lecture 13lecture 13 rc/rl circuits, time dependent op amp circuits 13.pdf · rc/rl circuits, time...
TRANSCRIPT
Lecture 13Lecture 13RC/RL Circuits, Time
Dependent Op Amp Circuits
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL CircuitsRL Circuits
The steps involved in solving simple circuits containing dcsimple circuits containing dc sources, resistances, and one energy storage elementenergy-storage element (inductance or capacitance) are:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
1 A l Ki hh ff’ t d lt1. Apply Kirchhoff’s current and voltage laws to write the circuit equation.
2. If the equation contains integrals, q g ,differentiate each term in the equation to produce a pure differential equationto produce a pure differential equation.
3 Assume a solution of the form K +3. Assume a solution of the form K1 + K2est.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
4. Substitute the solution into the differential equation to determine the values of K1 and s . (Alternatively, we can determine K1 by solving the circuit in steady state)
5. Use the initial conditions to determine the value of K2.
6. Write the final solution.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysisy
Fi d i(t) d th lt (t)Find i(t) and the voltage v(t)
i(t)= 0 for t < 0 since the switch is open prior to t = 0
Apply KVL around the loop:
0)()( =++− tvRtiVELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
0)()( =++ tvRtiVS
RL Transient Analysisy
0)()(S
tditvRtiV =++−
)(0)()(
SVdt
tdiLRti =+)()(
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
dt
RL Transient Analysisy
SVdt
tdiLRti =+)()( steKKtiTry 21)( +=
steKKtiTry 21)( +=SVtdiLRti =+)()(
Sst VesLKRKRK =++ )( 221
eKKtiTry 21)( +SVdt
LRti +)(
S)( 221
AVR
VKVRK SS 2
50100
11 =Ω
==→=LRssLKRK −
=→=+ 022
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RS 50Ω L22
RL Transient Analysisy
LtReKti /2)( −+= eKti 22)( +=
2220)0( 0 −=→+=+==+ KKeKi 2220)0( 222 −=→+=+== KKeKi
LtRti /22)( −
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
LtReti /22)( −=
RL Transient Analysisy
LD fiR
Define =τ
τ/22)( teti −−=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysisy
ττ // 100)22(50100)(50100)()( tt eetiRtiVtv −− ==== 100)22(50100)(50100)()( S eetiRtiVtv =−−=−=−=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysisy
T i t t t b i it hTransient starts by opening switch
Prior to t = 0 inductor acts as a short circuit so that:
v(t) = 0 for t < 0
i(t) = VS/R1 for t < 0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
( ) S 1
RL Transient Analysisy
Aft t 0 t i l t th h L d RAfter t = 0 current circulates through L and R, dissipating energy in the resistance R
/)()()( ftdi t τ
///
/
0)(
0)(0)()(
LKLRRKKL
tforKetiRtidt
tdiL
ttt
t >=→=+ −
τττ
τ
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
2
/// 0)(RLKeLRRKeKeL ttt =→=−=+− −−− τ
τττττ
RL Transient Analysisy
VV
1
0
1
)0(RVKKe
RVi SS =→==+
0)( / >= − tforeRVti tS τ
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
1R
RL Transient Analysisy
)( // ⎟⎞
⎜⎛ LVVdtdi 0)()( /
1
/
1
>−=⎟⎟⎠
⎞⎜⎜⎝
⎛== −− tfore
RLVe
RV
dtdL
dttdiLtv tstS ττ
τ00)( <= tfortv
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysiswith a Current Source
After the switch is opened, iR(0+) = 2A, IL(0+) = 0
Find v(t), iR(t), iL(t)( ), R( ), L( )
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysiswith a Current Source
After the switch is closed, iR(0+) = 2A, IL(0+) = 0
Atitiandtidt
tdiLtv LRRL 2)()()(10)()( =+==dt
( ))(210)(2 titdiL
L −=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
( ))(2102 tidt L
RL Transient Analysiswith a Current Source
)(tdi ( ))(210)(2 tidt
tdiL
L −=
10)(5)(=+ titdiL 10)(5 =+ ti
dt L
stL eKKtiTry −+= 21)(
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
L 21
RL Transient Analysiswith a Current Source
)(tdi 10)(5)(=+ ti
dttdi
LL
stL eKKtiTry −+= 21)(
i (0+) 0 K K
tt
iL(0+) = 0 → K1= -K2
5105)5(
10)(5
=→=+−
=−+−
−−
sKKes
KeKsKest
stst
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
)(
RL Transient Analysiswith a Current Source
= 2A= 2A
tL KeKti 5)( −−=L eti )(
In steady state the inductor becomes a h t i it i ( ) K 2A
teti 522)( −−=
short circuit → iL(∞) = K = 2A
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
L eti 22)( =
RL Transient Analysiswith a Current Source
dtdi )( ttL eedtd
dttdiLtv 55 20)22(2)()( −− =−==
ttLR eetiti 55 2)22(2)(2)( −− =−−=−=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysisy
Prior to t = 0 i(0) = 100V/100Ω = 1A
Find i(t), v(t)
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysisy
Prior to t = 0 i(0) = 100V/100Ω = 1A
After t = 0: 100)()(+ VRtitdiLAfter t = 0:
100)(200)(
100)( =+
itdi
VRtidt
L
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
100)(200)(=+ ti
dt
RL Transient Analysisy
tidt
tdi 100)(200)(=+
eKKesK
eKKtitrystst
st
100)(200
)(
212
21
=++−
+=−−
−
mssKeKs
eKKesKst 5200/1,200100200)200(
100)(200
12
212
===→=+−
++− τ
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RL Transient Analysisy
KKi 21 11)0( =+→=
AKAVKi
200
21 5.05.0200100)( =→=
Ω==∞
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
teti 2005.05.0)( −+=
RL Transient Analysisy
dtdi )( tt eedtd
dttdiLtv 200200 100)5.05.0()()( −− −=+==
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RC and RL Circuits with General Sources
)()()( tvtRid
tdiL t=+
)()()(
)()(
tvtitdiLdt
t
t
=+
)()()(
)(
tftxtdxR
tidtR
+
=+
τFirst order differential equation with constant Forcing
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
)()( tftxdt
=+τequation with constant coefficients
gfunction
RC and RL Circuits with GeneralRC and RL Circuits with General Sources
The general solution consists of two partsof two parts.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
The particular solution (also called theThe particular solution (also called the forced response) is any expression that
ti fi th tisatisfies the equation.
)(tdx )()()( tftxdt
tdx=+τ
In order to have a solution that satisfies the initial conditions, we must add theinitial conditions, we must add the complementary solution to the particular solution
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
solution.
The homogeneous equation is obtained byThe homogeneous equation is obtained by setting the forcing function to zero.
0)()(=+ tx
dttdxτ
The complementary solution (also called the natural response) is obtained bythe natural response) is obtained by solving the homogeneous equation.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Step-by-Step Solution
Circuits containing a resistance a sourceCircuits containing a resistance, a source,and an inductance (or a capacitance)
1. Write the circuit equation and reduce it to a first-order differential equationfirst order differential equation.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
2. Find a particular solution. The details of this pstep depend on the form of the forcing function.
3. Obtain the complete solution by adding the particular solution to the complementary solution xc=Ke-t/τ which contains the arbitrary constant K.
4. Use initial conditions to find the value of K.
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit with a Sinusoidal Source
)200sin(2)()( tCtqtRi =+
)200sin(2)0()(1)( tvtiC
tRit
c∫ =++
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
0C c∫
Transient Analysis of an RC Circuit with a Sinusoidal Source
Take the derivative:
)200cos(400)(1)( ttitdiR =+
Take the derivative:
)200(400)()(
)200cos(400)(
CitdiRC
ttiCdt
R +
:
)200cos(400)()(
solutionparticularaTry
tCtidt
RC =+
)200sin()200cos()(:
tBtAtisolutionparticularaTry
p +=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit with a Sinusoidal Source
)200(10400)()(105 63 ttitdi −−
:sin)200cos(10400)200sin()200cos()200cos()200sin(
)200cos(10400)()(105
6
63
tftffi ithtitxtBtAtBtA
txtidt
x
=+++−
=+
−
:cos0
:sin
termsfortscoefficientheequatingBABA
termsfortscoefficientheequating=→=+−
20010200
:10400
6
6
AA
SolvingxAB =+
−
−
)200sin(200)200cos(200)(20010200
200102006
6
tttiAxB
AxA
p +===
==− μ
μ
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
)()()(p
Transient Analysis of an RC Circuit with a Sinusoidal Source
The complementary solution is given by:
msFkRCKeti t μττ/ 5)1)(5()( − Ω msFkRCKetic μτ 5)1)(5()( =Ω===
The complete solution is given by the sum of the
AKettti t μτ/)200sin(200)200cos(200)( −++=
particular solution and the complementary solution:
AKettti μ)200sin(200)200cos(200)( ++
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit with a Sinusoidal Source
Initial conditions:
2sin(0+) = 0
vC(0+) = 1V
vR(0+) + vC(0+) = 0 → vR(0+) = -1V
i(0+) = vR/R = -1V/5000Ω = -200μA
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
( ) R μ
= 200cos(0)+200sin(0)+Ke0 = 200 + K → K= -400μA
Transient Analysis of an RC Circuit with a Sinusoidal Source
Attti t τ/400)200i (200)200(200)( −+
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Aettti t μτ/400)200sin(200)200cos(200)( −+=
Transient Analysis of an RC Circuit with a Sinusoidal Source
What happens if we replace the source with 2cos(200t) and the capacitor initially uncharged vc(0)=0?
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit with a Sinusoidal Source
What happens if we replace the source with 2cos(200t) and the capacitor initially uncharged vc(0)=0?
)(t
1
)200cos(2)()( tCtqtRi
t
∫
=+
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
)200cos(2)0()(1)(0
tvtiC
tRi c∫ =++
Transient Analysis of an RC Circuit with a Sinusoidal Source
Take the derivative:
)200sin(400)(1)( ttitdiR −=+
Take the derivative:
)200i (400)()(
)200sin(400)(
CitdiRC
ttiCdt
R +
:
)200sin(400)()(
solutionparticularaTry
tCtidt
RC −=+
)200sin()200cos()(:
tBtAtisolutionparticularaTry
p +=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit with a Sinusoidal Source
)200sin(10400)()(105 63 txtitdix =+ −−
:sin)200sin(10400)200sin()200cos()200cos()200sin(
)200sin(10400)(105
6
termsfortscoefficientheequatingtxtBtAtBtA
txtidt
x
−=+++−
−=+
−
:cos10400
:sin6
termsfortscoefficientheequatingxBA
termsfortscoefficientheequating
−=+− −
:0
6
SolvingAB
fffq g=+
)200sin(200)200cos(200)(20010200
200102006
6
tttiAxB
AxA
−−=
==−
−
μ
μ
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
)200sin(200)200cos(200)( ttti p −=
Transient Analysis of an RC Circuit with a Sinusoidal Source
The complementary solution is given by:
msFkRCKeti t μττ/ 5)1)(5()( − Ω msFkRCKetic μτ 5)1)(5()( =Ω===
The complete solution is given by the sum of the
AKettti t μτ/)200sin(200)200cos(200)( −+−=
particular solution and the complementary solution:
AKettti μ)200sin(200)200cos(200)( +
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit with a Sinusoidal Source
Initial conditions:
2cos(0+) = 2
vC(0+) = 0V
vR(0+) + vC(0+) = 2 → vR(0+) = 2
i(0+) = vR/R = 2V/5000Ω = 400μA
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
( ) R μ
= 200cos(0)-200sin(0)+Ke0 = 200 + K → K= 200μA
Transient Analysis of an RC Circuit with a Sinusoidal Source
Aettti t μτ/200)200sin(200)200cos(200)( −+−=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit with an Exponential Source
What happens if we replace the source with 10e-t and the capacitor is initially charged to vc(0)=5?
t)(
t
teCtqtRi −
∫
=+
1
10)()(
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
tc evti
CtRi −∫ =++ 10)0()(1)(
0
Transient Analysis of an RC Circuit with an Exponential Source
Take the derivative:
tetitdiR −−=+ 10)(1)(
Take the derivative:
tCtitdiRC
etiCdt
R
−+
+
10)()(
10)(
t
solutionparticularaTry
Cetidt
RC −=+
:
10)()(
tp Aeti
solutionparticularaTry−=)(
:
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
p
Transient Analysis of an RC Circuit with an Exponential Source
CeAeRCAe ttt 10−=+− −−−
CRCACRCAA
10)1(10
=−=−
AxCA
CRCA
μ20)102)(10(10
10)1(6
===
−=−−
ARC
A μ20121
=−
=−
=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit with an Exponential Source
The complementary solution is given by:
sFMRCKeti t μττ/ 2)2)(1()( − Ω sFMRCKetic μτ 2)2)(1()( =Ω===
The complete solution is given by the sum of the
AKeeti tt μ2/20)( −− +=
particular solution and the complementary solution:
AKeeti μ20)( +
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Transient Analysis of an RC Circuit with an Exponential Source
Initial conditions:
10e0+ = 10
vC(0+) = 5V
vR(0+) + vC(0+) = 10 → vR(0+) = 5
i(0+) = vR/R = 5V/1MΩ = 5μA
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
( ) R μ
= 20e0 + Ke0 = 20 + K → K= -15μA
Transient Analysis of an RC Circuit with an Exponential Source
Aeeti tt μ2/1520)( −−= Aeeti μ1520)( −=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Integrators and DifferentiatorsIntegrators and Differentiators
Integrators produce output voltages that areIntegrators produce output voltages that are proportional to the running time integral of the input voltages In a running time integralthe input voltages. In a running time integral, the upper limit of integration is t .
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
dtvdtiqvv
itt
in ∫∫11
d
dtvRC
dtiCC
vR
i
t
inincin
in
∫
∫∫−
====00
1
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
dtvRC
vvvv incoco ∫=−=→=−−0
10
t
( ) ( )dttvRC
tvt
o in1∫−=
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
RC 0∫
If R = 10 kΩ, C = 0.1μF → RC = 0.1 ms
( ) ( ) tdtvtvt
∫= 1000
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
( ) ( ) tdtvtvo ∫−=0
in1000
( ) ( ) mstfortdttdtvtvtt
o 10500051000100000
<<−=−=−= ∫∫ in
mstmsfordtdtms t
315510001
<<⎥⎥⎦
⎤
⎢⎢⎣
⎡−+−= ∫ ∫
t
ms
500010
0 1
+−=
⎥⎦⎢⎣∫ ∫
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.
Differentiator Circuit
dtdv
RCvR
vdt
dvC
dtdqi in
ooin
in −=→−
===0
ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, ©2008 Pearson Education, Inc.