Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege

[email protected] • ENGR-43_Lec-04a_1st_Order_Ckts.pptx1

Bruce Mayer, PE Engineering-43: Engineering Circuit Analysis

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

RC & RL 1st Order

Ckts



C&L Summary



Introduction Transient Circuits

In Circuits Which Contain Inductors & Capacitors, Currents & Voltages CanNOT Change Instantaneously

Even The Application, Or Removal, Of Constant Sources Creates Transient (Time-Dependent) Behavior



1st & 2nd Order Circuits

FIRST ORDER CIRCUITS• Circuits That Contain

ONE Energy Storing Element– Either a Capacitor or an Inductor

SECOND ORDER CIRCUITS• Circuits With TWO Energy Storing

Elements in ANY Combination



Circuits with L’s and/or C’s Conventional DC Analysis Using

Mathematical Models Requires The Determination of (a Set of) Equations That Represent the Circuit Response

Example; In Node Or Loop Analysis Of Resistive Circuits One Represents The Circuit By A Set Of Algebraic Equations

The Ckt The DC Math Model

Analysis

iv

G



Ckt w/ L’s & C’s cont.

When The Circuit Includes Inductors Or Capacitors The Models Become Linear Ordinary Differential Equations (ODEs)

Thus Need ODE Tools In Order To Analyze Circuits With Energy Storing Elements• Recall ODEs from

ENGR25• See Math-4 for

More Info on ODEs



First Order Circuit Analysis

A Method Based On Thévenin Will Be Developed To Derive Mathematical Models For Any Arbitrary Linear Circuit With One Energy Storing Element

This General Approach Can Be Simplified In Some Special Cases When The Form Of The Solution Can Be Known BeforeHand• Straight-Forward ParaMetric Solution



Basic Concept

Inductors & Capacitors Can Store Energy

Under Certain Conditions This Energy Can Be Released

RATE of Energy Storage/Release Depends on the parameters Of The Circuit Connected To The Terminals Of The Energy Storing Element

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Example: Flash Circuit

The Battery, VS,Charges the CapTo Prepare for aFlash

Moving the Switch to the Right “Triggers” The Flash• i.e., The Cap

Releases its Stored Energy to the Lamp

Say“Cheese”



Flash Ckt Transient Response The Voltage Across the Flash-Ckt Storage

Cap as a Function of TIME

Note That the Discharge Time (the Flash) is Much Less Than the Charge-Time



General Form of the Response Including the initial

conditions the model equation for the capacitor-voltage or the inductor-current will be shown to be of the form

xp(t) ANY Solution to the General ODE• Called the “Particular”

Solution

xc(t) The Solution to the General Eqn with f(t) =0• Called the “Complementary

Solution” or the “Natural” (unforced) Response

• i.e., xc is the Soln to the “Homogenous” Eqn

This is the General Eqn Now By Linear

Differential Eqn Theorem (SuperPosition) Let

tftxdt

tdx

0 txdt

tdx



1st Order Response Eqns Given xp and xc

the Total Solution to the ODE

txtxtx cp Consider the Case

Where the Forcing Function is a Constant• f(t) = A

Now Solve the ODE in Two Parts

For the Particular Soln, Notice that a CONSTANT Fits the Eqn:

0

txdt

tdx

Atxdt

tdx

cc

pp

0

so and

1

dt

tdx

Ktx

p

p



1st Order Response Eqns cont Sub Into the General

(Particular) Eqn xp and dxp/dt

AK

AK

1

1

or

0

Next, Divide the Homogeneous (RHS=0) Eqn by ∙xc(t) to yield

Next Separate the Variables & Integrate

1

tx

dttdx

c

c

tdtdxtx c

c

11

Recognize LHS as a Natural Log; so

const where

ln

c

ct

txc

Next Take “e” to The Power of the LHS & RHS



1st Order Response Eqns cont Then

tc

tcctc

eKtx

eeetx

2

Note that Units of TIME CONSTANT, , are Sec

Thus the Solution for a Constant Forcing Fcn

For This Solution Examine Extreme Cases• t =0• t → ∞

The Latter Case (K1) is Called the Steady-State Response• All Time-Dependent

Behavior has dissipated

/

21t

cp

eKKtx

txtxtx

121

210

KeKKtx

KKx



Effect of the Time ConstantTangent reaches x-axis in one

time constant

Decreases 63.2% after One Time Constant time

Drops to 1.8% after 4 Time Constants 0183.04 e



Large vs Small Time Constants Larger Time Constants Result in Longer

Decay Times• The Circuit has a Sluggish Response

Quick to Steady-State

Slow to Steady-State



1st Order Ckt Solution Plan

1. Use some form of KVL/KCL, and to develop an ODE

2. Isolate the “0th” order term which reveals the • Time Constant• Forcing Function

3. “EyeBall” the ODE to “Guesstimate the Particular Solution, or




4. CHECK that the particular Solution Satisfies the ODE

5. Separate the Variables in the Homogeneous Equation and Integrate to obtain the Complementary Solution or

6. A the particular and complementary solutions to obtain the total Solution;




The Solution Should include a Negative Exponential with an UnKnown PreFactor

7. Use the initial condition in the total solution to determine the Prefactor which completes the total solution

8. Check the total Solution for extreme cases:

• t = 0+ • t → ∞



Example 1st Order Ckt Soln

For the Ckt Below Find for



















Time Constant Example Charging a Cap

Use KCL at node-a

Now let• vC(t = 0 sec) = 0 V

• vS(t)= VS (a const)

ReArrange the KCL Eqn For the Homogenous Case where Vs = 0

Thus the Time Constant

vS

R S a

b

C

+

vc

_

dt

dvC C

S

SC

R

vv

S

S

S

CC

S

SCC

R

v

R

v

dt

dvC

R

vv

dt

dvC

0

SC

C

CRv

dtdv 1

SCR

t

cc

c eKtxtx

dttdx 2

1



Time Constant Example cont Charging a Cap

The Solution Can be shown to be

“Fully” Charged Criteria• vC >0.99VS OR

vS

R S a

b

C

+

vc

_

t

SSC eVVtv

)(

6.401.0ln

01.0

t

ore t

CRS tSC eVtv 1)(Or



Differential Eqn Approach Conditions for Using This Technique

• Circuit Contains ONE Energy Storing Device• The Circuit Has Only CONSTANT,

INDEPENDENT Sources• The Differential Equation For The Variable Of

Interest is SIMPLE To Obtain – Normally by Using Basic Analysis Tools; e.g., KCL, KVL,

Thevenin, Norton, etc.

• The INITIAL CONDITION For The Differential Equation is Known, Or Can Be Obtained Using STEADY STATE Analysis Prior to Switching– Based On: Cap is OPEN, Ind is SHORT



Example Given the RC Ckt At

Right with• Initial Condition (IC):

– v(0−) = VS/2

Find v(t) for t>0

Looks Like a Single E-Storage Ckt w/ a Constant Forcing Fcn• Assume Solution of Form

Model t>0 using KCL at v(t) after switch is made

)0();(

0,)(

211

21

xKKxK

teKKtxt

0

tdt

dvC

R

Vtv S

Find Time Constant; Put Eqn into Std Form• Multiply ODE by R

sVtvtdt

dvRC



Example cont Compare Std-Form with

Model• Const Force Fcn Model

Note: the SS condition is Often Called the “Final” Condition (FC)

In This Case the FC

Now Use IC to Find K2

1Kxdt

dx

• In This Case

sVtvtdt

dvRC )()(

Next Check Steady-State (SS) Condition• In SS the Time Derivative

goes to ZERO

s

s

V K

Vtvt

1 :confirms

)(



Example cont.2 At t = 0+

• The Model Solution

The Total/General Soln

Recall The IC: v(0−) = VS/2 = v(0+) for a Cap• Then

12

021

)0(

)0(

KvK

eKKv

2

2

)0(

2

2

12

S

SS

VK

VVK

KvK

RC

t

S eVtv 5.01)(

Check sS VeVv 5.05.01)0( 0

sS VeVv 5.01)(

or0,21

teKKtvt



Inductor Exmpl Find i(t) Given

• i(0−) = 0

Recognize Single E-Storage Ckt w/ a Constant Forcing Fcn• Assume Solution of Form

In This Case x→i

)0();(

0,)(

211

21

xKKxK

teKKtxt

To Find the ODE Use KVL for Single-Loop ckt

0)0()0( ii

)(ti

0;)( 21

teKKtit

Rv

Lv)(ti

KVL

)()( tdt

diLtRivvV LRS

Now Consider IC• By Physics, The Current

Thru an Inductor Can NOT Change instantaneously



Inductor cont Casting ODE in

Standard form

Recognize Time Const

Next Using IC (t = 0+)

)(ti

RL

Thus the ODE Solution

R

Vtit

dt

di

R

L S )()(

Also Note FC

RVtit s )(

RVK

RVK

KiK

S

S

2

2

12

0

)0(

tL

Rs

t

eR

VeKKti 1)( 21

RVK s1

Thus



Solution Process Summary

1. ReWrite ODE in Standard Form• Yields The Time-Constant,

2. Analyze The Steady-State Behavior• Finds The Final Condition Constant, K1

3. Use the Initial Condition• Gives The Exponential PreFactor, K2

4. Check: Is The Solution Consistent With the Extreme Cases

• t = 0+• t →



Solutions for f(t) ≠ Constant1. Use KVL or KCL to Write the ODE

2. Perform Math Operations to Obtain a CoEfficient of “1” for the “Zeroth” order Term. This yields an Eqn of the form

3. Find a Particular Solution, xp(t) to the FULL ODE above• This depends on f(t), and may require

“Educated Guessing”

) (yields tfxdt

dx



Solutions for f(t) ≠ Constant

4. Find the total solution by adding the COMPLEMENTARY Solution, xc(t) to previously determined xp(t). xc(t) takes the form:• The Total Solution

at this Point →

5. Use the IC at t=0+ to find K; e.g.; IC = 7

• Where M is just a NUMBER

tc Ketx

txKetx pt

MKxKe p 707 0



Capacitor Example

At t=0+ Apply KCL Assume a Solution of the Form for vc

C

1R

2R

)0();(

0,)(

211

21

CC

t

C

vKKvK

teKKtv

0)()(

0)(

21

21

cC

CC

vtdt

dvCRR

RR

vt

dt

dvC

For Ckt Below Find vo(t) for t>0 (note f(t) = const; 12V)



Cap Exmp cont Step1: By Inspection of

the ReGrouped KCL Eqn Recognize

Now Examine the Reln Between vo and vC

• a V-Divider

s

mFk

CRR

6.0

1.06

)( 21

)(3

1)(

42

2)( tvtvtv CCO

Step-2: Consider The Steady-State• In This Case After the

Switch Opens The Energy Stored in the Cap Will be Dissipated as HEAT by the Resistors

With xp = K1 = SSsoln

0)()( tvtv Co

0)(1 CvK



Cap Exmp cont Now The IC If the Switch is Closed for a Long Time before t =0, a

STEADY-STATE Condition Exists for NEGATIVE Times

vo(0−) by V-Divider Recall Reln Between vo and vC for t ≥ 0

Vv

VVv

O

O

3

8)0(

129

212

423

2)0(

VVvv

tvtv

CC

OC

8129

6)0()0(

)(3)(

Recall: Cap is OPEN to DC



Cap Exmp cont Step-3: Apply The IC

Now have All the Parameters needed To Write The Solution

)0,80)(

0,)(

6.0

21

tVetv

teKKtv

s

t

C

t

C

VvK

vK

vKK

C

C

C

8)0(

0)( and

)0(

2

1

21

Note: For f(t)=Const, puttingthe ODE in Std Form yields and K1 by Inspection:

0)()( 21 cC vt

dt

dvCRR

1K

Recall vo = (1/3)vC

0,V3

8)( 6.0

tetv s

t

o



Thevenin/Norton Techniques Obtain The Thevenin Voltage Across The Capacitor,

Or The Norton Current Through The Inductor

With This approach can Analyze a SINGLE-LOOP, or SINGLE-NODE Ckt to Find• Time Constant using RTH

• Steady-State Final Condition using vTH (if vTH a constant)

Circuitw ith

resistancesand

sources

I nductororCapacitor

a

b

R epresentation of an arbitrarycircuit w ith one storage elem ent

ThéveninVT H

R T H

I nductororCapacitor

a

b



Thevenin Models for ODE

KVL for Single Loop

VT H

R T H a

b

C

+

vc

_

Case 1.1Voltage across capacitor

ciRi

0 Rc iidt

dvCi C

c

TH

THCR R

vvi

0

TH

THCC

R

vv

dt

dvC

THCC

TH vvdt

dvCR

VT H

R T H a

b

L

iL

Case 1.2Current through inductor

Rv

Lv

THLR vvv LTHR iRv

dt

diLv L

L THLTHL viR

dt

diL

NTH

THL

L

TH

IR

vi

dt

di

R

L

KCL at node a



Find ODE Soln By Thevenin

Break Out the Energy Storage Device (C or L) as the “Load” for a Driving Circuit

Analyze the Driving Ckt to Arrive at it’s Thévenin (or Norton) Equivalent

ReAttach The C or L Load Use KCL or KVL to arrive at ODE Put The ODE in Standard Form



Find ODE Soln By Thevenin

Recognize the Solution Parameters For Capacitor

• = RTHC

• K1 (Final Condition) = vTH = vOC = xp

For Inductor• = L/RTH

• K1 (Final Condition) = vTH/RTH = iSC = iN = xp

In Both Cases Use the vC(0−) or iL(0−)Initial Condition to Find Parameter K2



Inductor Example

The Variable Of Interest Is The Inductor Current

The Thévenin model

Since this Ckt has a CONSTANT Forcing Function of 0V (the 24V source is switched OUT at t = 0), Then The Solution Is Of The Form

66

6

6

H3

V24

)(tiO

0t

0t;(t)iFind O

TH

THO

O

TH R

vi

dt

di

R

L

0;)( 21

teKKtit

O

Next Construct the Thévenin Equivalent for the Inductor “Driving Circuit”



Inductor Example cont The f(t)=const ODE in

Standard Form

s

H

R

L

R

v

TH

TH

TH

3.010

3

106666

0

The Solution Substituted into the ODE at t > 0

66

6

6V24

0t

Thevenin for t>0at inductor terminals

a

b

0;03.0 tidt

dis O

O

03.0

3.0 3.021

3.02

tt

eKKeK

0;)( 3.02

teKti

t

O

Combining the K2 terms shows shows K1 = 0, thus

From This Ckt Observe



Inductor Example cont.2

Now Find K2

Assume Switch closed for a Long Time Before t = 0• Inductor is SHORT to DC

Analyzing Ckt with 3H Ind as Short Reveals

• The Reader Should Verify the Above

0;3

16

36

24

2

1

6

24

tA

VViO

0;33.5)( 3.0

teAti s

t

O

Then the Entire Solution

66

6

6

H3

V24

)(tiO

0t

0t;(t)iFind O



Untangle to find iO(0−)

Untangle

Analyz

e

Be Faithful to Nodes



First Order Inductor Ciruit Transient Solution

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

-0.2 0.0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6 1.8

Time (s)

Cu

rren

t (A

)

i(t) (A)

file = Engr44_Lec_06-1_Last_example_Fall03..xls

0;333.5)( 3.0

teAti s

t

O

s

A

s

A

s

A

dt

di

s

i

dt

di

s

i

dt

di

s

i

dt

dii

dt

dis

t

O

O

t

OOO

OOO

O

78.1733

1016

103

316

103

0

103

3.003.0

0

0



WhiteBoard Work Let’s Work This

Problem

t= 0

24V 2

4 4

86 VC ( t)

(1 /4)H

+

-

Well, Maybe NEXT time…



WhiteBoard Work



Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 43

AppendixDE

Approach



Differential Eqn Approach cont Math Property

• When all Independent Sources Are CONSTANT, then for ANY variable y(t); i.e., v(t) or i(t), in The Circuit The Solution takes the Form

t

eKKty

21

The Solution Strategy• Use The DIFFERENTIAL EQUATION And The

FINAL & INITIAL Conditions To Find The Parameters K1 and K2



Differential Eqn Approach cont If the ODE for y is

Known to Take This Form

We Can Use This Structure to Find The Unknowns. If:

Then Sub Into ODE

Equating the TRANSIENT (exponential) and CONSTANT Terms Find

01

0

120

1 0

a

AK

a

aeKa

a t

0

01

)0( yy

Ayadt

dya

t

t

eK

dt

dy

teKKty

2

21 0,)(

AKaeKaa

AeKKaeK

a

t

tt

10201

2102

1



Differential Eqn Approach cont Up to Now

Next Use the Initial Condition

So Finally

If we Write the ODE in Proper form We can Determine By Inspection and K1

012

0

aa

t

eKa

Aty

102

021

00

21

0

or

)0(

)0(

KyK

yKK

yeKKy

yy

01

00

0

aa

t

ea

Ay

a

Aty

00

101 a

Ay

dt

dy

a

aAya

dt

dya

1K



Inductor Example

The Model for t>0 → KVL on single-loop ckt

Rewrite In Std Form

L

)(1 ti

Lv

S9

1

Recognize Time Const

0)()(18

21

1 titdt

di

)0();(

0,)(

12111

211

iKKiK

teKKtit

0)(126 11 ti

dt

diL

For The Ckt Shown Find i1(t) for t>0

Assume Solution of the Form



Inductor Example cont Examine Std-Form Eqn

to Find K1

For Initial Conditions Need the Inductor Current for t<0

Again Consider DC (Steady-State) Condition for t<0

The SS Ckt Prior to Switching• Recall An Inductor is a

SHORT to DC

0

0)()(9

1

1

11

K

titdt

di

)0(1 i

SoA

Vi 1

12

12)0(1



Inductor Example cont.1 Now Use Step-3 To

Find K2 from IC• Remember Current Thru

an Inductor Must be Time-Continuous

• Recall that K1 was zero

Construct From the Parameters The ODE Solution

2

2111

][1

)0()0(

KA

KKii

0],[)(

0],[)(9

1

91

1

tAeti

tAetit

t The Answer

0

Documents

Bruce Mayer, PE Licensed Electrical & Mechanical Engineer BMayer@ChabotCollege