2. Analogue Theory and Circuit Analysis 2.1 Steady-State (DC) Circuits 2.2 Time-Dependent Circuits DeSiaMorePowered by DeSiaMore1.

2. Analogue Theory and Circuit Analysis

2.1 Steady-State (DC) Circuits2.2 Time-Dependent Circuits

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Electrical systems have two main objectives:

• To gather, store, process, transport, and present information

• To distribute and convert energy between various forms


Electrical Engineering Subdivisions

• Communication systems

• Computer systems• Control systems• Electromagnetics

• Electronics• Power systems• Signal processing


Electrical Current

Electrical current is the time rate of flow of electrical charge through a conductor or circuit element. The units are amperes (A), which are equivalent to coulombs per second (C/s).


Electrical Current

t

t

tqdttitq

dt

tdqti

0

)()()(

)()(

0


Direct Current Alternating Current

When a current is constant with time, we say that we have direct current, abbreviated as dc. On the other hand, a current that varies with time, reversing direction periodically, is called alternating current, abbreviated as ac.


.


Voltages

The voltage associated with a circuit element is the energy transferred per unit of charge that flows through the element. The units of voltage are volts (V), which are equivalent to joules per coulomb (J/C).


Transients

The time-varying currents and voltages resulting from the sudden application of sources, usually due to switching, are called transients. By writing circuit equations, we obtain integrodifferential equations.


DC STEADY STATE

The steps in determining the forced response for RLC circuits with dc sources are:

1. Replace capacitances with open circuits.

2. Replace inductances with short circuits.

3. Solve the remaining circuit.


CAPACITANCE

cc

dvi C

dt

cv

ci

C [Farads]

CSS

dIn DC Steady State; 0

dti 0 Open Circuit


CAPACITANCE

Cvq

dt

dvCi

0

0

tqdttitqt

t

0

0

1tvdtti

Ctv

t

t


INDUCTANCE

LL

div L

dt

L [Henries]Lv

Li

LSS

dIn DC Steady State; 0

dtv 0 Short Circuit


INDUCTANCE

dt

diLtv

0

0

1tidttv

Lti

t

t

tLitw 2

2

1


SWITCHED CIRCUITS

• Circuits that Contain Switches

• Switches Open or Close at t = t0

• to = Switching Time

• Often choose to = 0

• Want to Find i’s and v’s in Circuit Before and After Switching Occurs

• i(to-), v(t0

-); i(to+), v(t0

+)

• Initial Conditions of CircuitDeSiaMore Powered by DeSiaMore 15

INITIAL CONDITIONS

• C’s and L’s Store Electrical Energy

• vC Cannot Change Instantaneously

• iL Cannot Change Instantaneously

• In DC Steady State; C => Open Circuit

• In DC Steady State; L => Short Circuit

• Use to Find i(to-), v(t0

-); i(to+), v(t0

+)

• Let’s do an ExampleDeSiaMore Powered by DeSiaMore 16

EXAMPLE

12 V2 4

2 1 F

Cv

1iSwitch Opens at t 0

Assume Switch has been Closed

for a long time before t 0

2v

1 v

2i

3 v 3i

+

Find Initial Conditions

i's and v's at t 0 and t 0

Ci


EXAMPLE

12 V

2

2 4

At t 0 :3i

3v (0 ) 0

3 v 1i

2i

1 2

12i (0 ) i (0 ) 3 A

2 2

1 v

1 2v (0 ) v (0 ) 3x2 6 V

2v

Cv

C 2 3v (0 ) v (0 ) v (0 ) 6 V

C

Open Ckt

Switch ClosedCi

Ci (0 ) 0 3i (0 ) 0

DC Steady State


EXAMPLE

12 V2

2

4

Cv

At t 0 :

C Cv (0 ) v (0 ) 6 V

1 F

1 1i (0 ) 0 v (0 )

2i

3 Ci i

2 3 C

6i (0 ) i (0 ) i (0 ) 1 A

4 2

2v

3 v

2v (0 ) 2x1 2 V 3v (0 ) 4x( 1) 4 V

1 v 1i

Switch Open


EXAMPLEInitial Conditions

t 0

1

2

3

C

1

2

3

C

i 3 A

i 3 A

i 0 A

i 0 A

v 6 V

v 6 V

v 0 V

v 6 V

t 0

1

2

3

C

1

2

3

C

i 0 A

i 1 A

i 1 A

i 1 A

v 0 V

v 2 V

v 4 V

v 6 V


1ST ORDER SWITCHED DC CIRCUITS

st

st

Will Look at 1 Order Circuits (Circuits with

1 C or 1 L) with Switched DC Inputs Tomorrow

Will Use Initial Conditions to Help Us Solve the

1 Order Differential Equation Relating the

st

Output to the Input

Today We Will Look at a 1 Order Circuit using

PSpice


ACTIVITY 13-1

100 V

R

20 nF Cv


ACTIVITY 13-1

• Charge a 20 nF Capacitor to 100 V thru a Variable Resistor, Rvar:

• Let’s Use a Switch that Closes at t = 0

• Rvar = 250k, 500k, 1 M

• Circuit File Has Been Run:• C:/Files/Desktop/CE-Studio/Circuits/act_5-

2.dat

• But Let’s Practice Using Schematics and Take a Quick LookDeSiaMore Powered by DeSiaMore 23

ACTIVITY 13-1

Circuit File

v 1 0 dc 100

R 1 2 {R}

C 2 0 20n ic=0

.param R=250k

.step param R list 250k 500k 1meg

.tran .1 .1 uic

.probe

.endDeSiaMore Powered by DeSiaMore 24

ACTIVITY 13-1

CPrint Graphs of v vs. time

Fill in Table for Activity 13-1

Hand In for Grading


Transient Behaviour Introduction Charging Capacitors and Energising Inductors Discharging Capacitors and De-energising Inductors Response of First-Order Systems Second-Order Systems Higher-Order Systems


Introduction So far we have looked at the behaviour of systems

in response to:– fixed DC signals– constant AC signals

We now turn our attention to the operation of circuits before they reach steady-state conditions– this is referred to as the transient response

We will begin by looking at simple RC and RL circuits


Charging Capacitors and Energising Inductors

• Capacitor Charging Consider the circuit shown here

– Applying Kirchhoff’s voltage law

– Now, in a capacitor

– which substituting gives

VviR

tv

Cidd

Vvtv

CR dd


The above is a first-order differential equation with constant coefficients

Assuming VC = 0 at t = 0, this can be solved to give

Since i = Cdv/dt this gives (assuming VC = 0 at t = 0)

– where I = V/R

)e1()e1( t-

CRt

-VVv

t-

CRt

-IIi ee


Thus both the voltage and current have an exponential form


• Inductor energising A similar analysis of this circuit gives

where I = V/R–

t-

LRt-

VVv ee

)e1()e1( t-

LRt-

IIi


Thus, again, both the voltage and current have an exponential form


Discharging Capacitors and De-energising Inductors

• Capacitor discharging Consider this circuit for

discharging a capacitor– At t = 0, VC = V

– From Kirchhoff’s voltage law

– giving

0viR

0dd vtv

CR


Solving this as before gives

– where I = V/R

–

t-

CRt

-VVv ee

t-

CRt

-IIi ee


In this case, both the voltage and the current take the form of decaying exponentials


• Inductor de-energising A similar analysis of this

circuit gives

– where I = V/R– see Section 18.3.1

for this analysis

t-

LRt-

VVv ee

t-

LRt-

IIi ee


And once again, both the voltage and the current take the form of decaying exponentials


A comparison of the four circuits


Response of First-Order Systems Initial and final value formulae

– increasing or decreasing exponential waveforms (for either voltage or current) are given by:

– where Vi and Ii are the initial values of the voltage and current– where Vf and If are the final values of the voltage and current– the first term in each case is the steady-state response– the second term represents the transient response– the combination gives the total response of the arrangement


• The input voltage to the following CR network undergoes a step change from 5 V to 10 V at time t = 0. Derive an expression for the resulting output voltage.

•


• Here the initial value is 5 V and the final value is 10 V. The time constant of the circuit equals CR = 10 103 20 10-6 = 0.2s. Therefore, from above, for t 0

•volts e510

e)105(10

e)(

2.0/

2.0/

/

t

t

tfif VVVv


• The nature of exponential curves


• Response of first-ordersystems to a squarewaveform


• Response of first-ordersystems to a squarewaveform of differentfrequencies


Key Points The charging or discharging of a capacitor, and the energising

and de-energising of an inductor, are each associated with exponential voltage and current waveforms

Circuits that contain resistance, and either capacitance or inductance, are termed first-order systems

The increasing or decreasing exponential waveforms of first-order systems can be described by the initial and final value formulae

Circuits that contain both capacitance and inductance are usually second-order systems. These are characterised by their undamped natural frequency and their damping factor


2. Analogue Theory and Circuit Analysis 2.1 Steady-State (DC) Circuits 2.2 Time-Dependent Circuits DeSiaMorePowered by DeSiaMore1.

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