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FIRST SEMESTER B.TECH DEGREE EXAMINATION,JANUARY 2016

COURSE CODE: BE101-04

COURSE NAME: INTRODUCTION TO ELECTRONICS ENGINEERING

PART A

(Answer all questions.Each question carries 2 marks)

1. Color band sequence on a resistor is yellow,violet,red and silver.What is it’s resistance value?

2. What is the working principle of a transformer? 3.A number 104 is written on the body of a ceramic capacitor.What is the value of capacitance?

4.Draw the piecewise linear model of diode.

5. Differentiate between zener and avalanche break down.

6. A silicon diode has reverse saturation current of 2.5 µA at 300K.Find forward voltage for a forward current of

10mA.

7.Discuss the role of bypass capacitor in a single stage RC coupled amplifier.

8. The widely used voltage amplifier configuration is CE , mention the reason.

9. Derive the relationship between α and β.

10. Draw the equivalent circuit of a UJT.

11. Write any four advantages of FET over BJT.

12. How FET functions as voltage variable resistor?

13. Why is the ripple of HWR higher than that of FWR?

14. Design a silicon diode clipper for transfer characteristics in figure below?

15. Assuming drop across the diode is .6 V, find the output voltage V0?

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16. Find the maximum and minimum value of zener diode current.

17. A CRO is set to a time base of .1ms/div with a 2 V/div amplitude.Sketch the display of a pulse signal

waveform with a frequency 1KHz and amplpitude 8V peak.

18.What is a precision and resolution of measuring instruments? 19. Compare an analog multimeter and a digital multimeter.

20. How testing of a diode is carried out?

PART B

Answer any 4 complete questions each having 10 marks

21. Differentiate the capacitors based on the types of dielectric used and explain their construction. (10)

22. A) Explain the constructional details of carbon composition resistors? (5)

B) Explain the formation of a potential barrier in a p-n junction and show the polarity of barrier potential. (5)

23. A) Explain the working of RC coupled amplifier with a neat circuit diagram. (6)

B) Explain the frequency response curve. (4)

24. A) With reference to the following circuit, draw the load line and mark Q point of a silicon transistor operating

in CE mode based on the following data( β= 80,Rs=47K RL=1K and neglect ICBO). (6)

B) Sketch the forward characteristics of a SCR. Explain the importance of Holding current in a SCR. (4)

25. With a neat diagram draw structure of n channel E-MOSFET and explain different regions of operation. (10)

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Answer any 2 complete questions each having 10 marks

26. A) Draw the circuit of a bridge rectifier and explain its working. (5)

B) Derive the expression for Vrms,Vdc,ripple factor and rectification efficiency, peak inverse voltage. (5)

27. A) With the help of suitable block diagram, discuss the working principle of the electronic device

which is used in the laboratories for generating the various standard waveforms. (5)

B) Draw the block diagram of DC power supply and list out the functions of each block. (5)

28. With neat schematic diagram, explain tne working of a CRO. List its applications. (10)

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FIRST SEMESTER B.TECH DEGREE EXAMINATION,JANUARY 2016

COURSE CODE: BE101-04

COURSE NAME: INTRODUCTION TO ELECTRONICS ENGINEERING

Answers

1.Color band sequence on a resistor is yellow,violet,red and silver.What is it’s resistance value?

Ans: 4.7K±10%

2. What is the working principle of a transformer?

Ans: The basic principle of a transformer is electromagnetic induction. When ac supply of voltage is connected to

primary winding, an alternating flux is set up in the core. This alternating flux when links with secondary winding,

an emf is induced in it and is called mutually induced emf. Although there is no electrical connection between

primary and secondary winding, but electrical power is transferred from primary circuit to the secondary circuit

through mutual flux.

3.A number 104 is written on the body of a ceramic capacitor.What is the value of capacitance?

Ans: 10,0000 pf = 100nf = o.1 µF

4.Draw the piecewise linear model of diode.

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5. Differentiate between zener and avalanche break down.

Ans:

Avalanche Break Down:

When a very large negative bias is applied to the p-n junction, sufficient energy is imparted to charge

carriers that reverse current can flow, well beyond the normal reverse saturation current. In addition, because of the

large electric field, electrons are energized to such levels that if they collide with other charge carriers at a lower

energy level, some of their energy is transferred to the carriers with low energy, and these can now contribute to the

reverse conduction process, as well. This process is called impact ionization. Now, these new carriers may also have

enough energy to energize other low energy electrons by impact ionization, so that once a sufficiently high reverse

bias is provided, this process of conduction takes place very much like an avalanche: a single electron can ionize

several others. This phenomenon is known as avalanche break down.

Zener Break Down:

When increasing reverse bias voltage across the junction, the electric field at the junction also increases.

This high electric field causes covalent bonds within the crystal to break. Thus a large number of charge carriers

become available. Thus a large current to flow through the junction, This phenomenon is called zener break down.

6. A silicon diode has reverse saturation current of 2.5 µA at 300K.Find forward voltage for a forward

current of 10mA.

Soln:

Given, reverse saturation current is = 2.5×10-6 A

Temperature in Kelvin,T= 300K

Diode forward current, i = 10×10-3A

Diode equation is given by

i = is [exp ( )-1]

Where,k = Boltzmann’s constant = 1.38 x 10-23 J/K

T= Temperature in Kelvin and

q = magnitude of charge on an electron = 1.6 x 10-19 C

After simplifying, we get;

Forward voltage, v = ln (1+ )

v= .071 volt= 71mv

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7.Discuss the role of bypass capacitor in a single stage RC coupled amplifier.

Ans: The emitter bypass capacitor CE,offers low reactance path to the signal.If it is not present ,the the voltage drop

across RE will reduce the effective voltage available across the base-emitter terminals (the input voltage) and thus

reduces the gain.

8. The widely used voltage amplifier configuration is CE ,mention the reason.

Ans: CE configuration is mainly used because its current, voltage and power gains are quite high and the ratio of

output impedance and input impedance are quite moderate.

9. Derive the relationship between α and β.

Soln:

We know that

Dividing numerator and denominator by

Therefore,

10. Draw the equivalent circuit of a UJT.

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11.Write any four advantages of FET over BJT.

Ans:

a. FET’s are less noisy compared to BJT’s.

b. FET exhibits high input resistance as compared to BJT.

c. Operation of FET is almost temperature independent.

d. FET is simpler to fabricate and small size as compared to BJT.

12.How FET functions as voltage variable resistor?

Ans: FET is a device that is usually operated in the constant-current portion of its output characteristics (saturation

region). But if it is operated on the region prior to pinch-off (Ohmic region where the VDS is small), it will behave as

a voltage-variable resistor. It is due to the fact that in this region drain-to-source resistance RDS can be controlled by

varying the bias voltage VGS. In such applications the FET is also referred to as a voltage-variable resistor.

13. Why is the ripple of HWR higher than that of FWR?

Ans: In the half wave rectifier, the output current in the load contains, in addition to dc component, ac

components of basic frequency equal to that of the input voltage frequency. The residual ac ripples is very

low in the output of a bridge rectifier. Therefore the ripple factor is high in the half wave rectifier than in

the full wave rectifier.

14. Design a silicon diode clipper for transfer characteristics in figure below?

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Solution

15. Assuming drop across the diode is .6 V, find the output voltage V0?

Ans: Here we take a 5 volt battery as DC source connected in series with the diode in such a way that negative

terminal of the battery is connected to the anode terminal of the diode. During the negative half cycle of the input

sine wave, the diode conducts and capacitor charges through diode and the DC source till (Vm -5) volts with positive

polarity at the right side of the capacitor. The charging of the capacitor is limited to (Vm -5) volts due to the presence

of the DC source. If we take the voltage drop across the diode is .6 volt.

The output is then represented as Vo= (Vm -5-.6)+ Vm sin t. Here the Vm is 10 volt. Therefore the output

voltage is Vo = (10-5-.6)+ 10 sin t = 14.4 volt.

16. Find the maximum and minimum value of zener diode current.

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Solution

The zener diode will conduct maximum current, when the input voltage is maximum. i.e. 120 V.

The total current is the current through 5KΩ resistor and is

Load Current

Maximum zener current

For minimum input voltage, zener current will be minimum. Therefore,

Minimum zener current

17. A CRO is set to a time base of .1ms/div with a 2 V/div amplitude.Sketch the display of a pulse

signal waveform with a frequency 1KHz and amplpitude 8V peak.

Ans:

1 KHz frequency means,Time period is 1/1×103 sec = 1 ms.

18.What is a precision and resolution of measuring instruments?

Ans: Precision is defined as a measure of the reproducibility of the measurement. That is, given fixed value of a

variable, precision is a measure of the degree to which successive measurements differ from one another.

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Resolution is defined as the smallest change in the measured value to which the instrument will respond.

19. Compare an analog multimeter and a digital multimeter.

Ans:

• Analog multimeters give the output as a reading on a scale against a pointer, while digital multimeter output is in

numerical form displayed on a LCD.

• Analog multimeters give a continuous output and carry a greater uncertainty in the measurement (about 3%), while

digital multimeter measurements have a far less uncertainty (about 0.5% or less). Digital multimeters are more

accurate than analog multimeters.

• Digital multimeters have a better range of measurements than analog multimeters.

• Analog multimeters are less costly while digital multimeters are expensive.

20. How testing of a diode is carried out?

Ans:

Insert the black test lead banana plug into the negative COM jack and the red test lead banana plug into the

positive diode jack.

Turn the function switch to the position.

Touch the test probes to the diode under test. Forward voltage will indicate .4 to .5V. Reverse voltage will

indicate “1”. Shorted devices will indicate near 0mV. Shorted devices will indicate near 0mV and an open

device will indicate “1” in both polarities.

PART B

21. Differentiate the capacitors based on the types of dielectric used and explain their construction.

Ans:

The dielectric material can be made from a number of insulating materials or combinations of these materials with

the most common types used being: air, paper, polyester, polypropylene, Mylar, ceramic, glass, oil, or a variety of

other materials.

Mica Capacitors

Mica capacitors are made from plates of Aluminium foil separated by sheets of mica. The plates are

connected to two electrodes. The mica capacitors have excellent characteristics under stress of temperature

variations and high voltage applications (~500 V). Available capacitances range from 5 to 10,000 pF. Its leakage

current is very small (Rleakage is about 1000 MW).

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Fig: Construction of a Mica Capacitor

Ceramic Capacitors

Ceramic capacitors are made in many shapes and sizes. A ceramic disc is coated on two sides with a metal,

such as copper or silver. These coatings act as two plates. After attaching tinned-wire leads, the entire unit is coated

with plastic and marked with its capacitance value—either using numerals or colour code. The colour coding is

similar to that used for resistances. Ceramic capacitors are very versatile. Their working voltage ranges from 3 V

(for use in transistors) up to 6000 V. The capacitance value ranges from 3 pF to about 3 mF. Ceramic capacitors

have a very low leakage current (Rleakage is about 1000 MW) and can be used in both dc and ac circuits.

Fig: construction of a Ceramic Capacitor

Paper Capacitors

This capacitor consists of two metal foils separated by strips of paper. This paper is impregnated with a dielectric

material such as wax, plastic or oil. Since paper can be rolled between two metal foils, it is possible to concentrate a

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large plate area in a small volume. Paper capacitors have capacitances ranging from 0.0005 mF to several mF, and

are rated from about 100 V to several thousand volts. They can be used for both dc and ac circuits. Its leakage

resistance is of the order of 100 MW.

Fig: Construction of a Paper Capacitor

Electrolytic Capacitors

An electrolytic capacitor consists of an aluminium-foil electrode which has an aluminium-oxide film

covering on one side. The aluminium plate serves as the positive plate and the oxide as the dielectric. The oxide is in

contact with a paper or gauze saturated with an electrolyte. The electrolyte forms the second plate (negative) of the

capacitor. Another layer of aluminium without the oxide coating is also provided for making electrical contact

between one of the terminals and the electrolyte. In most cases, the negative plate is directly connected to the

metallic container of the capacitor. The container then serves as the negative terminal for external connections.

22.A) Explain the constructional details of carbon composition resistors?

Ans:Carbon Composition Resistors are very commonly used low cost resistor. The construction of

carbon composition resistor is very simple. It is also commonly referred as carbon resistor. It is mainly

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made of carbon clay composition covered with a plastic case. The lead of the resistor is made of tinned

copper. The main advantages of these resistors are that they are easily available in local market in very

low cost and they are very durable too. But the main disadvantage is that they are very much temperature

sensitive. These resistors are available in wide range of values. It is available in as low as 1 Ω value and it

is also available in as high as 22 Mega Ω value. The tolerance range in resistance of carbon composition

resistor is of ± 5 to ± 20 %. Such resistor has a tendency of electric noise due to passage of electrical

current from one carbon particle to other. Where low cost is the main criteria of designing a circuit rather

than it's performance, these resistors are normally used.

22 B). Explain the formation of a potential barrier in a p-n junction and show the polarity of barrier

potential.

Ans:

When P-type and N-type semi conductor is suitably joined to an N-type semi conductor, PN junction is

formed. Such a PN junction is the basic building block on which the operation of all semi conductor devices

depends. The P-region has holes and acceptor ions and N-region has electrons and doner ions. Here electrons and

holes are mobile and ions are immobile. The electrons in the N-type material diffuse into the P-type and combine

with holes in P-type material, creating negatively charged ions in the P-type material nearby junction. Similarly

holes from P-type material diffuse into the N-type material and combine with electrons in the N-type material,

creating positively charged ions particularly in the region close to the junction in N-type material.

After a few recombination of electrons and holes, a narrow width of fixed positive charge on N-side of the

junction and fixed negative charge on P-side of the junction formed as shown in figure. This region is known as

depletion region. This region has immobile ions which are electrically charged, hence the region is also called space

charge region. Due to this region further diffusion is prevented, because now positive charge on N-side repels holes

to cross from P-type to N-type and negative charge on P-type repels electrons to enter from N-type to P-type. Thus

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barrier is setup against further movement of charge carriers and is called potential barrier or junction barrier. For

Silicon PN junction barrier potential is about .7 volt where as for Germanium, it is .3 volt.

23. A) Explain the working of RC coupled amplifier with a neat circuit diagram.

Ans:

Fig:R-C Coupled Amplifier

The figure shows the single stage Common emitter R-C Coupled amplifier circuit. Here both forward bias voltage at

emitter-base junction and reverse bias voltage at the collector-emitter junction are derived from the single supply

voltage VCC.

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The magnitude of these bias voltages are adjusted to operate the transistor in active region. In this arrangement base

is the input terminal and collector is the output terminal.

Working

When an input signal is applied in the emitter-base junction of the transistor,the signal is superimposed in

the d.c voltage at emitter-base junction.Therefore during the positive half-cycle of the input signal the forward bias

across the junction increases,because it is already positive with respect to ground.This increase in forward bias

increases the base current IB.Due to increase in base current the collector current also increases. In CE configuration

the corresponding increase in collector current will be ’β’ times the increase in IB. This increase in collector current

produces more voltage drop across the output terminal. During negative half-cycle of the input signal the forward

bias across emitter junction will be decreased and decreases base curent. This decrease in base current results in

corresponding (β-times) decrease in collector current.Consequently the drop across output terminals will be

decreased.It is clear that the collector current varies according to the input signal applied and variation is β-times to

that of input current variation.Due to this action an amplified form of input signal can be obtained at the output

terminal.

23 B) Explain the frequency response curve.

Ans:

The frquency response curve of a typical R-C Coupled amplifier is shown in figure.

At lower frequencies (below 50Hz) higher capacitive reactance of coupling capacitor allows very small part

of signal to pass from one stage to next and also because of higher reactance of emitter bypass capacitor CE,the

emitter resistor RE is not shunted. Thus the voltage gain falls off at low frequencies.

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In the mid frequency range (50Hz -20KHz),the voltage gain of the amplifier is constant.With increase in

frequency in this range ,the reactance of the coupling capacitor reduces thereby increasing the gain but at the same

time lower capacitive reactance causes higher loading resulting in lower voltage gain.Thus the two effect cancel

each other and uniform gain is obtained in mid frequency range.

At high frequencies (exceeding 20 KHz),gain of amplifier decreases with increase in frequency. At high

frequencies,the reactance of coupling capacitor become very small and it behaves as a short-circuit.This increases

the loading of the next stage and reduces the voltage gain. The other factor responsible for the reduction in gain at

higher frequencies is the presence of inter electrode capacitance Cbc between base and collector.It connects the

output with the input. Because of this, negative feedback takes place in the circuit and the gain decreases. This

feedback effect is more, when Cbc provides a path for higher frequency ac currents.

In the figure f1 and f2 are lower and upper cut-off frequencies respectively.The difference between upper

cut-off frequency(f2) and lower cut-off frequency(f1) is called the Bandwidth(BW).

Am is the maximum gain or mid frequency of the amplifier.It means at these frequencies ,the output voltage

is 1/ 2 times the maximum voltage.Since the power is proportional to the square of the voltage,the output power

at these cut-off frequencies become one half of the power at mid-frequencies.On dB scale this is equal to a reduction

in power by 3dB.For this reason these frequencies are also called 3 dB frequencies.

24. A) With reference to the following circuit, draw the load line and mark Q point of a silicon transistor

operating in CE mode based on the following data( β= 80,Rs=47K RL=1K and neglect ICBO).

Ans: Here the co-ordinates of A are obtained by writing Ic=0 in equation of Vcc.

Thus the co-ordinates of end ‘A’ are

= 20V

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The co-ordinates of end ‘B’ are obtained by writing in the equation

= 20mA

Thus the co-ordinates of end ‘B’ are

24 B) Sketch the forward characteristics of a SCR. Explain the importance of Holding current in a SCR.

V-I Characteristics of a SCR (Thyristor) is shown in figure.

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Fig: VI Characteristics of SCR

Importance of Holding current in a SCR

Holding current is the minimum current that should flow through an SCR to maintain it in the ON state, after it has

been switched ON. If the device current falls below the holding current, the SCR switches OFF. Hence the holding

current plays an important role in switching off the SCR.

25. With a neat diagram draw structure of n channel E-MOSFET and explain different regions of operation.

Ans:

Enhancement MOSFET (E-MOSFET):

The N-channel E-MOSFET consists of lightly doped P-substrate and two highly doped N-regions are

diffused into the P-type substrate. Here two N-regions act as Source and Drain. A thin layer of SiO2 is grown over

the surface of the structure with suitable cuts for the contacts to Drain and Source. A metallic layer such as

Aluminum is laid on the oxide layer between source and drain and also on the two cuts. This Aluminum layer forms

the gate of FET.With this structure a parallel plate capacitor is formed with the metal area of the gate and

semiconductor channel acting as the electrodes of the capacitor and oxide layer act as dielectric between the

electrodes.

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If the substrate is grounded and Positive voltage is applied at the gate, the positive charge on the gate

induces an equal negative charge on the substrate side between source and drain regions as shown in figure. Thus

electric field will be formed normal to SiO2 layer. As the positive voltage on the gate increases, the induced

negative charge in semiconductor also increases. Hence conductivity increases and current flows from source to

drain through the induced channel. It is observed that drain current has been enhanced on the application of positive

voltage. Therefore these types of FETs are called Enhancement MOSFET.

Fig: n-Channel E-MOSFET

26. A) Draw the circuit of a bridge rectifier and explain its working.

Ans:

Bridge Rectifier:

The circuit arrangement of the bridge rectifier is show in figure. It contains four diodes, but avoids the need

for a centre-taped transformer. During the positive half cycle of the secondary voltage, diode D2 and D3 will be

forward biased. At the same time diodes D1 and D4 are reverse biased. Therefore diodes D2 and D3 conduct and

current flows through D2, RL, D3 and transformer secondary. During the negative half cycle of the secondary voltage

diodes D1 and D4 will be forward biased and D2 and D3 will be reverse biased. Therefore the current flows through

D1, RL ,D4 and transformer secondary. Here in both cases the current flow through load resistor RL is in the same

direction. Hence DC voltage is obtained at the output. The wave forms are shown in fig (b).

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26. B) Derive the expression for Vrms,Vdc,ripple factor and rectification efficiency, peak inverse

voltage.

Ans:

DC output (Vdc):

The DC output voltage or a current of the rectifier is the average value of the output voltage or current.

Ie,

Let,

Alternating voltage appearing across the secondary winding of the transformer

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Current flowing through the load resistor

Diode forward resistance = 0

Here area =

=

= 2

Time period =

Now,

Voltage across RL,

=

While diode forward resistance is taken into account

DC voltage across RL can be written as

=

Since,

RMS Value (V rms):

When writing the rms value of the full wave rectifier both half cycles are to be considered. Hence the rms value of

the current is given by,

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Hence,

Ripple Factor:

= .482

Efficiency:

Irms = Im / √2

Vrms = Vm / √2

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( Where )

Peak inverse voltage (PIV)

When the secondary voltage is at its positive maximum value Vm diodes D2 and D3 are forward biased and

conduct. Conducting diodes have zero resistance and zero voltage drop across them, the points A and C in fig (a) are

in same potential. Similarly points B and D .Therefore the reverse voltage coming across diodes D1 and D4 will be

equal to the potential difference between the points A and D will be equal to Vm.

PIV = Vm

27. A) With the help of suitable block diagram, discuss the working principle of the electronic device

which is used in the laboratories for generating the various standard waveforms.

Ans:

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Function generator is the electronic device that has the capability of producing different types of waveforms as its

output signal. The most common output waveforms are sine-waves, triangular waves, square waves, and sawtooth

waves.

Block Diagram of a Function Generator

The block diagram of a function generator is given in the figure. In this instrument the frequency is

controlled by varying the magnitude of current that drives the integrator. This instrument provides different types of

waveforms (such as sinusoidal, triangular and square waves) as its output signal with a frequency range of 0.01 Hz

to 100 kHz.

The frequency controlled voltage regulates two current supply sources. Current supply source 1 supplies

constant current to the integrator whose output voltage rises linearly with time. An increase or decrease in the

current increases or reduces the slope of the output voltage and thus controls the frequency.

The voltage comparator multivibrator changes state at a predetermined maximum level, of the integrator

output voltage. This change cuts-off the current supply from supply source 1 and switches to the supply source 2.

The current supply source 2 supplies a reverse current to the integrator so that its output drops linearly with time.

When the output attains a predetermined level, the voltage comparator again changes state and switches on to the

current supply source. The output of the integrator is a triangular wave whose frequency depends on the current

supplied by the constant current supply sources. The comparator output provides a square wave of the same

frequency as output. The resistance diode network changes the slope of the triangular wave as its amplitude changes

and produces a sinusoidal wave with less than 1% distortion.

Fig: Block Diagram of a Function Generator

27. B) Draw the block diagram of DC power supply and list out the functions of each block.

The block diagram of a DC power supply is shown in figure.

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Fig:Block Diagram of a Power Supply

The first block in the DC power supply is a transformer. A transformer is a device which transforms high

voltage AC into low voltage AC or vice versa. Our goal is to convert high voltage AC into low voltage DC. So the

transformer that is used in power supply is step-down transformer, which steps down the input AC voltage.

The output of the transformer is given to the input of the rectifier.

The rectifier converts the input sinusoidal voltage to a unipolar output or pulsating DC. Full wave rectifiers are

the most commonly used rectifiers in power supply. The output of the rectifier is given to input of the filter circuit.

The output after being processed by full wave rectifier is not a pure DC. The output is a pulsating DC. The output

contains large fluctuations in voltages. The filter circuit reduces the variation in magnitude (ripples) of the rectifier

output. However, the output from the filter is still not a pure DC but filters removes the AC component in the

voltage to a considerable extent. This increases the average DC value of the output voltage.

The output of the filter is given to the voltage regulator circuit for stabilizing the magnitude of the DC output

voltage against the variations caused by changes in the load current and input voltage.

28. With neat schematic diagram, explain tne working of a CRO. List its applications.

Ans:

Working of a CRO :

An outline explanation of how an oscilloscope works can be given using the block diagram shown below:

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Fig: Block Diagram of CRO

A simple cathode ray oscilloscope consists of a cathode ray tube, a vertical amplifier, a time base,

a horizontal amplifier,a trigger circuit and a power supply.

Like a televison screen, the screen of an oscilloscope consists of a cathode ray tube. Although the size and

shape are different, the operating principle is the same. Inside the tube is a vacuum. The electron beam emitted by

the heated cathode at the rear end of the tube is accelerated and focused by one or more anodes, and strikes the front

of the tube, producing a bright spot on the phosphorescent screen.

The electron beam is bent, or deflected, by voltages applied to two sets of plates fixed in the tube. The

horizontal deflection plates, or X-plates produce side to side movement. As you can see, they are linked to a system

block called the time base. This produces a sawtooth waveform. During the rising phase of the sawtooth, the spot is

driven at a uniform rate from left to right across the front of the screen. During the falling phase, the electron beam

returns rapidly from right ot left, but the spot is 'blanked out' so that nothing appears on the screen.In this way, the

time base generates the X-axis of the V/t graph.

The slope of the rising phase varies with the frequency of the sawtooth and can be adjusted, using the

TIME/DIV control, to change the scale of the X-axis. Dividing the oscilloscope screen into squares allows the

horizontal scale to be expressed in seconds, milliseconds or microseconds per division (s/DIV, ms/DIV, µs/DIV).

Alternatively, if the squares are 1 cm apart, the scale may be given as s/cm, ms/cm or µs/cm.

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The signal to be displayed is connected to the input. The AC/DC switch is usually kept in the DC position

(switch closed) so that there is a direct connection to the Y-amplifier. In the AC position (switch open) a capacitor is

placed in the signal path. The capacitor blocks DC signals but allows AC signals to pass.

The Y-amplifier is linked in turn to a pair of Y-plates so that it provides the Y-axis of the the V/t graph. The

overall gain of the Y-amplifier can be adjusted, using the VOLTS/DIV control, so that the resulting display is

neither too small or too large, but fits the screen and can be seen clearly. The vertical scale is usually given in

V/DIV or mV/DIV.

The trigger circuit is used to delay the time base waveform so that the same section of the input signal is

displayed on the screen each time the spot moves across. The effect of this is to give a stable picture on the

oscilloscope screen, making it easier to measure and interpret the signal.

Changing the scales of the X-axis and Y-axis allows many different signals to be displayed. Sometimes, it

is also useful to be able to change the positions of the axes. This is possible using the X-POS and Y-POS controls.

For example, with no signal applied, the normal trace is a straight line across the centre of the screen. Adjusting Y-

POS allows the zero level on the Y-axis to be changed, moving the whole trace up or down on the screen to give an

effective display of signals like pulse waveforms which do not alternate between positive and negative values.

Applications of CRO

1. Measurement of voltage – Voltage waveform will be made on the oscilloscope screen. From the

screen of the cro, the voltage can be measured by seeing its amplitude variation on the screen.

2. Measurement of current – Current waveform will be read from the oscilloscope screen in the

similar way as told in above point. The peak to peak, maximum current value can be measured from

the screen.

3. Measurement of phase – Phase measurement in CRO can be done by the help of Lissajous

pattern figures. Lissajous figures can tell us about the phase difference between two signals.

Frequency can also be measured by this pattern figure.

4. Measurement of frequency – Frequency measurement in cathode ray oscilloscope can be made

with the help of measuring the time period of the signal to be measured.