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Electrical Sciences- I

Sudheer Hhsudheeer@gmail.com

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• Textbook:Fundamentals of Electrical Engineering, Leonard Bobrow, Oxford University Press 2nd edition 2005

• Reference Books:• Engineering circuit analysis, W.H.Hayt,

J.E. Kemmerly, McGraw Hill company, 6th Edition, 2000.

• Electronic Devices & Circuits, Millman & Halkias McGraw Hill, 2002.

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Reference Books:• Electrical Engineering: Principles and

Applications, Alan R. Hambley, Publisher, 2nd Edition 2003.

• Electrical & Electronic Technology, Hughes, Pearson education, eight edition 2003

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Lecture No. Learning Objective Topics to be covered (Ch./Sec./Text

Book)

1-2

To understand the concept of basic ckt. Elements

Introduction to Basic Circuit theory & Circuit elements 1.1

3-4

To understand the concept of basic electrical laws KVL & KCL 1.2, 1.3

5-6

To understand the concept of basic sources

Independent & Dependent sources 1.4,1.5

7-9

To understand the methods of ckt. Analysis

Mesh & Nodal Analysis, Ideal Operational Amplifier (OP-AMP) Applications 2.1,2.3,2.4

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10-11To understand the network theorems Thevenins & Nortons theorem 2.5

12-14

To understand the concept of basic theorems

Linearity, Superposition, Maximum power transfer theorems 2.5,2.6

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To solve non-series, non parallel circuit connections Star- Delta transformation R1/ 5.6

16-17

Learning characteristics of energy storage elements

Energy storage elements(Inductors & Capacitors) their relationships & their natural responses 3.1, 3.2, 3.3

18-21

To study forced and free response of a circuit

First order & second order System responses 3.4,3.5,3.6

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22-23To study basics of

semiconductorsSemiconductors: intrinsic and

doped; p-n junction 6.1,6.2

24-25

To study operation and characteristics of junction diodes

Junction Diode & its characteristics 6.3,6.4,6.5

26-27 Application of diodes Rectifier circuits & filtersR4/21.1 – 21.6

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Operation and application of Zener diodes

Zener Diode & its characteristics 6.6

29-30To study operation of

transistors Introduction to transistors 7.1

31-32

To study classification and characteristics of transistors

pnp and npn transistors and their characteristics & operation 7.2,7.3

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33-34To study operations of

FETSFETs, their operation &

characteristics 8.1

35-36

To study characteristics of MOSFETS

MOSFETs & its characteristics, CMOS its characteristics (No application of CMOS) 8.2,8.4

37-38To study working of

BJT Biasing the BJT 9.1

39To study A.C Model

of BJTA.C Model of BJT (Low

frequency model) 9.1

40-41Common emitter

characteristicsBJT Amplifier, common emitter

configuration 9.1

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Basic Electrical Quantities

• Basic quantities: current, voltage and power– Current: time rate of change of electric charge

I = dq/dt1 Amp = 1 Coulomb/sec

– Voltage: electromotive force or potential, V1 Volt = 1 Joule/Coulomb = 1 N·m/coulomb

– Power: P = I V1 Watt = 1 Volt·Amp = 1 Joule/sec

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Voltage, V

Voltage is the difference in energy level of a unit charge located at each of two points in a circuit, and therefore, represents the energy required to move the unit charge from one point to the other

Circuit Element(s)

+ –V(t)

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Current, I

• Normally we talk about the movement of positive charges although we know that, in general, in metallic conductors current results from electron motion (conventionally positive flow)

• The sign of the current indicates the direction of flow

• Types of current:– direct current (dc): batteries and some special

generators– alternating current (ac): household current which

varies with time

I(t)

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Default Sign Convention

• Passive sign convention : current should enter the positive voltage terminal

• Consequence for P = I V– Positive (+) Power: element absorbs power– Negative (-) Power: element supplies power

Circuit Element+ –

I

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BASIC ELEMENTS:BASIC ELEMENTS: 1. Voltage Sources1. Voltage Sources 2. Current Sources2. Current Sources 3. Resistors3. Resistors

LAWS:LAWS: 1. Ohm’s Law1. Ohm’s Law 2. 2. Kirchhoffs Voltage Law 3. Kirchhoffs Current Law

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1. Voltage Sources1. Voltage Sources

• A device that produces a voltage or potential difference of V volts across its terminals regardless of what is connected to it.

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Example 1.1The voltage produced by the voltage source shown above is described by v(t) =10e-tV.

Determine the value of the voltage at t=0s,1s,2s.

• At time t=0s =>V(t)=V(0)=10e-0=10 V;Then• t=1 s =>V(t) = V(1) =10e-1= 3.68 V• t=2 s =>V(t) = V(2) =10e-2= 1.35 V Finally,• t=3 s =>V(t) = V(3) =10e-3= 0.498 V

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2.Current Source

• Is a device that when connected to anything, will always move I amperes in the direction indicated by the arrow.

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Equivalent ideal current source:

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Resistors

• A resistor is a circuit element that dissipates electrical energy (usually as heat)

• Real-world devices that are modeled by resistors: incandescent light bulbs, heating elements (stoves, heaters, etc.), long wires

• Resistance is measured in Ohms (Ω)

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3.Resistance Resistance is mathematically, is given by:

R = l / Awhere is the resistivity (-m), l is the length (m), and A is the cross-sectional area (m2) of the conductor.

Note that is temperature dependent. The rate at which the resistance changes with temperature is called the temperature coefficient (). To determine the resistance when T1 changes to T2 , use:

R2 = R1 [1 + ( T2 - T1 )]

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Resistor Colour Code

Colour: Bk, Br, R, O, Y, Gn, Bl, V, Gr, W, Gl , S , No ColourBand 1: 1 2 3 4 5 6 7 8 9 Band 2: 0 1 2 3 4 5 6 7 8 9Band 3: 1 10 102 103 104 105 106 107 .1 .01 Band 4: 5% 10% 20%

Band

1 2 3 4 5

3

33

1

33

5

5533 5

4

2, Significant FiguresMultiplierToleranceReliability

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Models for Basic Electric Components

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Ohm’s Law

• There is a simple linear relationship between voltage, current and resistance.

Georg Ohm

V IR8/7/2012 ES-I 22

4.Ohm’s Law

Ohm’s Law states that current in a resistive circuit is directly proportional to its applied voltage and inversely proportional to its resistance.

In equation form:RV

I

A

E+

RV+

+

I

I (mA)

V

2 4

3

6

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Exercise 1.2For the circuit shown in fig, what value of R will result in v(t)=-2.5V?• Fig • Answer:

• V(t) = -2.5V• i(t) = -25 µA• R= V • I• R= -2.5• -25 x 10-6

•

= 100kΩ

25µA R+

-

i(t)

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Independent Sources

An independent source (voltage or current) may be DC (constant) or time-varying (AC), but does not depend on other voltages or currents in the circuit

+–

VoltageSource

CurrentSource

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Ohm’s Law

v(t) = i(t) R - or - V = I Rp(t) = i2(t) R = v2(t)/R [+ (absorbing)]

v(t)The

Rest of the

Circuit

R

i(t)+

–

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Open Circuit

• What if R = ?

• i(t) = v(t)/R = 0

v(t)The

Rest of the

Circuit

i(t)=0+

–i(t)=0

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Short Circuit

• What if R = 0 ?

• v(t) = R i(t) = 0

The Rest of

the Circuit

v(t)=0

i(t)+

–

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Series

Two elements are in series if the current that flows through one must also flow through the other.

R1 R2

Series

R1 R2

Not Series

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Parallel

Two elements are in parallel if they are connected between (share) the same two (distinct) end nodes.

Parallel Not Parallel

R1

R2

R1

R2

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Kirchhoff’s Laws

• Kirchhoff’s Current Law (KCL)– Algebraic sum of all currents entering a node

is zero– sum of currents entering node is equal to sum

of currents leaving node• Kirchhoff’s Voltage Law (KVL)

– Algebraic sum of voltages around any loop in a circuit is zero

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Kirchoff’s Current Law (KCL)

• The sum of all the currents in a node is equal to zero.

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KCL (Kirchhoff’s Current Law)

The Algebraic sum of currents entering the node is zero:

Analogy: mass flow at pipe junction

i1(t)

i2(t) i4(t)

i5(t)

i3(t)

n

jj ti

1

0)(

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KCL (Kirchhoff’s Current Law)

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Example 1.3Let us find the voltage v in the two node circuit given in Fig in which directions of i1,i2 and i3 and the polarity of v were chosen arbitrarily. the directions of the 2A and 13 A sources are given.)

1Ω 2 Ω 3 Ω2A13A

i3

i2 i1

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Current Division• If we know the current flowing into two parallel

resistors, we can find out how the current will divide up in one step.

• The value of the current through R1 is

i1 = iTOTAL R2 / (R1 + R2)

• The value of the current through R2 is

i2 = iTOTAL R1 / (R1 + R2)• Note that this differs slightly

from the voltage divisionformula for series resistors.

R1 R2

i1 i2

iTOTAL

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Kirchoff’s Voltage Law (KVL)

• The sum of the voltage differences around a circuit is equal to zero.

Gustav Kirchoff

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Kirchhoff’s Voltage Law

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Parallel Circuit

• Elements/branches are said to be parallel when they have only 2 nodes in common. The voltage across all parallel elements in a circuit will be the same.

• Voltage sources of different potentials should never be connected in parallel.

IT

E+

I1 I2

R1 R2 I3

R3RT

Ix = E / Rx; KCL: IT = I1 + I2 + I3 = E / RT

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KVL

E R

+-

+

-+

-VR

R

+ -V

V

1

2

3

2

3

1E

1

2+-

I

• Kirchhoff’s Voltage Law states that for a closed loop:V = 0, or Vrises = Vdrops

• The total resistance of n resistors in series is:RT = R1 + R2 + . . . + Rn

• The total power is: PT = P1 + P2 + . . . + Pn8/7/2012 ES-I 45

KVL

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Voltage Divider Rule

The voltage applied to a series circuit will be dropped across all the resistors in proportion to the magnitude of the individual resistors.

Vx = (Rx / RT) E

E R

+-

+

-+

-VR

R

+ -V

V

1

2

3

2

3

1

I

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Equivalent Resistance: Example I

20 30

20

8

7

a

b

12

20

8

7

a

b

20 20

7

a

b

10

7

a

b

17

a

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Equivalent Resistance: Example II

a

b

10 3

8

6

a

b

10 3

8

6

a

b

10

8

2

a

b

10 10

a

b

5

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INDEPENDENT SOURCES: 1. Voltage Sources 2. Current Sources

DEPENDENT SOURCES: 1. Voltage controlled Voltage source 2. Voltage controlled Current source 3. Current controlled Voltage source 4. Current controlled Current source

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VOLTAGE SOURCE

• is a two terminal network component with terminal voltage vab specified by a time function v(t) that is independent of the terminal current iab.

a b

Vab

V(t)

+ _i ab

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Current Source

• is a two terminal network component with terminal current iab specified by a time function i(t) that is independent of the terminal voltage vab

a b

Vab

i(t)

i ab

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REMEMBER

• The voltage across a voltage source is independent of the current through it, and

• The current through a current source is independent of the voltage across it.

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COMBINING SOURCES:

Series connection of voltage sources

Arbitrary Circuit

- ++-

a

b

v+

- v1

v2

SameArbitrary

Circuit

+--

a

b

v+

-v1+v2

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ArbitraryCircuit

i

b

a

i1 i2

Same ArbitraryCircuit

i

b

aI1+i2

COMBINING SOURCES:

Parallel connection of current sources

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Consider the following Figure Find v

v

i1

i2i3

2A 13A

+

-1Ω 2Ω 3Ω

After Combining the Parallel Current Sources

v +

-1Ω 2Ω 3Ω11A

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The Equivalent resistance of 3

116

611

6236

31

21

111

R

R

By Ohm’s Law we have that

Vv 611116

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Dependent Sources

• Some voltage (current) sources have their voltage (current) values varying with some other variables in the circuit

• Dependent sources generate voltages or currents proportional to other voltages or currents appearing in the circuit

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Current controlled Voltage source• is a network component that establishes

a voltage vcd between two points c and d in the circuit that is proportional to the current i in some branch of the circuit.

+ -

r i

c

d

+

-vcd

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Current controlled Current source

• is a network component that establishes a current icd in one branch of a circuit that is proportional to current i in some branch of the network.

βi

c

d

icd

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Voltage controlled Current source• is a network component that

establishes a current icd in a branch of the circuit that is proportional to the input voltage of the same circuit.

gmv

c

d

icd

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Voltage controlled Voltage source

• is a network component that establishes a voltage vcd between two points c and d in the circuit that is proportional to the input voltage v.

+ -

c

d

+

-vcdµvab

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POWER

tW

P

Power is defined as the rate of doing work or,equivalently, as the rate of transfer of energy.

(watts, W)

where W is the work (or energy) in joulesand t is in seconds.In terms of electrical quantities:

RVRIVIP

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Example Consider the circuit in fig. Let us determine the instantaneous power absorbed by each of the elements for the case that I=10 A.

+-

i1 i2 i3

I 6 V 1Ω 2Ω 3Ω

i

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Solution

• Since all the elements are connected in parallel,

• then the voltage across each of the elements is 6 V.

• Therefore by Ohm’s Law,

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AiAiAi 236;3

26;6

16

321

The power absorbed by 1Ω, 2 Ω and 3 Ω resistors are

;126;186;366

33

22

11

WipWipWip

Respectively;

For the total of;

WppppR 66121836321 Absorbed by the resistors.

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By KCL,

1123610 321 iiii

Wipv 66 Thus Power absorbed by Current source

Thus Power absorbed by voltage source

WpI 60106

Ai 1

Hence the total power absorbed in the circuit is

Wppp IVR 060666 8/7/2012 ES-I 67