Download Lecture 12

Lecture 12
First Order Transient Response
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Chapter 4
Transients
1. Solve first-order RC or RL circuits.
2. Understand the concepts of transient
response and steady-state response.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
3. Relate the transient response of first-order
circuits to the time constant.
4. Solve RLC circuits in dc steady-state
conditions.
5. Solve second-order circuits.
6. Relate the step response of a second-order
system to its natural frequency and damping
ratio.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transients
The time-varying currents and voltages
resulting from the sudden application of
sources, usually due to switching, are
called transients. By writing circuit
equations, we obtain integro-differential
equations.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Discharge of a Capacitance through
a Resistance
iC
iR
KCL at the top node of the
circuit:
q
dq
dv
C   q  Cv  iC 
C
v
dt
dt
vC (t )
iR 
R
dvC t  vC t 
C

0
dt
R
dvC t 
RC
 vC t   0
dt
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Discharge of a Capacitance through a
Resistance
dvC t 
dvC t 
1

v t 
RC
dt
 vC t   0
dt
RC
C
We need a function vC(t) that has the same form as it’s
derivative.
vC t   Ke
st
Substituting this in for vc(t)
RCKse  Ke  0
st
st
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Discharge of a Capacitance through a
Resistance
1
Solving for s: s 
RC
Substituting into vc(t):
Initial Condition:
vC 0    Vi
vC t   Ke
t RC
Full Solution:
vC t   Vi e
t RC
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Discharge of a Capacitance through a
Resistance
vC t   Ke
t RC
To find the unknown constant K, we need to use the
boundary conditions at t=0. At t=0 the capacitor is initially
charged to a voltage Vi and then discharges through the
resistor.
vC 0    Vi
vC t   Vi e
t RC
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Discharge of a Capacitance through a
Resistance
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Discharge of a Capacitance through
a Resistance
• The time interval t= RC is called the
time constant of the circuit
• RC circuits can be used for timing
applications (e.g. garage door light)
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Larry Light-Bulb Experiment
iR(t)
v(t ) Vi t / RC
iR (t ) 
 e
R
R
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Charging a Capacitance from a DC
Source through a Resistance
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Charging a Capacitance from a DC
Source through a Resistance
KCL at the node that joins
the resistor and the capacitor
dvC (t ) vC (t )  VS
C

0
dt
R
Current into the
dvc
capacitor:
C
dt
Current through the
resistor: vC (t )  VS
R
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Charging a Capacitance from a DC
Source through a Resistance
dvC (t ) vC (t )  VS
C

0
dt
R
Rearranging:
dvC (t )
RC
 vC (t )  VS
dt
This is a linear first-order differential equation with
contant coefficients.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Charging a Capacitance from a DC
Source through a Resistance
The boundary conditions are given by the fact that the
voltage across the capacitance cannot change
instantaneously:
vC (0)  vC (0)  0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Charging a Capacitance from a DC
Source through a Resistance
Try the solution:
vC (t )  K1  K 2e st
Substituting into the differential equation:
Gives:
dvC (t )
RC
 vC (t )  VS
dt
(1  RCs) K 2e st  K1  VS
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Charging a Capacitance from a DC
Source through a Resistance
(1  RCs) K 2e  K1  VS
st
For equality, the coefficient of est must be zero:
1
1  RCs  0  s 
RC
Which gives K1=VS
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Charging a Capacitance from a DC
Source through a Resistance
Substituting in for K1 and s:
vC (t )  K1  K 2e st  VS  K 2e t / RC
Evaluating at t=0 and remembering that vC(0+)=0
vC (0)  VS  K 2e0  VS  K 2  0  K 2  Vs
Substituting in for K2 gives:
vC (t )  K1  K 2e st  VS  VS e t / RC
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Charging a Capacitance from a DC
Source through a Resistance
vC t   Vs  Vs e
t 
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Charging a Capacitance from a DC
Source through a Resistance
If the initial slope is extended from t=0, when does it intersect the
final value VS?
vC t   Vs  Vs e t 
dvC VS t /
dvC

e

dt

dt

t 0
VS

A line with this slope
would intersect VS after a
time 
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Larry Light-Bulb Experiment
VS  i(t ) R  vC (t )  0
i(t ) R  VS  vC (t )  VS  (VS  VS e t )  VS e t / RC
VS e t / RC
i(t ) 
R
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady State
dvC (t )
iC (t )  C
dt
In steady state, the voltage is constant, so the
current through the capacitor is zero, so it behaves
as an open circuit.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady State
diL (t )
v L (t )  L
dt
In steady state, the current is constant, so the
voltage across and inductor is zero, so it behaves
as a short circuit.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady State
The steps in determining the forced response for
RLC circuits with dc sources are:
1. Replace capacitances with open circuits.
2. Replace inductances with short circuits.
3. Solve the remaining circuit.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady State
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady State
Find the steady state current ia and voltage va
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady State
• Open circuit the capacitor
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady State
va  ia R  (2 A)( 25)  50V
ia  2 A
• Short circuit the inductor
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady State
Find the steady state currents i1, i2, i3
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady State
• Open circuit the capacitor
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady State
10 
5
1A
1A
• Short circuit the inductor
• i1=20V/10=2A
• Current divider gives i2=i3=1A
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.