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Lecture 10

-

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

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Characteristics of Ideal OAmps

Infinite gain for the differential input signal

-

Infinite input impedances

Zero output impedance

Infinite bandwidth

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Real Versus Ideal Op Amp

Parameter Typical Range Ideal Values

Open-loop gain A 10^5-10^8

Input resistance, Ri 10^5 to 10^13

Output resistance, Ro 10 to 100 0

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741 Op-Amp

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741 Op-AmpCurrent Mirrors

Output Stage

DifferentialAmplifier

Class AAmplifier

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

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-CONSTRAINT

used with negative feedback, in which part of

opposition to the source signal.

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In a negative feedback system, the ideal op-

to force the differential input voltage and input

summing-point constraint.

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

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-analyzed by the following steps:

1. Verify that negative feedback is present.

2. Assume that the differential input voltageand the input current of the op amp are

forced to zero. (This is the summing-point

constraint.)

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The Basic Inverter

decreasesvvv xox

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Applying the Summing Point

Constraint

0 Rvvvvv

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

11211

22

RvRRRR inv ======

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Inverting Amplifier

1Ri

vZ

inin == 0

2 == outinout ZvR

RV

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Summing Amplifier

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Summing AmplifieriA

iF

iB

VVVV

==== BABAF

BB

AA

VVVVvv

RRRR

0

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

====

BA

Fo

BAFF

FRR

vRRRR

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Summing Amplifier

RAiA

== B

B

BA

A

A R

v

iR

v

i

V = 0

A

Vout

RfV

iout

iB

==

BAout

F

ou

F

ouout

vvv

RRi

RB

==

BA

BAF

BAout

vv

RRR

BAout RR

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

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Summing Amplifier

Input resistance seen by vA= RA

Input resistance seen by vB= RB

Since the output voltage does not depend on the load resistance

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

RL, the output impedance is zero.

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Exercise 14.2(a)

n 1, 2, x, oan vo:

11

1

1 mAk

V

R

v

i

in =

==

1011

1

2

12

kRv

mAii

o

==

11110

11

22

2

mAmAmAiiiiii

kkRRR

oxox

in

LL

o

====+

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

)1(

1

10)1)(10(

2

2 V

k

v

R

VkmAv ino

====

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Exercise 14.2(b)v

v

imAk

Vi

0

51

521

==

=

==

=

mAk

V

k

vi

k

51

5

1

01

3 =

=

=

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

Vvk

v

k

vvmAmAmAiii o

oo 1511

1055324 =

=

==+=+=

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Exercise 14.3

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Exercise 14.3

1

10vvo =

amplifier amplifier

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

212

12 24

10101020

101020 vv

kv

kk

kkkv

oout =

+

=

+

=

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With positive,

amps input and

increase in magnitude

voltage reaches one

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Non-inverting Amplifier

== 10 vvv ini

+==+=

2

1

2

1

21

21

11 1

Rv

vRvRvvRRv

o

ininoo

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

==

1Rvinv

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Non-inverting Amplifier

0 vin

=

=

1

vvR

outin

in

iin iout

Vin0 Vout

R1 R2 =

=

2

vvvii

R

outininoutin

ou

+=

21

R

v

R

v

R

v ininout

+= 21R

Rvv inout

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NONINVERTINGAMPLIFIERS

Under the ideal-op-am assum tion the

non- inverting

am lifier is an idealvoltage amplifier

havin infinite in ut

resistance and zeroout ut resistance.

21 Rv

A ov

+==

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

1in

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Voltage Follower

02Rvo

1 ==== Rvv

in

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Exercise 14.4

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Exercise 14.4

i1 = 0

i2 = 0

inv

inv

ininin vRivv =+=+

vo = vin

in

ii

R

vvi

==

=

= +

12

1

0

0

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

ininout vvRiv =+= 2

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Exercise 14.4

0

0

inino vvR

Rv ==

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Voltage-to-Current Converter

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Voltage-to-Current Converter

v

ino

Rvi =

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Exercise 14.6

vin

i1i1

v2

ii3

1= vi in

212

211

21111212

1

1

)()(

+

+=+=+=

RRvvvvv

RRR

vRRiRiRiv

inininin

in

1

21

1

221

1232

11

21

11111

213

1)(

++++=+=

=

===

R

RR

R

RvRR

R

vRivv

RRRRRRR

inino

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

2

1

2

1

2

1

2

1

2

1

2

1

2

1

2 311

++=++++=

R

R

R

R

R

R

R

R

R

R

R

R

R

R

v

v

i

o

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Amplifier design using op amps mainly

consists of selecting a suitable circuit

configuration and values for thefeedback resistors.

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If the resistances are too small, an

impractical amount of current and power

will be needed to operate the amplifier.

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Ver lar e resistance ma be unstable in

value and lead to stray coupling of undesired

signals.

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Example 14.3Want the voltage gain to be -10 5 percent:

Varying

resistance

Rv Need R >>R so that variabilit in

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.

1

=+

=RRv Ss RS is a small percentage change

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Example 14.3

o ge e ga n o , c oose 2 1 =

Since R1, RS, R2 can all vary, use 1% tolerance resistors:

R1 = 49.9k 499R2 = 499k 4.99k

ELECTRICAL ENGINEERING: PRINCIPLES AND APPLICATIONS, Fourth Edition, by Allan R. Hambley, 2008 Pearson Education, Inc.