance and discharges the
capacitor.
This shuts off the capacitor and starts the cycle over again.
35
1 2R R C
Lecture 8
Electric Circuits (Fall 2015) Pingqiang Zhou
Further Reading
• Refer to Ch. 7.9, [James W. Nillson 2014] and [Anant
Agarwal 2005] for more applications
Photoflash unit
Relay circuit
Automobile ignition circuit
Artificial pacemaker
Ramp and impulse responses
• Refer to Ch. 7.7 for first-order circuit with Op Amps.
36Lecture 8
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