Lecture 13Lecture 13 RC/RL Circuits, Time Dependent Op Amp Circuits 13.pdf · RC/RL Circuits, Time Dependent Op Amp Circuits ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth

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:


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.


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.


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


0)()(S

tditvRtiV =++−

)(0)()(

SVdt

tdiLRti =+)()(


dt


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


RS 50Ω L22


LtReKti /2)( −+= eKti 22)( +=

2220)0( 0 −=→+=+==+ KKeKi 2220)0( 222 −=→+=+== KKeKi

LtRti /22)( −


LtReti /22)( −=


LD fiR

Define =τ

τ/22)( teti −−=



ττ // 100)22(50100)(50100)()( tt eetiRtiVtv −− ==== 100)22(50100)(50100)()( S eetiRtiVtv =−−=−=−=



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


( ) S 1


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


2

/// 0)(RLKeLRRKeKeL ttt =→=−=+− −−− τ

τττττ


VV

1

0

1

)0(RVKKe

RVi SS =→==+

0)( / >= − tforeRVti tS τ


1R


)( // ⎟⎞

⎜⎛ LVVdtdi 0)()( /

1

/

1

>−=⎟⎟⎠

⎞⎜⎜⎝

⎛== −− tfore

RLVe

RV

dtdL

dttdiLtv tstS ττ

τ00)( <= tfortv


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( )



After the switch is closed, iR(0+) = 2A, IL(0+) = 0

Atitiandtidt

tdiLtv LRRL 2)()()(10)()( =+==dt

( ))(210)(2 titdiL

L −=


( ))(2102 tidt L


)(tdi ( ))(210)(2 tidt

tdiL

L −=

10)(5)(=+ titdiL 10)(5 =+ ti

dt L

stL eKKtiTry −+= 21)(


L 21


)(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


)(


= 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


L eti 22)( =


dtdi )( ttL eedtd

dttdiLtv 55 20)22(2)()( −− =−==

ttLR eetiti 55 2)22(2)(2)( −− =−−=−=



Prior to t = 0 i(0) = 100V/100Ω = 1A

Find i(t), v(t)



Prior to t = 0 i(0) = 100V/100Ω = 1A

After t = 0: 100)()(+ VRtitdiLAfter t = 0:

100)(200)(

100)( =+

itdi

VRtidt

L


100)(200)(=+ ti

dt


tidt

tdi 100)(200)(=+

eKKesK

eKKtitrystst

st

100)(200

)(

212

21

=++−

+=−−

−

mssKeKs

eKKesKst 5200/1,200100200)200(

100)(200

12

212

===→=+−

++− τ



KKi 21 11)0( =+→=

AKAVKi

200

21 5.05.0200100)( =→=

Ω==∞


teti 2005.05.0)( −+=


dtdi )( tt eedtd

dttdiLtv 200200 100)5.05.0()()( −− −=+==


RC and RL Circuits with General Sources

)()()( tvtRid

tdiL t=+

)()()(

)()(

tvtitdiLdt

t

t

=+

)()()(

)(

tftxtdxR

tidtR

+

=+

τFirst order differential equation with constant Forcing


)()( 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.


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


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.


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.


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.


Transient Analysis of an RC Circuit with a Sinusoidal Source

)200sin(2)()( tCtqtRi =+

)200sin(2)0()(1)( tvtiC

tRit

c∫ =++


0C c∫


Take the derivative:

)200cos(400)(1)( ttitdiR =+


)200(400)()(

)200cos(400)(

CitdiRC

ttiCdt

R +

:

)200cos(400)()(

solutionparticularaTry

tCtidt

RC =+

)200sin()200cos()(:

tBtAtisolutionparticularaTry

p +=



)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


)()()(p


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)( ++



Initial conditions:

2sin(0+) = 0

vC(0+) = 1V

vR(0+) + vC(0+) = 0 → vR(0+) = -1V

i(0+) = vR/R = -1V/5000Ω = -200μA


( ) R μ

= 200cos(0)+200sin(0)+Ke0 = 200 + K → K= -400μA


Attti t τ/400)200i (200)200(200)( −+


Aettti t μτ/400)200sin(200)200cos(200)( −+=


What happens if we replace the source with 2cos(200t) and the capacitor initially uncharged vc(0)=0?



What happens if we replace the source with 2cos(200t) and the capacitor initially uncharged vc(0)=0?

)(t

1

)200cos(2)()( tCtqtRi

t

∫

=+


)200cos(2)0()(1)(0

tvtiC

tRi c∫ =++



)200sin(400)(1)( ttitdiR −=+


)200i (400)()(

)200sin(400)(

CitdiRC

ttiCdt

R +

:

)200sin(400)()(


tCtidt

RC −=+

)200sin()200cos()(:

tBtAtisolutionparticularaTry

p +=



)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


)200sin(200)200cos(200)( ttti p −=



msFkRCKeti t μττ/ 5)1)(5()( − Ω msFkRCKetic μτ 5)1)(5()( =Ω===


AKettti t μτ/)200sin(200)200cos(200)( −+−=


AKettti μ)200sin(200)200cos(200)( +



Initial conditions:

2cos(0+) = 2

vC(0+) = 0V

vR(0+) + vC(0+) = 2 → vR(0+) = 2

i(0+) = vR/R = 2V/5000Ω = 400μA


( ) R μ

= 200cos(0)-200sin(0)+Ke0 = 200 + K → K= 200μA


Aettti t μτ/200)200sin(200)200cos(200)( −+−=


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)()(


tc evti

CtRi −∫ =++ 10)0()(1)(

0



tetitdiR −−=+ 10)(1)(


tCtitdiRC

etiCdt

R

−+

+

10)()(

10)(

t


Cetidt

RC −=+

:

10)()(

tp Aeti

solutionparticularaTry−=)(

:


p


CeAeRCAe ttt 10−=+− −−−

CRCACRCAA

10)1(10

=−=−

AxCA

CRCA

μ20)102)(10(10

10)1(6

===

−=−−

ARC

A μ20121

=−

=−

=




sFMRCKeti t μττ/ 2)2)(1()( − Ω sFMRCKetic μτ 2)2)(1()( =Ω===


AKeeti tt μ2/20)( −− +=


AKeeti μ20)( +



Initial conditions:

10e0+ = 10

vC(0+) = 5V

vR(0+) + vC(0+) = 10 → vR(0+) = 5

i(0+) = vR/R = 5V/1MΩ = 5μA


( ) R μ

= 20e0 + Ke0 = 20 + K → K= -15μA


Aeeti tt μ2/1520)( −−= Aeeti μ1520)( −=


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 .


dtvdtiqvv

itt

in ∫∫11

d

dtvRC

dtiCC

vR

i

t

inincin

in

∫

∫∫−

====00

1


dtvRC

vvvv incoco ∫=−=→=−−0

10

t

( ) ( )dttvRC

tvt

o in1∫−=


RC 0∫

If R = 10 kΩ, C = 0.1μF → RC = 0.1 ms

( ) ( ) tdtvtvt

∫= 1000


( ) ( ) tdtvtvo ∫−=0

in1000

( ) ( ) mstfortdttdtvtvtt

o 10500051000100000

<<−=−=−= ∫∫ in

mstmsfordtdtms t

315510001

<<⎥⎥⎦

⎤

⎢⎢⎣

⎡−+−= ∫ ∫

t

ms

500010

0 1

+−=

⎥⎦⎢⎣∫ ∫


Differentiator Circuit

dtdv

RCvR

vdt

dvC

dtdqi in

ooin

in −=→−

===0


Differentiator Circuit

d( )dt

dvRCtvoin−=


dt

Lecture 13Lecture 13 RC/RL Circuits, Time Dependent Op Amp Circuits 13.pdf · RC/RL Circuits, Time Dependent Op Amp Circuits ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth

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