lecture 12 first order transient response
TRANSCRIPT
Lecture 12Lecture 12First Order Transient Response
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Chapter 4Chapter 4Transients
1 S l fi t d RC RL i it1. Solve first-order RC or RL circuits.
2 U d t d th t f t i t2. 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.
TransientsTransients
The time-varying currents and voltages resulting from the sudden application ofresulting from the sudden application of sources, usually due to switching, are called transients. By writing circuit equations, we obtain integro-differential q gequations.
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 g ga Resistance
KCL at the top node of the circuit:iC iR
dtdvC
dtdqiCvq
vqC C ==→=→=
tv )(
( ) ( ) ( )R
tvi CR
)(=
( ) ( ) 0=+R
tvdt
tdvC CC ( ) ( ) 0=+ tvdt
tdvRC CC
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Rdt dt
Discharge of a Capacitance through a Resistance
( ) ( )tdv ( )tdv 1( ) ( ) 0=+ tvdt
tdvRC C
C ( ) ( )tvRCdt
tdvC
C 1−=
( )
We need a function vC(t) that has the same form as it’s derivative.
( ) stC Ketv =
S b i i hi i f ( )
0stst KRCK
Substituting this in for vc(t)
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
0=+ stst KeRCKse
Discharge of a Capacitance through a
1−Resistance
RCs 1=Solving for s:
( ) RCtC Ketv −=Substituting into vc(t): ( )g c( )
Initial Condition: Full Solution:
( ) RCteVtv −=( ) iC Vv =+0
Initial Condition: Full Solution:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
( ) iC eVtv =( ) iC Vv +0
Discharge of a Capacitance through a Resistance
( ) RCtC Ketv −=
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 thecharged to a voltage Vi and then discharges through the resistor.
( ) RCtiC eVtv −=( ) iC Vv =+0
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 g ga 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)applications (e.g. garage door light)
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Larry Light-Bulb Experimenty g
iR(t)
RCtiR e
RV
Rtvti /)()( −==
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
R RR
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
Current into the capacitor: dv
C cpdt
C c
Current through the i t )(resistor:
RVtv SC −)(
0)()(
=−
+R
Vtvdt
tdvC SCC
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Rdt
Charging a Capacitance from a DC Source through a Resistance
0)()(
=−
+R
Vtvdt
tdvC SCC
Rearranging:
SCC Vtvdt
tdvRC =+ )(
)(dt
This is a linear first-order differential equation with constant coefficients
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
constant coefficients.
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:instantaneously:
0)0()0( =−=+ CC vv
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
stC eKKtv 21)( +=Try the solution:
Substituting into the differential equation:
SCC Vtv
tdvRC =+ )(
)(
Substituting into the differential equation:
SC Vtvdt
RC + )(Gives:
st VKeKRCs =++ )1( SVKeKRCs =++ 12)1(
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
Sst VKeKRCs =++ 12)1(
For equality, the coefficient of est must be zero:
RCsRCs 101 −=→=+
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
RCtst eKVeKKtv /)( −++
Substituting in for K1 and s:
SC eKVeKKtv 221)( +=+=
Evaluating at t=0 and remembering that vC(0+)=0
sSSC VKKVeKVv −=→=+=+=+ 220
2 0)0(
RCtSS
stC eVVeKKtv /
21)( −−=+=
Substituting in for K2 gives:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
SSC eVVeKKtv 21)( +
Charging a Capacitance from a DC Source through a Resistance
( ) τtVV −
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
( ) τtssC eVVtv −=
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?
( )τ
τ
SCtSC
tssC
Vdt
dve
Vdt
dveVVtv
=→=
−=
−
−
/A line with this slope would intersect VS after a
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
ττ tdtdt =0S
time τ
Larry Light-Bulb Experimenty g
tRtiV 0)()( ++
eVeVVVtvVRti
tvRtiV
RCt
RCtS
tSSSCS
CS
/
/)()()(
0)()(−− =−−=−=
=++−
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.R
eVti
RCtS
/)(
−
=
DC Steady Statey
tdvC )(dt
tdvCti C
C)(
)( =
In steady state, the voltage is constant, so the current through the capacitor is zero, so it behaves as an open circuitas an open circuit.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady Statey
tdiL )(dt
tdiLtv LL
)()( =
In steady state, the current is constant, so the voltage across and inductor is zero, so it behaves as a short circuitas a short circuit.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady StateDC Steady State
The steps in determining the forced response for RLC circuits with dc sources are:
1. Replace capacitances with open circuits.
2 Replace ind ctances ith short circ its2. 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 Statey
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady Statey
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 Statey
• Open circuit the capacitor
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady Statey
VARiv aa 50)25)(2( =Ω==
Aia 2=
• Short circuit the inductor
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
DC Steady Statey
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 Statey
• Open circuit the capacitor
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.