FIRST ORDER TRANSIENT CIRCUITS

IN CIRCUITS WITH INDUCTORS AND CAPACITORS VOLTAGES AND CURRENTSCANNOT CHANGE INSTANTANEOUSLY. EVEN THE APPLICATION, OR REMOVAL, OF CONSTANT SOURCES CREATES ATRANSIENT BEHAVIOR

FIRST ORDER CIRCUITSCircuits that contain a single energy storing elements.Either a capacitor or an inductor

THE CONVENTIONAL ANALYSIS USING MATHEMATICAL MODELS REQUIRES THE DETERMINATION OF (A SET OF) EQUATIONS THAT REPRESENT THE CIRCUIT.ONCE THE MODEL IS OBTAINED ANALYSIS REQUIRES THE SOLUTION OF THE EQUATIONS FOR THE CASES REQUIRED.

FOR EXAMPLE IN NODE OR LOOP ANALYSIS OF RESISTIVE CIRCUITS ONE REPRESENTS THECIRCUIT BY A SET OF ALGEBRAIC EQUATIONS

WHEN THERE ARE INDUCTORS OR CAPACITORS THE MODELS BECOME LINEAR ORDINARY DIFFERENTIAL EQUATIONS (ODEs). HENCE, IN GENERAL, ONE NEEDS ALL THOSE TOOLS IN ORDER TO BE ABLE TO ANALYZE CIRCUITS WITH ENERGY STORING ELEMENTS.

ANALYSIS OF LINEAR CIRCUITS WITH INDUCTORS AND/OR CAPACITORS

THE GENERAL APPROACH CAN BE SIMPLIFIED IN SOME SPECIAL CASES WHEN THE FORMOF THE SOLUTION CAN BE KNOWN BEFOREHAND. THE ANALYSIS IN THESE CASES BECOMES A SIMPLE MATTER OF DETERMINING SOMEPARAMETERS.TWO SUCH CASES WILL BE DISCUSSED IN DETAIL FOR THE CASE OF CONSTANT SOURCES. ONE THAT ASSUMES THE AVAILABILITY OF THE DIFFERENTIAL EQUATION AND A SECOND THAT IS ENTIRELY BASED ON ELEMENTARY CIRCUIT ANALYSIS… BUT IT IS NORMALLY LONGER

A METHOD BASED ON THEVENIN WILL BE DEVELOPED TO DERIVE MATHEMATICAL MODELSFOR ANY ARBITRARY LINEAR CIRCUIT WITH ONE ENERGY STORING ELEMENT.

WE WILL ALSO DISCUSS THE PERFORMANCE OF LINEAR CIRCUITS TO OTHER SIMPLE INPUTS

THE MODEL

AN INTRODUCTIONINDUCTORS AND CAPACITORS CAN STORE ENERGY. UNDER SUITABLE CONDITIONS THIS ENERGYCAN BE RELEASED. THE RATE AT WHICH IT IS RELEASED WILL DEPEND ON THE PARAMETERSOF THE CIRCUIT CONNECTED TO THE TERMINALS OF THE ENERGY STORING ELEMENT

With the switch on the left the capacitor receivescharge from the battery.

Switch to the rightand the capacitordischarges throughthe lamp

dxxfetxetxet

tTH

xtt

)(1

)()(0

0

0

GENERAL RESPONSE: FIRST ORDER CIRCUITS

0)0(; xxfxdt

dxTH

Including the initial conditionsthe model for the capacitorvoltage or the inductor current will be shown to be of the form

Solving the differential equationusing integrating factors, one tries to convert the LHS into anexact derivative

t

TH efxdt

dx 1/*

TH

ttt

fexedt

dxe

11

TH

tt

fexedt

d

1

t

t0

dxxfetxetxt

t

TH

xttt

)(1

)()(0

0

0

0)0();()()( xxtftaxt

dt

dx

THIS EXPRESSION ALLOWS THE COMPUTATIONOF THE RESPONSE FOR ANY FORCING FUNCTION.WE WILL CONCENTRATE IN THE SPECIAL CASEWHEN THE RIGHT HAND SIDE IS CONSTANT

t

e

/*

circuit

the of speed reaction the on ninformatio

tsignifican provide to shown be willit

constant." time" the called is

.switchings sequentialstudy

to used be can expression general

The arbitrary. is , time, initial The ot

FIRST ORDER CIRCUITS WITHCONSTANT SOURCES

dxxfetxetxt

tTH

xttt

)(1

)()(0

0

0

0)0(; xxfxdt

dxTH

If the RHS is constant

dxef

txetxt

t

xtTH

tt

0

0

)()( 0

xtxt

eee

dxeef

txetxt

t

xtTH

tt

0

0

)()( 0

t

t

xtTH

tt

eef

txetx0

0

)()( 0

00

)()( 0

ttt

TH

tt

eeeftxetx

0

)()( 0

tt

THTH eftxftx

0tt The form of the solution is

021 ;)(0

tteKKtxtt

Any variable in the circuit is ofthe form

021 ;)(0

tteKKtytt

Only the values of the constants K_1, K_2 will change

TRANSIENT

TIMECONSTANT

EVOLUTION OF THE TRANSIENT AND INTERPRETATION OF THE TIME CONSTANT

A QUALITATIVE VIEW:THE SMALLER THE THE TIMECONSTANT THE FASTER THETRANSIENT DISAPPEARS

With less than 2% errortransient is zerobeyond this point

Drops 0.632 of initialvalue in one time constant

Tangent reaches x-axis in one time constant

CRTH

THE TIME CONSTANT

The following example illustratesthe physical meaning of timeconstant

vS

R S a

b

C

+

vc

_

Charging a capacitor

THCC

TH vvdt

dvCR

The model

0)0(, CSS vVv

Assume

The solution can be shown to be

t

SSC eVVtv

)(

CRTH

transientFor practical purposes thecapacitor is charged when thetransient is negligible

0067.0

0183.00498.0135.0

5

432

368.0

t

et

With less than 1%error the transientis negligible afterfive time constants

dt

dvC C

S

SC

R

vv

0

S

SCc

R

vv

dt

dvC

: KCL@a

1. THE CIRCUIT HAS ONLY CONSTANT INDEPENDENT SOURCES

THE DIFFERENTIAL EQUATION APPROACH

CIRCUITS WITH ONE ENERGY STORING ELEMENT

CONDITIONS

2. THE DIFFERENTIAL EQUATION FOR THE VARIABLE OF INTEREST IS SIMPLE TO OBTAIN. NORMALLY USING BASIC ANALYSIS TOOLS; e.g., KCL, KVL. . . OR THEVENIN

3. THE INITIAL CONDITION FOR THE DIFFERENTIAL EQUATION IS KNOWN, OR CAN BE OBTAINED USING STEADY STATE ANALYSIS

SOLUTION STRATEGY: USE THE DIFFERENTIAL EQUATION AND THE INITIAL CONDITIONS TO FIND THE PARAMETERS ,, 21 KK

( )

1 2

FACT: WHEN ALL INDEPENDENT SOURCES ARE CONSTANT

FOR ANY VARIABLE, ( ), IN THE CIRCUIT THE

SOLUTION IS OF THE FORM

( ) ,Ot t

O

y t

y t K K e t t

If the diff eq for y is knownin the form

Use the diff eq to find twomore equations by replacingthe form of solution into thedifferential equation

0

01

)0( yy

fyadt

dya

We can use thisinfo to find the unknowns

feKKaeK

att

2102

1

0110 a

fKfKa

0

120

1 0a

aeKa

at

t

eK

dt

dy 2

0,)( 21 teKKty

t

21)0( KKy

Use the initial condition to getone more equation

12 )0( KyK

SHORTCUT: WRITE DIFFERENTIAL EQ.IN NORMALIZED FORM WITH COEFFICIENTOF VARIABLE = 1.

00

101 a

fy

dt

dy

a

afya

dt

dya

1K

0),( tti FIND

0t FORKVL USE MODEL.

Rv

Lv)(ti

KVL

)()( tdt

diLtRivvV LRS

0)0()0()0(

0)0(0

iii

it

inductor

CONDITIONINITIAL

STEP 1R

Vtit

dt

di

R

L S )()( R

L

STEP 2 STEADY STATER

VKi S 1)(

STEP 3 INITIAL CONDITION

21)0( KKi

RLt

S eR

Vti 1)( :ANS

EXAMPLE

)0();(

0,)(

211

21

xKKxK

teKKtxt

1 2( ) , 0t

i t K K e t

0),( ttvO FIND

C

1R

2R

KCL USE 0.t FORMODEL

0)()(0)( 2121

cCCC vtdt

dvCRR

RR

vt

dt

dvC

sFCRR 6.0)10100)(106()( 6321 STEP 1

)(3

1)(

42

2)( tvtvtv CCO

STEP 2 0,)( 21 teKKtv

t

C 01 K

INITIAL CONDITIONS. CIRCUIT IN STEADY STATE t<0

)0(Cv V)12(9

6

][88)0( 221 VKKKvC STEP 3

0],[8)( 6.0

tVetvt

C

0],[3

8)( 6.0

tVetv

t

O

)(tvc DETERMINE

)0();(

0,)(

211

21

CC

t

C

vKKvK

teKKtv

USING THEVENIN TO OBTAIN MODELS

Obtain the voltage across the capacitor or the current through the inductor

Circuitw ith

resistancesand

sources

I nductororCapacitor

a

b

R epresentation of an arbitrarycircuit w ith one storage elem ent

Thevenin VTH

R TH

I nductororCapacitor

a

b

VTH

R TH a

b

C

+

vc

_

Case 1.1Voltage across capacitor

VTH

R TH a

b

L

iL

Case 1.2Current through inductor

KCL@ node a

ciRi0 Rc ii

dt

dvCi C

c

TH

THCR R

vvi

0

TH

THCC

R

vv

dt

dvC

THCC

TH vvdt

dvCR

Use KVL

Rv

Lv

THLR vvv

LTHR iRv

dt

diLv L

L

THLTHL viRdt

diL

TH

THL

L

TH R

vi

dt

di

R

LSCi