Digital Electronics by rajesh

7/25/2019 Digital Electronics by rajesh

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EN 202 Electronics Rajesh Gupta

Welcome

EN 202 Electronics


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My IntroductionProf. Rajesh Gupta

Dept. of Energy Science and Engineering

Office: Room No. 3. Near to Energy Sci and Engg. Office

Phone: 7837

Email: [email protected]


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Your Introduction

Name

Place

Background of electronics and electrical

What you expect from this course


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Objective of course

To give you a basic background of electronics

engineering, which is required for

Troubleshooting, understanding and making of

electrical/electronics circuits/instruments

Understanding of basic terminology of electronics

For laboratory experiments Minimum electronics knowledge which help in understanding

system in which electronics is one of the component


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Digital Electronics


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References A.P. Malvino and D.P. Leach, Digital Principles and

Applications, Tata McGraw Hill Edition

W.H. Gothmann, Digital Electronics An Introductionto Theory and Pratice, Prentice Hall of India PrivateLimited.

A. P. Malvino, J. A. Brown, Electronics : AnIntroduction to Microcomputers, Tata Mcgraw Hill.


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Logic Gates


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NOT Gate OR Inverter

Logic - Opposite of input


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AND Gate

Logic output is 0 if there is any input 0


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OR Gate



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NAND GateUniversal Gate



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NOR GateUniversal Gate



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XOR Gate

Logic output is 1 if there are odd number of 1s in input


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ICs of logic gates


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Problem Construct NOT, OR and AND function

by NAND Gate


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Problem Construct NOT, OR and AND function

by NOR Gate


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ProblemPropose an application based on digitalcircuit


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Acceptable input & output voltage

TTL Transistor-Transistor Logic


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Binary number system Why binary number system is required in digital electronics ?

Only two states are possible

Decimal Odometer

000, 001, 002, 003009, 010, 011..099, 100, 101 Binary Odometer

000, 001, 010, 011, 100, 101, 110, 111

Bit = X

Nibble= XXXX

Byte = XXXX XXXX

MSB LSB


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Weight of digits Weigh of digit in decimal system

Weight of digit in binary


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Problems A number 7 is required to electrically transmit

from one town to another town. What arepossible ways ?

Convert 10101 to decimal

Convert 55 to binary

Successive division


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Binary addition

Examples


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Binary subtraction0-0=0

1-0=1

1-1=0

10-1=1

Examples

407 100 (4) 1101 (13) 11001000

-328 -001 (1) -1010 (10) -01001011

------ --------- -------------- --------------

079 011 (3) 0011 (3) 01111101


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Binary Adder


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Half Adder


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Full Adder

A

B

C

FA

CARRY

SUM

A BC

CARRY

SUM


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Boolean Arithmetic- Based on logic gate


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OR and AND operation

Y= A+B Y= A.B


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NOT, NAND and NOR NOT

Y=

NANDY =

NOR

Y =

XOR Y = ??

AB

BA


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Basics of boolean algebra


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Boolean rules for simplification A+AB=A

BABAA

Truth tables should match


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Contd. A+BC=(A+B) (A+C)


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Circuit simplification example

Realize with less number of gates


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DeMorgan's Theorems


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Problem Solve

Ans BA


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Problems

1)CAB(CBAACAB

0ABAAB


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Converting truth tables into Boolean

expressions

Ans AB+BC+CA

Sum of product approach


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Contd.Product-Of-Sums approach


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Logic simplification with Karnaugh maps


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Contd.


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Contd.


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Larger 4-variable Karnaugh maps


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Contd.


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Contd.


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Contd.


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Sigma Notation


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Pi-Notation


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Don't Care cells in the Karnaugh map


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Problem

f= m (0,1,3,4,5,7,10,13)

f= m (1,2,3,5,6,7,8,12,13)

f= m(0,2,6,10,11,12,13)+d(3,4,5,14,15)


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FLIP-FLOP


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Sequential logic

Sequential logic, unlike combinational logic isnot only affected by the present inputs, but

also, by the prior history


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R-S Flip-flop or R-S latch


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Gated S-R latch


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D Flip-flop OR D latch


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D Flip-Flop Response


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Edge-triggered Response


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Edge trigger realization


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Edge triggered RS flip-flop


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J-K flip-flop


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Preset and Clear in flip-flop


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Counters


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Asynchronous counters


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Contd.


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Up and Down counter


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Propagation delay in asynchronous counter


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Contd.


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Application of counter


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Problem

Application of counters ?


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Shift Register


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Application

Shift registers produce a discrete delay of adigital signal or waveform

Very long shift registers served as digitalmemory


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Shift Register Types

Serial-in/serial-out

Serial-in/parallel-out

Parallel-in/serial-out

Parallel-in/parallel-out

Ring counter


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Serial-in/serial-out


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Parallel-in/serial-out


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Serial-in/parallel-out


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Parallel-in/Parallel-out


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Ring counter


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Serial-in, Serial-out shift register


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Parallel-in serial out


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Serial-in/parallel out


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Problem

Parallel-in parallel-out circuit ?


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Ring counter


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Digital Storage


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Why digital ?

The basic goal of digital memory is to provide a

means to store binary data: sequences of 1's and 0's

The most evident advantage of digital data storage isthe resistance to corruption

Magnetization method of storage

Digital data storage also complements digital

computation technology


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Random access and sequential access

Random access means that you can quickly and

precisely address a specific data location within the

device, and non-random (sequential) simply means

that you cannot

Examples

A vinyl record platter is an example of a

random-access device (CDs also)

Cassette tape is sequential


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Writing and Reading

The process of storing a piece of data to a memory

device is called writing

The process of retrieving data is called reading

ROM (read only memory)- Some devices do not allowfor the writing of new data, and are purchased "pre-

written.

Example: vinyl records

Read-write memory Memories allow reading andwriting

Example: Cassette audio and video tape


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Memory with moving parts: "Drives"

Paper tape

Magnetic tape (sequentialaccess,slow)

Magnetic storage drives

drum type (motor, R/W coil) Floppy disk (not reliable)

Hard drive

Compact disk (CD) Binary bits are "burned" into

the aluminum as pits by a high-power laser

Digital Video Disk (DVD)


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Modern nonmechanical memory

A very simple type of electronic memory isflip-flop


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ROM-Read-only memory

PROMs - Programmable Read-Only

Memory

The simplest type of ROM is that which uses tiny

"fuses" which can be selectively blown or left alone

to represent the two binary states

EPROM - Erasable Programmable Read-

Only Memory

Electrically-erasable (EEPROM)

Ultraviolet-erasable (UV/EPROM)


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Volatile and non-volatile memory

Volatile memory loose its data when power goes off

(e.g, RAM of computer)

Non Volatile memory retain data even without power(e.g. ROM, magnetic tapes)


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Memory Address

The location of this data within the storage device is

typically called the address, in a manner reminiscent

of the postal service.

the address in which certain data is stored can becalled up by means of parallel data lines in a

digital circuit

data is addressed in terms of an actual physical

location on the surface of some type of media

(e.g. tracksand sectorsof circular computer disks)


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Memory Array

For more storage,

many latches

arranged in a form ofarray where we can

selectively address

which one is reading

from or writing to.


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Memory Size

Number of bits

Generally memory size represented in bytes(1 byte = 8 bits)

Example 1.6 Gigabytes = 12.8 Giga bits

One kilobyte" = 1024 bytes (2 to the power of10) locations for data bytes (rather than exactly1000)

"64 kbyte" memory device actually holds 65,536bytes of data (2 to the 16th power)


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Digital computer

Main components

CPU: central procession unit

Memory

Input output device

Input Microprocessor OR CPU

Memory

Output


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Sections of CPU

Arithmetic and logic unit - to perform arithmetic operationssuch as addition and subtractions, logical operation (AND,OR,etc.)

Timing and control unit - control entire operation of a

computer. It acts as a brain. It also control all other devicesconnected to CPU

General purpose registers - for temporary storage of data andintermediate results while computer is making execution ofprogram

Accumulator It is a register which contain one the operandsand store results of most arithmetic and logical operations


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Electrical Theorem andComponents


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Thevenin's Theorem

Thevenin's Theorem is a way toreduce a network to an equivalentcircuit composed of a single voltagesource, series resistance, and seriesload

Useful in analyzing power systems and

other circuits where one particular

resistor in the circuit (called the "load"

resistor) is subject to change, and re-

calculation of the circuit is necessary

with each trial value of load resistance,

to determine voltage across it andcurrent through it.


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Contd.


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Contd.

2. Find the Thevenin resistance by

removing all power sources in the

original circuit (voltage sources

shorted and current sources open)

and calculating total resistancebetween the open connection

points.


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Contd.

3. Draw the Thevenin equivalent

circuit, with the Thevenin

voltage source in series with

the Thevenin resistance. The

load resistor re-attachesbetween the two open points

of the equivalent circuit.

4. Analyze voltage and current for

the load resistor following the

rules for series circuits.


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Norton's Theorem

Norton's Theorem is a way toreduce a network to an equivalentcircuit composed of a single currentsource, parallel resistance, andparallel load.

Useful in analyzing power systems andother circuits where one particular

resistor in the circuit (called the "load"

resistor) is subject to change, and re-

calculation of the circuit is necessary

with each trial value of load resistance,

to determine voltage across it and

current through it.


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Steps to follow for Norton's Theorem

1. Find the Norton source current by

removing the load resistor from the

original circuit and calculating

current through a short (wire)

jumping across the open connection

points where the load resistor usedto be

Designate R2 as the "load" resistor

Determining voltage and current across

R2


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Contd.


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Contd.

2. Find the Norton resistance by

removing all power sources in the

original circuit (voltage sources

shorted and current sources open)

and calculating total resistancebetween the open connection

points.


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Thevenin-Norton equivalencies


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Electrical Components


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Capacitors and calculus


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Contd.


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Contd.


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Factors affecting capacitance

PLATE AREA

PLATE SPACING

DIELECTRICMATERIAL


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Series and parallel capacitors


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Practical considerations

Working voltage

Polarity

Equivalent circuit

Physical Size


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Inductors

The ability of an inductor to store energy in the form of amagnetic field is called inductance. It is measured in the unit ofthe Henry(H).

Inductors used to be commonly known by another term: choke.In large power applications, they are sometimes referred to asreactors.

Inductors react against changes in current by dropping voltagein the polarity necessary to oppose the change.

When an inductor is faced with an increasing current, it acts asa load

When an inductor is faced with a decreasing current, it acts as asource


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Inductors and calculus


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Contd.


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Contd.


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Factors affecting inductance

TURNS IN THE COIL

COIL AREA

COIL LENGTH

CORE MATERIAL


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Series and parallel inductors


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Practical considerations

Rated current

Equivalent circuit

Inductor size

Interference


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Analog Electronics


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References

A.P. Malvino, Eletronic Principles, Tata McGraw-Hill Publishing Company Limited, New Delhi

N.N. Bhargava, D.C. Kulshreshtha, S.C. Gupta,Basic Electronics and Linear Circuits, TataMcGraw-Hill Publishing Company Limited, New Delhi


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Diodes


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n-type semiconductor

Legends :Free electron (Negative Charge)

Hole (Positive Charge)Immobile ion (Positive Charge)

Fifthelectron

+5

P

+4 +4 +4

+4

+4+4+4

+4


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p-type semiconductor

Legends :Hole (Positive Charge)Electron (Negative Charge)Immobile ion (Negative Charge)

+4

Hole

Incomplete bond

+4 +4

+4

+4

+4

+4+4

+3

+4 +4 +4

+4 +3 +4

+4 +4 +4


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p-n junction

P - type N - type N - typeP - type Electric field

Depletion region


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p-n junction forward biased

P - type N - type

Depletionregion


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p-n junction reverse biased

Depletion region

P - type N - type


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Introduction

Schematic symbol

Stripe marks cathode

Real component appearance

Anode Cathode

Flow permitted

Flow permitted

+ -

- +

Hydraulic check valve

+

-

-

+

Diode operation

Current permittedDiode is forward-biased

Current prohibitedDiode is reverse-biased


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Contd.

DI

SI

e

q

VN

k

T

diode current

saturation current

Eulers Constant(2.71828.)

Charge of electron(1.6 As)1910

Voltage across the diodeNon-ideality coefficient(typ.between 1 and 2)

Boltzmanns constant (1.38) )2310

Junction temperature in kelvin

)1( NkTqV

SD eII

forward

reverse-bias

reverse

Breakdown!

0.7

DI

DV

V

forward-bias

)1( 026.0/ DVSD eII


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Summary

A diodeis an electrical component acting as a one-way valve forcurrent.

When voltage is applied across a diode in such a way that the diodeallows current, the diode is said to be forward-biased.

When voltage is applied across a diode in such a way that the diode

prohibits current, the diode is said to be reverse-biased. The voltage dropped across a conducting, forward-biased diode is

called the forward voltage. Forward voltage for a diode varies onlyslightly for changes in forward current and temperature, and is fixedprincipally by the chemical composition of the P-N junction.

Silicon diodes have a forward voltage of approximately 0.7 volts.

Germanium diodes have a forward voltage of approximately 0.3 volts. The maximum reverse-bias voltage that a diode can withstand without

"breaking down" is called the Peak Inverse Voltage, or PIVrating.


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Problem

Find out Application of Diodes


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Rectifier Circuits

Rectificationis the conversion of alternating current

(AC) to direct current (DC).

A half-waverectifier is a circuit that allows only one half-cycle of

the AC voltage waveform to be applied to the load, resulting inone non-alternating polarity across it. The resulting DCdelivered to the load "pulsates" significantly.

A full-waverectifier is a circuit that converts both half-cycles ofthe AC voltage waveform to an unbroken series of voltagepulses of the same polarity. The resulting DC delivered to the

load doesn't "pulsate" as much.


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Half-wave Rectifier

Bright

Dim

AC Voltage source ~V A

Half-wave rectifier circuit

~

V A

+

-LoadAC Voltage

source


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Full-wave Rectifier

V A

+

Full-wave rectifier circuit (center-tap design)

~ Load

-

AC Voltagesource

V A

V A

+~

-

V A

+

-

+

-

V A

+~

V A

-

+

-

+


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Full wave bridge rectifier

V A

Full-wave rectifier circuit (bridge design)

V A

ACVoltagesource

~

Load

+

-

V AV A

~+

-

+

-

V AV A

~+

-

+

-


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Diode ratings

In addition to forward voltage drop (Vf) and peak inversevoltage (PIV), there are many other ratings of diodes

Maximum DC reverse voltage = VRor VDC The maximumamount of voltage the diode can withstand in reverse-bias modeon a continual basis. Ideally, this figure would be infinite.

Maximum reverse current = IR the amount of currentthrough the diode in reverse-biasoperation, with the maximumrated inverse voltage applied (VDC). Sometimes referred to as

leakage current. Ideally, this figure would be zero, as a perfectdiode would block all current when reverse-biased. In reality, itis very small compared to the maximum forward current.


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Contd.

Maximum (average) forward current = IF(AV) The maximumaverage amount of current the diode is able to conduct in forward biasmode. This is fundamentally a thermal limitation: how much heatcan the PN junction handle, given that dissipation power is equal tocurrent (I) multiplied by voltage (V or E) and forward voltage isdependent upon both current and junction temperature. Ideally, thisfigure would be infinite.

Maximum (peak or surge) forward current = IFSM or if(surge) Themaximum peak amount of current the diode is able to conduct inforward bias mode. Again, this rating is limited by the diodejunction's thermal capacity, and is usually much higher than the

average current rating due to thermal inertia (the fact that it takes afinite amount of time for the diode to reach maximum temperature fora given current). Ideally, this figure would be infinite.


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Contd.

Maximum total dissipation = PD The amount of power (in watts)allowable for the diode to dissipate, given the dissipation (P=IE) ofdiode current multiplied by diode voltage drop, and also the dissipation(P=I2R) of diode current squared multiplied by bulk resistance.Fundamentally limited by the diode's thermal capacity (abilityto tolerate high temperatures).

Operating junction temperature = TJ The maximum allowabletemperature for the diode's PN junction, usually given in degreesCelsius (oC). Heat is the "Achilles' heel" of semiconductor devices: theymustbe kept cool to function properly and give long service life.


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Contd.

Typical junction capacitance = CJ the typical amount ofcapacitance intrinsic to the junction, due to the depletion region actingas a dielectric separating the anode and cathode connections. This isusually a very small figure, measured in the range of picofarads (pF).

Reverse recovery time = trr the amount of time it takes for a diodeto "turn off" when the voltage across it alternates from forward-bias toreverse-bias polarity. Ideally, this figure would be zero


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Power Supply

Transformer Rectifier Filter Regulator

AC mains

Regulated dc output


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Need of Filter

Full-wave rectifier output is not smooth, it has lot ofripples

In order to minimize these ripples, a filter is required

to smooth the out of full wave rectifier.


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Shunt Capacitor Filter

Shunt capacitor is simplestand cheapest type filter

Connect a large value ofcapacitor across load

Block DC, allow AC to follow

Rate of discharge dependon RLC

Large C give less ripples

Upper limit of C depends oncurrent handling rating of

diodes

Powermains

Full-waverectifier C

Filter Load

+

- LR o

2 3 40

mV

dcV

B D

A C

t

2 3 40

mV

dcV

B D

A C F

E

t

o

o


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Conduction angle of diode

2 3 40

mV

dcV

B D

A C

t

2 3 40

mV

dcV

B D

A C

t

o

o E

F


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Problem

Load current = 10 mA

Capacitance= 470 F

Line frequency = 50 Hz

Find ripple voltage of full wave rectifier and half waverectifier


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Need of Regulator

Output of filter also have some ripples, to make itmore smooth, regulator is required.

Zener diode is one of the simplest type of regulator


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C d


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Contd.

45 V224 mV

447.76 A

44.776 V

100 K

447.76 A 447.76 A

447.76 A

0A loadR

500

45 V12.6 V

32.4 mA

32.4 V

1 K

32.4 mA 25.2 mA

25.2 mA

7.2 mA

loadR500

45 V

224 mV447.76 A

44.776 V

100 K447.76 A

loadR

500

447.76 A 447.76 A

S h ttk di d


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Schottky diodes

Schottky diodesare constructed of a metal-to-N junction rather thana P-N semiconductor junction.

Schottky diodes are characterized by fast switching times (low reverse-recovery time), low forward voltage drop (typically 0.25 to 0.4 volts for ametal-silicon junction), and low junction capacitance. This makes them

well suited for high-frequency applications. In terms of forward voltage drop (VF), reverse-recovery time (trr), and

junction capacitance (CJ), Schottky diodes are closer to ideal than theaverage "rectifying" diode. Unfortunately, though, Schottky diodestypically have lower forward current (IF) and reverse voltage(VRRMand VDC) ratings than rectifying diodes and are thus unsuitablefor applications involving substantial amounts of power.

T l di d


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Tunnel diodes

Tunnel diodesexploit a strangequantum phenomenon called resonanttunnelingto provide interesting forward-bias characteristics having peakcurrent (IP) Valley current (IV).

Able to transition between peak andvalley current levels very quickly,"switching" between high and low statesof conduction much faster than evenSchottky diodes.

Tunnel diode characteristics are alsorelatively unaffected by changes intemperature.

Forwardcurrent

Forward voltage

Tunnel diodeAnode

Cathode

PI

VI

PV VV

Li ht itti di d


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Light-emitting diodes

Some semiconductor junctions, composed ofspecial chemical combinations, emitradiant energy within the spectrum of visiblelight as the electrons transition in energylevels.

Simply put, thesejunctions glow whenforward biased. A diode intentionallydesigned to glow like a lamp is called a light-emitting diode, or LED.

Diodes made from a combination of theelements gallium, arsenic, and phosphorus(called gallium-arsenide-phosphide) glow bright

red, and are some of the most common LEDsmanufactured

Light-emitting diode (LED)

Cathode

Anode

V t Di d


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Varactor Diode

It is operated reverse-biasedso no current flows through it,but since the width of thedepletion region varies withthe applied bias voltage, the

capacitance of the diode can bemade to vary.

Varactors are commonly used involtage-controlled oscillators

or

Symbol

C

maxC

minC

Reverse voltage


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P iti Li it


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Positive Limiter

PV

PV PV

R

LR0 0

~

Bi d li it


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Biased limiter

0 0

PV

PVPV

+

-

R

LRV

V+0.7

~

C bi ti li it


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Combination limiter

0

7.01V

7.02V

R

1D

1V

2D

2V LR+

--

+PV

PV

0~

Ci it t ti li it


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Circuit protection limiter

inV

outV1 2

1N914

+5 V

inV outV


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Transistors

Introduction


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Introduction

The invention of the bipolartransistor in 1948 ushered in arevolution in electronics.

Bipolar transistors consist ofeither a P-N-P or an N-P-Nsemiconductor "sandwich"

structure. The three leads of a bipolar

transistor are called the Emitter,Base, and Collector.

Difference between PNP andNPN transistor is the properbiasing of junctions. Currentdirections and voltage polaritiesfor each type of transistor areexactly opposite

NN

Emitter Collector

Base

P

NP

Emitter CollectorBase

cE

B

E c

BB

cE

E c

B

P

Transistor Mode of Operation


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Transistor Mode of Operation

Biasing an NPN transistor for active operation

Condition Emitter junction Collector junction Region ofoperation

FR Forward-biased Reverse-biased Active

FF Forward-biased Forward-biased Saturation

RR Reverse-biased Reverse-biased Cutoff

RF Reverse-biased Forward-biased Inverted

BaseEmitter Collector

- + - +

E C

Spacecharge

Spacecharge

B

EEV

CCV1S 2S

N P N

O l itt j ti f d bi d


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Only emitter junction forward biased

Larger current flow

~ 99 % of total current carried by electrons (moving fromemitter to base)

Emitter current and base current very large (IE=IB), IC=0

N NP

Reducedbarrier

CE

B

B

I

EI

EEV

+-

O l ll t j ti bi d


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Only collector junction reverse biased

Very small current flow (minority carrier current temperaturedependent current) called collector leakage current ICBO

ICBO signifies current between collector and base when thirdterminal (emitter) is open

+-

Electron

Hole

B

CCV

N P N

E C

BI

CI


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Working of transistor


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Working of transistor

Ratio of no. of electrons arriving at collector to no. of emitted electrons isknow as base transportation factor (typically ~ 0.99)

No. of electrons (like 3) and holes (like 7) crossing the E-B junction is muchmore than the no. of electrons (like 5) and holes (like 8) crossing the C-Bjunction. The difference of these two currents is base current.

E C7

6

5

1234

B

EI

+-

EEV

+-

CCV

B

BI CI

N NP

C d


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Contd.

Collector current is less than emitter current

A part or emitter current consists of holes that do not contribute tocollector current

Not all the electrons injected into the base are successful inreaching collector.

Ratio of collector current to emitter current is typically 0.99 denotedby dc

Collector current made up of two parts

Fraction of emitter current which reaches the collector

Reverse leakage current ICO

Total current equationCOEdcC III

BCE III

Problem


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Problem

An NPN transistor has of dc 0.98 and a collectorleakage current of 1 uA. Calculate the collector andbase current, when emitter current is 1 mA.

Transistor amplifying action


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Transistor amplifying action

Transfer + resistor = transistor

5 k

E C

~

Ei Ci

+

-

CCV

B

+

- -

+

EEV

EB CBLR os

40inR kRo 500

mAIe 5.040

1020 3

mA5.0II ec LcO RIV

V5.2)105()105.0( 33

S

O

V

V

A

1251020

5.23

Different configurations of transistor


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Different configurations of transistor

Figure shows three configuration from ac point of view.

In all configurations, emitter-base junction is always forward-biased and collector base junction is always reverse-biased.

~+

+

--

B

C

E

LR

E C

B~

+

-

+

-

s

LRs ~

+

+

--

B

CLR

s

E

OO

O

Transistor characteristics


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Transistor characteristics

Static characteristic curves to relate currentand voltage in a transistor. Input characteristic

Output characteristic

Common Base Input characteristics


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Common-Base Input characteristics

.ConstVCB

E

EBi

ir

mA mA

CCVVV B

E

1R 2R

Ei Ci

++

+

+

-

-

- -

EEV

SR

EB CB

C

Common Base output characteristics


179/360


Common-Base output characteristics

High output resistance can be good current source

Saturation region - collector current not remain same with change in

emitter current Cut-off region- collector current is not zero even emitter current is zero

due to leakage current ICB0 OR ICO

.ConstIEC

CBo

ir

E

Cfb

i

i

orh

.ConstVCB

Summary of C-B characteristics


180/360


Summary of C-B characteristics

Current gain slightly less then unity (~0.98)

Dynamic input resistance very low (~ 20 ohm)

Dynamic output resistance very high (~ 1 M ohm)

Leakage current very low (~ 0.02 uA for Si transistor)

C-E Configuration


181/360


C-E Configuration

Mostly work in active region CEOBdcC III

dc

dcdc

1

dc

CBOCEO

II

1

CBOBCdcC IIII )(

CBOBdcCdc III )1(

CBO

dc

B

dc

dcC I

1

1I

1I

C

CCV

+

-

N

NP+E

B

BBV

+

BBV

-

-

CCV+

B

E

C

EI

CIBI

-


182/360

C-E input characteristics


183/360


C E input characteristics

V

A

mA

V

BBV

CCV

1R BEV E

CEV 2R

B

BI

CI

C

+ -

+

+

+-

--

.ConstVCE B

BE

i ir


184/360

Comparison between CB and CE


185/360


Comparison between CB and CE

Leakage current lead to Thermal Runway

Parameters Common baseConfiguration

Common emitterConfiguration

1. Input dynamic resistance

2. Output dynamicresistance

3. Current gain

4. Leakage current

Very low (20 )

Very high (1 M)

Less than unity (0.98)

Very small (5 A for Ge, 1 A

for Si)

Low (1 k)

High (10 k)

High (100)

Very large (500 A for Ge, 20

A for Si)

Problems


186/360


Problems

When emitter current of transistor changed by 1mA,its collector current changed by 0.995 mA. Calculate

CB current gain

CE current gain

The DC current gain of a transistor in CEconfiguration is 100. Find DC current gain in CBconfiguration.

Why CE configuration widely used


187/360


Why CE configuration widely used

A good amplifier stage is one which has high input resistance and lowoutput resistance

Current gain is more in CE configuration

Amplifierstage 1

Amplifierstage 2

1A 2A 3A

1B 2B 3B

SRLR

S ~

Common-collector configuration


188/360


Common collector configuration

Emitter current as a function of base current

B

C

EBBV

LR

CCV

CBOEdcC

CBE

III

III

BE

CLR

~

+

-

oS

Contd


189/360


Contd.

CBOEdcBE IIII

CBOBEdc III )1(

CBO

dc

B

dc

E III

1

1

1

1

11

1

dc

dc

CBOdcBdcE III )1()1(

)1(

)1(

dc

B

E

BdcE

I

I

II


190/360

Transistor data sheet


191/360


Transistor data sheet

Important parameters

Maximum power dissipation

Maximum allowable voltage

Current gain

Max frequency of operation

Basic CE amplifier circuit


192/360


Basic CE amplifier circuit

+

-

B

C

E+

-

CR

BR

s

o

CCVBBV

1CC

2CC


193/360

Amplification and Q-point


194/360


Amplification and Q point

13750110125 AAi=

beb

cec

VI

VI

poweracinput

poweracOutputgainPoweriii ,)(

1101020

9.41.7,)( 3 BECEAgainVotageii

12510)4060(

10)8.43.7(,)(

6

3

B

Ci

i

iAgainCurrenti

Selection of Q-point


195/360


Selection of Q point


196/360


Logic Gate Using Transistor

Current regulator


197/360


Current regulator

Transistors function as currentregulators by allowing a small current tocontrol a larger current. The amount ofcurrent allowed between collector andemitter is primarily determined by theamount of current moving between base

and emitter. In order for a transistor to properly

function as a current regulator, thecontrolling (base) current and thecontrolled (collector) currents must begoing in the proper directions: meshingadditively at the emitter and goingagainstthe emitter arrow symbol.

BC

E

BC

E

= Small, Controlling Current

= large, Controlled Current

Transistor as a switch


198/360


a s sto as a s tc

Transistor's collector current isproportionally limited by its basecurrent, it can be used as a sort ofcurrent-controlled switch. Arelatively small flow of electronssent through the base of the

transistor has the ability to exertcontrol over a much larger flow ofelectrons through the collector

When a transistor has zero currentthrough it, it is said to be in a stateof cutoff

When a transistor has maximumcurrent through it, it is said to bein a state of saturation.

Switch

NPNtransistor

PNPtransistor

Switch

Inverter


199/360


0 input = open switch = output 1

1 input = close circuit = output 0

BR

CR

CCV

inV

outV

B

BEinB

R

VVI

BdcC II

CCCCout RIVV

C

CC

R

V

CI

CCVCEV

Open switch

Closed switch

CCV

CR

outV

CCV

CR

outVC

E

C

E


200/360

OR Gate


201/360


C

1Q 2Q 3QA

B

CCV

AND Gate


202/360


1QA

2Q 3Q

C

B

CCV


203/360

XOR Gate


204/360



205/360


Transistor Biasing Circuit

Stabilization of Q-point


206/360


Q p

Requirement of biasing circuit Operating point in the centre of active region

Stabilization of collector current against temperature variations

Making operating point independent of transistor parameters

Different biasing techniques used for achieving these points

T T

Temperaturecontinues toincrease

CI

CBOI

CEOI

Fixed Bias


207/360


CECCCC VRIV

CEOBC III

C

CCSatC

R

VI )(

C

CCCR

VI

B

BEBBB

R

VVI

BR CR

CCV

BR CR

CCVBBV

CR

CCV

BBV

BR

Steps for calculating Q-point in fixed bias


208/360


p g Q p

1. Calculate base voltage, in case VBE is known, usemore accurate calculation.

2. Calculate collector current from base current, makesure its not greater than Ic(sat)

3. Calculate collector-emitter voltage

Problem


209/360


Calculate the Q-point for the circuit given in figure

BR CR

50

VVCC

9

300 k 2 k

Problem


210/360


Calculate Q-point in circuit. Given RC=1 k ohm and RB=100 k ohm

If transistor is replaced by another unit of beta=150 insteadof 60. Determine its new Q-point.

BR CR

VVCC 10

60

AC125

Assignment (1 Mark)


211/360


In a given circuit, a supply of 6 V and aload resistance of 1 k ohm is used.

Find the value of resistance RB so thata germanium transistor with =20 andICBO=2uA draws an IC of 1 mA.

What Ic is drawn if the transistorparameters change to =25 and ICBO =10 uA due to rise in temperature ?

BR

V6

1 k

Fixed bias features


212/360


Advantages Very simple

very few component

Easy to fix Q point by changing RB

Limitations Thermal runway

Strongly dependent Q-point

CI

Leads to thermalrunway

CI

T T T CEOI CEOI

CBOI

CBOI

Collector to base bias circuit


213/360


BC

BECCCCB

BEBBCCCCC

BEBBBCCCC

RR

VRIVI

VIRRIRV

VRIIIRV

)(

)(

)(

BC

BECEB

RR

VVI

CBOI

CEOI

T

Rising tendency is checked

CI CI

BI

CEV

-

BR

CR

+

CCV

BEV

+

-

+

- CI

BI

CEV

)( BC II

Contd.


214/360


IB ~ Vcc/(RB+Rc)

Shift in Q-point due to change in is not much asit occurs in case of fixed bias

BCBBECC

BEBBCBCCC

IRRVV

VIRRIRV

])1([

)(

CCCCCBCCCCE

CECCBCC

RIVRIIVV

VRIIV

)(

0)(

Features of collector to base bias circuit


215/360


Advantages

Check thermal runway

Q-point less dependent on value

Limitations

Base resistance also provide AC feedback, thatreduce voltage gain

Problem


216/360


Calculate the minimum and maximum collectorcurrent in the given circuit, if varied with in thelimit indicated.

BR

CR

200 k

2 k

20 V

50


217/360


B

BEEECCB

EEBEBBCC

R

VRIVI

RIVIRV

)(

B

EECCB

R

RIVI

)(

CI

CBOI

CI

CEOIT

BI

EERI

EIRising tendency ischecked

CI CI

BI

EERI

EI

Rising tendency ischecked

+

-

-

BR CR

ER+

CI

EI

BEV

+

-

CCV

Contd.


218/360


EBBEBBCC RI)1(VRIV

EB

CC

EB

BECCB

RR

V

RR

VVI

)1(

)/(

BE

CC

EB

CCBC

RR

V

RR

VII

CECCCCE

EECECCCC

IRRVV

RIVRIV

)(

Features of bias circuit with emitter resistance


219/360


Advantages

Provide good stabilization of Q-pointagainst temperature and variation

Limitations

Emitter resistance provide a feedback

causes reduction in voltage gain. For getting very good stabilization

For large RE, high DC source is required

For small RB, low DC source is required Problem: Calculate values

of 3 currents

ER BR>>

100

1 M 2 k

1 k EC+ +

- -

10 v

BR CR

ER

Voltage divider biasing circuit


220/360


1R

A

2R ER

CCV

B

CR

B

CCV

CR

ER

ATHR

BEVTHV

EI

BI

21

21

RR

RR

RTH

CCVRR

R

21

2

CECCCCE IRRVV )(

EEBETHBTH IRVRIV

EBBETHBTH RIVRIV )1(

BETHETHB VV]R)1(R[I

ETH

TH

ETH

BETHB

RR

V

RR

VVI

)1(

2R

A

1R

B

Sourceshortage

1R 2R

A

B

Problem


221/360


Find Q point

60

40 k 5 k

5 k 1 k EC+

-

+12 v

+

-


222/360


Amplifiers

DC Behavior


223/360


1R cR

CCV

2R ER

EECECCCC RIVRIV

)( ECCCE RRIV

)RR(

VV)RR(

1I

EC

CCCE

EC

C

Input and Output phase


224/360


AC Behavior


225/360


+

-

~+

-

1R

2R

CR

O

+

+

-

-++

-

-

CCV

CC

CC

+ +

--

iER

EC

OR

OR

+

-

~

+

-

1R 2R

CR

O

i

AC equivalent

Transistor Equivalent Circuit


226/360


Input

Output

B

E

bi

ir

C

E

bi or

Contd.


227/360


C

E

bi or

cibi

C

B

+

-

+

-b

ir

h-Parameter


228/360


Manufacture specify characteristics of a transistor interms of h(hybrid) parameters

Hybrid is used with these parameters because they

are a mixture of constants having different units

It becomes popular because they can be measureeasily

Transistor as two-port network


229/360


1 2

3

+ +

- -1 2

1i 2i

2121111 hih

2221212 hihi

= Forward Current ratio (with output shorted) = fh

1

111

ih

02

= Input impedance (with output shorted) = ih

1

221

i

ih

02

2

112

h

01 i= Reverse voltage ratio (with input open) = rh

2

222

ih

01 i oh= Output admittance (with input open) =

Hybrid equivalent circuit


230/360


i bi

E

C

+

-

B

~~

+

- O-

+

ci

bi

ieh

fehoeh

reh C

,rh iie

,feh

,/1 ooe rh

dynamic input resistance

current amplification factor

dynamic output resistance.

ieh 1 k4105.2 reh

50fehoeh 25 s (or, 1/ oeh 40 k)

Amplifier analysis


231/360


OC

OCOCac

RR

RRRRR

||

B C

i ~ ~

+

-

+

-

E

1R 2R cR OR Oci+

-

bi

ieh

reh C feh bi oeh

i Obi

E

C

+

-

~

B

-

+

biieh feh

acR

Contd.


232/360


ieiein hhRRZ |||| 21

acacoeo RRhZ ||)/1(

AAA ip

b

ci

i

i

CurrentInput

CurrentOutputAgainCurrent ,

fe

b

bfeh

i

ih

iA

ieb

acbfe

hi

Rih

VoltageInput

VoltageOutputAgainVoltage

,

ie

acfe

h

Rh

0180i

ac

r

RA

Problem


233/360


Current Gain 150, Rin=2 kohm

Calculate voltage gain andinput impedance

~

15 V

75 K 4.7 K

7.5 K 1.2 K

12 K

15 F15 F

100 F

SO

+

+

--

Multi-stage Amplifiers


234/360


1

210P

PLog

A =12

1

1

21

n

o

n

n

ss

o

1A nn AAA 12 =

s ~

ins 11A 2A nA2 1n on

Numbers of bels =

Contd.


235/360


Number of dB = 10 Number of bels = 10

Gain in dB = 20

1

210logV

V

dBn2dB1dBdB A...AAA

1

210logP

P

Why dB


236/360


Simple addition of gain

Permits us to denote very small and very large value

Our hearing power in logarithmic

Coupling of two stages


237/360


Minimum loss of signal

Should not affect biasing of other stage

Typical Couplings RC coupling

Transformer coupling

Direct coupling

RC coupling


238/360


Widely used

Makes DC biasingindependent

Not good for lowfrequency applications

-

O

+

+

-

+ + + +

- - - -

cR

cR

1R1R

2R 2RER E

REC EC

LR

CC

CCCC

CCV

Transformer coupling


239/360


DC isolation provided bytransformer

Bigger in size

Does not amplify signals ofdifferent frequency equally

Suited for power amplifiersand tuned voltage amplifiers

CCV

O

1R1R

2R 2RER

SC

+ ++

+

--

- - EC

EC

S

ERSC

Direct coupling


240/360


Required at very low frequency

Affect biasing of other stage (consider thiswhile designing)

Frequency response of Amplifier


241/360



242/360

At high f


243/360


~

cRcR 1

R

CCV

CCCC

ECER2RECER

2R

1R

+

+++ +

-- -

- -

+

-

CC

s o

LR

1wC

1wC2wC

2wC

bcCbcC

beC

ceCceC

beC

Contd.


244/360


Bandwidth


245/360


mm A2/1A707.0

Effect on band width with no. of stages


246/360


Bandwidth decreases with increase in no. ofstages

Because greater no. of capacitor in the circuit

Voltage gain

Upper and lower cut of frequency

n

mAA )(

1

)12(

11

/1ff

n

f22f )12( n/1

Two-stage RC-coupled amplifier


247/360


CCV

-

O

++

- + + + +

- - - -

1R 1R

2R2RER ER

EC EC

LR

CCCCCC

1CR 2CR

S

-

++

-

1R 2R 2CR

S

O1CR 1R 2R

LR

Contd.


248/360


A1 is always less than A2, becauseof lower Rac1 due to loading effect

21

2121||

RR

RRRRRB

ie

acfe

h

RhA

2

2

LC

LCLCac

RRRRRRR

2

222 ||

ie

acfe

h

RhA

11

21 AAA m

-

+

1B 2B1bi 1C 2C

acR1CR

2bi

Sieh

ieh1bie ih 2bie

ih

E

Problem


249/360


Calculate

input impedance

output impedance

voltage gain

both transistor

hfe= 120

hie=1.1k ohm ~ 5.6k

56 k 6.8 k 56 k 3.3 k

S

0.5k

5.6k 0.5

k

2.2k

O

+

-

VVCC 20

5F5F

500F

500F

+ ++ +

- - - --

+

5F

Distortion in Amplifiers


250/360


When wave shape of the output is not an exact replica ofinput wave

Caused by Reactive component and non-linear characteristic of transistor

Frequency distortion Caused by electrode capacitance and other reactive components

Contd.


251/360


Phase distortion Delay introduce by the amplifier is different for various

frequency

Reactive components of the circuit are responsible for thisdistortion

Harmonic distortion Output contain new frequency components that are not

present in the input. New frequency are harmonics ofpresent in input

Happen due to non-linearity in the dynamic transfercharacteristic curves

Contd.


252/360


AMPLIFIERInput Output

i o

1f 2f 1f 2f 22 f12 f 13f 23ff f


253/360


Operational-Amplifiers

Introduction


254/360


The operational amplifier is most useful singledevice in analog electronic circuitry.

With only few external components, it can perform awide variety of analog signal processing tasks.

One key to the usefulness of these little circuits is inthe engineering principle of feedback, particularlynegativefeedback, which constitutes the foundationof almost all automatic control processes.

Single-ended


255/360


A "shorthand" symbol for anelectronic amplifier is a triangle, thewide end signifying the input sideand the narrow end signifying theoutput.

To facilitate true AC output from anamplifier, we use a splitor dualpower supply, with two DC voltagesources connected in series withthe middle point grounded, giving apositive voltage to ground (+V) anda negative voltage to ground (-V).

General amplifier circuit symbol

input output

SUPPLYV

SUPPLYV

input output+V

-V

inputV15 V

15 V

loadR

+

+

-

-

Differential amplifiers


256/360


Most amplifiers have one input and one output. Differentialamplifiershave two inputs and one output, the output signal beingproportional to the difference in signals between the two inputs.

The voltage output of a differential amplifier is determined by thefollowing equation: Vout = AV(Vnoninv - Vinv)

Differential amplifier

output

SUPPLYV

SUPPLYV

-

+

1input

2input

0 0 0 0 1 2.5 7 3 -3 -2

0 1 2.5 7 0 0 0 3 3 -7

0 4 10 28 -4 -10 -28 0 24 -20

1)( Input

2)( InputOutput

Voltage output equation = )( 12 InputInputAV Vout

OR))(Input)(Input(VAoutV

The "operational" amplifier


257/360


High-gain differential amplifiers came to be known as operationalamplifiers, or op-amps, because of their application in analogcomputers' mathematical operations.

Op-amphave extremely high voltage gain (AV = 200,000 or more).

Long before computers were built to electronically perform calculationsby employing voltages and currents to represent numerical quantities.

Op-amps typically have very high input impedances and fairly lowoutput impedances.

dtdvCic

Where,

ci = Instantaneous current through capacitorC = Capacitance in farads

dt

dv= Rate of change of voltage over time

dtdvmF

Where,

F = Force applied to object

m = Mass of object

dt

dv = Rate of change of velocity over time

Op-amp electrical model


258/360


1V

2V

+

-

inr

NONINVERTING

INVERTING

)( 21 VVA

OutVOutr

Comparator


259/360


+

-

CCV

inV

EEV

outV

outV

inV

0

satV

satV

Moving trip point


260/360


outV

satV

inV

satVrefV

EEV

+

-

OutV

CCV

CCV

inV

1R

2R BYC

EEV

+

-

OutV

CCV

EEV

inV

1R

2R BY

C

outV

satV

inV

satV

refV

CCVRR

R

21

2ref

Schmitt Trigger


261/360


Comparator contain noise, output may beerratic when input voltage is near to trip point

Noise causes output to jump back and forth betweenlow and high states

Noise triggering can be avoided by schmitt trigger

Contd.


262/360


Difference between UTP and LTP is called hysteresis,required to prevent false triggering due to noise

21

2

RR

RB

EEV

+

-

OutV

CCV

inV

2R 1R

satBVref

satBVref

satsat BVLTPand,BVUTP

outV

inV

satV

satV

satBV satBV

Moving trip point of schmitt trigger


263/360


1R2R

+

-

EEV

CCV

inV

outV

3RCCV

1R

2R

+

-inV

outV

+3R

- CCVRR

R

32

2

outV

inV

satV

satV

LTP UTP

CCcen V

RR

R

32

2

321

32

R||RR

R||R

B

satcen BVUTP

satcen BVLTP

Problem


264/360


Find UTP and LTP

Given

Vcc= 12 V, Vee=-12V, R2=R3=2 k ohm,R1= 100 k ohm

Sine to square waveform


265/360


outV

inV

satV

satV

LTP UTP

-

+

CCV

EEV

1R

2R

00

UTP

LTP

0

satV

satV0

Relaxation oscillator

TOWARD tV


266/360


EEV

+

-

OutV

CCV

2R 1R

R

C

B

BRCT

1

11n2

signaloutputofperiodT

resistancefeedbackRecapacitancC

fraction,feedbackB

UTP

LTP

0

satV

satV

0OUTPUT

CAPACITOR

TOWARD sat

T

Counter based A/D convertor


267/360


ClockGate andcontrol

Start

Counter

Level amplifiers

Binary ladder

Comp.Ref.voltage

Analoginputvoltage

Nlines

Nlines

Nlines

Digital output

OP-AMP ICs


268/360


Most popular 741Typical 8-pin DIP op-ampIntegrated circuit

8 7 6 5

1 2 3 4

- +

Offsetnull

Offset null

-V

Output+VNo ConnectionDual op-amp in 8-pin DIP

+

8 7 6 5

1 2 3 4

+

-

-

+V

-V

Comparator

T l


269/360


To compare two voltages Op-amps are used as signal comparators, operating in full

cutoff or saturation mode depending on which input(inverting or non-inverting) has the greatest voltage.

+V

-V

LED

-

+

inV

Pulse width modulation by comparator

O t li ti i l idth d l t d i d b


270/360


One comparator application is pulse-width modulator, and is made bycomparing a sine-wave AC signal against a DC reference voltage. As the DCreference voltage is adjusted, the square-wave output of the comparatorchanges its duty cycle (positive versus negative times). Thus, the DC referencevoltage controls, or modulatesthe pulse width of the output voltage.

inV outV

+V

-V

inV outV

-+

~

+V

-V

inV outV

-+~

inV outV

Analog to digital convertorSimple bargraph driver circuit

+V


271/360


Vin

+V

-

+

-

+

-

+

-

+

1LED

2LED

3LED

4LED

Analog to digital convertor


272/360


Analog input voltage

1A

2A

0A

Digitaloutputs

02

12

22

9318

1C

2C

3C

4C

5C

6C

7C

4/3V

8/5V

8/7V

2/V

8/3V

4/V

8/V

Low

1C 2C 3C 4C 5C 6C 7C

Low LowLow Low Low Low

Low Low Low Low Low LowHigh

021222

Binary outputComparator for level

Input voltage

0 0 0

0 0 1

LowLowLowLowHigh High Low 0 1 0

0 1 1

1 0 0

1 0 1

1 1

1 1 1

0

High

High

High

High

High

High High

High High High

High

High

High

High High High

High High High High

High High High High

Low

Low

Low Low Low

Low

Low

Low

Low

Low

High

8to0 V/

4to8 V/V/

83to4 V/V/

2to83 V/V/

85to2 V/V/

43to85 V/V/

87to43 V/V/

VV/ to87

Negative feedback

Connecting the output of an op amp to its inverting ( ) input is called negative


273/360


Connecting the output of an op-amp to its inverting (-) input is called negativefeedback.

When the output of an op-amp is directlyconnected to its inverting (-) input, avoltage followerwill be created. Whatever signal voltage is impressed upon thenoninverting (+) input will be seen on the output.

An op-amp with negative feedback will try to drive its output voltage to whateverlevel necessary so that the differential voltage between the two inputs is practically

zero. The higher the op-amp differential gain, the closer that differential voltage willbe to zero.

-

+inV out

V

-

+

The effects of negative feedback

29.99985V

5.999970000149999V

6 V

Divided feedback

N i ti A lifi


274/360


Non-inverting AmplifierA negative-feedback op-amp circuit with the input signal going to thenoninverting (+) input is called a noninverting amplifier. The outputvoltage will be the same polarity as the input. Voltage gain is given bythe following equation: AV = (R2/R1) + 1

-

+

Vin

1R 2R

1K 1K

VoutVout

Contd.

Inverting Amplifier


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Inverting AmplifierA negative-feedback op-amp circuit with the input signal goingto the "bottom" of the resistive voltage divider, with thenoninverting (+) input grounded, is called an inverting amplifier.Its output voltage will be the opposite polarity of the input.

Voltage gain is given by the following equation: AV = R2/R1

-

+

Vin

All voltage figures shown in reference to ground

0 V

1K 1K

1R 2R

0 V

Vout

Average circuit

Passive averager Circuit VVV


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Passive averager Circuit

outV

1V 2V 3V

1R

2R

3R

With equal value resistors:

outV 3321 VVV

321

3

3

2

2

1

1

111

RRR

R

V

R

V

R

V

321

3

3

2

2

1

1

111

RRR

R

V

R

V

R

V

1V 2V 3V

1R 2R 3R

outV

Summer circuit

A summer circuit is one that sums or adds multiple analog voltage


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A summer circuit is one that sums, or adds, multiple analog voltagesignals together. There are two basic varieties of op-amp summercircuits: noninverting and inverting.

Summer circuits are quite useful in analog computer design.

33 321

VVVVout

321 VVVVout )( 321 VVVVout

-

+

R

RR

R

0 V

0 V

1V

2V

3V

1I

2I

3I

321 III

outV

-

+outV

R1V

3V

R

R

2V

1 k 2 k

Differentiator circuits

Differentiator produces a voltage output proportional to the


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Differentiatorproduces a voltage output proportional to theinput voltage's rate of change.

Applications: analog computation, process control

-

+

0 V

0 V

Differentiator

0 V

RC

inV

outV

CChangingDCVoltage

dt

dvCi

dt

dvRCV inout

Integrator circuits

i t t d lt t t ti l t th


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integratorproduces a voltage output proportional to the

product (multiplication) of the input voltage and time

Applications: analog computation, process control

-

+

0 V

0 V

Integrator

0 V

R C

inV

outV

RC

V

dt

dvinout

OR

cdtRC

VV

tin

out 0

Where,C=Output voltage at start time (t=0)

Voltage-to-current signal conversion

Voltage signals are relatively easy to produce directly from transducer


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Voltage signals are relatively easy to produce directly from transducerdevices, whereas accurate current signals are not.

DC current signals are often used in preference to DC voltage signalsas analog representations of physical quantities. Current in a seriescircuit is absolutely equal at all points in that circuit regardless ofwiring resistance, whereas voltage in a parallel-connected circuit mayvary from end to end because of wire resistance.

Current independent of load resistance

-

+

250 4 to 20 mA

+

-

4 to 20 mA

inV 1 to 5 volt signal range

loadR-

+

Slew rate

Maximum rate of output voltage change


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Maximum rate of output voltage change

dt

dvCi C

i

dt

dv

c

out

C

I

dt

dv max sVpFA

dt

dvout

/2

30

60

Slew rate distortion

+10 V+10 VRSSlope


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+10 V

0

-10 V

-10 V

Frequency response of Op-amp


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1

1

10 100 1 10 100 1

10

100

1000

10,000

70,700100,000

unityf

ZMHZkHZH

OLf

OPEN-LOO

P

VOLTAGE

GAIN

FREQUENCY

Directly coupled amplifier

Gain bandwidth product constant

Op-amp Models and circuits

+V


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A simple operationalamplifier made fromdiscrete components

+V

-V

Input(+)

(-)Input

Output

1Q 2Q

3Q4Q

5Q 6Q


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555 Timer

555 Timer

Versatile IC have so many application


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+

+

-

-

S

R

Q

Q

CCV

5K

5K

5K

RESET

GROUND

OUTPUT

TRIGGER

CONTROL

THRESHOLD

DISCHARGE

4

5

6

8

1

2

7

3

Versatile IC, have so many applicat

Digital Electronics by rajesh

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