introduction electronics

8/14/2019 Introduction Electronics

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Introduction to

Electronic CircuitsAn informal self-study course

Written by Prof. C.K. Michael [email protected]

CONTENTS

1 Introduction

1.1 Outline of Objectives

Some basic concepts of analogue signals are introduced in this section. You will learn the basic functions of an analogue electronic system, viz. transformation and generation osignals of continuous nature. Here, signal transformation refers to all kinds of manipulationsuch as filtering (selecting signals of a certain range of frequencies) and amplification(enlarging the magnitude of a signal). Signal generation in the context of analogue electronics

can be taken to mean the construction of a signal with a specified waveform, as in the case oa sinusoidal wave generator. This section prepares you for the study of electronic amplifiersand oscillators in the later sections.

1.2 Signals

A. The concept of signals

For the purpose of this introductory course, it suffices to view signals as voltages that

change in time in a particular manner. A sinusoidal voltage, for instance, is a signal.

The definitions of frequency, amplitude and phase, are central to the description of a

signal. The diagram below illustrates the meaning of these terms. (See Fig. 1.)

1. Introduction2. The bipolar junction transistor (BJT)3. Amplifiers4. Operational amplifiers5. Feedback and oscillators

Page 1 of 43Introduction to Electronic Circuits

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Fig. 1: Sinusoidal signal

The mathematical expression for this sinusoidal signal is

v(t) =A sin (2 f t+ ).

The units of measurement should be noted. In particular, your attention is drawn to the

unit of frequency which is cycle per secondorhertz. It is also important to distinguish

between peak values and root-mean-square values for the description of amplitudes o

signals, as studied in an earlier section in the course.

Remarks:

A common misunderstanding is worth noting: the unit of frequency is NOT per

second, as may be wrongly concluded from the dimension of frequency. In fact it is

cycle per second! The consequence of this misunderstanding can be serious. 1 cycle

actually equals 2radian, with the radian being the truly dimensionless unit. Thus,

writing 50 Hz as 50 per secondwould result in an unacceptable error! Alternatively,

you may use angular frequency which has the unit of per secondor radian per

second.

B. Comparing signals

Before introducing the concept of amplification, it is important to explain how two

signals of the same frequency can be compared. The simplest comparison is to take

the ratio of the amplitudes of two signals, leading to the concept ofgain. Moreover, it

is important to mention that since the magnitude of gain is usually quite large, it is

better to employ a "compressed" scale. This brings up the unit decibels (dB). Whentwo voltages of amplitudes V1 and V2 are compared, their ratio (gain), in dB, is given

by the formula:

dB

Another type of comparison between two signals of the same frequency is the phase

difference. This can be expressed in either degree or radian.

Phase difference = 2

- 1

where 1 and 2 are the phase angles of the two signals relative to a fixed referencesine wave.

Class discussions:

What is the amount of gain or attenuation in dB when a signal is reduced by half in

power or by a factor of 0.7071 in amplitude? (ans: 3dB)

C. What is analogue electronics?

At this point, it is worth mentioning, as a prelude to the study of analogue electronics,

that the subject deals with signals of continuous nature. (At a later stage, you willcontrast this with digital electronics which deals with signals of discrete nature.)




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Remarks:

For the purpose of these introductory notes, it suffices to regard analogue electronics

as systems that eithertransform orgenerate continuous signals. Here, the function o

transforming signals may include filtering, amplification and phase shifting, and the

generation of signals refers in particular to the creation of signals of certain specified

waveforms, as in the case of oscillators.

1.3 Interim Review and Recapitulations

Upon completing this section, you should be well aware of the characteristic features o

analogue electronics. More specifically, you should be able to

1. describe a sinusoidal signal (or more generally periodic signals) in terms of frequency,

amplitude and phase;

2. compare two signals in terms of gain and phase difference;

3. name the basic roles of analogue electronics.




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2 The Bipolar Junction Transistor (BJT)


The Bipolar Junction Transistor (BJT) is introduced here as a crucial electronic device for the

construction of amplifiers. The basic characteristics of BJT will be discussed, with minimal

reference to semiconductor physics. The simplest application of a transistor as a switch bringsup two operating conditions of a transistor, namely, cut-off and saturation. Without involving

complex mathematical expressions, the input, transfer and output characteristics will be

explained for active transistor operation.

2.2 The Transistor Model and Operation

A. Basic transistor model

A BJT is viewed as a three-terminal device available in two flavours: npn and pnp.

Fig. 2: Simple models for BJTs

At this point, you should note the following properties of an npn transistor (withreverse polarities for a pnp):

Property 1: The collector must be more positive than the emitter.

Property 2: The base-emitter and base-collector junctions behave like diodes, as

shown above.

Property 3: Collector current is roughly proportional to base current , and

can be written as , where is the current gain of the

transistor and is typically about 100. Both and flow to the

emitter. An important observation should be made here: The collectorcurrent is not due to forward conduction of the base-collector diode;

that diode is in fact reverse-biased. This situation may be viewed as

"transistor action".




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Remarks:

1. The npn transistor should be the initial focus of discussion in order to avoid

obscuring the essentials. You may forget about pnp for the time being. When you

have mastered the npn transistor, you should be able to extrapolate the ideas to the pnp

transistor by merely reversing signs and polarities.

2. Property 3 gives the transistor its usefulness: a small current flowing into the

base controls a much larger current flowing into the collector.

3. You should be warned that is NOT a "good" transistor parameter because itsvalue can vary from 50 to 250 for different specimens of the same transistor type.

Hence, any circuit that relies on a particular value is a bad circuit. Details will bepostponed to a later time when DC biasing is studied.

4. Property 2 implies that an operating transistor has . (Here, polarities are given for the npn transistor. They should be reversed for the pnp

transistor.)

Discussions:

The following question is interesting. Is the collector current a result of diode

conduction? The answer is "no". The collector-base diode is normally reverse-biased,

and the collector current does not vary much with the collector-emitter voltage. In

most cases of active operation, the collector current is fairly constant, in opposition to

the usual V-I relationship of a diode junction.

Fig. 3: Transistor as a switch

B. Cut-off and saturation: application as a switch

Use of a BJT in switching is typically illustrated by Fig. 3. This will conveniently

explain the concepts of cut-off and saturation.

Since this is probably the first transistor circuit you have ever met, it is worthwhile to

study this circuit in depth.

Situation 1: The idea can be put in a very simple way. Referring to Fig. 3, the

collector current should be times the base current. Also, collectorcurrent causes voltage drop across the lamp (assuming that the lamp is a

resistance). Proportionality between collector current and base current




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is maintained only if the voltage drop across the lamp is less than the total available

voltage (10V). In this circuit, the max collector current is 100 mA. So,

if the base current exceeds 100/ mA, the collector current is saturatedat 100 mA.

In this situation, since the collector-emitter voltage is less than a diode

drop, the collector-base diode is not reverse biased, invalidating the

"transistor action" stated in Property 3. The transistor is said to be in

saturation.

Situation 2: As shown in Fig. 3, the base is effectively grounded. Since both base-

emitter and base-collector diodes are not conducting, the base current is

zero. The collector current is zero, according to Property 3. The lamp is

off.

In this situation, the transistor is said to be cut-off, having no base

current to control the collector current.

Remarks:The case of cut-off is usually more easily understood by you than the case o

saturation. Indeed, you will find it confusing when trying to determine the validity o

Property 3. Here, a simple rule will help you understand this idea: Property 3 is true

only if Property 1 is true.

C. BJT characteristics for active operation

There are many possible views of an npn transistor. The usual view is to consider the

base-emitter terminals as input terminals, and the collector-emitter terminals as output

terminals. This view will conveniently give the various voltage-current relationships.

1. From Property 2, the input characteristic resembles that of a simple diode, i.e.,

exponential function. It may be helpful to give the characteristic curve o

as shown in Fig. 2.4.

2. From Property 3 and the input characteristic, the transfer characteristic is

likewise an exponential function relating and . (Mathematicalexpression may be optionally shown.)

3. Regarding the output characteristic, it is important to point out that is

independent of . The curve of versus is simply a flat straight lineunder this condition.

The above characteristics remain valid when the transistor is in active operation.

You may put together the previously described saturation and cut-off conditions to

obtain a complete picture of transistor operation.

Remarks:

Instead of a detailed examination of the mathematical expressions, you should have a

strong intuitive feeling about the characteristics.

1. The base current increases exponentially with . Typically, remains

zero until rises to about 0.6 V. Thereafter increases very rapidly. It

may be assumed that stays around 0.6 V for all practical values of .




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Fig. 4: Transistor characteristics

2. Although it may NOT be necessary to mention the characteristic equation, it is important to draw your attention to the fact that

the incremental change of is proportional to the incremental change of ,

i.e., , where the proportionality constant k is numerically equal

to the slope of the transfer characterisric curve (i.e., versus ) measured at

a given . This incremental transfer characteristic is probably the MOSTIMPORTANT transistor characteristic. You are advised to memorize it.

3. The term incremental transconductance orsmall-signal transconductance, ,may be introduced here as an important transistor parameter. Here, the formula

= may be helpful, where 25 mV at roomtemperature.

Class discussions:

Some calculations of the value of may improve understanding of its meaning.

Here, a difficult (perhaps new) concept has to be explained. While is the ratio o

the change of to the change of , its value depends on the actual value of the .

For example, try to calculate how much would fluctuate if there is a 10 mV

fluctuation of , given that the value of is set at 10 mA. Now, can you see thepossibility of signal amplification?




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At this point, you should be well aware of the basic characteristics of an npn transistor (forget

about pnp if you find it hard to swap polarities). These include:

1. an intuitive model of a BJT, i.e., the three properties;

2. operations in cut-off and saturation;

3. application as a switch;

4. active operation, with (a) base-emitter voltage more or less kept at 0.6 V; (b) base

current proportional to collector current; (c) the change of collector current

proportional to the change of base-emitter voltage; and (d) collector current

independent of the collector-emitter voltage.




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3 Amplifiers


In this section, you will be introduced to the concept of amplification and the basic technique

of constructing an amplifier using a BJT. The main transistor circuit to be studied is the

common-emitter (CE) amplifier. A simple graphical technique is used to explain theoperation of a CE amplifier and to deduce the voltage gain. The factors affecting the gain

value will be considered using the graphical technique.

3.2 Amplification: Concepts and Implementation

A. The basic idea of amplification

The idea of amplification is introduced here as a gain of amplitude when a signal

passes through an assembly of electronic devices known as an amplifier.

Alternatively, an amplifier can be considered as an electronic circuit with an input portto which a signal enters, and an output port from which an enlarged signal emerges.

The diagram shown in Fig. 5 may help explain this notion.

Fig. 5: Voltage amplifier

Two aspects of practical amplification should be highlighted. The first aspect is, o

course, the enlargement (reduction) of signal amplitude. The second aspect is the

phase shift of the output signal relative to the input signal. It is important here to point

out that a practical amplifier does not give the same gain and phase shift for signals o

all frequencies. Usually, a practical amplifier maintains a fixed gain and phase shift

for a range of frequencies only. For example, an amplifier with a first-order low-pass

cut-off should be adequate for this purpose. Your attention should be drawn to the

gain and phase shift variation as frequency increases.

1. The use of semi-log scale is recommended. The x-axis is logf, and the y-axis

is the gain in dB. Real figures may be given to clarify the usage. For instance,

if Hz corresponds to on the x-axis, then Hz corresponds to onthe x-axis as shown in Fig. 6.

2. The amplifier maintains a gain ofA dB for low-frequency signals up to Hz.

Thereafter the gain starts to drop. Here, is called the cut-off frequency orthe bandwidth of the amplifier.

3. At , the gain is exactly 3dB below the intended low-frequency gainwhich is A dB. Therefore, the output signal amplitude at this frequency is

reduced by a factor 0.7071 in comparison with the intended amplitude at low

frequencies. In terms of power, the output power at this frequency is reduced to




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half the power for low frequencies.

Fig. 6: First-order frequency response of amplifier

4. The gain continues to drop, beyond the cut-off frequency, at a rate of 20 dB

every ten-fold rise in frequency. See Fig. 6.

5. The phase shift between the input signal and output signal increases (actually

getting more negative) as frequency increases. At the cut-off frequency, theoutput signal is exactly 45 deg lagging the input voltage, i.e., phase shift = -45

deg. At about 10 times the cut-off frequency, the phase shift stabilizes to -90

deg.

Remarks:

1. The study of the amplifier concept is centered around voltage signals. This

should be sufficient for the purpose of studying transistor amplifiers later on in the

course.

2. Details of frequency response of electronic systems may be omitted for the

time being. It suffices for you, at this stage, to appreciate the usual drop of gain andphase shift in most practical amplifiers as frequency increases.

Class discussions:

Some of you may question the universality of the gain-phase features mentioned

above. To clear up ambiguity, it is worth pointing out that some amplifiers do exhibit

higher-order responses in which the gain drops off more rapidly and the phase shift

increases to a larger extent.

B. Concept of loading




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It is appropriate at this point to discuss the importance of loading effects. The

phenomenon can be well motivated by the requirements of a "good" voltage amplifier

which draws almost no current from the input signal and can supply very high current

to the load. This concept is best explained in terms of input and output resistances.

The input resistance can be viewed as the equivalent load of the amplifier as seen by

the input signal. If the output of the amplifier is regarded as a voltage source

supplying an enlarged version of the input to a load, then the output resistance can beviewed as an internal resistance of this voltage source. A "good" voltage amplifier will

need to have a large input resistance and a small output resistance. This can be

explained using a simple voltage divider concept. The model shown in Fig. 2.7 is

convenient for explaining this situation.

This model can also be used to present theMaximum Power Transfer Theorem. The

theorem states that maximum power transfer takes place when the load resistance is

chosen to have the same value as the output resistance. Mathematical derivation is

optional.

Fig. 7: Amplifier model

Remarks:

1. A simple way to illustrate the Maximum Power Transfer Theorem is to derive

the expression for the power dissipated in the load and to examine how this power

varies with the load resistance. The derivation involves first finding the current and

then using the formula . Upon differentiating the power with respect to the load

resistance, the maximum power transfer can be shown to occur when .

2. The Maximum Power Transfer Theorem must be treated with care. It may give

a wrong impression that the output resistance should be designed to equal the load

resistance. In fact, maximum power transfer takes place when the output resistance iszero. This seems to contradict the theorem, but it certainly makes sense to have zero

output resistance in order to maximize the output. So, what is wrong with the

theorem? In fact, nothing is wrong. The theorem assumes that the output resistance

cannot be changed and suggests the value of the load resistance for achieving

maximum power transfer. You will benefit from a careful reading of the statement o

the Maximum Power Transfer Theorem and a thorough examination of its meaning.

Class discussions:

You can appreciate loading effects easily if you try to calculate the actual input voltageof the amplifier when are 50 and 1k respectively and to repeat the same

calculation using 50 and 100 instead. Likewise, you can try to calculate the actual




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output voltage when are 100k and 50 respectively and to repeat the

calculation with 100 and 50 instead.

C. Basic operating principles of transistor amplifiers

The operation of transistor amplifiers is unarguably an important topic of Analogue

Electronics. It will be helpful for you to get a quick glimpse at the whole construction

process before probing into details. The following "golden trio" will serve this purpose

well.

1. Before the transistor can be made to amplify signals, it must be given a set o

DC currents and voltages around its terminals. This is called biasing.

2. Once the transistor is biased, its transfer characteristic is exploited to do

amplification. Basically, when the transistor is in active operation and has a

certain fixed , a small change of will give a wide swing of about thefixed value.

3. If a resistor is connected to the collector, the voltage across this resistor will

swing too, according to the swing of . This gives voltage amplification.

D. Biasing a BJT

The process of biasing a transistor can be viewed as an assignment of a set of DC

values for , in such a way that active operation is maintained, i.e.,ensuring the transistor is neither cut-off nor in saturation. The specific objectives are:

1. to set collector current to a desired value; and

2. to set collector voltage to about (for max voltage swing).

Fig. 8: Biasing arrangements

Biasing is best illustrated with the two examples shown in Fig. 8. The objective, for

both arrangements, is to work out the values of the resistances.

It is important to understand the design procedure and the difference of the two biasing

methods.

Circuit (a): For this circuit, the objective is to find and , in order to achieve thetarget and . Normally, you can work out the design by assuming

. With little difficulty, can be calculated for any given




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and . Finally, can be chosen to make equal to .

Circuit (b): Initially, and are chosen to satisfy the equation:

, where is the desired collector

current. Finally, is chosen to give = . Note that the choice

of and is flexible (infinite possibilities).

Remarks:

1. Circuit (a) is considered "bad" because the exact value must be known.

(Quite impossible!) Hence, calculated values are highly unreliable if differs greatlyfrom the assumed value. You are warned to refrain from using this biasing method. In

fact, nobody really uses it in practice, unless there is a compelling reason to save

resistors (e.g., lower cost).

2. Circuit (b) is better because and can be chosen to make negligible

so that the design equation is . A simple and practicalapproach is to choose and in the range of a few k or less such that is at

least ten times bigger than . Then, and essentially form a voltage divider,

and the ratio of to will determine . With the desired value of in mind,

calculate the value of to make equal to about . Finally, fine tune either

or until is really equal to . A numerical example is helpful here.

For instance, if = 10 V, choose = 9.4 k and = 0.6 k initially to give

0.6 V. If the desired is 10 mA, then set = 5V/10mA = 500 to target

= 5 V. Finally, adjust slightly until is really 5V.3. Attention should be drawn to some rules of thumb in the choice of voltage

values as well as the collector current values (about one to tens of mA).

Discussions:

To probe further, consider the following questions:

1. What is the error in the calculated for circuit (a) if the actual value differsfrom the assumed value by 50%?

2. Why should be set around ? (Hint: consider the largest possible swing

of .)

3. What modification will make the design almost independent of the value?(Hint: consider putting a resistor between the emitter and ground.)

E. Principle of amplification (simple graphical approach)

After the transistor is biased, its collector current and voltage are fixed, and the circuit

is ready for amplifying signals.

The amplifier configuration to be considered is the common-emitter (CE)configuration. Basically, input signal goes into the base, and output is taken from the

collector, with the emitter serving as a shared reference.

A simple graphical consideration is recommended for explaining the circuit operation.




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This approach can eliminate unnecessarily complex algebra and is very easy to understand.

1. The description may start with a simple and comfortable statement: The

resistor must obey Ohms law. So, , which means

.

Fig. 9: Graphical illustration of small-signal amplification

Hence, the value of must lie along a straight line having a slope of 1/

as shown in Fig. 9. But is also related to by the transistor outputcharacteristic. The intersection of this characteristic with the straight line

stated earlier will give the biasing values and .

2. You can now easily visualize the amplification process by considering a small

swing in . This small swing causes to swing widely. (Refreshyour memory here about the transfer characteristic.) See Fig. 9.

3. At this point you should look at . Guided by the graph, you should easilyconvince yourselves that the gain (the ratio of the output to input swings, i.e.,

) is equal to , where is the fixed incremental trans-conductance as studied previously.

Remarks:

This part requires some conceptual thinking, and it is worth spending more time on.The amplitude of the input voltage applied to the base-emitter should not normally

exceed one-tenth of the biased value, i.e., less than 60 mV. Too large an input swing

will cause nonlinear operation and hence distortion. There is also a limit to the output

swing. If is biased at 5 V, then the maximum swing is 5 V (actually about 4.8 Vif the saturation voltage of 0.2 V is taken into consideration).

Finally the dependence of the gain on both the biasing current and the collector

resistance should be noted.

Class discussions:

How can the small input signal be injected to the base? Direct connection of the inputvoltage signal across the base and emitter will surely destroy the biasing condition!

Can you explain the use of coupling capacitors to inject and extract the signals?




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Upon completing this section, you should be well aware of the principle of operation of a

common-emitter amplifier. In particular, you should be well acquainted with the following:

1. The role of an amplifier;

2. The practical characteristics of an amplifier in relation to gain, bandwidth and phase

shift;

3. The loading effects and the requirements on input and output resistances of a practical

amplifier;

4. The maximum power transfer theorem;

5. The procedure for biasing a transistor amplifier in the common-emitter configuration;

6. The operating principle of small-signal amplification in terms of a simple graphical

interpretation; and

7. The incremental or small-signal gain expression.




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4 Operational Amplifiers


The operational amplifier (op-amp) is introduced in this section as an important device which

finds applications in a variety of electronic systems. Some essential properties of the ideal op-

amp will be discussed in detail. These properties form the basis on which the operations osome selected op-amp circuits are to be explained. These op-amp circuits also provide an

excellent starting point for the study of the feedback concept.

4.2 Ideal Op-amps and Applications

A. The operational amplifier (Op-amp) once over lightly

Before commencing a formal discussion of op-amp circuits, it is worthwhile to explain

to you, in very general terms, what the op-amp is and how it is used. In particular, the

following facts should be mentioned:

1. The op-amp is an assembly of electronic circuits whose function is to provide

very high voltage gain (typically to ). Normally, details of theanatomy of an op-amp are not needed for understanding its basic properties.

However, you may note that the very high gain of an op-amp is actually

provided by a common-emitter stage which you have studied previously. So, at

least, you shouldn't find the op-amp a complete stranger!

2. Op-amps are now available in literally hundreds of different types, from

different manufacturers, with the universal symbol shown in Fig. 10.

Fig. 10: Operational amplifier

3. An op-amp typically has two input terminals, one output terminal, one ground

or reference terminal, and power supply terminals. (The ground and power

supply terminals are frequently not shown in circuit diagrams.)

4. Op-amps always come in Integrated Circuit (IC) packages.

5. When the (+) input goes more positive than the (-) input, the output goespositive, and vice versa.

6. Op-amps have enormous voltage gain, and they are hardly ever used without




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feedback. Up to this point, you have probably never heard of the term "feedback".

While detailed discussion of feedback is left to a later time, it may simply be

regarded, for the time being, as a way in which the output terminal is connected

to either or both of the input terminals.

7. By appropriately connecting the output terminal with the input terminals, i.e.,

feedback, many useful analogue functions can be realized using op-amps.

8. The supply voltages set the maximum and minimum voltage levels that the

output can take.

Fig. 11: Op-amp IC top-view

Remarks:

1. Do not worry too much at this stage about the detailed theory of feedback. Theplan of attack, as will become clear, is to make use of op-amp circuits to illustrate the

difficult concept of feedback. This kind of reverse treatment often works well,

especially when you have only minimal mathematical background.

2. In many electronics text-books, the op-amp is often introduced as an ideal

infinite-gain voltage amplifier, with all real physical aspects either omitted or

postponed to a later time. Although this ideal op-amp model proves to be extremely

useful in presenting the essence of op-amp circuits, it often leaves in our mind the

questions of how op-amps are actually used and what they really are. In case you feel

unsatisfied with the naive "triangle" that keeps being claimed as an infinite-gain

amplifier, think of it as a real circuit.3. The op-amp symbol may need further explanation. It is natural for you to

assume that the (+) terminal should be kept positive with respect to the (-) terminal.

This is, however, not the case! This point will become apparent when negative

feedback is discussed at a later time. The (+) and (-) symbols in fact tell the relative

phase of the output. Therefore, in order to avoid confusion, let's use the words non-

inverting and inverting, rather than "plus" and "minus".

Discussions

Without feedback, an op-amp is already useful for one application. Can you guess

what it is? (Ans: comparator)

B. The ideal op-amp model and two golden rules




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The op-amp is not a very complicated device to understand. For most practical

purposes, the op-amp behaves reasonably close to an ideal (hypothetical) amplifier

which possesses the following properties:

Property 1: The gain is infinite.

Property 2: The output resistance is zero.

Property 3: The input resistance is infinite.

The above three properties are easy to remember, but they are not very easy to use in

practical situations. However, with a bit of re-wording, these properties can be re-

written as two useful rules which can be easily applied to predict actual circuit

behaviour. Here they are:

Golden Rule 1: The output attempts to take whatever value necessary to make the

voltage difference between the input terminals zero.

(This rule is a direct consequence of Properties 1 and 2.)

Golden Rule 2: The inputs draw no current.

(This rule is a direct consequence of Property 3.)

You are advised to memorize these two rules.

Remarks:

1. Properties 2 and 3 in fact correspond to freedom of loading effects (Recall

Section 2.2.2B.), making the ideal op-amp a perfect voltage amplifier.

2. You should be reminded that Golden Rule 1 is NOT applicable when there is

no feedback! However, since op-amps are hardly ever used without feedback, the

Golden Rules usually apply.

3. It does not matter if you cannot grasp Golden Rule 1 immediately. Later on, acouple of examples will clear things up.

Discussions:

You may be asked to consider a simple voltage follower circuit in the light of the two

Golden Rules. Referring to Fig. 12, you should easily work out a relationship between

the output and the non-inverting input.

Fig. 12: Voltage follower

C. Basic op-amp circuits

Some of you may now be eager to apply the Golden Rules to the analysis of op-ampcircuits. The following simple op-amp circuits are recommended for class illustration,

because they are very popular op-amp circuits and are simple enough to allow clear

step-by-step illustrations of the application of the Golden Rules.




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Discussions:

You will get a better idea of what the circuit actually does by finding the value (or the

form) of the output voltage for different values (or forms) of the input voltage. For

example, assuming = 10k and = 100k, work out the output voltage if theinput voltage is

(i) -0.3 V (DC);

(ii) 0.5 V (DC);

(iii) 0.4 sin 20 t V (AC); and

(iv) 0.1 + 0.4 sin 20 t V (AC with DC offset).

For case (iv), you should also try to sketch the waveforms of the input and output

voltages.

2. Non-inverting amplifier

The non-inverting amplifier is shown in Fig. 14. From Golden Rule 1, the output

attempts to make the voltage at point A equal to the input. Also, since no current isdrawn into the inverting input of the op-amp (Golden Rule 2), the voltage divider

formula directly gives the relation , which means

Fig. 14: Non-inverting amplifier

The input resistance is infinitely large since no current is drawn into the non-inverting

input.

Remarks:

1. It is interesting to compare the input resistance of the non-inverting amplifier

with that of the inverting amplifier. Obviously, the non-inverting amplifier is more

desirable because of its freedom from loading the input voltage source.

2. The non-inverting amplifier is suitable for both DC and AC. Unlike the case o

the inverting amplifier, there is neither sign reversal for DC signals nor phase reversal

for AC signals.

3. Voltage follower

You may have already come across the voltage follower in a previous exercise. If not,

you should try finding out how this circuit works.




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Fig. 15: Voltage follower

Discussions:

Apart from deriving the voltage gain, you should attempt to calculate the input

resistance of the voltage follower. The result of calculating the input resistance should

give a clue to the usefulness of the voltage follower in practical situations. It is

recommended that you try to explain, in the light of loading effects, why the voltage

follower is useful. The model shown in Fig. 16 is convenient for this purpose.

Suppose the objective is to supply the load with the voltage . Unfortunately, the

source has an internal resistance (or output resistance) equal to 500. You shouldreadily see that

(i) without the voltage follower (Fig. 16 (a)), the output is only able to get

two-third of ; and

(ii) with the voltage follower (Fig. 16 (b)), the output is equal to due tofreedom of loading effects.

Fig. 16: Illustration of loading effects




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4. Current source

The op-amp can also be used to provide a constant or adjustable current for a given

load resistor. This application is simple and should be quite easily understood by you

who have only basic understanding of op-amps. It should be reasonably

straightforward to derive the value of the current being supplied to the load in Fig.

17 (a). Essentially, Golden Rule 1 ensures that the voltage acrossR is equal to ,

and Golden Rule 2 forces to be equal to the current flowing in R. Therefore, thefollowing equation holds:

In other words, the current that is being supplied to the load is determined by and

R. Note that the load current is actually supplied by the 12V source and the op-amp

simply plays the role of providing a fixed reference voltage acrossR.

(a)

(b)




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Fig. 17: Current source

Remarks:

It should be stressed that the example shown in Fig. 17 (a) represents an arrangement

for supplying current to a "floating" or "ungrounded" load. In practice, one of the

terminals of the load is often connected to ground, and in that case, the current source

of Fig. 17 (a) is not suitable.

Discussions:

Can you derive an alternative configuration that can supply current to a grounded load?

A possible arrangement is shown in Fig. 17 (b) in which the pnp transistor supplies to a

grounded load a current, either fixed or adjustable by R. You should easily verify that

the current being supplied is given by

5. Differential amplifier

The op-amp has many interesting applications apart from providing voltage gain as in

the cases of inverting and non-inverting amplifiers. For example, the circuit shown in

Fig. 18 can be used to measure the difference between two voltages.

Fig. 18: Differential amplifier

Discussions:

As a class exercise, you may be asked to prove that the output voltage of the op-amp

circuit, shown in Fig. 18, is given by the following formula:

What happens if the resistances are not perfectly matched? Suppose the actual values

of the resistances are as shown in Fig. 19. What would then happen to the output

voltage?




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Discussions:

It is interesting to consider the circuit shown in Fig. 21 and to explore its potential

application.

Fig. 21: Digital-to-analogue converter

You may be prompted to calculate the output voltage for each of the followingcombinations of input voltages:

The use of this circuit in converting a 4-bit binary number to an analogue voltage (i.e.,

digital-to-analogue conversion) should be apparent.

7. Comparator

It is quite common to use op-amps to compare two signals and to determine which one

is larger. The required circuit is simple. The op-amp alone will do!

Fig. 22: Comparator

The ability to compare signals can be attributed to the high voltage gain. You may use

the following simple argument to explain how the comparator works. Since the

voltage gain typically exceeds 100,000, the inputs must be within a fraction of a

millivolt in order to prevent the output from swinging all the way to extreme positiveor negative. It is assumed that the supply voltages are +10 V and -10 V and that the

gain is 100,000.

0V 0V 0V 1V

0V 0V 1V 0V

0V 0V 1V 1V

... ... ... ...

1V 1V 0V 1V

1V 1V 1V 0V1V 1V 1V 1V




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1. If is larger than by more than 0.0001 V, the output will swing to +10 V.

2. If is larger than by more than 0.0001 V, the output will swing to -10 V.

Remarks:

The operation of the op-amp without feedback is entirely different from that withfeedback. Essentially, in the absence of feedback, there is no way to make the voltage

at the inverting input equal to that at the non-inverting input, no matter how hard the

output tries to achieve so. Here, Golden Rule 1 fails to describe what happens!

Your attention should be drawn to the exact wording of Golden Rule 1:

"The output attempts to take whatever value necessary to make the voltage

difference between the input terminals zero."

The output only attempts to do what the rule says, but it can fail to do so.

Discussions:

Now, you should discover that the op-amp is never a perfect comparator, although it is

good enough for most practical purposes. Given the gain and the supply voltage, you

should be able to work out the resolution of the comparator. For instance, if the supply

voltage is 10 V and the gain is 100,000, the comparator works well for signals that

differ by more than about 10V/100000 = 0.1 mV.

8. Schmitt trigger

A classic example of the use of the comparator action is the Schmitt trigger, as shown

in Fig. 23. This circuit may prove to be difficult to understand, and should be studied

only if you have gained enough familiarity with the earlier op-amp circuits.

Fig. 23: Schmitt trigger

You may study this circuit based on the comparator action of the op-amp. The Golden

Rules should not be used here. A simple way to tell the story is to begin with the

assumption that the op-amp operates as a comparator. Suppose the output is at the

positive extreme, say, +10 V. The voltage at point A is, therefore, equal to

. Obviously, the input can be anything below this value. Otherwisethe output would have swung to the negative extreme. If changes to a value which

is higher than , the output will swing to the negative extreme, say, -10




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V. Interestingly, the output wont swing back to +10 V when the input comes down

and passes again. This is because the voltage at point A has become

, and the output remains at the negative extreme unless goes down

below the voltage at point A. So, only if goes below , the outputswings back to +10 V. This circuit has an interesting ability of memorizing the history

of the output!The behaviour of the Schmitt trigger of Fig. 23 can be summarized by a sequence o

events that takes place when the input goes up and down, above and below the

threshold values of and .

1. The output is at +10 V, and the input is lower than .

2. The output remains at +10 V until the input goes up to , at whichthe output swings to -10 V.

3. The output remains at -10 V until the input goes down to , at

which the output swings to +10 V.The waveforms shown in Fig. 24 illustrate this sequence of events.

Remarks:

1. The Schmitt trigger operates entirely differently from the non-inverting

amplifier although it looks pretty much like a non-inverting amplifier. In fact, there is

one important difference. In the Schmitt trigger, the feedback is connected to the non-

inverting input, rather than the inverting input.

Fig. 24: Waveforms from Schmitt trigger

2. To avoid obscuring the essentials, you are recommended not to consider

positive and negative feedback at this stage. Indeed, detailed explanation of the circuit

in terms of feedback is not necessary here. You should readily understand the circuit

operation in terms of comparator action.

3. The terms lower trip pointand upper trip pointare often used to describe the




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two threshold levels for the Schmitt trigger.

4. The term hysteresis is often used to describe the operation of the Schmitt

trigger. This refers to the apparent memorizing ability of the circuit. This particular

viewpoint may, however, be omitted to avoid confusion.

Discussions:

To probe further, you may try to work out the operation of the Schmitt trigger shown

in Fig. 25. In particular, find the upper trip point and the lower trip point of this

Schmitt trigger.

Fig. 25: Schmitt trigger example


Upon completing this section, you should be well aware of the properties of the operational

amplifier and the principles of operation of various op-amp circuits. In particular you should

be

1. able to apply some simple rules to analyze op-amp circuits;

2. familiar with some common op-amp circuits such as the inverting and non-inverting

amplifiers; and

3. aware of the use of feedback in the construction of many op-amp circuits.




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5 Feedback


The concept of feedback is formally studied in this section. The emphasis is on the systematic

analysis of feedback amplifiers based on a general feedback configuration. You will learn in

this section how negative feedback stabilizes gain and how positive feedback causesoscillation. Furthermore, some practical oscillator circuits are used to illustrate how negative

feedback can become positive feedback by deliberately introducing phase shift along the

feedback path.

5.2 Feedback Theory and Applications

A. General feedback configuration

A convenient starting point for discussing feedback is the general negative feedback

configuration shown in Fig. 26.

Fig. 26: Feedback configuration

The following general procedure of analysis should be reasonably easy to understand.

1. The overall feedback amplifier is composed of an amplifier of gain A and a

feedback network that couples the output with the input. For convenience, the

amplifier of gain A, as shown in Fig. 26, is called the basic amplifier. (Note

that in some text-books this amplifier is referred to as the open-loop amplifier.)

2. Without the feedback network, the gain of the amplifier is A which is usually

called the open-loop gain, i.e.,

Open-loop gain =A

3. When feedback is connected as shown in Fig. 26, the input to the overallfeedback amplifier is substracted by a fraction of the output before being

applied to the input of the basic amplifier.

4. The input to the basic amplifier is , i.e.,

5. The output of the basic amplifier is, therefore, equal to , i.e.,




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6. From the above equation, an explicit expression for can be easily obtained.

7. The gain of the overall feedback amplifier is , which is usually

called the closed-loop gain of the feedback amplifier.

Closed-loop gain =

Remarks:

You should be aware of the usual terminology:

Open-loop gain: A

Closed-loop gain: G =

Loop gain: T=

Also, the feedback network is sometimes called the beta network.

Class discussions:

Think about the implication of the feedback action in the light of the above gain

formula. Specifically, ifA is infinitely large and is a finite number, what valuewould G take?

B. Effects of feedback --- Why is feedback so useful?

Some important desirable effects of feedback should be highlighted here. In particular

the following are most significant:

1. Predictability of gain:

The closed-loop gain G is . When the open-loop gain A is very

large such that , the closed-loop gain G is approximately ,regardless of the exact value ofA.

2. Increased input resistance (in the case of voltage amplifier):

Since only a portion of the input voltage is applied across the basic amplifier,

the ratio of the input voltage to the input current becomes bigger compared to

that of the basic amplifier. The simple model shown in Fig. 27 (a) can be used

to explain this. Essentially, comparison is made between the input resistance o

the basic amplifier, , and that of the feedback amplifier, .

Referring to Fig. 27 (a), is given by

Also, the input resistance of the feedback amplifier is




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Clearly is larger than by a factor of , i.e.,

This is definitely an advantage since loading effect on the input voltage source

is significantly reduced. The equivalent model for the entire feedback amplifier

is shown in Fig. 27 (b).

Fig. 27: Feedback amplifier model

3. Reduced output resistance (in the case of voltage amplifier):

Reduction of the output resistance can be appreciated using the same model

shown in Fig. 27 (a). Comparison is again made between the output resistance o

the basic amplifier, , and that of the feedback amplifier, .

Referring to the model of Fig. 27 (a), and assuming that the feedback network




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does not draw current from the output, the output resistance of the basic amplifier, ,is given by Ohms law as

Referring to the model of Fig. 27 (b), the output resistance of the feedback

amplifier, , is given by . Also, since G = A/(1+A), we

can rewrite as

Now, combining the above equation with the previous expression of leadsdirectly to the following relation:

Clearly, the feedback amplifier's output resistance is less than the basic

amplifiers by a factor of , i.e.,

Again this is an advantage since the amplifier has much less internal voltage drop

as explained previously in Section 2.2.

Remarks:

1. The analysis of feedback, as outlined above, is based on the voltage amplifier,

for which the input is a voltage, the output is a voltage, and the feedback network

substracts a fraction of the output voltage from the input voltage. Everything is

voltage! For other types of amplifiers, the analysis is generally the same but the

conclusion could be different. For the purpose of this introductory course, the voltage

amplifier suffices to illustrate the basic principles.

2. The above analysis of output resistance has assumed that the feedback network

does not draw any current from the output. However, in practice, the feedback

network often draws current from the output. For example, in the non-inverting

amplifier shown in Fig. 28, the feedback network is a simple voltage divider which is

composed of two resistors. In this case, the output resistance would contain an

additional parallel resistance which is equal to . Such an intuition leads to the

following formula for .

If you find it hard to apply intuitive reasoning in this situation, try to prove the result

formally. This may not be necessary but would certainly be beneficial. The following

class exercise (or homework) may be attempted.




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Fig. 28: Illustration of loading effects (optional)

Discussions:

The non-inverting amplifier is an excellent example for illustration of the effects o

negative feedback. Referring to Fig. 28, a portion of the amplifier output is substracted

from the input, cancelling part of the input and making the gain much less than what itused to be without feedback. You may attempt to verify the following:

1. The value of is ;

2. The closed-loop gain is approximately (i.e., );

3. The input resistance is ; and

(Caution: Careless algebra often gives !)

4. The output resistance is .

You may try to prove these results without resort to intuitive reasoning. While the

proofs of the first three results are straightforward, the fourth result may be a bit tricky

to get. The derivation of the output resistance may be proceeded as follows.

The gist of the problem originates from the non-zero current drawn by the feedback

network. This current is . Therefore, the expression for is

Again, using = and G = A/(1+A), the output resistance of thefeedback amplifier is obtained, after a few lines of algebra, as




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You may find this result surprisingly simple! Indeed, as can be intuitively reasoned,

the feedback network essentially imposes an additional resistance, , inparallel with the feedback amplifiers output resistance.

C. Negative and positive feedback

So far, the effects of negative feedback have been discussed in rather quantitatve

terms. At this point, some intuitive viewpoints may help you appreciate the subtleties

of the feedback concept.

1. Negative feedback is the process of coupling the output with the input so as to

cancel some of the input. The result is a much smaller closed-loop gain incomparison with the original open-loop gain. In exchange, the gain becomes a

lot more predictable in the sense that it depends only on the feedback network

and is unaffected by any of the undesirable characteristics of the basic amplifier

such as nonlinearity and distortion.

For example, the op-amp normally has a gain of 100,000, but the closed-loop

gain of a non-inverting amplifier can be less than 100. In this case, negative

feedback "eats" up almost all the open-loop gain!

2. Positive feedbackis also possible. Instead of cancelling out some of the input,

the feedback network actually adds a portion of the output to the input, before

feeding into the basic amplifier. This makes the closed-loop amplifier gaineven bigger! In practice, the output cannot take the value it is supposed to take

(too large!), and it could at best swing to either the positive extreme or the

negative extreme (limited by the power supply). This is exactly how the

Schmitt trigger works.

3. Negative feedback can become positive feedback if the feedback network has

the ability to revert the sign of the output voltage prior to subtracting it from the

input. In the case of AC, if the feedback network is able to introduce a phase

shift of , the intended negative feedback will again become positivefeedback.

Remarks:

1. You may recall the Schmitt trigger where the positive feedback is present in the

circuit.

2. Positive feedback is usually regarded as unstable configuration because it tends

to give an endless boost to the input signal. Therefore, all practical configurations o

positive feedback are accompanied by some kinds of gain limiters (or amplitude

limiters) in order to prohibit the growth of the voltage in the feedback loop. Many

oscillator circuits are based on the combined use of positive feedback and gain limiters

as will be discussed later.

Discussions:

The closed-loop gain of a feedback amplifier depends only upon the feedback




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network. What would happen to the closed-loop gain if the feedback network itself is

frequency dependent? There are two possible areas of discussion: (1) construction o

filters; and (2) hazard of unwanted oscillation.

D. How can oscillation start and be sustained?

By now, you should have a vague idea of how positive feedback can cause oscillation.The simplified models of Fig. 29 will help you grasp the idea more firmly.

Fig. 29: Oscillation by feedback

The model of Fig. 29 (a) is a simple positive feedback configuration whereas that o

Fig. 29 (b) is a negative feedback with an additional phase shift of . Thediscussion starts with two assumptions.

Assumption 1: Signals of all frequencies exist in the loop at the very beginning.Assumption 2: Only the signal at a selected frequency is allowed to go around the

feedback loop.

For both models of Fig. 29, oscillation will start when the output is sent back to the

input and gets re-amplified by the amplifier provided is larger than 1 for theselected frequency. The process goes on continuously, and very soon the output will

go wildly towards infinity! In practice the output saturates to the extreme value limited

by the power supply.

Now, a dilemma arises. With less gain the oscillation will stop, and with more gain the

output will saturate! This seems to be a rather difficult problem. A practical solution isto reduce below 1 once oscillation starts, and keeps changing slightly aboveand below 1 continuously and "quickly" so as to sustain oscillation. Of course, this

must be done automatically by the circuit itself. A class exercise will clear this up

later.

In summary, two major processes contribute to sustained oscillation:

1. Set > 1 to start oscillation by positive feedback, or negative feedback with

a phase shift of ;

2. ControlA to sustain oscillation.




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Fig. 30: Oscillator illustrative example

Remarks:

Assumption 1 can be easily justified because "the world is noisy," and nothing can be

perfectly protected from the invasion of noise. So, signals of all frequencies already

exist in the loop.

Assumption 2 can be satisfied artificially by designing an appropriate feedback

network. In fact, oscillator design deals partly with the choice of the feedback

network, and some specific examples will be discussed later.

Discussions:

Consider the oscillator shown conceptually in Fig. 30. For the time being, discussions

need not involve actual electronic components. The necessary concept can be well

explained at the block diagram level. Suppose the feedback network has the following

properties:

1. For signals at 100kHz, = 1/3 and phase shift = 0; and

2. For signals at all other frequencies, is less than 1/3 and phase shift is within

.

At what frequency will positive feedback occur? What value ofA is needed to start the

oscillation?

E. Oscillators

You should now be ready to meet some oscillator circuits, e.g., the Wien-bridge

oscillator and the phase shift oscillator.

Before probing into details of these oscillator circuits, some general features should be

mentioned.

1. Positive feedback:

The basic structure should resemble the model shown in either Fig. 29 (a) or




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Fig. 29 (b).

2. Frequency selection:

The feedback network provides a gain of and a phase shift which are

frequency dependent. The value of the gain and the amount of phase shiftare chosen such that oscillation occurs only at a certain (desired) frequency.

Wien-bridge oscillator

The Wien-bridge oscillator is shown in Fig. 31.

You should be aware that this circuit corresponds directly to the model of Fig. 29 (a),

with the basic amplifier being the familiar non-inverting amplifier. The gain A is

therefore given by

Perhaps the most difficult part is the feedback network. The following two propertiesregarding the feedback network are needed in order to understand the circuit operation:

Fig. 31: Wien-bridge oscillator

1. For signals at 1/(2 RC) Hz, = 1/3 and phase shift = 0; and

2. For signals at all other frequencies, is less than 1/3 and phase shift is within

.

Therefore, if the basic amplifier's gain is larger than 3, positive feedback occurs at 1/2

RCHz, i.e.,

Oscillating frequency = Hz

The condition for oscillation to start is




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5. For extreme low and high frequencies, approaches zero. Intuitively, there

must be a frequency at which is maximum. At this frequency, both the seriesand parallel RC portions should be half way between short circuit and open

circuit.

6. The RC portion is half way between short and open when neither component

dominates. This happens when the frequency is at exactly 1/(2 RC) Hz.

7. Clearly, the value is expected to be maximum at 1/(2 RC) Hz.

8. To explain the phase-shift property, the approximate networks shown in Fig. 33

should be adequate. Here, you should be told that when a capacitor and a

resistor are connected in series, the capacitor voltage is always behind the

resistor voltage by . At very low frequencies (Fig. 33 (a)), should be

nearly ahead of because the capacitor voltage is nearly equal to .

Similarly, at very high frequencies (Fig. 33 (b)), should be nearly

behind because the resistor voltage is nearly equal to .

9. Intuitively, there must be a frequency at which and are just in phase.Again, this must be the point where the RC portions are half way between short

and open. This occurs at exactly 1/(2 RC) Hz.

Discussions:

You may realize that the story is only half told. The Wien-bridge oscillator can start

oscillating when , and do so at a frequency of 1/(2 RC) Hz. However, sincenothing is there to limit the gain, the op-amp will saturate very quickly. Now, you

should try to complete the story by considering the method for limiting the gain. The

arrangement shown in Fig. 34 may be used for this purpose. Here, the incandescant

lamp's resistance has a positive temperature coefficient. You should be able to work

out its operation without much difficulty.

Fig. 34: Practical Wien-bridge oscillator with amplitude limiting

Phase-shift oscillator

The phase-shift oscillator is shown in Fig. 35. You should be aware that this circuit




40/43

corresponds directly to the model of Fig. 29 (b), with the basic amplifier being the

familiar inverting amplifier. The gainA is therefore given by

The negative sign is removed because it has already been accounted for in the negative

feedback configuration as shown in Fig. 29 (b).

A similar story may be told here: The feedback network selectively produces a phase

shift of for signals at a chosen frequency. Furthermore, at this chosen frequency,

the value of should be larger than 1 in order to start oscillation. Of course, anarrangement for limiting the gain or amplitude is needed.

The feedback network consists of three RC ladder sections. For the sake of analytical

simplicity, all resistors are identical and all capacitors are identical as shown in Fig.

36. Note that in Fig. 35 one of the resistors in the feedback network is R'. This is

merely to make up for the loading effect of the inverting amplifier's input resistance.

The parallel combination of andR'is stillR as shown in Fig. 36.

Neglecting this loading effect is a very common error!

Fig. 35: Phase-shift oscillator

The important parameters to remember are the frequency at which the phase shift is

exactly , , and the attenuation at this frequency, .

Hz and

Using the same argument as in the case of the Wien-bridge oscillator, the condition foroscillation to start is:




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The oscillation frequency is Hz.

Fig. 36: Phase-shift network

Remarks:

The following properties of the feedback network are important in the operation of the

oscillator.

1. It produces phase shift for signals of all frequencies, but to different extents.

2. It produces less phase shift for signals of higher frequencies, and at the same

time it attenuates less at higher frequencies, i.e., gets larger as frequencyincreases.

3. This three-RC ladder network can produce a maximum phase shift of .

(Note: Each RC section can give a maximum phase shift of . So, two RCsections can give a maximum of , etc. However, in this type of RCladders, maximum phase shift occurs at a very low frequency. Therefore, at

least three RC sections are needed to give a phase shift of at a finitefrequency.)

You may note that the feedback network is a high-pass filter.

Discussions:

The basic amplifiers in the previous two examples are based on op-amps. It is in fact

possible to use transistor amplifiers such as the common-emitter amplifier. Aworthwhile exercise is to modify the phase-shift oscillator using a transistor amplifier

as the basic amplifier. Fig. 37 shows one possibility.

Due to the loading effect of the CE amplifier, the analysis is more complicated than the

one employing an op-amp inverting amplifier. Only an approximate analysis should be

attempted. The following questions will guide the analysis.

1. Why is one of the resistors in the feedback network replaced by 25k? (Hint:

consider the loading effects of the 100k and 20k.)

2. What is the biasing collector current of the transistor? (Hint: the voltage at

emitter is 1.4V since the voltage at base is 2V.)

3. What is the voltage gain of the basic amplifier? (Hint: the 280 can be ignoredwhen calculating gain since the big capacitor has surely shorted it out at the

signal frequency.)




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4. Will oscillation start?

5. What is the oscillation frequency?

Fig. 37: Phase-shift oscillator using transistor amplifier


Upon completing this section, you should be able to explain the feedback principle and the

effects of negative feedback. In addition, you should be able to explain how positive feedback

causes oscillation and to derive the condition for oscillation for some simple oscillator

circuits. Specifically, you should be familiar with the following:

1. the general feedback structure;

2. derivation of the formula for the closed-loop gain of a feedback amplifier;

3. the desirable effects of negative feedback;

4. the principle of sustained oscillation; and

5. two specific oscillator circuits, namely, the Wien-bridge oscillator and the phase-shiftoscillator.




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Written July 1998; updated 23 September 2003. Copyrighted materials.


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