Slide008 Op-Amp Intro

8/22/2019 Slide008 Op-Amp Intro

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EE 4623: Analog Integrated Electronics

Operational Amplifier Applications


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Objectives

Students should be able to: Explain basic amplifier terms including:

Gain, Linear, Distortion, Decibel, Saturation, Biasing, Models,Cascade, Coupling Capacitor, Transconductance, Open- andClosed-Circuit Models.

Calculate gains for amplifier configurations. Understand tradeoffs inherent in amplifier designs.

Identify the four amplifier types.

Understand how frequency response relates to

amplifiers.Analyze op-amp simple circuits and effect of a finite

open-loop gain .

Understand the effects of offset voltage and offsetcurrent in an Op-Amp.


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Amplifiers

o iv A v

Amplifiers are 2-portnetworks:

input port

output port

A is called the amplifiergain.

If the gain is constant, we call this a l inearamplifier.

iv ovA

iv ov


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Amplifier Gain in Decibels

Amplifier gain is expressed in decibels (dB) Originally it was expressed as Bels (named afterAlexander Graham Bell), but these proved to be ofinsufficient size so we multiply Bels by 10 decibels.

Decibels are a log-based ratio and are thereforedimensionless.

Purpose: We want to measure the ratio of some valuerelative to another (e.g. sound power in a stereoamplifier).

Derivation of dB


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Derivation of Decibels

Ratio of power of interest (call it p1) to some other

reference power (say,p2):

However, these values are generally quite huge and

tend to be logarithmically related; thus, creation of

the Bel:

1

2

p

p

1

2

logp

Belp


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Derivation of Decibels (Contd.)

However, as mentioned, the Bel is a bit too small,

so lets multiply it by 10 and call it a decibel (10 x

Bel = 1 dB).

Which gives us the decibel expression for power:

1

210logpower

p

decibel p


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In-Class Exercise: dB for Voltage

Given the following equation for expressing

power in decibels, write a similar equation

expressing voltage in decibels:

1

210logpower

p

decibel p


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dB for Voltage

First, lets relate voltage to power:

2

/p vii v r

p v r


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In-Class Exercise: dB for Voltage

Upon substitution:

Which gives us the decibel expression for

vol tage:

22 2

1 1 1

2 2

2 2 2

10log 10log 10logv r v v

v r v v

1

2

20logvoltagevdecibelv


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Lets Draw some Physical

Conclusions

if dB is positive, then v1 > v2, the signal is

amplified.

if dB is negative, then v1 < v2, the signal is

attenuated.

if dB is 0, then v1 = v2.


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Signal Amplification

A is the amplifiergain.

The premise is that this is a l inearamplifier.

Amplifiers that are not linear exhibit nonlinear

distortion.

)()( tAvtv io


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Gain (in Decibels)

Power Gain,Ap

Voltage Gain,Av

Current Gain,Ai


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The Meaning of Active in EEWhat makes an amplifier an act ivecircuit component?

Active components are devices that add intelligence in someway to a signal that passes through it (in contrast to passivecomponents which consume but do not produce energy).Passive components are incapable of power gain.

Active components can also switch the flow of current and arecapable ofgain. Examples: Transistors, diodes, and vacuumtubes.


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Amplifier Saturation

What is the range ofvI?

L+andLare the positive, and

negative saturation levels,

respectively.

In order to avoid output saturation,

the input must be kept within the

linear range of operation:

I

v v

L LvA A


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Nonlinearity and Biasing We want to operate in the linear

region. What have we added tothe amplifier (below) to achievethe transfer characteristic(right)?

By adding a DC offset to the

input, we bias the input voltageto operate in this region.


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Biasing

The bias point is also referredto as the

quiescent point (Q-point).

DC bias point.

operating point.

The time-varying input signal

vi (t)is superimposed on the

DC bias voltage VI to createthe total instantaneous

input,vI(t)

( ) ( )I I iv t V v t


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Coupling Capacitors

You will be using coupling capacitors rather heavily so itis good to get an idea about what they are accomplishing!


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Symbol Convention

Explain the nomenclature on the following graph:


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Closed- and Open-Circuit

Amplifiers


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Open-Circuit Amplifier Model

What is the relationship ofvo to vi in the open-circuit voltageamplifier model? Here, R

iaccounts for the fact that the amp draws an

input current (ii) from ______. R

oaccounts for changes in output

voltage as the amp is called on to supply output current (io) to _____.

Avois the open-circuit (voltage) gain factor (i.e., it is the gain of the

unloaded amplifier)


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Closed-Circuit Amplifier Model

What is the relationship ofvo

to viin the following closed-

circuit

model? What is Rs? Finally, what is the relationship ofv

oto

vs?


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Procedure/Solution

oL

L

si

ivo

s

o

si

isi

oL

Lvo

i

ov

Lo

oL

Livoo

RR

R

RR

RA

v

v

RR

Rvv

RR

RA

v

vA

RRRR

RvAv


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Cascaded Amplifiers


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Cascaded Amplifiers

Calculate the voltage gains of each stage assuming thenext stage is the load of the previous.


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vi1/vs = 0.909 (i.e., a lossof gain at the input)

Av1 vi2/vi1= 9.9

Av2 vi3/vi2= 90.9

Av3 vL/vi3 = 0.909 (Voltage gain of output stage)

Av vL/vi1 =Av1 Av2 Av3 = 818 = ______dB (Total gain)

Gain (Source-to-Load): vL/vs = (vL/vi1)(vi1/vs) = 818 0.909 =_____


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about

Cascaded Amplifiers

To avoid losing signal strength at the input (where the signal isusually very small), the first stage has a relatively high inputresistance. The tradeoff is a moderate voltage gain.

The second stage doesnt need to have as high an inputresistance, and performs the bulk of the voltage gain.

The third stage provides no voltage gain, but functions as a

bu ffer amp li f ier, with a large input resistance and very smalloutput resistance (even lower than RL). This allows connectionto a small 100 load.


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Questions

What would the overall voltage gain be without stage 3?Without stage 2? Without stage 1?

IfRL varies from 10 to 1,000 , what is the range of thecorresponding voltage gain?

Answer: Range is 409 Av 810.


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Other Amplifier Types

Input signals of interest can be voltages or currents.

Some transducers have very high output resistances and can bebetter modeled as current sources.

Output signals can be either as well.

The most popular is the voltage amplifier, but it is just onetype of model. Other are:

Current amplifiers Transconductance amplifiers Transresistance amplifier


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These models are unilateral(i.e., signal flow isuniderectional from input to output). Most real amplifiersshow some reverse transmission (generally undesirable ).

Our transistor models are generally of thetransconductance type.


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Operational Amplifiers: Ideal

The Inverting Configuration


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Circuit Symbol


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741 Op Amp (Pin-Out Diagrams)

Ch t i ti /P ti


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Characteristics/Properties

(Ideal Op-Amp)

Property No.1: Infinite input impedance (Zin)

Input impedance is the ratio of input voltage to input current (Zin = Vin/Iin)

As Iin0Zin.

High-grade op-amps can have input impedance in the T range.

Some low-grade op-amps (on the other hand) can have mA input currents.


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Ideal Properties (Contd)

Property No.2: Zero output impedance (Zout0)

The ideal op-amp acts as a perfect internal voltage source with no internalresistance.

Real-life op-amps have some internal resistance which then is in series with

an external load, thus reducing the output voltage available to the load. Example:

Real-life op-amps have output-impedance in the 100-20 range.


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Property No.3: Infinite Open-Loop Gain (A vo l)

Open-loop gain is the gain of the op-amp without positive or

negative feedback.

Ideally infinite (typical values range from 20,000 to 200,000 inreal-life devices; normally feedback is applied around the op-ampso that the gain of the overall circuit is defined and kept to a figure

which is more usable).

Open-loop gain falls very rapidly with increasing frequency.


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Property No.4: Zero Common-Mode Gain (what does this mean?!?)

An ideal op-amp only amplifies the voltage di f ferencebetween its two

inputs.

If these inputs were to be shorted together (thus ensuring zero potential

difference between them), there should be no change in output voltage

for any amount of voltage applied between the two shorted inputs and

ground.

Example:


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Common-Mode Gain (Meaning)

Voltage that is common between either of the inputs and ground (as Vcommon-

mode is in this case) is called common-mode voltage.As we vary this

common voltage, the perfect op-amp's output voltage should hold absolutely

steady (i.e., no observable change in output for any arbitrary change in

common-mode input). This translates to a common-mode voltage gain of

zero.

The performance of a real op-amp in this regard is most commonly measured

in terms of its differential voltage gain (i.e., how much it amplifies the

difference between two input voltages) versus its common-mode voltage gain

(i.e., how much it amplifies a common-mode voltage). The ratio of the former

to the latter is called the common-mode rejection ratio (CMRR):

)(

)(

ModeCommonv

alDifferentiv

A

ACMRR


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Common-Mode Rejection Ratio

An ideal op-amp, withAv(Common-Mode) = 0 would have an infinite CMRR.

Real-life op-amps have high CMRRs (the ubiquitous 741 having somethingaround 70 dB, which works out to a little over 3,000 in terms of a ratio).

A note about feedback compensation

Because the CMRR in a typical op-amp is so high, common-mode gain is usuallynot a great concern in circuits where the op-amp is being used with negativefeedback.

If the common-mode input voltage of an op-amp circuit were to suddenly change(thus producing a corresponding change in the output due to common-modegain), that change in output would be quickly corrected as negative feedback andAvolworked to bring the system back to equilibrium.

Conclusion: A change might be seen at the output, but it would be a lot smaller

than what you might expect!


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Characteristics (Contd)

Property No.5: Infinite Bandwidth (What does this mean?!?)

The ideal op-amp will amplify all signals from DC to the highest

AC frequencies.

In real op-amps, the bandwidth is rather limited; this limitation is

specified by the Gain-Bandwidth product (GB), which = the

frequency where the amplifier gain unity.

Some op-amps (such as the 741 family) have very limited

bandwidth of up to a few KHz.


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Characteristics (Contd)

Property No.6: Zero Output Offset (What does this mean?!?)

The output offset is the output voltage of an amplifier when bothinputs are grounded, i.e.;

Ideal op-amp: Zero output offset.

Real op-amps: Finite amount of output offset voltage


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The Inverting Configuration


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Closed-Loop Gain

Determine the closed-loop voltage gain vo/vI


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(Solution)

1

2

21211

111

11

12

0

0

0

R

R

v

v

RiRivv

R

v

R

v

R

vvi

A

v

vve

evA

I

o

o

III

vol

o

in

in

ovol


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Synthesis of the Feedback Resistor

Design an inverting op-amp where the closed-loop gain is =10 and the total resistance is

150K.


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(Solution)

2

1

2 1

2 1

1 1

1

2

10

10

150

10 150

13.63

136.3

o

I

v R

v R

R R

R R K

R R K

R K

R K


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Synthesis of the Feedback Resistor

Design an inverting op-amp where the closed-loop gain is =10 and the total resistance is

150K.


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(Solution)

k3.136

;k63.13

k15010

k150:But

10

10

2

1

11

12

12

1

2

R

R

RR

RR

RR

R

R

v

v

I

o


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Closed-Loop Gain II

Determine the closed-loop gain vo/vi


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(Solution)

1 2

1

2 2

2 2 23

3 3 1 3

24 2 3

1 1 3

4 4

22 2 4 4 2 4

1 1 1 3

2 4 2 4 2 4 2

1 1 1 3 1 1 3

2 4 2

1 1

0

1

1

I

x

xI

II

o x

I Io I

o I I

o

i

vi iR

v i R

v i R Ri v

R R R R

v Ri i i v

R R R

v v i R

v v Rv i R i R R v R

R R R R

R R R R R R Rv v v

R R R R R R R

v R R R

v R R

3

R


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The Very Real Effects of Finite

Open-Loop Gain


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Recall the Ideal Op-Amp Model:


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In Reality, v1 0

The Effect of Finite Open Loop


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The Effect of Finite Open-Loop

Gain

012 vol

oin

A

vvve

111

11

R

Av ol

vv

R

Av ol

vv

R

vvi

oI

oI

I

21

:

RiA

vv

KVL

vol

oo

21

RR

A

vv

A

v vol

oI

vol

o

vol

volI

o

AR

R

ARR

RR

v

v

1

2

12

12

1

make...thereforeshouldThus,/)/1(1

/

???Why


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What is a Weighted Summer?

Simply put, it is an op-amp configured to produce the weighted sum

of two or more voltage inputs


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Analysis of a Weighted Summer

Use superposition to analyze the following weighted summer (i.e.,

write an expression forvo in terms of the input voltages and input andfeedback resistors).


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Solution

n

n

fff

o vR

Rv

R

Rv

R

Rv ...2

2

1

1

n

nn

R

vi

R

vi

R

vi ...,,

2

2

2

1

1

1

nf iiiii ...21

f

o

R

vi

0:Where

fo iRv


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The Non-Inverting Configuration

How to derive the closed-loop gain (vo/vI) for this

configuration:


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Closed-Loop Gain Derivation

111

11

0

:lawsOhm'

R

v

R

v

R

vvi III

2

1

21 0

:KVL

RRvvv

Rivv

IIo

Io

vol

I

ovcl

A

RR

vvA

:Observe

1

:Thus

1

2


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Design Exercise

Design a non-inverting op-amp where the gain

is +10 and the total resistance is 150k. Sketch

the final circuit.


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Solution

1 2

1

2

1

2

150k

1 10

15k

135k

R R

RR

R

R


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The Difference Amplifier


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What is a Difference Amplifier?

A Difference amplifier combines features of the inverting amplifier

and the non-inverting amplifier. It is the complement of the summing

amplifier and allows the subtraction of two voltages or, as a special

case, the cancellation of a signal common to the two inputs.


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Use superposition

set v1= 0, solve forvo (i.e.,

this is a non-invertingamp)

set v2= 0, solve forvo (i.e.,

this is an inverting amp)


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Result of setting v1= 0 and then solving forvo (non-

inverting configuration):

2

43

4

1

22 1 v

RR

R

R

Rvo

43

42

RR

Rv


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Result of setting v2= 0 and then solving forvo (inverting

configuration):

1

1

21 v

R

Rvo


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Add the two results

vo = vo1 +vo2

1

21

43

4

1

22 1

R

Rv

RR

R

R

Rvvo

Hint on the Shunt Resistor


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Hint on the Shunt Resistor

Observation

WithoutR2, what

happens if there is a

tiny dc voltage?

WithoutR2, what

happens if there is atiny dc input offset

current?


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Examples and Problems

Examples: 1.2, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 1.10, 1,11, 1.12, 1,13,

Problems: 1.8, 1.11, 1.16, 1.18, 1.19, 1.25, 1.26 1.31, 1.43, 1.44


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Assignment No: 04

Problems: 1.16, 1.18, 1.19, 1.25, 1.26 1.31, 1.43, 1.44

Last date of submission: 07-05-2013

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