suggested answers electronics

8/9/2019 Suggested Answers Electronics

1/24

December 2008

a) (i) Q 1 is cmos p- channel while Q2 is cmos n- channel

input Switch position Q1 Q2 output0V 1 ON OFF 15

15V 2 OFF ON 0V

(b)

2 a

Gain- bandproduct = product of open loop gain and bandwidth

.

With voltage gain of 1000 and bandwidth of 1000,

Gain- bandwidth = 1000 1000 = 10 6


2/24

b. i The circuit is a low- pass active filter

ii mid- band gain ( Av ) = 13

2

R

R= 1

100

56= 1.56

iii cut- off frequency =RC2

1=

9310100102.22

1

= 723.4Hz

3 a.i hfe is the ratio of the collector current to the base current at constant of collector- emitter voltage.

i.e hfe =B

C

I

I@ Constant Vce

ii hie is the ratio of the base- emitter voltage to the base current at constant of base emitter

voltage. i.e hie =b

be

I

V

@ Constant Vbe

b i

ii Voltage gain = hfeIN

OUT

R

R

Rin = hie = 1K= 100; Rout =oeh

1// R

L=

1000001.0

1000001.0

=9.9910

3

Voltage gain =1000

1099.980 3= 799.2

Voltage gain=IN

OUT

V

V

Vout = Voltage gain Vin = 310102.799 = 7.992V

4

a


3/24

N- Channel JFET comprises of a channel of n- type material surrounded by material of theopposite polarity. The ends of the channel( in which conduction takes place ) form electrode

known as the source and drain. The effective width of the channel is controlled by a charge placed

on the third electrode called gate.

OPERATION

There are two distinct regions of operation, both of which have useful application.For low values of drain- source voltage, the drain current is progressively increased as the gate-

source voltage is made less negative. The operation region is known as the triode region

(amplification)

For high values of drain- source voltage, the drain current is progressively reduced as the gate-

source voltage is made more negative. At a certain value of gate- source voltage, drain current

falls to zero and the dvice is said to be cut-off( switching)

b) Limiting conditions of transistor as a switch are;i Saturation

ii Cut-off

c) Vin = BBRI mAIC 4= ; 90=feh

B

INB

I

VR =

B

Cfe

I

Ih =

mAh

II

fe

CB 04.0

90

1043

=

==

=

==

=

KRB 12504.0

5000

1004.0

5

3

CCECCC I

VVR =

@ Saturation VVCE 0 == KRC 5.2004.0

10

5 a Effects of negative feedback

1 Less harmonic

2 Increase bandwidth3 Highly stabilized gain

b) i Voltage- series feedback

CECCCC VRIV =


4/24

ii Practical circuit which utilizes this type of feedback is emitter- follower ( common- collector

configuration )

c)gainopenedloopfeedback

gainopenedloopedgainclosedloop

=1

32.3310000003.01

100000 =

=

6 a

1. The gain must be infinite

2. The overall loop voltage gain must be greater than one

3. Must occur at a single frequency

b) Tunned- LC Oscillator

When the supply is first switched ON a transient current is developed in the tunned circuit as the

collector current rises to its quiescent value. This transient current initials natural oscillations in the tank

circuit (LC) . These natural oscillations induce a small e.m.f into L2 by mutual induction which causes

corresponding variation in base current. These variations in base current are amplified times and

appear in the collector circuit. Part of this amplified energy is used to meet losses taking place in the

oscillatory circuit and the balance is radiated out in form of electromagnetic wave.

ii frequency of oscillation = =

=

KLC

6.791041012

1

2

1

93

7 a

INPUT OUTPUT

A B BA.

0 0 1

0 1 1

1 0 1

1 1 0

BACBACBA ..... C CBA.

)()( CCABCCAB

AB BA = )( BBA = A

c)

d) Latch is needed to momentarily store information electronically.


5/24

8a

i Class A- 50%ii Class AB- 70%

iii Class B- 78.5%

b) The complementry pair is made use of in push- pull class B amplifier. One NPN and one PNPtransistor. The NPN is forward biased during positive half- cycle allowing current to flow while

PNP is reverse biased, so it cut-off. The PNP is forward biased during negative half- cycle

allowing current to flow while NPN is reverse biased. During this operation, crossover distortion

occurs as show below.

Crossover distortion can be eliminated by applying slight forward bias to each emitter diode. i.e locating

the Q- point of each transistor slightly above cu-off so that each one operates for more than one half cycle.

9

a)

i Channel gain ; It is used for the adjustment of pattern width and height.

ii Time base ; This controls the speed at which the beam is swept across the screen from left to right. Itdoes this by setting the sweep time of the time base generator.

b) i Where two or more waveforms are to be display simultaneously

ii Where waveshape needed to be measured and displayed with equal accuracy.

c) Expected voltage drop across resistor 5kwhen voltmeter is not connected is given by

V9900.91055000

5000=

=

With the voltmeter connected; The reading of the voltmeter 9875.91054000

4000=

= V

Error in the measured reading = 9.9900- 9.9875= 0.0025V

10

a)

Phototransistor is light- sensitive transistor and similar to an ordinary bipolar junction transistor except that

it has no connection to the base terminal. Its operation is based on the photodiode that exists at the


6/24

collector-base junction . Instead of the base current, the input to the transistor is provided in the form of

light.

OPERATION

When light is incident on the collector- base junction, a base current is produced which is directly

proportional to the light intensity. When there no incident light on the collector base junction, there is a

small thermally generated collector- to- emitter leakage current which in this case, is called dark current

and is in the nanoamps range.

b)

In a fibre- optical communication system, information signal which is an incoming electrical signal is turninto light using a device called an emitter. The emitter may be a LED ( light emitter diode ) or laser diode.

The light travels down the core . It remains in the core and never leaves it ( never entering the cladding ) .

At the receiver side, the light signal, which made it through the fibre- optic cable is returned to an electricalsignal by a detector. Detector might be a PIN diode or an APD ( Avalanche photo diode ). The light send

down the fibre- optic cable corresponds to an electromagnetic wave with a frequency in the range of 1014

to 1015 Hz. This system is capable of sending information at rates of 10 14 bits/s.

c) When the intensity of the incident light on the light depend resistor increases, the resistance of the LDR

decreases, current flow through LDRand not to the base of the transistor . Thus TR2 is OFF and no

output. When the intensity of the incident light on the LDRdecreases, in such a way that the resistance ofthe LDRis greater than R1 , current flow to the base ofTR2 and turn it ON. Thus, there is output.

June 2008


7/24

3 a) i hfe is the ratio of the collector current to the base current at constant of collector- emitter voltage.

i.e hfe =B

C

I

I@ Constant Vce

ii hie is the ratio of the base- emitter voltage to the base current at constant of base emitter voltage. i.e

hie =b

be

I

V@ Constant Vbe

b)


OUT

R

R

Rin = hie// RB

= 4.249100000250

100000250=

; Rout =

oeh

1// R

L=

==

K7.17.1666

200010000

200010000

Voltage gain =4.249

107.1120 3= 817.96

Voltage gain=IN

OUT

VV

Vout = Voltage gain Vin = 3101096.817

=8.2V

4. a)

i Class A The transistor is biased that output current flows for the full- cycle of the input signal .

That is the transistor remains forward biased through out the input cycle. Its condition angle is 3600 . The maximum efficiency of the amplifier is 50%.

ii Class B The transistor is biased and the amplitude of the input signals are such that output

current flows for only half- cycle (180 0 ) of the input signal.

The maximum efficiency of the amplifier is 78.5%.

iii Class C The transistor is biased and signals amplitudes are such that output current flows for

appreciably less than half- cycle of the input signal i.e 120 0 or 150 0 angle codition.

b) Distortion can occur in a class A amplifier if a large signal is applied to it which will

eventually cause a shift in Q- point to non-linear regions near saturation . This in turn

results to distortion.5 a)

Values of characteristics of a typical operational amplifier are ;

i Input resistance is high ( infinite )

ii Output resistance is low

iii Open loop gain is finite

iv Wide bandwidth


8/24

b)

i High pass active filter

ii Cut-off frequency =KHzHz

RC592.16.1591

101010102

1

2

1

93==

=

iii Mid- band voltage gain = 101010

10100

3

3

1

2=

=

R

R

6

a) Feedback factor ( ) is the ratio of the portion of output that is fed back into the input.

b)

i Less harmonic

ii Increase bandwidthiii Highly stabilized gain

c)

i The RC phase- shift oscillator uses an amplifier and RC network to provide feedback . The

amplifier is to provide phase reversal and amplified the oscillating signal while RC network is

used to control the phase of the feedback signal. This enable the oscillator to become sensitive to a

signal of only one particular frequency

ii

Resonant frequency (fo ) HzRCRC 6.5271022106.5

065.0065.0

62

193=

==

7

a)

i OR GATE ii NAND GATE

INPUT OUTPUT

A B BA.

0 0 1

0 1 1

1 0 1

1 1 0

INPUT OUTPUT

A B A B0 0 0

0 1 1

1 0 1

1 1 1


9/24

b)

Assuming A is high, A will be low since the output of an inverter is the complement of the input . The

low input to NAND gate 1 cause the Q output to go high. This high Q output is also fed to the input of

NAND gate 2 . The other input to NAND gate 2 B is high , with both input to gate 2 high, the output goes

low. The low Q output is also fed to NAND gate 1 to be used as the LATCH signal . If A goes low while

this condition exists, there will be no change to the output because the flip-flop would be in the LATCH

condition both A and B low.

c)

F= CBACBACAB

ii ABCAACBCABABCCBACBACABF == )(

CBABCBCCABABCCBCAB === )(

8 a b)

i Class A- 50% Linear audio amplifier

ii Class B- 78.5% push- pull amplifier

c) The complementry pair is made use of in push- pull class B amplifier. One NPN and one

PNP transistor. The NPN is forward biased during positive half- cycle allowing current to flowwhile PNP is reverse biased, so it cut-off. The PNP is forward biased during negative half- cycle

allowing current to flow while NPN is reverse biased. During this operation, crossover distortion

occurs as show below.

9 a)


ii Time base ; This controls the speed at which the beam is swept across the screen from

left to right. It does this by setting the sweep time of the time base generator.


10/24

iii Synch/trigger; These controls allow the scope to triggered from various selected

sources. Triggering causes the trace to being its sweep across the screen.

b)

i It produces a visual indication of waveformii The frequency, period and phase angle of the waveform(s) can be determined

iii It can display two or more waveforms simultaneously

c)

c) Expected voltage drop across resistor 5kwhen voltmeter is not connected is given by

V612200000

1000000==

The resistance of the voltmeter at 10V range is= 10000/V 10V= 100000

With the voltmeter connected; The reading of the voltmeter V41210050

50=

=

Error in the measured reading = 6V-4V= 2V

10 a

i Construction

Optical- isolators are designed to electrically isolate one circuit from another which allowing one circuit to

control the other .The purpose of isolation is to provide protection from high- voltage transients surge

voltages and low-level electrical noise that could possible result in an erroneous output or damage to thedevice. An optical isolator consists of a light source such as light emitting diode (LED ) and a photo

detector such as phototransistor

Operation

When LED is forward biased, the light produced by it is transferred to the phototransistor which is turned

ON thereby producing current through the external load.

ii Application

They are used in applications where a low-level input voltage is required to latch a high voltage relay foractivating some kind of electromechanical device

b)


11/24

i The resistance of a light dependent resistor is affected by the amount of light falling on its surface. The

higher the intensity of the light , the lower the resistance of the light dependent resistor.

ii The T1 is a phototransistor which is turned ON by the ray of light descending on it.

This causes current to flow through to transistor T2 which is also turned ON by producing current throughthe relay coil to the external load.

December 2009

1 a


12/24

i

INPUT OUTPUT

A B Q1 Q2 Q3 Q4

0 0 ON ON OFF OFF VDD

0 1 ON OFF OFF ON 0V

1 0 OFF ON ON OFF 0V1 1 OFF OFF ON ON 0V

The operation of the circuit is described as per the above truth table shown above

ii The logic function is NOR gate

b)

Time constant ( )= sCR 00001.0101010139==

Period(T)=s

f001.0

100

11==

CR T

2 a

N- Channel JFET comprises of a channel of n- type material surrounded by material of the opposite

polarity. The ends of the channel( in which conduction takes place ) form electrode known as the sourceand drain. The effective width of the channel is controlled by a charge placed on the third electrode called

gate.

OPERATION

There are two distinct regions of operation, both of which have useful application.

For low values of drain- source voltage, the drain current is progressively increased as the gate-


(amplification)

For high values of drain- source voltage, the drain current is progressively reduced as the gate-

source voltage is made more negative. At a certain value of gate- source voltage, drain current

falls to zero and the dvice is said to be cut-off( switching)

b)

i The biasing conditions are achieved by the use of voltage- divider consisting of R1 & R2

ii The circuit is stabilized against temperature variation by the emitter resistor R4

3 a


13/24

i hfe is the ratio of the collector current to the base current at constant of collector- emitter voltage. i.e hfe

=B

C

I

I@ Constant Vce

ii hie is the ratio of the base- emitter voltage to the base current at constant of base emitter voltage. i.e

hie =b

be

I

V@ Constant Vbe

b)


OUT

R

R NB: hie=t5K is assumed to be an

erro from the examiner. Hence, the value of Rb is considered as the Rin

Rin = hie// RB

K68 ; Rout =oeh

1// R

L/ Rc=

==

K7.11.1651

200003.1803

200003.1803

Voltage gain = 3

3

1068

107.1100

= 2.5

Voltage gain=IN

OUT

V

V

Vout = Voltage gain Vin = 310105.2 =25mV

4 a

b)

The reasons why the gain falls off at the lower and higher frequencies is th presence of

capacitances, some of which are connected in series along the signal path and some in parallel.

c).i Increase bandwidth


14/24

ii Highly stabilized gain

iii Less harmonic distortion

5 a

First- order high- pass active filter

The filter allows signals of frequency greater than the cutt-off to pass while blocking signals of

frequency below the cutt-off.The function of this filter is to differentiate the signal.The presence of an operational amplifier

gives room for infinite gain but does not alter the function of the circuit.

b)

i Class B The transistor is biased and the amplitude of the input signals are such that output



ii Class C The transistor is biased and signals amplitudes are such that output current flows for

appreciably less than half- cycle of the input signal i.e 120 0 or 150 0 angle condition.

6a

1. The gain must be infinite

2. The overall loop voltage gain must be greater than one

3. Must occur at a single frequency

c) Tunned- LC Oscillator

When the supply is first switched ON a transient current is developed in the tunned circuit as the

collector current rises to its quiescent value. This transient current initials natural oscillations in the tank






15/24


16/24

then recombined at the output . This arrangement gives excellent efficiency but suffers from the

drawback that there is a small mismatch at the joins between the two halves of the signal.

iii Crossover distortion can be eliminated by applying slight forward biased to each emitter transistor . In

effect, it means locating the Q- point of each transistor slightly above cut-off so that each one operates formore than one half-cycle. This resulted to class AB operation.

9 a


ii Time base ; This controls the speed at which the beam is swept across the screen from

left to right. It does this by setting the sweep time of the time base generator

b)

i When the frequency, period and phase angle of a waveform is to be determined

ii When simultaneous display of two or more waveforms is required .

iii When a visual indication of the waveform of a quantity is needed.

c)

Limitation of the instrumentThe instrument disturbing the circuit

Instrument aging

10 a

Like all other diode, light-emitter diodes are constructed of a p-layer and n-layer between which a depletion

layer is formed. If a diode is operated in the forward direction , and if the doping of the two layers is

roughly the same, the same number of electrons will travel from the n-layer to the p-layer as the number ofholes migrating from the p-layer to the n-layer. In the case of LED, however, the n-layer is very heavily

doped, whereas the p-layer is less heavily doped.

Operation

When the LED is operated in a forward direction the current through the depletion layer is almost

completely carried by electrons. The electrons reaching the p-layer recombine with the holes present

there.Energy is liberated ,which , depending on the material

Used, is emitted to some extent as visible light or as infra-red radiation.


17/24

b)

An optical-fibre is a glass or plastic fibre designed to guide light along its length

It is a cylindrical dielectric waveguide that transmit light along its axis by the process oftotal internal

reflection

c)

In a fibre- optical communication system, information signal which is an incoming electrical signal is turninto light using a device called an emitter. The emitter may be a LED ( light emitter diode ) or laser diode.


At the receiver side, the light signal, which made it through the fibre- optic cable is returned to an electrical

signal by a detector. Detector might be a PIN diode or an APD ( Avalanche photo diode ). The light send



June 2009

1 a

i Differential amplifier

ii Single-ended and Double-ended operations

iii The typical voltage gains for single-ended and double ended operation of differe ntial amplifier are

ONE (1) and TWO (2) respectively.

b)

i


18/24

input voltage(Vin) tt == 1002sin10sin10

output voltage(Vout)

= ttdtdCR = 200cos10200106.1101200sin10 36

The amplitude of the output waveform= V54.10)01.0142.3200cos(6.10 =

The phase shift between the input and the output waveform is given by;

03.601.0200 === t

2 a

.

ii The primary cause of gain variation in amplifiers is the presence of capacitances, some of which are

connected in series along the signal path and some in parallel

At low frequencies, the series connected capacitors offer relatively large reactance thereby dropping off a

large part of the input signal.At high frequencies, the reactance of the internal transistor capacitance and stray capacitances drops.

b)

BBBCC VRIV = 1 ---------- (1) EEBEBB RIVV = ---- (2) EEE RIV = --- (3)

Using Voltage- Divider Rule

VVRR

RV

CCBB 941.2910681033

1033

33

3

21

2=

=

=

From (1) mARVVI BBCCB 0891.0

1068

941.29

3

1

=

=

=

CE II

4RIV CE = ---- (4) mAR

VI EC 2

101

2

3

4

=

==

CCC IV = ECEC VVR


19/24

Collector Voltage

V

VRIVV ECCCCCE

6.2)2102.2002.0(9

)(

3==

=

3 a

N- Channel JFET comprises of a channel of n- type material surrounded by material of the opposite

polarity. The ends of the channel ( in which conduction takes place ) form electrode known as the source

and drain. The effective width of the channel is controlled by a charge placed on the third electrode called

gate.

OPERATION

There are two distinct regions of operation, both of which have useful application.For low values of drain- source voltage, the drain current is progressively increased as the gate-


(amplification)

For high values of drain- source voltage, the drain current is progressively reduced as the gate-source voltage is made more negative. At a certain value of gate- source voltage, drain current

falls to zero and the device is said to be cut-off ( switching)

b)

INGS VV = ----- (1) OUTGSmO RVgV = ----- (2)

== KRY

DS

OS

501

=

= KRR LD 4.3109.28

109.325//3

6

Voltage Gain 16102.310533====

GS

OUTGSm

IN

OUT

V

RVg

V

V

The output voltage VVnVoltageGaiV INOUT 85.016)( ===

4a

LD

OS

OUT RRY

R ////1

=

=

= KROUT 2.3104.53

104.31050

3

33


20/24

i Input resistance is high ( infinite )

ii Output resistance is low

b)

The function of the operational amplifier is to allow signals whose frequency is less than cut-off frequency

to pass while blocking the signals with frequency above cut-off. It integrates the input signals.

ii with

TCR

c)

i Class A The transistor is biased that output current flows for the full- cycle of the input signal .

That is the transistor remains forward biased through out the input cycle. Its condition angle is 3600 . The maximum efficiency of the amplifier is 50%.

ii Class B The transistor is biased and the amplitude of the input signals are such that output



iii Class C The transistor is biased and signals amplitudes are such that output current flows for

appreciably less than half- cycle of the input signal i.e 120 0 or 150 0 angle condition.

5 a

i Voltage- series or shunt derived series and current- series or series derived series fed feedback.

ii The effects of current-series fed feedback on the input and output impedance is that both the impedances

increased.

While the effect of voltage-series fed feedback is that the input impedance increased but the output

impedance decreased.

b)

.i Increase bandwidth

ii Highly stabilized gain

iii Less harmonic distortion

iv Reduced noise


21/24

6 a

d) Tunned- LC Oscillator

When the supply is first switched ON a transient current is developed in the tunned circuit as thecollector current rises to its quiescent value. This transient current initials natural oscillations in the tank





ii frequency of oscillation = KHzLC

6.791041012

1

2

1

93=

=

b)

Two advantages of using crystal control in oscillators are

i For high frequency stability

ii Frequency can be adjusted

7 a

EX-OR gate

The gate will produce a logic 1 output whenever either one of the input is at logic 1 and the other is at

logic 0

Boolean Expression

= BABA

b)

i

ABCBCACBACABF =

ii BCACBAABBCACBACCABF == )(

= BCAACABBCACBABCACBBA == )()(

BCACABBAACAB == )(

INPUT OUTPUT

A B A B0 0 0

0 1 1

1 0 1

1 1 0


22/24

8 a

i Class A- 50%

ii Class B78.5%

b)i Class B push- pull amplifier

ii mAR

VI

L

CC

C 77.1108.6

12

3=

==

iii

A practical circuit using class B element is the complementary pair or push- pull arrangement . The

complementary devices are used to amplify the opposite halves of the input signal which is then

recombined at the output . This arrangement gives excellent efficiency but suffers from the drawback

that there is a small mismatch at the joins between the two halves of the signal.

Crossover distortion can be eliminated by applying slight forward biased to each emitter transistor . In

effect, it means locating the Q- point of each transistor slightly above cut-off so that each one operates for

more than one half-cycle. This resulted to class AB operation.

iv


23/24

9 a

i Frequency of4

10

1014

13

31=

=

W ii Amplitude of

mVW 10101013

2==

iii phase difference between0

2190=andWW

b)i Resolution is the size or magnitude of each piece. It is the analog equivalent weight of the least

significant bit

ii Resolution Vscaleinputfulln

9.112

20

125.3

=

=

= NB: n= Number of digit

10 a

i Basic function of photovoltaic is to convert light energy into electrical energy.

ii Basic function of photoconductive is to create resistance which varies inversely with intensity of light

that falls upon it.

b)

Phototransistor is light- sensitive transistor and similar to an ordinary bipolar junction transistor except

that it has no connection to the base terminal. Its operation is based on the photodiode that exists at the

collector-base junction . Instead of the base current, the input to the transistor is provided in the form of

light.

OPERATION

When light is incident on the collector- base junction, a base current is produced which is directly

proportional to the light intensity. When there no incident light on the collector base junction, there is a


24/24

small thermally generated collector- to- emitter leakage current which in this case, is called dark current

and is in the nanoamps range.

c)

An optical-fibre is a glass or plastic fibre designed to guide light along its lengthIt is a cylindrical dielectric waveguide that transmit light along its axis by the process oftotal internal

reflection

In a fibre- optical communication system, information signal which is an incoming electrical signal is turn

into light using a device called an emitter. The emitter may be a LED ( light emitter diode ) or laser diode.


At the receiver side, the light signal, which made it through the fibre- optic cable is returned to an electrical

signal by a detector. Detector might be a PIN diode or an APD ( Avalanche photo diode ). The light send



suggested answers electronics

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