L02 Kvs Circuit Fundamentals Full

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Introduction to electronics IIT KanpurLectures

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ESc201: Introduction to ElectronicsESc201: Introduction to Electronics

Circuit Fundamentalsquick

A RA RecapDr. K. V. SrivastavaDr. K. V. Srivastava

Dept. of Electrical Engineering

IIT KIIT Kanpur1

ConceptsConcepts

• Charge Current Voltage Power and EnergyCharge, Current, Voltage, Power, and Energy

• Ohm’s Law

C• KCL

• KVL

2

Electrical Current

Th ti t f fl f l t i l hThe time rate of flow of electrical charge

– The units are amperes (A), which are equivalent to p ( ) qcoulombs per second (C/s)

André-Marie Ampère1775-1836

3

Flow of electrons through a wire or other electrical conductor i i t tgives rise to current

• Electrons are negatively charged particles

The charge per electron is -1.602×10-19 C

g y g

I

1016 electrons flow per secondp

How much current flows?

19 1631.6 10 10 1.6 10

1QI At

−−− × ×

= = =− ×4

I

Current has a magnitude and a direction

I

1016 electrons flow per second

Direction of current flow is opposite to direction of electron flow

Large number of electrons have to flow for appreciable current

Exercise 1: For q(t) = 2 - 2e-100t, t > 0 and q(t) = 0 for t < 0, fi d i(t)find i(t)

5

Direct Current (DC) & Alternating Current (AC)

When current is constant with time we say that we haveWhen current is constant with time, we say that we have direct current, abbreviated as DC.

~

O th th h d t th t i ith ti i di tiOn the other hand, a current that varies with time, reversing direction periodically, is called alternating current, abbreviated as AC

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VoltageVoltage difference is a Source of current flow

Units of Voltage: Volts (V)

Alessandro Giuseppe Antonio Anastasio Volta 1745-1827

12V 12V

0V0V12V

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Voltage Sources

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DC and AC voltages

12V V V+ −− =

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Electrical Systems are made of Voltage sources, wires and avariety of electrical elements

Resistor Capacitor Inductor

Diode

TransformerTransistor

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Current flows in a loop

El t i l t ll d l t i l i itElectrical systems are called electrical circuits

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Connection of several circuit elements in closed paths by

Electrical Circuit

Connection of several circuit elements in closed paths by conductors

Before we learn how to analyze and design circuits, we mustBefore we learn how to analyze and design circuits, we must become familiar with some basic circuit elements.

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Resistance

( ) ( )v t R i t= ×( ) ( )v t R i t= ×

Ohm’s law

The constant R is called the resistance of the component and

Ohm s law

The constant, R, is called the resistance of the component and is measured in units of Ohm (Ω)

RResistor Symbol:

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Conductance

( ) ( )v t R i t= ×( ) ( )v t R i t×

( )t( )( ) ( )v ti t G v tR

= = ×

G = 1/R is called conductance and its unit Ernst Werner von Siemens

1816-1892is Siemens (S)

1816 1892

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vR =iiiGv

=

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Resistance Related to Physical Parameters

LRA

ρ= ×

Resistance is affected by the dimensions and geometry ofResistance is affected by the dimensions and geometry of the resistor as well as the particular material used

ρ is the resistivity of the material in ohm meters [Ω-m]ρ is the resistivity of the material in ohm meters [Ω m]– Conductors (Aluminum, Carbon, Copper, Gold)– Insulators (Glass, Teflon)

Semiconductors (Silicon)– Semiconductors (Silicon)

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Any electrical element which obeys ohms law can be modeled as a resistor

Can we model an electric bulb as a resistor?17

Electrical Bulb

Even though characteristics are non-linear, over a certain grange, the bulb can be thought of as a resistor

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Power and Energy

12VQ

X

0VThe charge loses energy = Q x 12 Joules

0V

This energy is taken from the voltage source and delivered t th i it l tto the circuit element

19

12VI Q Q I t×

X

I QIt

= Q I t= ×

X

In 1 second charge equal to Q = I flows0V

In 1 second, charge equal to Q = I flows through the element X

Every time a charge q goes from 12V to 0V it transfers energy qx12 J to the element X

Total Energy transferred in 1 second = I x 12 J

Power = Energy/time P = I x 12 Watts

Joules/second = watts20

P = I x V

A charge of 1 coulomb receives or delivers an energy of 1 joule in moving through a voltage of 1 volt.

dwv =d

j g g g

dqdtdqi =

(( )) ( )dw dw dqP td d d

v t i t= = =dt

2td

(( )) ( )dt dq dt

2

1

( ) ( )t

t

d wP t w p t d td t

= ⇒ = ∫1t

21

V1

Power

I1 2( )P V V I= − ×

X

V2

If V1 > V2 then P is positive and it means that power is being delivered to the electrical element X

If V1 < V2 then P is negative and it means that power is being extracted from the electrical element X. X is a source of power !X is a source of power !

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Note on direction of current

X2A -2AX

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12V

Examples

X

1A P = ?

X1 2( )

(12 6 ) 1 6P V V I

W= − ×= − × =

6V( )

12V1A

P = ?

1 2( )P V V I= − ×X

1 2( )(12 6 ) 1 6W= − × − = −

6V24

Power is supplied by element X instead of dissipation

6V P = ?

X

6V1A

1 2( )P V V I= − ×X 1 2( )(6 12 ) 1 6V V

W= − × − =12V

power is being delivered to the electrical element Xelectrical element X

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There is only one battery in the circuit. Can you find which element is a battery?y

A battery is a source of power, so Power dissipated is negativeA battery is a source of power, so Power dissipated is negative

Answer is C

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Power dissipated in a Resistor

+ i v i R= × viR

=R

-

v

P i

R- P v i= ×

22vPR

=2P i R= ×

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Circuit Analysis

R3R1

R2

R3

R4VS

R1

IX

What is current in R2 ?Procedure:

Use Kirchhoff's voltage law (KVL) and Kirchhoff's Current law(KCL) to transform the circuit into a set of equations whose( ) qsolution gives the required voltage or current value

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Engineering Analysis

Real-life SystemReal life System

Abstract Model

Mathematicalproblem

2 2v v

1 2

v vPR R

= +29

Nodes and loops

Node: A point where 2 or more circuit elements are connected.

R2

R3

R4VS

R1

IX

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A loop is formed by tracing a closed path through circuitelements without passing through any intermediate node morep g g ythan once

R2

R3

R4VS

R1

IX2 R4VS IX

This is not a valid loop !

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Kirchhoff's Current Law (KCL)

Sum of currents entering a node is equal to sum of currents leaving a node

1 2 3i i i+ =

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Kirchhoff's Current Law (KCL)N

Net current entering a node is zero1

0ji =∑

Current entering a node is considered positive and current leaving a node is considered as negative

0i i i+1 2 3 0i i i+ − =

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3 4i i=3 4

1 3 0ai+ − = 1 3 2 0bi+ + − =

4ai A= 2bi A= − 34

1 3 4 0i

8i A

1 3 4 0ci+ + + =

8ci A= −35

Example

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The sum of currents entering/leaving a closed surface is zero

KCL: More general formulation

The sum of currents entering/leaving a closed surface is zero.

i i2i1

R2

R3

R4VS

R1

IX

i3 i4

1 2 3 4 0i i i i+ + − =37

Series Circuit

Two elements are connected in series if there is no otherTwo elements are connected in series if there is no otherelement connected to the node joining them

A, B and C are in series

The elements have the same current going through them

a b ci i i= =38

A and B are in series E, F and G are in series

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Th l b i f th lt l f

Kirchhoff's Voltage Law (KVL)

The algebraic sum of the voltages equals zero for any closed path (loop) in an electrical circuit

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Example

41

L oop3 : 0e d b av v v v− + − + =

KVL and Conservation of Energy

V Th h l Q (V V ) J lV1

Q

The charge loses energy = Q x (V1-V2) Joules

X

V22

Energy gained Energy lost

KVL: law of conservation of Energy 42

Parallel CircuitsTwo elements are connected in parallel if both ends of one pelement are connected directly to corresponding ends of the other

A and B are connected in parallel

D E and F are connected in parallelD, E and F are connected in parallel

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The voltage across parallel elements are equal (both g p q (magnitude and polarity)

a b cv v v= = −

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Example

3 5 0 8c cv v V− − + = ⇒ =

( 10 ) 0 2c e ev v v V− − − + = ⇒ = −( )c e e

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Use KVL , KCL and Ohm’s law to solve the given problem

+ v1 -

+

V

i1

V2

-

i2

1 2 1 20x xv v v v v v− + + = ⇒ = +

1 1 5v i= × 2 2 1 0v i= ×1 1 5v i × 2 2 1 0v i ×46

Next Problem: Find currents i1 and i2

i11 2xv v v= +

i

1 1 5v i= ×

i22 2 10v i= ×

( 2 ) 5v i i= + ×1 2( 2 ) 5xv i i= + ×

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Use ohm’s law : v = I x R

I1 = ?

+

5V

+

5V1A

+

5V0.5A

5V

-

5V

-

5V

-

Apply KCL at the indicated node

1 10 .5 1 1 0 2 .5i i A− − − = ⇒ = 1 1 5 12 .5v i V= × =

1 2 12 .5 5 17 .5xv v v V= + = + =48

Independent Sources

12V12V

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Dependent (Controlled) Voltage Sources

+

-Vx2Vx

+- Ix

3Ix+-

Current-controlledVoltage-controlledVoltage source

Very useful in constructing circuit models for real-world devices such as transistors and amplifiers

Voltage sourceVoltage source

devices such as transistors and amplifiers

For a voltage controlled source: V = K1Vx , K i i t ith itK1 is a gain parameter with no units

For a current controlled source: V = K2Ix, 2 xK2 is a gain parameter with units [V/A]

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Dependent (Controlled ) Current Sources

+Vx2Vx Ix

3Ix-

Current-controlledVoltage-controlled

Very useful in constructing circuit models for real-world f

Current-controlledcurrent sourcecurrent source

devices such as transistors and amplifiers

For a voltage controlled source: I = K3Vx,g 3 x,K3 is a gain parameter with units [A/V]

For a current controlled source: I = K IFor a current controlled source: I = K4Ix, K4 is a gain parameter with no units

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SummaryCurrent: The time rate of flow of electrical charge

Voltage difference is the Source of current flow

Units of Voltage: Volts (V)g

– The units are amperes (A), which are equivalent to coulombs per second (C/s)

Units of Voltage: Volts (V)

V1 IPower( )Direction of current flow is opposite to direction of electron flow

X

V2

I1 2( )P V V I= − ×

2

1

( ) ( )t

t

d wP t w p t d td t

= ⇒ = ∫Ohm’s law V2

( ) ( )v t R i t= ×

( )( ) ( )v ti t G v tR

= = ×

Kirchhoff's Current Law (KCL)Sum of currents entering a node is equal to sum of currents leaving a node

R+

-v

i 2vPR

=

2P i R= ×currents leaving a node

1 2 3i i i+ =

Kirchhoff's Voltage Law (KVL)Two elements are connected in series ifthere is no other element connected tothe node joining them. Same currentflows The algebraic sum of the voltages

equals zero for any closed path (loop) in an electrical circuit

Two elements are connected in parallel if both ends of one element are connected directly to corresponding ends of the other. Same voltage

flows

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