Ch 5 - Ch 6

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5.1

ELEC166 Chap5.doc 02/01/06

Chapter 5

Electronic Systems

5.1 The Black-Box Approach to Signal Processing

For anyone without specialised knowledge, a circuit diagram just does not containthe right sort of information. There is too much detail, and the old adage about not seeingthe wood for the trees applies. We need a diagram that concentrates on ends rather thanon means, on signals rather than on arrangements of components. Such a visual aidis called a block diagram.

What we see in a block diagram are predominantly blocks. Each block may show

signals in and signals out, along with a name for each block, and often some sketches ofsignal waveforms. Power supplies are usually ignored, as well as individual components.Sometimes one block can represent a single chip; sometimes one block summarises actionsof more complicated circuitry. Of course one cannot build a circuit from a block diagramalone. But the operation of systems is much more easily understood using block diagrams,as unnecessary details have been left out.

Source

Preamplifier

Power amplifier

Loudspeaker

Controls

Figure 5.1 A simple block diagram for an audio circuit.

A simple block diagram is shown in Figure 5.1. In this system an input signalfrom a source (tape, CD, radio) is taken and the signal amplitude (voltage) amplified.Other controls may modify the signal and then apower amplifieris used to drivea loudspeaker. All this is basically described in words that accompany each block.Amplification means making a copy of a waveform, and the copy is usually larger.An amplifier block is very likely to be labelled with its gain expressed in decibels.

In another example, in a certain audio system low-level signals from three microphonesare mixed together by a mixer which gives a low-level signal out. This mixed signalis applied to a power amplifier which provides a high-level signal to drive a loudspeaker.The power-supply unit converts the AC mains supply voltage to a DC voltage suitable forthe system.

Exercise 5.1 On the basis of the above description draw the block diagram for the abovesystem.

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5.2


The block diagram provides only very limited knowledge of the system's operation.However, with even this limited knowledge we can apply some commonsense logicif there is a need to trace a fault. If there is no output from the loudspeaker, but we canmeasure an output from the mixer, then the loudspeaker or the power amplifier is faulty.There would be no point in changing the mixer.

In summary, electronic systems are usually composed of several basic functionalcomponents connected in such a way that an output signal from one becomes the inputsignal to another.

5.2 Putting Black Boxes Together

The output signal from the entire system is usually derived from the overall input,but it may not be a faithful copy it may have some of its attributes changed. We will nowanalyse, one by one, which attributes can change and in what way.

In any linear system, at any given frequency, there are essentially two signal parametersthat may change: the signal phase and the signal amplitude. The exact amount of thesechanges may vary depending on the signal's frequency. The signal amplitude is justone expression of the system behaviour, being a consequence of properties of each ofthe individual black boxes which are its subcomponents.

We will now discuss the signal amplitude and what exactly will happen to it if wepass the signal through a system built of several black boxes, each with its own frequencyresponse. Note that, in the case when the gain is less than one, it is called attenuation.This is relevant, because many electronic components, for example some filters, attenuaterather than amplify.

For a start we will concentrate on a signal consisting of a single frequencyf= 700 Hzwith an amplitude of 1 V. The signal is processed by two black boxes, withfrequency responses as in Figure 5.2. After having passed through Box 1 the signal

amplitude is 1 V 61 = 61 V, where the gain of 61 was read from Figure 5.2.This signal now becomes the input to Box 2, and we calculate that the output amplitude is

61 V 12 = 732 V. In summary the output amplitude is 1 V 61 12 = 732 V.

Note that 61 12 is the product of the gains of the two boxes at 700 Hz.

Now we consider a more complicated signal, which is made up of 10 different frequencies,of 1000 Hz, 2000 Hz, 3000 Hz, 4000 Hz, ..., 10000 Hz. The procedure for each of theseindividual frequencies is the same as that for a single frequency. We have to read offthe gain of the first box at that frequency, then read off the gain of the second box,then multiply the two. The product of the two gains and the input amplitude is equal tothe output amplitude.

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5.3


Gain curve for Box 1

0

1

2

3

4

5

6

7

8

9

10

11

100 1000 10000

Frequency (Hz)

Gain

Gain curve for Box 2

0.0

0.5

1.0

1.5

100 1000 10000

Frequency (Hz)

Gain

Figure 5.2 Frequency responses of the two black boxes.

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5.4


Gain Curve for Box 1

0

1

2

3

4

5

6

7

8

9

10

11

100 1000 10000

Frequency (Hz)

G

ain

Gain Curve for Box 2

0.0

0.5

1.0

1.5

100 1000 10000

Frequency (Hz)

Gain

Gain Curve for System

0123456789

101112131415

100 1000 10000

Frequency (Hz)

Gain

Figure 5.3 The frequency response of the two black boxes is calculatedby multiplying the two individual frequency responses point by point.

If a signal is even more complicated and contains a continuous range of frequencies,then the procedure is the same it is just a bit harder to carry out. We still need tomultiply the gains of each of the two boxes, but at all frequencies. In practice we canmultiply the two gains at a large number of frequency points (Figure 5.3). We end up witha lot of points showing the product of the two gain curves. Now we draw a smooth linethrough these points, a procedure which is only approximate, but is often close enough.

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5.5


Now we discuss what might happen to the signal phase if the signal is sent throughthe two boxes. Again we start with a simple signal, at a single frequency of 700 Hz,and we assume that at t = 0 its phase is zero as well. The first box can, in principle,

shift the phase by 1 . The value of this phase shift depends what is inside the box.

If there are resistors only, then 1 will be zero. With capacitors and inductors inside,

the phase shift becomes generally non-zero. If this phase-shifted signal is sent to the inputof the second box, an additional phase shift 2 will be applied. Once again, the value

depends on what is inside the second box. So the net effect is that the signal on the output

is shifted in phase by 1 + 2 with respect to the input signal.

As earlier, we will now examine a more complicated signal made up of two frequencies,say 1000 Hz and 2000 Hz. Both frequency components will be phase shifted after havingpassed through the two boxes, but the phase shift may be different for each of these twofrequency components. As a result of this phase shift, the appearance of the output signalwhen viewed on an oscilloscope may be very different from the input signal.

The same approach underlies the phase shift of a signal made up of very many frequencycomponents, or a whole range of them. Each individual frequency component will beshifted by a different amount. However nearby frequencies will nearly always havesimilar phase shifts; in electronics major changes rarely occur very suddenly.

We have examined what happens to signal amplitude and phase if the system is made upof two black boxes. It is easy to see what will happen if we have, not two boxes, but more.Once again all gains are multiplied for each frequency separately and all phase shifts areadded up.

5.2.1 Voltage and Current Limits

In our earlier introduction, linear circuits were described as having an output proportionalto the input. However in many electronic circuits this linear relationship holds only withincertain limits. For example the circuit may have a gain of, say, 1000, but only up toan output voltage of 10 V. If we apply AC signals with an amplitude of less than 10 mV,the circuit behaves linearly. If, however, the input signal starts to exceed 10 mVthen the output may hold the value of 10 V, until the time that the input signal becomesless than 10 mV again.

This type of circuit behaviour is called clippingand is illustrated in Figure 5.4. Sometimesclipping may be designed into circuits as a safety feature. Clipping may refer to the signalvoltage or to the signal current or to both.

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5.6


Input

voltage

Output

voltage

Time

Time

Figure 5.4 Clipping.

Only some circuits are designed to clip; others may, for example, simply distorta signal.This is illustrated in Figure 5.5. In this figure, the output signal, instead of beingjust a sinusoidal waveform, has its maxima rounded off and its minima sharpened.This kind of distortion in the case of audio signals can be clearly heard.

Input

voltage

Output

voltage

Time

Time

Figure 5.5 Distortion of output due to non-linearity.

The sine wave is heard as a sound of a single frequency so a clear, pure note.The distorted sinusoid if we work out its frequency spectrum contains some higherfrequencies, so it would be heard at the same pitch with some higher notes in thebackground. (Because they are at multiples of the input frequency, they change the soundof the note rather than its apparent frequency compare a flute and a saxophoneboth playing middle C.)

Clipping and distortion are often signs that we are trying to use the system in a way thatit has not been designed for. In these circumstances we should reduce the input signalto bring the system back to the region of linear operation. If we are applying too large

an input signal, we are subjecting our system to an overload. In some cases this may causedamage.

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5.7


5.3 Loading Effects

Loading effects are best seen with the example of a simple battery. We have all seenmanifestations of loading effects when starting a car at night while the headlights are on.When we turn the starter key the lights go dim, which shows that for a brief while when the starter motor is powered by the car battery and draws huge currents from it

the voltage across the battery terminals decreases. This lowered voltage is due to loading.Loading in this case means drawing a large current from the battery by connectinga small resistance across it.

We will now show how to calculate how much the battery voltage will decrease.

Example 5.1 A battery, nominally 12V and with an internal resistance of 1 , is connected

to an external load of 2 . Calculate the voltage on the battery terminalswith the load connected.

Solution

1 ohm

2 ohms

Internalresistance

Battery

A

B

A

B

2 ohms12 V

Terminals

Figure 5.6 Batteries have internal resistance that cannot be removedfrom the circuit.

The layout of this circuit is shown in Figure 5.6 (right), and the real-life behaviourmay be estimated using the equivalent circuit for the battery as shown in Figure 5.6(left) . We have to calculate the voltage on the battery terminals, that is between pointA and point B.

First we calculate the total resistance Rtotwhich is 1 + 2 = 3 .Therefore the current is 12 V / 3 = 4 A .

The voltage between A and B is 12 V 1 4 A = 8 V. So, with the load, the voltageof the battery is lower.

Note that, if the external load has a relatively high resistance, say 100 , then the voltagedrop would be barely noticeable. (It is suggested that students repeat the same calculationfor this load.) Note also that, unlike with heavy objects, where we say that a heavier loadhas more weight, a heavier electrical load actually has less resistance.

Although a battery has been used in this example, any voltage source (for examplethe output of an amplifier) is susceptible to loading effects of this sort.

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5.8


5.4 Examples

Problem 5.1 A typical problem on clipping may read as follows: The voltage gainof a certain 'black box' for weak signals is shown in Figure 5.7. The voltage gain is20 dB between 0 Hz and 10 kHz. A -3dB point is at 20 kHz. The box is designedto clip signals at vO= 5 V. What is the maximum peak amplitude of the input signal

atf= 10 kHz and atf= 20 kHz, such that the output signal is not distorted?

0

2

4

6

8

10

12

14

16

18

20

0 5 10 15 20 25 30

Frequency (kHz)

Voltagegain(dB)

-3dB point

Figure 5.7 The gain curve in the problem on clipping.

SolutionCase a) f= 10 kHz.

The voltage gain at 10 kHz is 20 dB; we now have to calculate Vo/ Vi .

20 log10 (Vo / Vi) = 20

so Vo/ Vi= 10 .

Now we work out the maximum input amplitude Vi, max .

Maximum Vo = 5 V = 10 Vi, max

and so

Vi, max = 05 V .

We now estimate the maximum input voltage at 20 kHz. Note that the solutionassumes that the limited frequency response and the clipping levels are two isolated

phenomena. In a simple transistor amplifier this may be not the case.

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5.9


Case b) f= 20 kHz.

At the -3dB point the voltage gain is 20 dB 3 dB = 17 dB. Now we can calculateVo / Vi.

log10 (Vo / Vi) = 17/20 = 085

and so

Vo / Vi = 71 .

Now we work out the maximum Vi, max .

Maximum Vo = 5 V = 71 Vi, max

Vi, max = 07 V .

Problem 5.2 The following is a typical problem on the system bandwidth of an opticalscanner, designed to read a pattern of regular stripes on labels (rather like a barcodescanner).

Box 1Preamp.

Box 2Filter

CounterInput

Laserscanner

Output

Figure 5.8 A block diagram of the electronics of the scanner system.

Response of Box 1

20

25

30

35

40

45

50

55

60

0 20 40 60 80 100 120 140

Frequency (kHz)

VoltageGain(dB)

Response of Box 2

-60

-55

-50

-45

-40

-35

-30

-25

-20

0 2 4 6 8 10

Frequency (kHz)

Voltagegain(dB)

Figure 5.9 The voltage gain curves for the boxes in the barcode scanner system.

A block diagram of a certain system is shown in Figure 5.8. The voltage gain curvesof the two boxes are shown in Figure 5.9.

The input signal to the electronics is a regular series of rectangular pulses with aperiod Tof 1 msec. This counter recognises the pattern if the output signal containsthe fundamental frequency and at least the fifth harmonic of the fundamentalfrequency, substantially unchanged. Will this system be able to recognise thepatterns?

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5.10


SolutionFirst we calculate the voltage gain curve for the combined system of two boxes. Thisis done by multiplying the voltage gain curves point by point at each frequency. Ifthe gains are given in dB we have to add the gains at each frequency, point to point.We get the graph as in Figure 5.10.

Combined Response

0

5

10

15

20

25

30

35

40

0 2 4 6 8 10

Frequency (kHz)

Voltagegain(dB)

60 dB 20 dB = 40 dB

Figure 5.10 The voltage gain curve for the barcode scanner system.

Since the stream of input pulses is repetitive (periodic), its spectrum consists of afundamental frequency component and harmonics. We calculate the fundamental

frequency of the signal as

f= 1/T= 1/1msec. = 1 kHz .

From figure 5.10 we can see that the combined gain of the system is relativelyconstant up to a frequency of about 5 kHz, the fifth harmonic of the fundamentalfrequency. Thus the signal will be recognised satisfactorily.

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5.11


Problem 5.3 A typical loading problem may, for example, look at the issue of delivery ofpower to the load.

A black box has a voltage gain of 60 dB (atf= 50 Hz) and its output resistance

(often called impedance) is 10 . The peak amplitude of the input signal

(atf= 50 Hz) is 1 mV. A load resistor of 2 is connected across the output terminals.

a) Calculate the peak amplitude of the output voltage, Vload , between the output

terminals.

b) Calculate the peak power and the average power delivered to the load.Assume that there is effectively no capacitance and no inductance betweenthe output terminals. In these circumstances there is no phase shift betweenthe output voltage and the output current, so we can use our previous formulasfor the peak and average power.

c) How would you change the load resistance so that the power (peak or average)delivered to the load is maximised?

Solution

a) First we calculate Vo /Vi . We have

20 log10 (Vo /Vi ) = 60 ,

solog10 (Vo /Vi ) = 3

andVo = 1000 Vi .

This means that we would have a voltage Vo = 1 mV 1000 = 1 V between

the open-circuit output terminals (with no load resistance connected).But, with a load, the situation is different. We can draw the circuit diagram as inFigure 5.11. Note that Figure 5.11 is the diagram of an AC voltage divider.We can calculate the peak amplitude of the output voltage:

Vload = 1 V 2 /(2 + 10 ) = 017 V .

1 mV 1000

10

Rload

Figure 5.11 The circuit diagram for calculating

the power delivered to the load.

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5.12


b) The peak current amplitude Iload can be calculated by looking at the same

diagram:

Iload = 1 V /(10 + 2 ) = 1 V/12 = 008333 A .

The average power is

Pave= VloadIload /2 = 0007 W .

The peak power is

Ppeak = VloadIload = 0014 W .

c) In order to see if we can increase the power delivered to the load, we will try

a couple of alternative load resistors. We will try Rload = 10 and Rload= 40 .

For Rload= 10

Vload = 1 V 10 /(10 + 10 ) = 05 VIload = 1 V /(10 + 10 ) = 1 V/20 = 005 A .


Pave = VloadIload /2 = 05 V 005 A /2 = 00125 W .

This is more than the average power delivered to the 2 resistor discussedpreviously.

For Rload= 40

Vload = 1 V 40 /(40 + 10 ) = 1 V 40 /50 = 08 V

Iload = 1 V /(40 + 10 ) = 1V/50 = 002 A .


Pave = VloadIload /2 = 08 V 002 A /2 = 0008 W .

This is less than for 10 .

Conclusion: Rload = 10

gives the highest power. The value of 10

was chosendeliberately. In fact it can be mathematically proven that the maximum poweris delivered to the load if the load resistance is made equal to the output resistance.This condition is also called impedance matching.

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6.1

ELEC166 chap6.doc 02/01/06

Chapter 6

AmplifiersThere are many instances when one needs to make an electrical signal bigger(to amplify it) or to control the flow of current in one circuit with a signal in another circuit to have an electrically controlled switch. Amplifiers are often used for analogue signalssuch as speech and music. Electronically controlled switches are used primarily for digitalsignals; such switches are found in computers. The difference between an amplifierand an electronically controlled switch is that an amplifier output must mirror preciselyevery small change in the input level, while a switch might produce only two differentoutput voltages, high or low, depending whether the input is in a high or a low range.

Amplifiers are needed in a variety of situations. A typical example might be a telephone

conversation with a distant country. Our voice can be converted to voltage signals througha microphone, but the amplitude remains low, and because of circuit resistance thisvoltage would be too weak to be detected at the overseas destination. Thereforewe need amplifiers to restore the amplitude, and generally speaking more than one

a whole series of them, that maintain the voltage level1.

Playing a CD also requires amplifiers. The tiny marks on a CD are converted into electricalsignals by moving the disk through an infrared beam produced by a small laser;the reflected beam is allowed to fall on a light detector where it produces currentscoded with music information. These currents are extremely weak and must be boosted

before the music signal can be extracted. After this we need amplifiers to increase thesignal amplitude and to produce sufficient power to drive loudspeakers.

Amplifiers in circuit diagrams are represented by several different symbols as inFigure 6.1. Gain = 10 means that the output signal is ten times the input signal.

Amplifiers are identified as being current amplifiers or voltage amplifiers2. As the namesuggests, voltage amplifiers amplify voltages and may keep the current the same or evenreduce it. Current amplifiers amplify currents and may keep the voltage the same.Most amplifiers amplify AC input signals, but there are also DC amplifiers which amplifyboth the DC and the AC part of a signal. The + and - terminals indicate invertingand

non-invertinginputs (gain + or -), but generally if we swap the polarity there may beno problem (for example in audio amplifiers).

1 Note that this sort of communication is now carried out by converting the speechto digital ('on-off') signals, which are much more resistant to degradation; however,so-called repeater amplifiers are still required along the way.

2 It is possible to build amplifiers which have a voltage input and a current output,and vice versa. These are referred to as transconductance and transresistanceamplifiers.

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6.2


i in ioutGain = 10

vin

voutvin

vout

iin iout

+

-

+

-

Figure 6.1 Symbols used to denote various amplifiers.

6.1 Features of Amplifiers

6.1.1 Single-Ended and Differential Amplifiers

Note that in Figure 6.1 the amplifiers appear to have two different styles of inputs(and outputs). Some symbols indicate that the signal is fed into the amplifier throughone terminal only; likewise the output seems to be generated at one terminal only.

If the amplifier on the circuit diagram has only one input and one output, it means that theinput signal is applied between the input point on the amplifier and the earth orground;

likewise the output is taken between the output point and ground3. The groundedterminal/wire is simply not shown, and we only see the other terminal whichenters the amplifier. The same applies to the output: the output signal is generatedbetween the single terminal shown on the diagram and the earth. This kind of operationis called single-ended.

Other amplifiers have two distinct inputs (in addition to any grounded terminals,which are not shown). They are capable of dealing with voltage signals from ungrounded

sources, i.e. sources with two outputs at non-zero potentials. The input voltage for such anamplifier is then equal to the voltage difference between the amplifier input terminals.Because of that, this amplifier is referred to as a differential amplifier. Similarly,the amplifier output may also have the two terminals at a non-zero potential.

3 Ground in this case means a common connection, which usually is connected tothe actual earth.Also not shown are the power-supply connections; this supply is responsible forproviding the increased voltage or power.

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6.3


Differential amplifiers are extremely useful, for example if two voltage signals v1 and v2

coming from two inputs may both have a large voltage offset but a small difference v

v2 = v1 + v

and we are interested in this small difference v only. By connecting these inputs

to a differential amplifier, we would get an output proportional to v only.

Note that a differential amplifier can always be used in a single-ended way(i.e. with one terminal grounded) but not the other way round.

It is quite common for an amplifier to invertthe signal in addition to increasingits amplitude. In this case a positive input would give rise to a negative output andvice versa. Such amplifiers are called invertingamplifiers; if they do not invert, they arecalled noninverting.

6.1.2 Effects of Input and Output Resistance,Ri,Ro

We have already looked at the effects of loading on voltage sources. Let us nowlook at this a little more closely in the case of amplifiers. This is most easily understoodwhen we discuss the case of voltage amplifiers. When we are connecting an amplifierto a signal source the signal source can be visualised as a Thvenin equivalent circuit,made up of an ideal voltage source and a series resistor. This circuit is connected tothe amplifier input terminals. The signal flowing through to the input terminals is affectedby the resistance 'looking into' these terminals; this resistance is called the input resistanceRi . Ideally, this resistance should be infinite for a voltage amplifier and thus leave

the input signal unchanged; however, this is a real world, not an ideal one (Figure 6.2).Likewise, we need to visualise the two output terminals of an amplifier as presenting

an output resistanceRo . Ideally, the output resistance should be zero, but it never is.

ii io

Ri

R S Ro

LR

vi

Source

v'i

Amplifier Load

a.

R S

vi

v'o

b.

RL=Aov vi ovvi

=Ao

v vi

i'oii

Figure 6.2 A signal source is connected to an amplifier and the amplifierto a load. (a. ideal amplifier; b. real amplifier)

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6.4


On the input side the presence of Rimeans that there is an input current ii given by

ii = vi / Ri .

The larger Ri is, the more nearly ideal is the input stage of the amplifier, i.e. ii is smaller.

On the output side the presence of Ro means that, for output current io , the output voltage

v'o is not just

vo = gain vi ,

but is given by

v'o = vo - io Ro .

The smaller Ro is, the more ideal is the output of the amplifier; i.e., it behaves more nearly

like an ideal voltage source.

6.1.3 How to measureRiandRo

We can measure the output resistance Ro by keeping vi constant and observing v'o first

for a very light load (RL >> Ro ), when the output is just v'o1 = vo . Then a smaller load

RL

is connected and we observe the new value of the amplifier output voltage v'o2 .

These two values (v'o1and v'o2) allow us to deduce the output resistance Ro because

v'o2= voRL / (RL + Ro) = v'o1 RL / (RL + Ro )

from which Ro can be calculated. The latter formula is that of a potential divider.

Likewise we can measure the input resistance Ri in the following way. The observed

quantity is the output voltage v'o in the presence of a load RL :

v'o = gain viRL / (RL + Ro ) = Kvi .

First we drive the amplifier from a source v'i of low source resistance (RS

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6.5


6.1.4 Worked Examples Involving Input and Output Resistance

Problems related to the input and output resistances of amplifiers are often based onvarious potential divider circuits.

Problem 6.1

With a certain input signal the open-circuit voltage in an amplifier is 10 volts.When we connect a load of 1 ohm across the output, the voltage drops to 8 volts.Calculate the output resistance, Ro , of this amplifier.

SolutionWe draw a circuit diagram (Figure 6.3) describing the situation with the load.

1

Ro

10 V 8 V

Figure 6.3 The circuit diagram to calculate the output resistanceif the voltage is 10 V into an open circuit but 8 V into a load.

In this circuit we have

10 V 11

Ro +

= 8 V .

We simplify and get

10 = 8 + 8 Ro

and so finally

Ro = 025 .

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6.6


Problem 6.2An ideal signal source (with zero source resistance) was connected tothe amplifier input. The amplifier output measured across the load was 1 V.

When we connected the same signal source in series with a 100k resistor,

the voltage at the amplifier output dropped to 09 V. What is the value of the inputresistance, R

i, of this amplifier?

Solution

The circuit diagram at the input with the 100k resistor is shown in Figure 6.4.

R i

100 k

vs

Amplifier

Figure 6.4 The circuit diagram to calculate the input resistanceif the output is 1 V with low source resistance but becomes

09 V with the 100k source resistance.

When an ideal source is connected across the input, we get

vs

K= 1 V .

When an ideal source with a 100k resistor in series is connected across the input,we get

vs RiRi

k

+ 100 K= 09 V .

We can now substitute for vs K in the second equation:

Ri

Ri k+ 100 1 V= 09 V

orRi

Ri k+ 100 = 09

and so

Ri = 09 Ri + 90 k

01 Ri = 90 k

Ri = 900 k .

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6.7


Problem 6.3This problem describes real electronic components in a realistic situation.

We have a microphone with a source resistance of 1 M, which gives a voltage signalof 1 mV. We also have an amplifier with the following specifications.

Voltage gain = 1000, input resistance = 2 M , output resistance = 1 . Additionally,

we plan to connect a loudspeaker with a resistance of 3 . By what factor will ourvoltage signal be amplified?

Hint: In the amplifier specifications it is assumed that you use an ideal sourceand measure your output voltage without any load (open circuit).In these circumstances the gain is 1000, but in the real circuit the gain will be lower.

Solution

Source Amplifier Load

3 1 M

2 M1

V

Figure 6.5 The circuit diagram to calculate the overall gainof the amplifier.

We draw the circuit diagram of the system (Figure 6.5). The open-circuit voltage Vof the amplifier is

V= (1 mV 2 M / (2 M + 1 M)) 1000 .

The voltage VL that appears across the load is

VL = V 3 / (1 + 3 ) .

Putting the two equations together we get

VL = 1 mV 2

2 1

M

M M

+ 1000

3

1 3

+= 1 V

2

3

3

4= 05 V .

Thus the final gain is

05 V / 1 mV = 500 .

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6.8


6.2 DC Amplifiers

A DC amplifier must obviously amplify DC voltages and voltages which have a DC part,for example Figure 4.12 (b). There must be no low-frequency cut-off. Such an amplifierwill also amplify AC signals, but only up to the high-frequency cut-off point,fh , so that

its bandwidth is alsofh - 0 =fh . To summarise this part, both DC and AC amplifiersare capable of amplifying AC signals, but only a DC amplifier will amplify DC signals.

Example 6.1An amplifier has a gain curve as in Figure 6.6. Derive from the graph:i) the high-frequency cut-off pointfh ;

ii) the low-frequency cut-off pointfl ;

iii) the bandwidth f.

2 3 41 5

20

10

0

Gain(dB)

flog

17

Figure 6.6 A gain curve of an amplifier.

SolutionWe should use the definition given in the text for the high-frequency and thelow-frequency cut-off points, where the gain drops by 3 dB.

The maximum gain is 20 dB, thus the cut-off point is at the point where gain drops to

20 - 3 = 17 dB. This gives logfh = 425 ,fh = 17783 Hz , and logfl = 20,fl = 100 Hz .

The bandwidth is then given by

f = fh -fl = 17683 Hz 177 kHz .

There is one important feature of DC amplifiers namely they have a tendency to driftin time. Their gain may remain constant, but the output voltage slowly acquires an offset.This may happen over any period of time minutes, hours or days. To compensatefor this, DC amplifiers may have a facility for removing the additional constant offset,by adding a small voltage to the input. This offset may need to be readjusted as time

goes by.

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6.9


6.3 Power Amplifiers

The primary purpose of power amplifiers is to deliver a substantial amount of power

(volts amps) to a load (for example, a loudspeaker, lamp, heater or motor), and they areoften required to deliver large currents, perhaps many amperes. They may or may nothave considerable voltage gain. Power amplifiers are often massive devices, and efficiencymight be an important consideration.

Problem 6.4A transistor radio has an output amplifier system with a frequency response from

150 Hz to 60 kHz.

i) Is this a DC or an AC amplifier?ii) Calculate the bandwidth of the amplifier. Is it adequate compared to

the frequency range of sounds that we can hear?

iii) Explain why the small bandwidth give poor-quality sound.

Problem 6.5A pH-meter is an instrument that enables us to quantitatively distinguish acidsfrom bases. Its block diagram is shown in Figure 6.7. We need to use two amplifiersA and B as shown in the diagram. The input to A from the pH electrode is a voltage

signal which changes by 590 mV for a unit change on the pH scale. The output

from A also changes by 590 mV per one pH unit.

A B

Meter

Electrode

Figure 6.7 Block diagram of a pH-meter.

The output from B operates the display meter and is a voltage signal which changesby 1 volt for a unit change on a pH scale.

(i) State which type of amplifier should be used for A and B. You should choosefrom the following options: AC amplifier, DC amplifier, buffer amplifier,power amplifier.

(ii) Based on the information in the text, estimate the voltage gain for each ofthe amplifiers A and B.

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6.10


6.4 Feedback

6.4.1 Feedback in General Terms

Consider the two following examples of a heater system.

1) The heater receives an input from a timer that keeps it switched on for five minutesand then switches it off for five minutes. The whole system repeats this cycleuntil turned off.

2) The heater receives an input from a thermistor(temperature-sensitive resistor);it switches on when the temperature falls below 20 degrees Celsius and turns offwhen the temperature rises above 23 degrees Celsius.

Once the heater in example 1 is turned on, is it affected by the surroundings?The answer is no regardless of the surrounding temperature, the heater will go onheating the room in five-minute intervals. This is an example of an open-loop systemwhere the system has no control over its output. The system will continue to producean output until it is turned off at the power supply. In the second example the heater isdirectly affected by the surroundings it is heating up. When the room gets too hot,the heater turns off; when the room is too cool the heater turns on. This is an example ofa closed-loop system, where the system takes information about its output and feeds backthis information into the input. It has a closed loop between the output and inputsince the temperature-sensing input is affected by the heat the system is producing.Thefeedback loop means that the system can check its own performance.

Exercise 6.1In the above two examples, which will heat the room most effectively and why?Think about wasting heat and the time needed to heat up the room. Will the timeof year affect which is the most efficient?

An open-loop system cannot control its own output since it has no feedback loop betweenthe output and input. It cannot pass information back to the input.

In summary, a closed-loop system can control its own output because it has a feedbackloop between the output and input. It uses this to pass information back to the inputthat will affect the output. A closed-loop system checks its output.

Exercise 6.2Draw a block diagram of the two heater systems described above.

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6.11


6.4.2 Feedback in Amplifiers

A

iv

sv

vf

vo

Summer Amplifier

Feedback

Figure 6.8 A block diagram of a feedback amplifier.

Feedback is commonly used in amplifiers, and its characteristic feature is that we have

a closed loop. Figure 6.8 shows an amplifier, a feedback block and an additional elementcalled a summerdesigned to add (or subtract, as in the example above) all or a part ofthe output signal to the input. We can easily see that, in Figure 6.8, vi affects vo , vo affects

vf , vf affects vi , which is where we started from. The feedback loop lends itself to

mathematical analysis and it can be shown that the gain of the feedback amplifier isstrongly related to the 'strength' of the feedback. Let's see how it is done.

The amplifier has a function of amplifying the input by a factor A , so

vo =Avi .

The feedback block takes the output signal and passes only a fraction of it, indicated by

vf = vo .

We normally require to be very stable, and it often consists of a simple potential divider.Finally the third element the summer has the function of subtraction:

vi = vs - vf .

Sometimes the summer is a separate block, but often it is incorporated with an amplifierin the form of a differential amplifier.

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6.12


Often, the value ofA is very large and the circuit operates as follows. Sincevi = vo /A

then vi must be very small, and very close to 0 volts. So we can treat it as a virtual earth.

Sincevs - vf = vi

andvi 0

we have

vsvf = vo

so that finally

vo/ vs 1/

and we find that the gain vo / vs does not depend on A only on ! This is quite

advantageous, becauseA depends on properties of elements inside the amplifier,mostly transistors, and they may not be very stable, whereas a potential divider gives

a very stable value of .Note that the circuit includes a closed loop, and that it includes three different gains.The summer has a gain of +1, the amplifier has a gain of A , and the feedback block has

a gain of . The total gain around the closed loop is called the loop gain and is equal to

A.

Now we will show that the feedback leads to increasedgain stability. To do thatwe need to first derive an expression for the closed-loop gain vo/ vs in terms of

the amplifier gainA and the feedback . We have

vo=Avi =A (vs - vo )

so

vo =Avs -Avo .

Thereforevovs

A

AA

=+

=

+1

1

1

.

If 1/A is much less than then this becomes

vo/ vs 1 / ,as we have shown above. Thus, provided that the gainA is high enough, the ratio ofthe input to the output is (although much smaller thanA ) almost independent ofA .

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6.13


Example 6.2

Suppose thatA = 104 , = 001 . Then

vo / vs = 104 /(1 + 104 10-2) = 9901 .

We note that, for this circuit, 1/A = 10-4 is indeed much less than = 001 .

Now let us imagine that something happened to our amplifier and the gain A wasnow halved that is,A = 5000 . The ratio of output to input is now

vo / vs = 5000 /(1 + 5000 10-2) = 9804 .

This ratio has changed by 1% only. So we have good stability of gain, but we haveless gain. The gain of our amplifier A used to be 10000; now it is near 99, so it hasbeen reduced by a factor of 101. At the same time the gain stability has improved bythe same factor of 101.

6.4.3 Other Advantages of Feedback

We have just seen that negative feedback can compensate for a drop inA caused by,for example, aging of components. We can then expect that it can also compensate fora drop in A caused by a limited frequency response. Figure 6.9 shows such an improvedfrequency response. The 3dB frequencies, f HF' and f LF' , of an amplifier with a feedback

loop are related to the 3dB frequencies of an amplifier without feedbackfHFandfLF

(for the simplest of amplifiers) through

f HF' =fHF (1 +A)

f LF' =fLF/ (1 +A) .

open loop

Gain

Gain

(dB)

Frequency (log scale)

closed loop

Gain /(1+ )A A

HFf 'LFff 'LF HFf

A

Figure 6.9 Improved frequency response of an amplifier with

negative feedback.

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6.14


We can see that the bandwidth is improved by a factor (1 +A) , the same factor by whichthe gain is reduced. Feedback amplifiers have also other benefits, such as reduceddistortion of signals. Distortion is due to nonlinearities (which we may think of as a gainthat varies with the signal amplitude), and in feedback amplifiers the effect of these

nonlinearities is reduced by the same factor (1 +A) .

We learned that an important feature of amplifiers is their input and output resistanceRinand Rout . In ideal voltage amplifiers the input resistance is infinite and the output

resistance is zero. In real amplifiers, however, both are finite, although the input resistancemay be high and the output resistance low. These resistances are further improvedif the amplifier is placed in a feedback loop. (We may think of them as causinggain variations dependent on the source and load resistances.) It can be shown that

the input resistance may be increased by a factor of (1 + A ) and the output resistancemay be reduced by the same factor. So both of these resistances are improved, i.e. becomecloser to the ideal.

Feedback seems to be a wonderful remedy for all sorts of amplifier problems and it lookslike a miraculous fix. While it is a very useful method it does have some potentialproblems. These have to do with phase shifts. The amplifier itself may contributea considerable phase shift. The feedback block which may be made up of resistors,as with a potential divider does not introduce phase shifts by itself but, combined withthe amplifier and connecting cables, which always have a non-zero capacitance, it will leadto non-zero phase shifts. These become more and more pronounced at higher frequencies.If the extra phase shift approaches 180 degrees then, instead of subtracting a fraction ofa signal in a feedback loop, we will now be effectively addinga fraction of the signal.

The feedback becomes positive. Immediately all the advantages of feedback will turn intodisadvantages and the whole amplifier may even start to oscillate, giving output withoutany input. One remedy is to limit the frequency response of the basic amplifier.This can be done with the aid of filters, discussed in Section 3.6.

Example 6.3A voltage amplifier A has the performance as listed in Table 6.1. It is proposed

to reduce the gain to 50 using negative feedback with consisting of a potential

divider of total resistance about 10 k to 20 k. We will (i) calculate suitablecomponent values and (ii) predict the performance of the resulting feedback

amplifier.

Mid-band gain +1000

Gain tolerance 20%

Frequency response Flat 50 Hz to 5 kHz (-3 dB)

Input resistance Rin 50 k

Output resistance Rout 100

Distortion 5% with output 20 V p-p

Table 6.1 Parameters of the amplifier A .

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6.15

Answeri) The gain with feedback applied is

Af=A /(1 +A)

so we have

50 = 1000 / (1 + 1000 )

which gives1 + 1000 = 20therefore

= 0019 .

If the potential divider has R1 connected to the amplifier output and R2 to ground,

we must have

R2 / (R2 + R1 ) = 0019

0019 R1 = 0981 R2

R1

= 516 R2

.

Getting this exact ratio requires non-standard resistance values. Two standard valuesthat would be close enough to be satisfactory are

R1 = 20 k

R2 = 390

for which

R1 = 5128 R2

R1 + R2 = 2039 k .

ii) The gain has been reduced by a factor

1 +A= 20 .

Quantity Previous value New value

Gain tolerance 20% 1%

fLF 50 Hz 25 Hz

fHF 5 kHz 100 kHz

Rin 50 k 1 M

Rout 100

5

Distortion 5% 025%

Table 6.2 Parameters of the amplifier A without and with the feedback.

Therefore we predict a variety of other amplifier properties which will also

be improved by a factor 1 +Aas in Table 6.2. In view of the predictedgain tolerance, we would be well advised to choose R1 and R2 carefully, using

025%-tolerance resistors. Note that an amplifier which was barely adequateas an audio amplifier (telephone quality) in terms of frequency response has been

converted by feedback into a quite acceptable one.

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