ct1 unit 4

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Unit 4: Amplifiers 4-1

AMPLIFIERS

4

By the end of this unit you should be able to:

1. Define or explain the meaning of the following terms:

. ac amplifier

. bandwidth

. buffer amplifier

. closed-loop gain. comparator

. current gain

. dc amplifier

. decibel

. equivalent circuit

. gain-bandwidth product

. hysterisis in a comparator

. ideal op-amp

. input resistance

. inverting amplifier

. negative feedback

. non-inverting amplifier

. open-loop gain

. operational amplifier

. output resistance

. power gain

. saturation

. summing amplifier

. tuned amplifier

. virtual ground/earth

. voltage follower

. voltage gain

2. Understand the basic operation of amplifier circuits and perform

simple calculations on voltage gain, current gain and power gain inan amplifier, given the appropriate data.

3. Understand the operation of operational amplifier circuits for the

amplification, summation, and comparison of voltage signals.

4. Calculate the voltage gain and input resistance of simple inverting,

non-inverting and voltage - follower operational amplifier circuits,

given the appropriate data.

5. Calculate the output of simple operational amplifier summing

circuits, given the appropriate data.

6. Sketch the output of operational amplifier comparator circuits for

given input signals.

OBJECTIVES

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4-2 Communications Technology 1

7. Design simple inverting, non-inverting and summing operational

amplifier circuits given the appropriate data.

INTRODUCTION

An amplifier is a very signifcant components in a telecommunications

system. Its basic function is to take an electronic signal of small

amplitude and create a similar version of it with a much greater

amplitude, so that it is less corruptible by noise that naturally occurswithin any communications system.

In the last Unit we studied the bipolar junction transistor (BJT) and the

metal-oxide-semiconductor field-effect transistor (MOSFET) and

described how these devices could provide signal amplification. We

also considered simple, complete, single-stage amplifier circuits based

on both devices. It is possible to extend such amplifier circuits to have

several stages, and consequently greater signal gain, as well as other

desirable amplifier characteristics such as high input resistance and low

output resistance. Such amplifiers can be manufactured on a tiny piece

of silicon and supplied as single integrated circuit (IC) package.

AMPLIFIER

EQUIVALENT CIRCUITS

To understand or use amplifiers, we do not need to have a detailed

knowledge of the complete circuitry inside the amplifier, but rather we

can consider the amplifier as a black box, which can be represented by

the equivalent network shown in Figure 4.1

IiIo

RiVi

+

-

Vo

+

-

Ro

Gvo Vi

+

-

Figure 4.1

Equivalent Circuit of a

Voltage Amplifier

When the voltage signal to be amplified, Vi

is connected to the input

terminals of the amplifier, a current Iiflows. From Ohms law, the ratio

Ri= V

i/ I

iis referred to as the input resistance of the amplifier. Thus

the input side of the amplifier can be modelled as a resistance Ri

comnnected between the two input terminals. The ability of the

amplifier to produce an amplified signal requires a source of controlled

energy and this is represented by the voltage generator Gvo

Vi, where G

vo

denotes the open-circuit voltage gain of the amplifier, ie the gain when

there is no load connected to the amplifier output terminals. In practice,

when the amplifier is connected to a load (other electronic circuitry), the

voltage gain will be reduced slightly from its open-circuit value. This ismodelled by the inclusion of the component R

owhich represents the

output resistance of the amplifier. In the open-circuit condition the

cuurent Io

= 0 and the output voltage Vo

= Gvo

Vi. When the amplifier

output is connected to other electronic circuitry, the current Iois not zero

and some of the amplified voltage will be lost across the output

resistance Ro, just as we saw how some voltage is lost across the internal

resistance of a battery in Unit 1.

In most practical amplifiers the input resistance is relatively high and the

output resistance is relatively low. Ideally Riwould be infinitely large,

thereby reducing the input current and the input power to zero, and Ro

would be zero minimising the output power loss in Ro.

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Figure 4.2 shows an amplifier, represented by its equivalent circuit with

an input signal source Vs

and internal resistance Rs

driving a resistive

load RL. The circuit may be easily analysed as follows.

Ii Io

RiVi

+

-

Vo

+

-

Ro

Gvo Vi

+

-

Rs

Vs RL

Figure 4.2

Equivalent Circuit of aVoltage Amplifierwith Input

Signal Source and Output

Load

IV

R RV I R

R

R RVi

s

s ii i i

i

s is= +

= =+

and .

Similarly

V RR R

G VoL

o Lvo i= +

.

Therefore the overall voltage gain Gv

is

GV

V

R

R RG

R

R Rv

o

s

i

s ivo

L

o L

= =+ +

. .

The output signal current is

IG V

R R

G I R

R Ro

vo i

o L

vo i i

o L

=+

=+

Therefore the current gain Giis

GI

I

G R

R Ri

o

i

vo i

o L

= =+

The power gain Gp

is given by

GV R

V R

I R

I R

V I

V IG Gp

o L

i i

o L

i i

o o

i iv i= = = = =

signal power into load

signal power into amplifier

2

2

2

2

/

/

The voltage gain of an amplifier is often expressed in logarithmic units

called decibels, or dBs ie

G GV

VvdB

vo

s

= =20 2010 10log log

In practice a number of amplifiers may be connected in cascade as shown

in Figure 4.3 for two amplifiers in which the output of the first amplifier

is connected to the input of the second. Thus the second amplifier acts

as a load for the first. The overall voltage gain Gvo

is

GV

V

V

V

V

Vvo

o

i

o

i

o

o

= = 11

.

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Figure 4.3

Amplifiers in CascadeVi Vo1 Vo

Amplifier 1 Amplifier 2

Expressing this in decibels gives

GV

V

V

V

V

V

V

VvodB o

i

o

o

o

i

o

o

=

= +20 20 2010

1

110

110

1

log . log log

Thus the overall voltage gain in decibels is equal to the sum of the

individual voltage gains in decibels. This is a very useful result and can

be extended to any number of amplifiers in cascade.

SAQ 1An integrated circuit amplifier has an open-circuit voltage

gain of 60dB, an input resistance of 90k and an outputresistance of 50. Its input is connected to a 1mV (rms)signal source with an internal resistance of 10k and itsoutput to a 450 resistive load. Determine the rms value ofthe output voltage across the load.

FREQUENCY

RESPONSE

So far we have been assuming that the gain of an amplifier is constant no

matter what the frequency of the input signal is. Unfortunately practical

amplifiers do not amplify all input signal frequencies to the same degree.

This often results from a natural practical limitation of the amplifier butmay also be due to the fact that the amplifier has been specifically

designed to amplify a small range of frequencies only. The general

shape of the frequency response of an ampifier which uses capacitors at

the input and output of the amplifier is shown in Figure 4.4(a). Such

amplifiers are referred to as capacitively coupled or ac amplifiers (see,

for example, Figures 3.24 and 3.29). For low-frequency input signals,

the presence of the input and output coupling capacitors causes the

amplifier frequency response to fall off as shown. The frequency at

which the voltage gain of the amplifier has fallen to 0.707Gv(-3 decibels)

is referred to as the low-frequency cut-off of the amplifier fL. At high

frequencies the amplifier gain decreases also due to internal transistor

capacitances. At the high-frequency end, the freqency at which the gain

is down 3dBs on its nominal value is referred to as the high-frequency

cut-off fH. The frequency difference f

H- f

Lis referrred to as the bandwith

of the amplifier and defines the range of input signal frequencies over

which the amplifier can provide the specified gain.

It is possible to construct an amplifier which doesnt use input and output

coupling capacitors. Consequently such an amplifier doesnt have a

low-frequency cut-off and for low-frequency input signals the gain

remains constant right down to 0 Hz, as shown in Figure 4.4(b). Another

common type of amplifer is the so-called tuned amplifier which is

common in radio circuits and is designed to amplify a very narrow range

of frequencies only. A typical tuned amplifier response is shown in

Figure 4.4(c).

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Figure 4.4

Frequency Response of an

AC, DC and Tuned

Amplifiers

NEGATIVE FEEDBACK

Another important technique that is used in amplifier design is negative

feedback. A detailed study of this topic is beyond the scope of this unit,

but basically feedback is a process whereby a signal (voltage or current)

is taken at the output side of th amplifier and a fraction of it is fed back

and subtracted from a signal on the input side of the amplifier. This

process is used mainly to alter the gain of the amplifer but also alters its

input resistance, its output resistance and its frequency response.

Figure 4.5 shows a basic amplifier with an open-loop (non-feedback)

voltage gain Gv = A in which feedback is applied such that a fraction of the output voltage across the load is fed back and subtracted from the

input voltage. The circuit may be easily analysed as follows.

G =Av

Amplifier

Vi

Io

RL Vo

+

-

+

-

Vin

+

-Vf+

FeedbackNetwork

Figure 4.5

Amplifier with Negative

Feedback

The feedback signal Vfis

V V

V V V V V

V AV A V V

V A V AV

GV

V

A

A

f o

in i f i o

o in i o

o o i

vfo

i

=

= + = +

= = +

= = =

( )

the voltage gain with feedback1

1 10 10 10 10 102 3 4 5

bandwidth of dc amplifier

bandwidth of ac amplifier

3db

1 10 10 10 10 102 3 4 5

bandwidth

of tuned

amplifier

(c) tuned amplifier

Frequency (Hz)

(log scale)

(a) ac amplifier

(b) dc amplifier

gain

Gv

0.707Gv

Frequency (Hz)(log scale)

Gain (db)db

Gv

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If the magnitude of (1 - A) is greater than 1 then the magnitude of Gvf

is less than that of A and the feedback is said to be negative, otherwise

the feedback would be positive and the amplifier output could potentially

oscillate, ie generate a sinusoid at a particular frequency . The easiest

way to ensure the feedback is negative is to use an amplifier which

provides a 180phase shift between input and output, which means that

A will be negative and the magnitude of (1 - A) is guaranteed to begreater than unity. Thus in this case the gain with feedback can be written

as

GV

V

A

AA

vfo

i

= =+

=+1

1

1

If the term 1/A is much smaller than the quantity , as is generally thecase, then the gain with feedback is given approximately by:

Gvf 1

This is an important result because it indictes that in a feedback amplifier

the gain is determined essentially by the feedback network and not the

amplifier itself. We will see examples of this in the next section.

To conclude this section on negative feedback lets consider its effect on

the frequency response of an amplifier. Figure 4.6 shows the typical

response of an ac amplifer without negative feedback. Its gain is Gv

=

A and its upper and lower frequencies are fH

and fL

respectively. When

feedback is applied the gain falls to a value Gvf

which is much smaller

than Gv= A, and the characteristic becomes much flatter. It follows that

the new lower and upper cut-off frequencies fLfand fHfare further apartand the bandwidth of the feedback amplifier is increased. In fact the

product of gain and bandwidth ( the gain-bandwidth product ) in an

amplifier is a constant. Thus if b denotes the bandwith of the original

amplifier and bfdenotes the bandwidth of the feedback amplifier then we

can say that

G b G bv vf f =

Figure 4.6

Effect of Negative Feedback

on Amplifier Frequency

Responsebandwidth with feedback

Frequency (Hz)

bandwidth without feedback

b

Gain

G = Av

0.707A

Gvf0.707Gvf

fLf

fL

bf

fHf

Gain withFeedback

Gain withoutFeedback

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OPERATIONAL

AMPLIFIERS

There are certain types of amplifier which have the general capability of

amplifying the difference between two input signals. When such an

amplifier is manufactured on a tiny piece of silicon and supplied as a

single integrated circuit (IC) package, it is generally referred to as an

operational amplifier or op-amp. The term operational derives from

the fact that an op-amp can electronically perform such mathematical

operations as addition, multiplication and integration.

Fortunately, it is not necessary to understand the detailed circuitry inside

an op-amp in order to analyse and design circuits which contain op-amp

devices. We may therefore treat the op-amp as a black box, as shown

in Figure 4.7.

+ V

- VS

S

vo

v2

v1

-

+

Inverting Input

Non-Inverting Input

Positive Supply

Negative Supply

Figure 4.7

Circuit Symbol for an

Operational Amplifier

Firstly for an op-amp to work two DC power supplies, VS, must beconnected to the device. These supplies are typically in the range 5Vto 15V. To simplify circuit diagrams containing op-amps, the powersupply connections are often omitted; but it is important to remember

that they must be connected for the amplifier to function.

There are two separate input terminals: the non-inverting input terminal(marked +) and the inverting input terminal (marked -). When a

voltage V1 is applied to the non-inverting input, an amplified voltage Vo= +AV1 appears at the output terminal, where A is the voltage gain of the

amplifier. The + sign indicates that the input and output voltages are in

phase.

If a voltage V2 is applied to the inverting input, an amplified voltage Vo= -AV2 appears at the output terminal. The - sign indicates that the input

and output voltages are 180 out of phase (antiphase).

If voltages V1 and V2 are applied simultaneously then the output voltage

Vo is given by the expression:

Vo = +AV1 - AV2 = A(V1 - V2 )

That is, the input signal V2 is subtracted from the input signal V1 and the

result of the subtraction is amplified. This signal differencing and

amplification function is very useful in many practical applications. For

example, when an electrocardiogram (ecg), which is a signal indicatiing

the electrical activity of the heart, is acquired by placing a pair of

electrodes on the human body (chest), one electrode is the reference

electrode and the ecg signal potential at the other electrode is with

respect to this reference electrode. The ecg signal is very small, of the

order only of a few microvolts, and requires a high level of amplification.

Unfortunately both electrodes pick up high levels of other electrical

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activity in the human body which appears as high amplitude-noise

signals on both electrodes, thus masking the desired signal on the ecg

electrode. This is where an op-amp can prove extremely useful. The

reference electrode is fed to the inverting input of the op-amp and the ecg

signal electrode is fed to the non-inverting input. Since the noise on both

electrodes is virtually the same, the differencing action of the op-amp

causes the noise to be cancelled and the small ecg signal is amplified bythe high gain of the op-amp.

We can draw a black box equivalent circuit of an op-amp as shown in

Figure 4.8 and summarise its main properties as follows:

. A very high voltage gain, called the open-loop gain. This is

typically in the range 105 - 106 at DC and low frequencies, but

decreases with frequency as shown in Figure 4.9.

. A very high input resistance Ri between the inverting and non-

inverting input terminals, which is typically of the order 1012.

. A very low output resistance Ro which is typically 100.

V1

V2

Vd

Vo

Ri

Ro

AVd

+

_

Vd = V1 - V2

Figure 4.8

Black Box Representation

of an Op-Amp

1

10

0

210

310

410

510

610

102

103

104

105

106

107

108

10

Frequency (Hz)

Open-Loop

Voltage Gain

Figure 4.9

Open-loop voltage gain as a

function of frequency

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THE IDEAL OP-AMP

The above properties of an op-amp - high voltage gain, high input

resistance and low output resistance - imply that it is close to being an

ideal amplifying device. In fact, the analysis and design of op-amp

circuits is much simplified if ideal devices are assumed. The properties

of an ideal op-amp are:

. Infinite open-loop gain (A = ).

. Infinite input resistance (Ri =). That is, the op-amp does not drawany current from any source connected to it.

. Zero output resistance (Ro = 0). That is, the output acts as an idealvoltage source.

The equivalent circuit of an ideal op-amp is shown in Figure 4.10.

-

+

v2 v1

v0

= A (v1

- v2)

Figure 4.10

Equivalent circuit of an ideal

op-amp

INVERTING

AMPLIFIER

The very high gain of an op-amp (typically 105 - 106) is far too high for

many applications and some method of controlling the gain is required.

This is achieved using negative feedback as described above, in which

resistors are connected around the op-amp in such a way as to reduce the

gain. Consider the circuit shown in Figure 4.11.

Figure 4.11

Op-Amp inverting amplifier

With the resistors R1 and R2 connected, the gain G = Vo/Vin is called the

closed-loop gain and will always be less than the open-loop gain, A,

which is the gain without any resistors connected. V1 and V2 denote the

values of the voltages at the non-inverting and inverting input terminals

of the op-amp. Since the non-inverting input terminal is connected to 0V

or ground, V1 = 0.

Vo = - AV2

where A is the open-loop gain of the amplifier. If the DC supply voltages

VS applied to the op-amp are 15V, and the op-amp gain A = 105, then

the maximum AC output voltage swing VO

(max) from the amplifier is

30V and the maximum input voltage swing at point P, V2 (max) is given

by:

R 2

R1P v2

v1

-

+

v0

v0

v in

-R 2

R 1=

v in

i 2

i 1

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VV

AV Vo2 5

30

10300 150(max)

(max)= = = ( )

This is a very small voltage. In fact, if we assume an ideal op-amp in

which A = , then V2 = 0 and the point P is virtually at ground or earthpotential. Thus, the point P is often referred to as a virtual earth or

virtual ground.

The current i1 through R1 is given by:

iV V

R

V

RVin in1

2

1 12 0=

= =, since

Since the input resistance of the op-amp is so high, a negligibly small

current flows into the inverting input terminal and therefore all of the

current i which flows through R1 also flows through R2.

Vo = V2 - i2 R2

= -i1R2, since V2 = 0 and i1 = i2

The gain G is given by:

GV

V

i R

i R

R

R

o

in

= =

= 1 21 1

2

1

The gain of the circuit is therefore simply the ratio of the two resistors

connected around the op-amp. The minus sign indicates that a 180o

phase shift exists between the input voltage Vin and the output voltage,Vo. Hence the term inverting amplifier.

The input resistance Rin of the amplifier is given by:

RV

i

V

V RRin

in in

in

= = =1 1

1/

SAQ 2An inverting operational amplifier circuit has R2 = 33 k andR1 = 3.3 k. Calculate the voltage gain of the amplifier.

SAQ 3Design an operational amplifier circuit which has a gain of -

100 and an input resistance of 1 k.

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NON-INVERTING

AMPLIFIER

Consider the circuit of Figure 4.12, which is similar to the previous

circuit, except in that the input voltage source is now directly connected

to the non-inverting input terminal and the input end of the resistor is

connected to ground.

Figure 4.12

Op-amp non-inverting

amplifier

An expression for the voltage gain of the circuit can again be calculated,

by assuming that no current flows into the amplifier input terminals and

that the voltage between the inverting and non-inverting input terminals

is zero.

For zero voltage between the inverting and non-inverting input terminals,

the voltage at point P must equal the input voltage V in.

iV

R

V V iR V V

RR V

R

R

GV

V

R

R

in

o in inin

in

o

in

=

= + = + = +

= = +

1

21

22

1

2

1

1

1

Because the gain is positive, the output voltage is in phase with the input

voltage and hence the name non-inverting amplifier. In this case, the

input resistance Rin of the amplifier is equal to the very high input

resistance of the op-amp, between the non-inverting and inverting

terminals.

SAQ 4A non-inverting operational amplifier circuit has R1 = 1 kand R2 = 10 k. Calculate the voltage gain of the amplifier.

SAQ 5Design an operational amplifier circuit with a gain of +50, the

smallest resistor used being 1 k.

VOLTAGE FOLLOWER

(BUFFER) AMPLIFIER

Consider the simple circuit of Figure 4.13. Again assuming an ideal op-

amp, with zero voltage difference between the inverting and non-

inverting input terminals, we can say that

V VV

Vo in

o

in

= =or 1

R2

R1

P

v2v1

-

+

v0

v0

v in

R 2

R 1=

vin

i

i

(1 + )

v0

v in

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-

+

v0

vin

Figure 4.13

Voltage Follower or Buffer

Amplifier

Thus this circuit provides a gain of +1 and the output voltage simply

follows the input voltage - hence the term voltage follower. The

input resistance of the circuit is the high differential input of the

operational amplifier. This property of the circuit, together with the low

output resistance of the op-amp, makes it useful for converting a high -

impedance source to a low - impedance one. When it is used in this way

it is often referred to as a buffer amplifier.

SUMMING AMPLIFIER

It is often required to have a circuit which can add two or more voltages

together. Consider the circuit of Figure 4.14 which is capable of adding

two voltage signals V1 and V2 and amplifying the result. Assuming an

ideal op-amp.

iV

Ri

v

Ri

v

R

o

F1

12

2= = =, and

R

RRF

P-

+

ii1

i2

v2

v1

v0

Figure 4.14

Summing Amplifier Circuit

From Kirchoffs Laws:

i i i i i i

V

R

V

R

V

R

VR

RV V

o

F

oF

+ + = = +

= +

= +

1 2 1 2

1 2

1 2

0, ( )

( )

or

The output signal, V0, is the sum of the input signal V1 and V2 amplified

by the factor - RF/R. More than two signals can be summed by having

additional input resistors, R. It is also possible to multiply the input

signals by different gain factors by having unequal input resistors R and

hence different gain factors RF/R.

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SAQ 6In the circuit of Figure 4.14, R = 10 k, RF = 10 k. Derivean expression for the output voltage for the following input

voltages.

(a) V1 = 2 sin (400t), V2 = 3 sin (400t)

(b) V1 = + 3V (dc), V2 = 4 sin (200t)

SAQ 7Design an op-amp summing amplifier which adds three input

signals V1, V2 and V3 such that V0 = -(V1+2V2+4V3). The

smallest resistor to be used is 1 k.

OP-AMP VOLTAGE

COMPARATOR

Consider an operational amplifier connected in the open-loopconfiguration with input signals V1 and V2 connected to the inverting

and non-inverting inputs as shown in Figure 4.15. From our previous

study we know that the output voltage in this configuration is given by

V0 = A(V1 - V2)

where A is the open-loop gain of the amplifier which normally has a

value in the range 105 - 106. The voltage difference between the inputs

V1 and V2 is amplified and appears at the output. Since the value of A

is so large a voltage difference of only a few hundred microvolts between

V1 and V2 causes the amplifier output to saturate, i.e. have a value close

to the supply voltage - either + VS or - VS. When V1 is slightly greater

than V2, then V0 + VS and when V1 is slightly less than V2, V0 - VS. Thus the circuit can be used as a comparator to determine whether

V1 is greater than or less than V2.

Suppose in a voltage comparator V1 is a triangular waveform and V2 is

a dc voltage as shown in Figure 4.16(a). The output voltage waveform

is as shown in Figure 4.16(b).

When the instantaneous value of the triangular signal, V1, is less than the

d.c signal, V2, the output signal has value close to - VS and when V1 is

greater than V2, the output signal has a value close to + VS. Thus V0repeatedly switches between approximately -VS and +VS producing a

pulse waveform. The mark - to - space ratio, ie the ratio of the high-levelvoltage to low-level voltage in each period, of the output waveform

depends on the switching threshold which is determined by the value of

the d.c voltage V2.

-

+

v2

v0

v1

+Vs

-VsFigure 4.15

Op-Amp VoltageComparator Circuit

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Figure 4.16

Comparator Input and

Output Signals

(a) Comparator

Input Signalst

t-vs

+vs

vin

v1

v2

(b) Comparator

Output Signals

v0

SAQ 8

In a simple voltage comparator circuit the power supplyvoltages are 15V. Input signal V1 is a triangular voltagewaveform with an amplitude of 10V and a period of 4

seconds. Calculate the mark-to-space ratio when V1 is a dc

voltage of:

(i) 0V

(ii) -5V

(iii) +5V

(iv) +12V

A disadvantage of the simple comparator configurations above is that

their behaviour is erratic in the presence of noise. If the input voltage has

superimposed noise, and has an amplitude close to Vref

the output may

be forced to change states several times. This erratic behavior may be

suppressed by introducing hysteresis into the comparator. Hysteresis is

a phenomenon in which the transition point is different when switching

from the high-to-low state as compared with switching from the low-to-

high state. Comparators with hysteresis are often referred to as Schmitt

triggers. Figure 4.17 indicates the circuitry of an inverting Schmitt

Trigger.

R1

R2

Vo

Vo

Vt-Vt

Vi

Vi _

V+Sat

V_

Sat

(a) (b)

Figure 4.17

(a) Circuitry of a Schmitt

Trigger,

(b) Output characteristic.

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Resistors R1and R

2form a voltage divider, establishing a voltage at the

noninverting terminal Vt, proportional to V

o.

VR

R RVt sat = +

2

1 2

when, Vo=+V

sat, the noninverting input = V

t

Vo=-V

sat, the noninverting input = -V

t

When Viis negative, V

o= +V

satand V+ = V

t. In order for the output to

change signs, Vimust exceed the value of V

t. To return to the previous

state the input voltage must reach a value of Vi< -V

t, causing the output

to return to +Vsat

. By choosing a sufficiently large value of Vtthe effects

of noise at the transition points may be minimised. Care however must

be taken to ensure that Vtdoes not become too large, in which case the

accuracy may be degraded for certain applications. The rectangle on the

input-output curve is referred to as a hysteresis loop. Note, it is

important to label the curve with arrows to indicate directions of change.

It is also possible to develop a non-inverting Schmitt trigger, which is

based on the same principles as above; however this circuit will have a

slightly different hysteresis loop.

SAQ 9Design an inverting Schmitt Trigger having thresholds close

to +/- 50mV based on supply voltages to the op-amp of +/-

15V. The exact value of Vtis not critical in this application.

Assume op-amp saturation voltages of +/- 14V.

SUMMARY

1. An amplifier is an electronic device for increasing the amplitude of

a signal, and may be characterised by its voltage gain, its input

resistance, its output resistance, and its frequency response.

2. Negative feedback is used to alter the gain of an amplifier but also

affects its input resistance, its output resistance and its frequency

response.

3. An operational amplifier is a high gain, BJT, FET or BJT/FET

amplifier with a high input resistance and low output resistance. It

is normally supplied as a single integrated (IC) package.

4. An ideal operational amplifier has infinite gain, infinite input

resistance and zero output resistance.

5. The voltage gain G of the inverting amplifier circuit of Figure 4.11

is given by the expression:

GR

R= 2

1

The input resistance Rin

= R1. The input and output voltages are out

of phase by 180o.

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6. The voltage gain G of the non-inverting amplifier circuit of Figure

4.12 is given by the expression:

GR

R= +1 2

1

The input resistance Rin

= . The input and output voltages are inphase.

7. A voltage follower or buffer amplifier has a high input impedance,

low output impedance and a gain of +1.

8. A summing amplifier circuit may be used to add two or more

voltages together - in different ratios if required.

9. A voltagecomparator indicates whether the instantaneous value

of one input signal is greater than or less than that of another.

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ANSWERS TO SAQSSAQ 1

GV

VG anti

GV

V

R

R RG

R

R R

V G V mV V

vodB o

ivo

vo

s

i

s ivo

L

o L

o v s

= = = =

= =+ +

=+

+= =

= = =

20 60 3 1000

90

10 901000

450

50 4500 9 1000 0 9 810

810 1 0 81

10 10log log

. .

. .

.

(rms)

SAQ 2

Voltage Gain GR

R= =

= 21

3

3

33 10

3 3 1010

.

SAQ 3

Input Resistance Rin = R1

Rin = 1 k => R1 = 1 k

Voltage Gain GR

R= = 2

1

100

or R2 = 100 x R1 = 100 x 1 k = 100 k.

Therefore, the circuit is as shown in Figure 4.18.

Figure 4.18

OP-AMP Circuit for SAQ 3

-

+

v0

vin

100k

1k

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SAQ 4

Voltage Gain GR

R= +

= +=

1 110

1112

1

SAQ 5

Voltage Gain

or

GR

R

R

RR R

= +

=

= =

1 50

49 49

2

1

2

12 1

Let R1 = 1 k

R2 = 49 x R1 = 49 x 1 k = 49 k.

The circuit is shown in Figure 4.19.

Figure 4.19

OP-AMP Circuit for SAQ 5

SAQ 6

(a)

VR

RV V t t t o

F= +( ) = + = 1 210

102 400 3 400 5 400( sin sin ) sin

(b) V t to = + = 1 3 4 200 3 4 200( sin ) sin

-

+

v0

vin

49k

1k

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SAQ 7

The basic summing circuit is shown in Figure 4.20.

Let R3 = 1 k => RF = 4 k.

R2 = 2 k and R1 = 4 k.

RF

-

+

v1

v0

R1

R2

R3

v2

v3

Figure 4.20

SAQ 8

Consider the sketch of Figure 4.21.

The mark -to-space ratios are as follows:

(i) 1 : 1

(ii)T T

4

3

4

1 3: :=

(iii)3

4 43 1

T T: :=

(iv) Output voltage always low ( -15V)

Voltage(volts)

+15

+10

+5

0

-5

-10

-15

T/4 T/2 3T/4 T-5V

0V

+5V

+12V

V 2

V 1

t

Figure 4.21

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SAQ 9

VR

R RV

R

R R

R R R R R R R

R R k k

t sat =+

=+

+ = = =

= =

2

1 2

3 2

1 2

1 2 2 1 2 1 2

2 1

50 10 14

0 05 14 0 05 13 95 279

100 27 9 28

.

. ( ) .

.

.

Let and ( )

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