Ch4_First Order Circuits

51
Chapter 4: First Order Circuits BEE1133 : Circuit Analysis I

Transcript of Ch4_First Order Circuits

Page 1: Ch4_First Order Circuits

Chapter 4: First Order Circuits

BEE1133 : Circuit Analysis I

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First Order Circuits : Syllabus

4.1 Introduction to energy storage elements4.2 The source-free RC circuit4.3 The source-free RL circuit4.4 Unit step function4.5 Step response of an RC circuit4.6 Step response of an RL circuit

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First Order Circuits:Lesson Outcomes

• Understand the characteristics of capacitor and inductor

• Differentiate the behaviour of first-order circuit with dc circuit

• Demonstrate the behaviour of natural responses of source-free RC and RL circuits

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Energy Storage Elements :• Capacitor and Inductor

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Energy Storage Elements: Capacitor• A capacitor is a passive element that stores energy

in its electric field.• It consists of two conducting plates separated by

an insulator.• Capacitance is the ratio of the charge on one plate

of a capacitor to the voltage difference between the two plates

• Measured in farads (F)• A capacitor is an open circuit to dc.• The voltage on a capacitor cannot change abruptly.

v

qC

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Energy Storage Elements: Capacitor• The capacitance value depends on the physical

dimensions of the capacitor:

• Current voltage relationship of the capacitor

• Energy stored in the capacitor

d

AC

dt

dvCic

t

t

c tvidtC

v0

)(1

0

2

2

1Cvwc

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Series and Parallel Capacitors• Series connection of capacitors

• Parallel connection of capacitors

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Energy Storage Elements: Inductor• An inductor is a passive element that stores energy

in its magnetic field.• It consists of a coil of conducting wire.• Inductance is the property whereby an inductor

exhibits opposition to the change of current flowing through it.

• Measured in henrys (H)• An inductor acts like a short circuit to dc.• The current through an inductor cannot change instantaneously.

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Energy Storage Elements: Inductor• The inductance value depends on the physical

dimensions and construction of the inductor:

• Current voltage relationship of the inductor

• Energy stored in the inductor

AN

L2

dt

diLvL

t

t

L tidttvL

i0

)()(1

0

2

2

1LiwL

where is the permeability of the core

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Series and Parallel Inductors• Series connection of inductors

• Parallel connection of inductors

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Energy Storage Elements: Summary

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Series and Parallel Capacitor/InductorsExercise 1Find equivalent capacitance, Ceq for circuit below.

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Series and Parallel Capacitor/InductorsExercise 2Find equivalent inductance, Leq for circuit below.

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Source free RC circuits :• Natural response

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Source free RC circuits• Source free RC circuit occurs when its DC source is

suddenly disconnected.• The energy already stored in the capacitor is

released to the resistors.

A source free RC circuit

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• A first-order circuit is characterized by a first-order differential equation.

• Apply Kirchhoff’s voltage laws to circuit above results in algebraic equations.

• Apply the laws to RC circuits to produce differential equations.

Capacitor law Ohm law

0 R

v

dt

dvC0 RC ii

By KCL

Source free RC circuits

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Source free RC circuits• Voltage response of the RC circuit is solved below:

• This is a 1st order differential equation (only 1st derivative of v is involved:

• Integrating both sides

• Taking power of e (V0 is initial voltage at t =0)

0R

v

dt

dvC 0

RC

v

dt

dvor

dtRCv

dv 1

ARC

tv lnln

RC

t

A

vlnbecomes

RCtAetv )( orRCteVtv 0)(

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• The natural response of a circuit refers to the behavior (i.e voltages & currents) of the circuit itself, with no external sources of excitation.

• The time constant of a circuit is the time required for the response to decay by a factor of 1/e or 36.8% of its initial value.

• v decays faster for small t and slower for large t.

CRTime constantDecays more slowly

Decays faster

Source free RC circuits

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• The key to working with a source-free RC circuit is finding:

1. The initial voltage v(0) = V0 across the capacitor.

2. The time constant = RC.

/0)( teVtv CRwhere

Source free RC circuits

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Source free RC circuitsExample 1Refer to RC circuit below. Let vc(0) = 30 V. For t 0,

determine vc , vx , io , and wc

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Source free RC circuitsSolution for Example 1• Since the voltage across capacitor cannot change

instantly, the voltage across capacitor at t = 0- is the same at t = 0:

• Solve for the time constant

• Find Req to solve for

VVvc 30)0( 0

CReq

1286//12eqR

s43

112

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Source free RC circuitsSolution for Example 1 (cont…)• Now solve for vc and wc

• Now solve for vx and io

tttcc eeevtv 25.04 3030)0()(

JCvw cc 150303

1

2

1)0(

2

1)0( 22

tttxx eeevtv 25.025.0 10

84

430)0()(

ttto eee

vti 25.025.08 5.2

8

20

8)(

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Source free RC circuitsExample 2If the switch opens at t = 0, find v(t) and wc(0) for t 0

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Source free RC circuitsSolution for Example 2• Voltage across capacitor at t = 0- is the same at t = 0.

To calculate v(0), capacitor is open (due to dc source)

• Solve for the time constant

• Find Req to solve for

Vvc 82436

324

)4//12(6

4//12)0(

CReq

34//12eqR

s5.06

13

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Source free RC circuitsSolution for Example 2 (cont…)• Now solve for vc and wc

tttcc eeevtv 25.0 88)0()(

JCvw cc 33.586

1

2

1)0(

2

1)0( 22

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Source free RC circuitsExercise 3If the switch opens at t = 0, find :i) , vc(t), wc(0)and iR(t) for t 0

ii) Value of vc at t = 3 sec.

t = 0

+VC

_

iR10

2

5

7 +_20 V

12

1 F

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Source free RC circuitsExercise 4If the switch opens at t = 0, find :i) , vc(t), wc(0)and iR(t) for t 0

ii) Value of vc at t = 4 sec.

t = 0

+VC

_

iR

+_15 V

5

10 10

5

F5

1

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Source free RL circuits :• Natural Response

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• A first-order RL circuit consists of a inductor L (or its equivalent) and a resistor (or its equivalent)

0 RL vv 0 iRdt

diL

Inductors law Ohms law

Source free RL circuits

By KVL

• Apply Kirchhoff’s voltage laws to circuit above.• Apply the laws to RL circuits to produce

differential equations.

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Source free RL circuits• Current response of the RL circuit is solved below:

• This is a 1st order differential equation (only 1st derivative of i is involved:

• After Integrating both sides

• Taking power of e (V0 is initial voltage at t =0)

or

L

Rt

I

ti

0

)(lnor

LRteIti 0)(

0 iRdt

diL 0 i

L

R

dt

di

dtL

R

i

di

tti

I

dtL

R

i

di

0

)(

0

0 Iln i(t)ln 0 L

Rt

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• The time constant of a circuit is the time required for the response to decay by a factor of 1/e or 36.8% of its initial value.

• i(t) decays faster for small t and slower for large t.

/0)( teIti

R

L

A general form representing a RL

where

Source free RL circuits

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/0)( teIti

R

L

A RL source-free circuit

where /0)( teVtv RC

A RC source-free circuit

where

Comparison between a RL and RC circuit

Source free RL circuits

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• The key to working with a source-free RL circuit is finding:

1. The initial voltage i(0) = I0 through the inductor.

2. The time constant = L/R.

/0)( teIti

R

Lwhere

Source free RL circuits

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Source free RL circuitsExample 3If the switch opens at t = 0, find iL(t) and wL(0) for t 0

25 V +_ 6 6

2 3 H

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Source free RL circuitsSolution for Example 3• Current through inductor at t = 0- is the same at t = 0.

To calculate i(0), inductor is open (due to dc source)

• Solve for the time constant

• Find Req to solve for

AiL 5.26

125

32

3

6

125

)6//6(2

6//6)0(

eqRL

1266eqR

s25.012/3

2

6 6 6 6

3 H

25 V

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Source free RL circuitsSolution for Example 3 (cont…)• Now solve for iL and wL

tttLL eeeiti 425.0 5.25.2)0()(

JLiw cL 375.95.232

1)0(

2

1)0( 22

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Source free RL circuitsExercise 5If the switch opens at t = 0, find :i) , iL(t), wL(0)and iR(t) for t 0

ii) Value of iL at t = 5 sec.

t = 0

15 A 5 10

10

5

iL(t) 5 H

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Source free RL circuitsExercise 6If the switch opens at t = 0, find :i) , iL(t), wL(0)and iR(t) for t 0

ii) Value of iL at t = 5 sec.

t = 0iL(t)

15 A10 10 5 5 H

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First Order Circuits:Lesson Outcomes

• Understand the concept of singularity functions and relate with application in electric circuits

• Formulate the step response of an RC and RL circuits

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Unit step function

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Unit Step Function• Unit step function is one type of singularity function,

useful in circuit analysis.• It can be used as a good approximation to the

switching signals.• Three most widely used singularity functions are unit

step, unit impulse and unit ramp functions.

0

0

,1

,0)(

t

ttu

• The unit step function u(t) is 0 for negative values of t and 1 for positive values of t.

• In mathematical terms:

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Unit Step Function• Below are the images for delayed unit step

function (a) and advanced (b) by t0

0

00 ,1

,0)(

tt

ttttu

0

00 ,1

,0)(

tt

ttttu

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Step Response of an RC circuit

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Step Response of an RC circuit• The step response of a circuit is its behavior when

the excitation is the step function (a voltage or a current source).

• In other words, it is the response of the circuit due to a sudden application of dc voltage or current source.

• For finding the step response of an RC circuit, we break the v(t) into natural response and forced response:

Complete response = natural response + forced responsestored energy independent source

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Step Response of an RC circuit• In mathematical form:

• Where

• A forced response is produced by the circuit when an external force is applied.

• A natural response refers to the behavior of the circuit itself with no external source of excitation.

• The natural response eventually dies out along with the transient component of the forced response.

fn vvv t

n eVv 0

)1( tsf eVv

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Step Response of an RC circuit• Expending the equation for complete response:

• If the switch changes position at time, t = t0 (with time delay), the response becomes:

fn vvv )1(0

ts

t eVeVv t

sst eVVeV 0

ts

ts eVeVV 0

ttss evvveVVV ))()0(()()( 0

)(00

0))()(()()( ttevtVvtv

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Step Response of an RC circuitExample 4Refer to RC circuit below. Assume the switch has

been open for a long time and is closed at t =0. Calculate v(t) at t = 0.5 s

+_

_+

2 6

5 V10 V 1/3 F+v_

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Step Response of an RC circuitSolution for Example 4For t < 0, switch is closed. Capacitor act as o/c. v (0-) = 10 V = v(0).

At t = 0, switch is closed. = ReqC = (2//6)*1/3 = 3/2*1/3 = 0.5 s

v() =

+__+

2 6

5 V10 V 1/3 F+v_

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Step Response of an RL circuit

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Conclusion

• Nodal

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The End