Download Response of First-Order RL and RC Circuits

Chapter 7
Engr228
Circuit Analysis
Dr Curtis Nelson
Chapter 7 Objectives
• Be able to determine the natural response of both RL and
RC circuits;
• Be able to determine the step response of both RL and RC
circuits.
Engr228 - Chapter 7, Nilsson 10E
1
Characteristics of R, L, C
• Resistors resist current flow;
• Inductors resist change of current;
• Capacitors resist change of voltage.
L and C are considered “dynamic” elements because
their V-I characteristics are a function of time.
Review: DC Characteristics of an Inductor
i(t)
i(t)
1V
1Ω
L
vL (t ) = L
1V
1Ω
diL (t )
dt
When iL(t) is constant, diL(t) = 0 thus, vL(t) = 0.
In other words, the inductor can be replaced with a short circuit.
Engr228 - Chapter 7, Nilsson 10E
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Review: DC Characteristics of a Capacitor
i(t)
1V
i(t)
1Ω
C
iC (t ) = C
1V
1Ω
dvC (t )
dt
When vC(t) is constant, dvC(t) = 0 thus, iC(t) = 0.
In other words, the capacitor can be replaced with an open circuit.
Types of Responses
• Transient Response, natural response, homogeneous
solution (e.g. oscillation, temporary position change)
– Fades to zero over time;
– Resists change.
• Forced Response, steady-state response, particular solution
(e.g. permanent position change)
– Follows the input;
– Independent of time passed.
Engr228 - Chapter 7, Nilsson 10E
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Pendulum Example
I am holding a ball with a rope attached. What is the movement
of the ball if I move my hand to another point?
Two movements:
1. Oscillation.
2. Forced position change.
Mechanical Analogue
Forced response
Engr228 - Chapter 7, Nilsson 10E
Natural response
at a different time
4
Transient Response
• RL Circuit
First-order differential equation
Chapter 7
• RC Circuit
Second-order differential equation
Chapter 8
• RLC Circuit
Source Free RL Circuits
i(t)
R
+
+
L
-
Inductor L has energy stored
so that the initial current is I0.
Similar to a pendulum that is
at a height h (potential energy
is nonzero).
height
Engr228 - Chapter 7, Nilsson 10E
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Source Free RL Circuits
i(t)
-
+
L
-
R
+
Ri (t ) + L
di (t )
=0
dt
di (t ) R
+ i (t ) = 0
dt
L
There are 2 ways to solve first-order differential equations.
Solving Source Free RL Circuits
Method 1: Assume solution is of the form i(t ) = Ae st
where A and s are the constants that we wish to solve for.
Substitute i(t ) = Ae st in the equation
Ase st +
(s +
R st
Ae = 0
L
R
) Ae st = 0
L
s=−
R
L
i (t ) = Ae
Engr228 - Chapter 7, Nilsson 10E
di (t ) R
+ i(t ) = 0
dt
L
R
− t
L
6
Solving Source Free RL Circuits - continued
Initial condition: i(0) = I 0
i (t ) = Ae
from
R
− t
L
I 0 = Ae 0
I0 = A
Therefore
i (t ) = I 0 e
R
− t
L
Solving Source Free RL Circuits
Method 2: Direct integration
di (t ) R
+ i (t ) = 0
dt
L
di (t )
R
= − i (t )
dt
L
di (t )
R
= − dt
i (t )
L
i (t )
∫
I0
i (t )
ln i (t ) I
0
t
R
=− t
L 0
ln i (t ) − ln I 0 = −
i (t ) = I 0 e
R
(t − 0)
L
R
− t
L
t
di (t )
R
= ∫ − dt
i (t ) 0 L
Engr228 - Chapter 7, Nilsson 10E
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Time Constant
The ratio L/R is called the time constant and is denoted
by the symbol τ (tau).
τ=
L
R
Units: seconds
One time constant is defined as the amount of time
required for the output to go from 100% to 36.8%.
i(t ) = I 0e
R
− t
L
= I 0e
−
t
τ
e −1 = 0.368
Time Constant Graph
Engr228 - Chapter 7, Nilsson 10E
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1st Order Response Observations
• The voltage on a capacitor or the current through an inductor is
the same prior to and after a switch at t = 0 seconds because
these quantities cannot change instantaneously.
• Resistor voltage (or current) prior to the switch v(0-) can be
different from the voltage (or current) after the switch v(0+).
• All voltages and all currents in an RC or RL circuit follow the
same natural response e-t/τ.
General RL Circuits
The time constant of a single-inductor circuit will be τ = L/Req
where Req is the resistance seen by the inductor.
Example: Req=R3+R4+R1R2 / (R1+R2)
Engr228 - Chapter 7, Nilsson 10E
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Example: RL with a Switch
Find the voltage v(t) at t = 200 ms.
v(t) = -12.99 volts at t = 200 ms
Source-Free RC Circuits
As you might expect, source-free RC circuits are an analogue
of source-free RL circuits. The derivation for the capacitor
voltage is now a node equation rather than a loop equation.
Engr228 - Chapter 7, Nilsson 10E
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Source-Free RC Circuits
• To be consistent with the direction of
assigned voltage, we write
v(t )
dv (t )
+C
=0
R
dt
dv (t ) v(t )
+
=0
dt
RC
• Comparing the last equation to that of the RL circuit, it
becomes obvious that the form of the solution is the same
with the time constant τ = RC rather than L/R
di (t ) R
+ i(t ) = 0
dt
L
Comparison Between Source-free RL and RC
• Transient response equations:
iL (t ) = I L 0e
R
− t
L
v C (t ) = VC 0e
−
= I L 0e
t
RC
−
t
τ
= VC 0e
−
t
τ
• Time constants:
L
R
τ C = RC
τL =
Engr228 - Chapter 7, Nilsson 10E
11
RC Natural Response
The time constant is τ = RC
for a first-order RC circuit.
General RC Circuits
The time constant of a single-capacitor circuit will be τ = ReqC
where Req is the resistance seen by the capacitor.
Example: Req=R2+R1R3 / (R1+R3)
Engr228 - Chapter 7, Nilsson 10E
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The Source Free RC Circuit
Find the voltage v(t) at t = 200µS.
v(t) = 321mV at t = 200µS
Textbook Problem 8.22 Hayt 7E
(a) Find vC(t) for all time in the circuit below.
(b) At what time is vC = 0.1vC(0)?
vC (t ) = vC (0)e −125t
vC(0) = 192V
vC = 0.1vC(0) when t = 18.42mS
Engr228 - Chapter 7, Nilsson 10E
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Driven RL and RC Circuits
• Many RL and RC circuits are driven by a DC or an AC source.
The complete response of a driven RL or RC source is the sum
of a transient response and a forced response:
i (t ) = tran (t ) + forced (t )
The Unit Step Function u(t)
• A Step Function is often used to drive circuits.
• The forcing function of 1u(t) represents a function that has
zero value up until t = 0, and then a value of 1 forever after.
Engr228 - Chapter 7, Nilsson 10E
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Step Functions
t=0
t=0
R
1V
1V
R
v(t)
v(t)
1V
1V
0V
0V
t
t
Unit Step function
Example
I
1V
1Ω
I = 1A
I
L
2V
1Ω
L
I = 2A
Suppose the voltage source changes abruptly from 1V to 2V.
Does the current change abruptly as well?
Engr228 - Chapter 7, Nilsson 10E
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Voltage
Example - continued
2V
1V
time
Current
2A
1A
time
Transient Response + Forced Response
Complete Response
• The differential equation for the circuit above now becomes
di (t )
= vS (t )
dt
• The transient part of the complete solution is determined by
setting the forcing function vS(t) = 0
Ri (t ) + L
Ri (t ) + L
Engr228 - Chapter 7, Nilsson 10E
di (t )
=0
dt
16
Complete Solution
• The complete solution is solved by assuming i(t) is of the
form:
i(t)
k1 + k 2 e
−
t
τ
R
+
u(t)
AC
di (t )
Ri (t ) + L
= u (t )
dt
L
-
• As shown on the following pages through direct integration,
substituting the solution above into the circuit equation yields
R
R
− t⎞
− t
V⎛
i(t ) = ⎜⎜1 − e L ⎟⎟ + i(0)e L
R⎝
⎠
Solution – Direct Integration
For t > 0 (letting V = u)
di (t )
Ri (t ) + L
=V
dt
di (t )
L
= V − Ri (t )
dt
Ldi(t )
= dt
V − Ri (t )
Ldi(t )
∫ V − Ri (t ) = ∫ dt
L
− ln(V − Ri (t )) = t + c1
R
Engr228 - Chapter 7, Nilsson 10E
L
ln(V − Ri (t )) = t + c1
R
R
R
ln(V − Ri (t )) = − t − c1
L
L
−
V − Ri (t ) = e
V − Ri (t ) = e
R
− t
L
R
− t
L
Ri (t ) = V − c2 e
⋅e
R
− c1
L
⋅ c2
R
− t
L
V c2 − RL t
i (t ) = − e
R R
17
Direct Integration - continued
V c2 − RL t
i(t ) = − e
R R
We can find c2 from the initial condition i(0)
V c2
− ⋅1
R R
c 2 = V − i ( 0) R
i ( 0) =
V V − i(0) R − RL t
e
Therefore, we have i (t ) = −
R
R
R
i(t ) =
R
− t⎞
− t
V⎛
⎜1 − e L ⎟ + i(0)e L
⎟
R ⎜⎝
⎠
Review: Solving First-Order Circuits
• Start by finding the initial current through the inductor L or
the initial voltage across the capacitor C;
• Assume that the desired response is of the form
k1 + k 2 e
−
t
τ
• Find the time constant τ;
• Solve for k1 and k2 using initial conditions and the status of
the circuit as time approaches infinity;
• After solving for the inductor current (or capacitor voltage),
find other values that the problem may ask for.
Engr228 - Chapter 7, Nilsson 10E
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Example
The switch is in the position shown for a long time before t = 0.
Find i(t).
t=0
R=1Ω
i(t)
1V
L=1H
2V
i (t ) = k1 + k 2 e
−
t
τ
Time constant τ = 1 sec
Example - continued
i (t ) = k1 + k 2 e
At t = 0, i(0) = 2 A
2 = k1 + k2
At t = ∞, i(∞) = 1 A
1 = k1 + 0
−
t
τ
Therefore, k1 = 1, k2 = 1
The answer is:
Engr228 - Chapter 7, Nilsson 10E
i (t ) = 1 + e −t
19
Example Graph
i (t )
2A
1A
i(t ) = 2
i(t ) = 1 + e −t
Example: RL Circuit with Step Input
Find i(t)
i(t)=25+25(1-e-t/2)u(t) A
Engr228 - Chapter 7, Nilsson 10E
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Driven RC Circuits (Part 1 of 2)
Find Vc (t) and i(t)
vC(t)=20 + 80e-t/1.2 V and i(t)=0.1 + 0.4e−t/1.2 A
Driven RC Circuits (Part 2 of 2)
vC=20 + 80e-t/1.2 V
i=0.1 + 0.4e−t/1.2 A
Engr228 - Chapter 7, Nilsson 10E
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Chapter 7 Summary
• Showed how to determine the natural response of both RL
and RC circuits;
• Showed how to determine the step response of both RL
and RC circuits.
Engr228 - Chapter 7, Nilsson 10E
22