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Department of Electronics and Communication Engineering Electronic Circuit Analysis Lab

INTRODUCTION

Engineering: A profession in which a knowledge of mathematical, natural and social sciences gained by study, experience, and practice is applied with judgment to develop ways to utilize economically the materials and forces of nature for benefit of society.

Circuit: An interconnection of electrical devices in which there is at least one closed path in which current may flow.

Passive and active elements: An active element is defined as an element which is capable of furnishing an average power greater than zero to some external device, where the average is taken over an infinity time interval. Ideal sources are active elements, and an operational amplifier is also an active device.

Passive element is defined as an element that cannot supply an average power that is greater than zero over an infinite time interval. The resistor, capacitor and inductor are passive elements. The resistor is a passive element because the energy it received is usually transformed into heat. Both the inductor and capacitor are also passive elements because which are capable of storing finite amounts of energy, but they cannot provide an unlimited amount of energy or finite average power over an infinity time interval

ResistorsResistor: is a component which opposes flow of current. The voltage and current relation in resistor is given by the Ohm’s law V= IR. The resistance of a conductor of length l and cross sectional area A is given by R = l /A, where is the resistivity of the conductor.

When N number of resistors is connected in series the equivalent or total resistance (R) is the sum of the individual resistances. R = R1 +R2 + …….. +RN .

When N resistors are connected in parallel the equivalent or total resistance (R) is given by 1/R = 1/R 1

+ 1/R2 + …. 1/RN

Tolerance: Tolerance is the range of resistance values which enables a user to use a single resistor within its range. This is necessary because resistors cannot manufactured for all values. Let tolerance be x%, and the resistance be R. Then this resistor can be used in the range R+x%.Color coding:

The resistance value of the resistor can be found by the color bands on a resistor. The first two color bands give the first two digits of the resistor. The third color gives the power of 10, multiplied with the first two digits or the number of zeros followed by the first two digits. The fourth band gives the tolerance value.

RESISTOR WITH FOUR COLOUR BANDS

Band 1 Band 2 Band 3Band 4

COLOUR BAND VALUES Black Brown Red Orange Yellow Green Blue Violet Gray White 0 1 2 3 4 5 6 7 8 9Power rating of a resistor: It indicates the maximum power that a resistor can handle.

Maharaj Vijayaram Gajapathi Raj College of Engineering, Vizianagaram 1


While we design the circuit, the maximum current (which should not exceed) that can flow through the resistor Imax is given by

Variable resistors:There are two types of variable resistors, continuous and discrete variable resistors

Potentiometers:In potentiometers the variation of resistance is in continuous manner. It has three leads. The maximum resistance can be obtained by connecting the extreme leads to the circuit. In this case the resistance can not be varied though the knob is turned. Variable resistance can be obtained by connecting center lead with any one of the extreme leads. In this case the variation in the resistance can be obtained by turning the knob. Decade Resistance Box:In DRB, the variation in resistance is in discrete manner. The required resistance is achieved by locating the knob at the required numerical positions.

The total resistance of DRB = sum of the product of the knob position numbers and the multiplication factors.

Rheostat: Rheostat is a continuously varying resistor, whose resistance can be varied linearly. Rheostats have very high power ratings. These are used where precise value of resistance is required. Capacitor

A capacitor is used for storing charge. It consists of two conducting surfaces separated by a thin insulating layer of a very large resistance. If we assume that this resistance is infinite, then equal and opposite charges placed on the capacitor plates can never re-combine, at least by any path between the elements. Capacitance of the capacitor is given by the equation C = ЄA/d and is expressed in Farads Where, ‘A’ is the cross sectional area of the plates.

‘d’ is the distance between plates ‘Є’ is the permittivity constant of the insulating material between the plates.

For Air and Vacuum, Є = Є0 = 8.84 PF/meter We define capacitance C by voltage current relationship by

i = C dv/dt The reactance offered by the capacitor is given by Xc = 1/2ΠfC, so that it acts as an open circuit to DC.

When the capacitors are connected in series then the equivalent or total capacitance is 1/C = 1/C1 + 1/C2 + ………. + 1/CN.

When the capacitors are connected in parallel then the equivalent or total capacitance is C = C1+C2+ ……………. +CN.

Finite amount of energy (½ CV2) can be stored in a capacitor even if the current through the capacitor is zero such as when the voltage across it is constant.

It is impossible to change the voltage across the capacitor by finite amount in zero time. For this requires an infinite current through the capacitor.Inductors

The ability of conductor to produce voltage is known as inductance. A current carrying conductor produces a magnetic field and that changing magnetic field will induce a voltage. The inductance is expressed in Henries. We define the inductance L by the voltage current relationship

V = L (di/dt)


L

V

i

C

+ -Vi


Physical inductor may be constructed by winding a length of wire into coil. The inductance of this coil is proportional to the square of the number of complete turns made by conductor out of which it is formed. For example, an inductor or coil, that has the form of a long helix of very small pitch is found to have an inductance of µN2A/S.

Where A is the cross-sectional area, S is the Axial length of the coil, and µ (mu) is the constant of the material inside the helix called the permeability, µ = µo = 410-7 H / m2

The reactance offered by the inductor is given by XL = 2f L, where f is the applied signal frequency, thus inductor acts as a short circuit to DC.When N inductors are connected in parallel then the equivalent or total inductance is 1/L = 1/L 1 + 1/L2 + ………. + 1/LN.

When N inductors are connected in series then the equivalent or total capacitance is L = L 1+L2+ ……………. +LN.

An infinite amount of energy can be stored in an inductor even the voltage across the inductor is zero, such when the current through it is constant.

It is possible to change the current through the inductor by finite amount in zero time. For this requires an infinity voltage across the inductor.

D.C. Power supply.The available power supplies in this laboratory are single channel and dual channel power supplies.

Dual power supply can be used as two independent voltage/ current sources, where as a single voltage/current source can be obtained form a single channel power supply.

The DC power supply can be used as either voltage source or current source. The voltage which is obtained can be varied by varying the knobs coarse and fine. Coarse knob is used to get the large range variation, where as the fine knob is used to get small variation in voltage. The digital display it self indicates the voltage drawn form the source.

To observe the voltage from channel 1 or channel 2 on single digital display, the selector switch is to be placed in the respective position. When we connect the DC power supply as DC current source the current which is drawn from the source can be varied by varying the current knob.

To observe the current from channel 1 or channel 2 on single digital display, the selector switch is to be placed in the respective position. Cathode Ray Oscilloscope.

The Cathode Ray Oscilloscope (CRO) is an instrument which provides visual presentation of any waveform applied to its input terminals, while the multi-meter provides numerical information about applied signal. The CRO allows the actual form of the waveform to be displayed. The cathode ray tube (CRT) is the heart of the CRO, providing visual display of an input signal waveform.The CRT contains four basic parts.1. An electron gun to produce a stream of electrons.2. Focusing and accelerating elements to provide a well defined beam of electrons 3. Horizontal and vertical deflection plates to control the path of the electron beam.4. An evacuated glass envelop with a phosphorescent screen which glows visibly when struck by the

electron beam.



The voltage to be measured is applied to the vertical defection plates, where as the continuous sweep voltage is applied to the horizontal deflection plates.

Measurement using calibrated CRO: The oscilloscope tube face has been calibrated to compute the amplitude and time measurement.

Amplitude measurement:The amplitude of the voltage can be measured by using the scale-setting of the scope (volts/div knob.)

and the signal measured off the face of the scope.The peak-to-peak amplitude = No. divisions between peak – to –peak of the signal volts/division

knob multiplication factor.

Time measurement: The horizontal scale of the scope can be used to measure the time of the applied signal. The time

period of the periodic signal = No. of divisions during one cycle time/division knob multiplication factor.

Frequency measurement: The measurement of the repetitive waveform-period can be used to calculate the signal frequency.

Since the frequency is the reciprocal of the time period. f=1/T

Measurement of phase and frequency (Lissajous Patterns)The patterns that appear on the screen of a CRT, when sinusoidal voltages are simultaneously applied

to horizontal and vertical plates, are known as Lissajous figures. When two sinusoidal voltages of equal frequency which are in phase with each other are the horizontal and vertical deflection plates, the pattern appearing on the screen is a straight line as is clear from fig below.

ey

(voltage applied to y-plates, frequency = f)

(Voltage applied to x-plates, frequency = f)

Fig 1.2 LISSAJOUS PATTERNS

When two equal voltages of equal frequency but with 900 or 2700 phase shift displacement are applied to a CRO, the trace on the CRO is a circle. An ellipse is formed if the two voltages are not equal and /or out of phase.

When two equal voltages of equal frequency but with a phase shift ‘’(not equal to 00 or 900 )are applied to a CRO, we obtain an ellipse regardless of the two amplitudes of the applied voltages, the ellipse provides a simple means of finding phase difference ‘’ between two voltages.


1

0 2 4

3

t (0,2,4)

1

3

1

0

2

4

3

t

ex


= sin-1 y1 / y2 = sin-1 x1 / x2.

Y Major Axis.

y1 y2

X x1

x2

Fig: 1.3 PHASE SHIFT MEASUREMENT

Y

Major axis

Y2 Y1

X X1

X2




If the major axis of the ellipse lies in the first and third quadrants as in fig:1, the phase angle is either between 00 and 900 or 2700 and 3600. when the major axis of the ellipse lies in second and fourth quadrants as in Fig:2, the phase angle is either between 900 and 1800 or between 1800 and 2700.

Bread board

A bread board has been designed for easy circuit connection. The figure below represents a bread board.

I

II

III

IV

Bread board connection diagram

The upper and the lower parts of the bread board (I&IV) are similar and are generally used for ground connections. The middle parts (II&III) are similar. Part I consists of 100 holes in four groups with insulation in between succeeding horizontal and vertical 25-hole groups. All 25 holes in each group are short circuited.

Part II of the bread board has 5 rows with 65 holes in each row. All holes are short circuited horizontally and insulated vertically.

Digital multimeter

Digital instrument are generally used to measure the parameters of the interest in a laboratory are 1) voltage 2) current 3) power 4) frequency 5) Logic.The basic building block of a Digital instrument is shown below.

In digital multi-meters the measured quantity is displayed numerically instead of deflection, as in conventional analog meters.

3 ½ digit:

In our laboratory we use 3 ½ digit multi-meters. ‘3’ indicates 3 digit display on DMM. For example a 3 bit display on DMM for a 0 to 1V range will indicate values 0 to 999mV with a smallest increment of 1mV. Normally a fourth digit capable of indicating 0 or 1 (hence called half digit) is placed to the left. This permits the digital meter to read values above 999 up to 1999.


A/D converterSignal

Processing Digital display


Half digit 0 or 1 full digit 0-9

Insulators, semiconductors and metals.

A very poor conductor of electricity is called an insulator. An excellent conductor is a metal. A substance whose conductivity lies between an insulator and a conductor is known as a semiconductor.

The forbidden energy gap is relatively high (approx. 6 eV) for an insulator, where as it is relatively small (approx. 1eV) for a semiconductor. In a metal the conduction and valance band overlap each other.

Semiconductors are of two types1. Intrinsic semiconductor2. Extrinsic semiconductor

The semiconductor in pure form is known as intrinsic semiconductor. The conduction level of the intrinsic semiconductor can be increased by increasing the temperature. The semiconductor is thus said to have –ve temperature co-efficient of resistance. Negative temperature coefficient of resistance indicates the decrease in resistance with increase in temperature (increase in conductance with increase in temperature).

The conductance of the intrinsic semiconductor can also be increased by adding some impure atoms to it. This process is called doping. The resultant semiconductor is called extrinsic semiconductor.

The extrinsic semiconductors are divided into two types depending upon the impure atoms added. If trivalent (acceptor) atoms (Boron, Gallium, Indium) are added to the intrinsic semiconductor, p-

type semiconductor is formed. If pentavalent (donor) atoms (Arsenic, Antimony, Phosphorus) are added to the intrinsic semiconductor, n-type semiconductor is formed. The majority carriers in n-type and p-type semiconductors are electrons and holes respectively, holes being the absence of electrons. The minority carriers in n-type and p-types semiconductors are holes and electrons respectively.

The current in semiconductor is due to two distinct phenomenons:

1. Drift current, where carriers drift in an electric field (this conduction current is also available in metals)-

2. Diffusion current, where carriers diffuse if a concentration gradient exists (a phenomenon which does not exist in metals).

The commonly available semiconductors materials are Silicon and Germanium and the difference between these two are given below.

Silicon :



1. high PIV rating (upto 1000V)2. higher current rating3. wider temperature range (upto 2000 C)

Germanium:1. Lower PIV rating (up to 400V)2. Lower current rating3. Lower temperature range (1000C)

The advantage of Germanium over Silicon is the lower forward bias voltage required to reach the region of upward swing ,0.3V for Germanium and 0.7V for Silicon.On the other hand, the reverse saturation current for silicon is less than that of germanium, Is (Si) : Is (Ge) = 1 : 103

which makes it advantageous to use Silicon diodes at higher temperatures.

Experiment No : 1

FREQUENCY MEASURMENT USING LISSAJOUS FIGURES

AIM: To measure the frequency using Lissajous figures.

APPARATUS: 1. Function Generator 2No.2. CRO 1No.

THEORY:Cathode Ray Oscilloscope.

The Cathode Ray Oscilloscope (CRO) is an instrument which provides visual presentation of any waveform applied to its input terminals, while the multi-meter provides numerical information about applied signal. The CRO allows the actual form of the waveform to be displayed. The cathode ray tube (CRT) is the heart of the CRO, providing visual display of an input signal waveform. The CRT contains four basic parts.1. An electron gun to produce a stream of electrons.2. Focusing and accelerating elements to provide a well defined beam of electrons 3. Horizontal and vertical deflection plates to control the path of the electron beam.4. An evacuated glass envelop with a phosphorescent screen which glows visibly when struck by the electron beam. The voltage to be measured is applied to the vertical defection plates, where as the continuous sweep voltage is applied to the horizontal deflection plates.

Measurement using calibrated CRO: The oscilloscope tube face has been calibrated to compute the amplitude and time measurement.

Amplitude measurement:The amplitude of the voltage can be measured by using the scale-setting of the scope (volts/div knob.)

and the signal measured off the face of the scope.The peak-to-peak amplitude = No. divisions between peak – to –peak of the signal volts/division

knob multiplication factor.



Time measurement: The horizontal scale of the scope can be used to measure the time of the applied signal. The time

period of the periodic signal = No. of divisions during one cycle time/division knob multiplication factor.

Frequency measurement: The measurement of the repetitive waveform-period can be used to calculate the signal frequency.

Since the frequency is the reciprocal of the time period. f=1/T

Measurement of phase and frequency (Lissajous Patterns)The patterns that appear on the screen of a CRT, when sinusoidal voltages are simultaneously applied

to horizontal and vertical plates, are known as Lissajous figures. When two sinusoidal voltages of equal frequency which are in phase with each other are the horizontal and vertical deflection plates, the pattern appearing on the screen is a straight line as is clear from fig below.

(voltage applied to y-plates, frequency = f)

(Voltage applied to x-plates, frequency = f)

Fig 1.2 LISSAJOUS PATTERNS

When two equal voltages of equal frequency but with 900 or 2700 phase shift displacement are applied to a CRO, the trace on the CRO is a circle. An ellipse is formed if the two voltages are not equal and /or out of phase.

When two equal voltages of equal frequency but with a phase shift ‘’(not equal to 00 or 900 )are applied to a CRO, we obtain an ellipse regardless of the two amplitudes of the applied voltages, the ellipse provides a simple means of finding phase difference ‘’ between two voltages.

= sin-1 y1 / y2 = sin-1 x1 / x2.


1

0 2 4

3

t (0,2,4)

1

3

1

0

2

4

3

t

ex


Y Major Axis.

y1 y2

X x1

x2


Y

Major axis

Y2 Y1

X X1

X2


If the major axis of the ellipse lies in the first and third quadrants as in fig:1, the phase angle is either between 00 and 900 or 2700 and 3600. when the major axis of the ellipse lies in second and fourth quadrants as in Fig:2, the phase angle is either between 900 and 1800 or between 1800 and 2700.

A wave is applied to Y-plates and another known wave with variable frequency is applied across X-plates. The known frequency is varied till a suitable stationary pattern is obtained on the screen. Knowing this frequency and pattern, the frequency of the unknown wave is found out.

PROCEDURE:

1. Two sinusoidal waveforms are applied to the CRO in XY mode.



2. A known frequency fH is applied to horizontal plates and an unknown frequency fV is applied to the vertical plates.3. The unknown frequency is varied until a stable stationary waveform is obtained on the CRO.4. The unknown frequency is calculated using the formula: fVfH = No. of cuts on the horizontal line No. of cuts on the vertical line

OBSERVATIONS:

If the two sine waves are in phase but the frequency of the horizontal sine wave is twice the frequency of the vertical sine wave the pattern is shown in fig(a).

If the sine wave 90 degrees out-of-phase with the frequency of the horizontal sine wave three times the frequency of the vertical sine wave the pattern is shown in fig(b).

If the sine wave 90 degrees out-of-phase and the frequency of the vertical sine wave three times the frequency of the horizontal sine wave the pattern is shown in fig(c).

(a) fVfH =2:1 (b) fVfH =3:1 (c) fVfH =1:3

RESULTS

QUESTIONS:

1. Define Lissajous pattern.

2. What are the applications of CRO?



Experiment No : 2

P-N JUNCTION DIODE CHARACTERISTICS

AIM: 1. To plot the V-I characteristics of the given semiconductor diode, both for forward and reverse bias.2. To find out the cut-in voltage.3. To calculate the dynamic and static resistances of the diode.

APPARATUS: 1. Power supply 0-30V 1No.2. Voltmeters 0-1V 1 No.

0-30V 1 No.3. Ammeters 0-10mA 1 No.

0-500A 1 No.

COMPONENTS: 4. Semiconductor diode 1N4007 1No.5. Resistor 470 1No

THEORY:

p -n junction diode:

If donor impurities are introduced into one side and acceptors into the other side of a single crystal of a semiconductor, a p-n junction is formed.

Junction Acceptor ion Donor ion

Hole Electron


NP

V

+ - V

+ -

PN JUNCTION DIODE UNDER FORWARD BIAS


p-type n-typeP N JUNCTION DIODE

The region uncovered by +ve and –ve ions is called the depletion region, the space charge region or the transition region. The thickness of the region is the order of the wavelength of the visible light (0.05 micron). When p-n junction is formed, the concentration of the holes in p side is much greater than that in the n-side, a very large hole-diffusion current tends to flow across the junction from the p to n material. Hence an electric field must buildup across the junction in such a direction that the hole drift current will tend to flow across the junction from n to p side in order to counter balance the diffusion current. This equilibrium condition of zero resultant hole current results at potential barrier Vo. The numerical value for Vo is of the order of magnitude of a few tenths of a Volt.

The p-n junction can be operated in two regions

1) forward bias 2) reverse bias

Forward bias:A forward bias or ‘on’ condition is established by applying the +ve potential to the p-type material

and –ve potential to the n-type material as shown in the figure.

When the voltage difference between the p and n points is greater the VZ then the diode is forward bias otherwise reverse biased.

In forward bias the height of the potential barrier at the junction will be lowered by the applied forward voltage V. The holes cross the junction from p in to the n type region and become a minority current in the P side. Similarly the electrons cross the junction in reverse direction and become the minority carriers in p side. Holes traveling from left to right constitute a current in the same direction as the electrons moving from right to left. Hence the resultant current crossing the junction is the sum of the hole and electron minority current. Reverse bias:

A reverse bias or ‘off’ condition is established by applying the –ve potential to p-type material and +ve potential to n-type material as shown in the figure below

In the reverse bias both the holes p-type and the electrons in n-type will move away from the junction. The height of the potential barrier increases. This increase in the barrier serves to reduce the flow of majority carriers. Hence zero current results. However the minority carriers are uninfluenced by the increase in height


NP

V

- + V

- +

PN JUNCTION DIODE UNDER REVERSE BIAS


of the barrier. So a small current will flow due to these minority carriers and is called the reverse saturation current.

Volt-ampere characteristics:

I

The characteristics of the semiconductor diode can be defined by the following equation for forward and reverse bias regions.

ID = IS (e VD

/ ηVT – 1 )

Where VD = diode voltage, ID = diode current IS = reverse saturation current or a scale current which is a function of donor and acceptor impurity concentration, diode temperature and area of the junction etc.,


Ge Si

Cut-in-Voltage V

VZ

I


Η = emission co-efficient (empirical constant). This empirical constant accounts for any recombination of hole and electrons, which may occur when the carriers diffuse across the depletion region of the forward biased p-n junction.

The value of η lies in the range of ‘1’ to ‘2’ and depending upon the1. Size of the diode2. The semiconductor used to make it.3. The magnitude of the forward current.4. The value of Is.In general the η value for silicon is ‘2’ and for germanium is ‘1’

VT = Volt equivalent temperature KT/QK = Boltzmann constant = 1.38 × 10 – 23 j / 0 KT = Temperature in 0KQ = charge = 1.603 × 10 – 19 coulombs.

If V is positive, I is flowing from p to n side and the diode is in the forward bias. When the diode is reverse-biased and V is several times VT, ID - Io. The reverse current is therefore constant independent of applied reverse bias. At a reverse biasing voltage VZ, a large reverse current flows and the diode is said to be in break down region.Cut-in voltage (offset, break point or threshold voltage) (V):

The cut-in voltage is defined as the voltage across the diode below which the current is very small ( say less than 1% max rated value), and beyond V the current raises very rapidly.

The temperature dependence of the V-I characteristics The reverse saturation current Io doubles for every 100 raise in temperature. If IO = IO1 at T = T1 then at

temperature T, Io is given by Io(T) = IO1 × 2(T-T1)/10. The diode voltage decreases with the temperature and is given as dv / dt = -2.5mv / 0CDynamic resistance η VT / IDiode equivalent circuits:

CIRCUIT DIAGRAM:


470Ω

0-30V V+

0-1V

A+0-10mA

470Ω

0-30V V+

0-30V

A+0-500uA

1N4007 1N4007

Fig-1FORWARD BIAS

Fig -2REVERSE BIAS

VA

B

A

B

Rf Forward resistance(in the order of 100Ω)

Diode

A

B

Rr Forward resistance(in the order of MΩ)

Forward bias Reverse bias

Ir

Vr

I A

If

Vf

I

V

I (mA)

V (volts)


PROCEDURE

Forward Bias:

1. Connect the circuit as shown in the fig 12. Vary the supply voltage gradually, starting from zero. Increase the applied voltage and note the voltmeter reading(V)3. For each 0.1 V step in V note the corresponding forward current (I) till V becomes say 0.7 V. I should not exceed 10mA.4. Tabulate the result and draw the V-I characteristics under forward bias conditions.

Reverse Bias:

1. Connect the circuit as shown in the fig(2)2. Measure the current(reverse current) and voltage by increasing the voltage 1V steps. Do up to 20V.3. Tabulate the results and plot the reverse bias characteristics.4. Here the reverse current will be in micro amperes. 5. The reciprocal of the slope of these curves for both the conditions gives the resistance for forward and reverse bias conditions.

OBSERVATIONS:


S.No V(volts) I (mA)

MODEL GRAPHS

V(volts)




Forward bias characteristics Reveres bias characteristics

RESULTS:1. Cut-in voltage = 2. Dynamic forward resistance = Vf / If

3. Static forward resistance = V/I 4. Dynamic reverse resistance = Vr / Ir

QUESTIONS:

1. Draw the piece wise linear V-I characteristics of a p-n diode, what is the circuit model for ON state and OFF state?2. What is Hole? How does it contribute to conduction?3. What is the depletion region?4. How diode acts as a switch?5. What parameter in a germanium diode differs from those in a silicon diode?

Experiment No : 3

ZENER DIODE CHARACTERISTICS

AIM: 1. To plot the V-I characteristics of the given Zener diode, both for forward and reverse bias. 2. To study the performance of a Zener diode as a voltage regulator.

APPARATUS:

1. Power supply 0-30V 1 No. 2. Ammeters 0-10mA 1 No.

0-100mA 1 No.3. Voltmeters 0-1V 1 No.

0-10V 1 No.

COMPONENTS :

1. Zener diode BZ 6.2 V 1No.2. Resistor 470 1No.



THEORY:

Diodes which have adequate power-dissipating capabilities to operate in the break-down region are commonly called Zener diodes. These devises are employed as voltage regulators.

The location of the Zener region can be controlled by varying the doping levels. An increase in the doping produces an increase in the number of added impurities. Further, this will decrease the Zener potential. Zener diodes are available in the Zener potential range of 1.8 V to 200V with power rating from ¼ to 50W. Silicon is usually preferred in the manufacturing of Zener diodes because of its higher temperature and current handling capabilities.

The equivalent circuit of the Zener diode in the Zener region is given below.

Vz Vz Rz

Complete Approximated

Two processes which produce the breakdown region are Avalanche multiplication and Zener breakdown which are explained below.

Avalanche breakdown

The thermally generated electrons and holes acquire sufficient energy from the applied potential to produce new carriers by removing the valance electrons from their bonds. These new carriers in turn produce additional carriers again through the process of disrupting bonds. This cumulative process is referred as avalanche breakdown. Avalanche multiplication involves when the reference voltage is above ‘6V’. The temperature co-efficient is positive (% change in reference voltage per centigrade degree change in diode temperature)

A junction with broad depletion layer, and therefore low field intensity will break down by the avalanche mechanism.Zener breakdown

This process initiates breakdown through a direct rupture of the bonds because of the existence of a strong electric field. Zener breakdown involves when the reference voltage is below ‘6V’. The temperature co-efficient is negative.

A junction having a narrow depletion layer width and high field intensity will breakdown by the Zener mechanism.

The networks employing Zener diodes can be analyzed by replacing the Zener diode equivalent circuits (‘on’ and ‘off’ states)


VV VZVZ

‘ON’ stateV>VZ

‘OFF’ stateVZ >V>0V


CIRCUIT DIAGRAMS:

PROCEDURE:

Forward Bias:

1. Connect the circuit as shown in fig(1)2. Vary the supply voltage gradually, starting from zero. Increase the supply voltage and note the voltmeter reading (V) for each 0.1V step in V, note the corresponding forward current (I) till V becomes say 0.7 V. I should not exceed 10mA. 3. Tabulate the results and draw the V-I characteristics under forward bias conditions.

Reverse Bias:1. Connect the circuit as shown in fig(2)2. Increase the supply voltage suitably, to read Iz in steps of 5mA, starting from zero upto say 40mA, note the corresponding values of Vz.3. Tabulate the results and draw the V-I characteristics under reverse bias conditions.

OBSERVATIONS:





470Ω

0-30V V+

0-1V

A+0-10mA

470Ω

0-30V V+

0-10V

A+0-100mA

6Z2 6Z2

Fig – 1FORWARD BIAS

Fig - 2REVERSE BIAS

V (volts)

I (mA)

V (volts)

I (mA)


MODEL GRAPHS

RESULTS:

Cut-in voltage =Break down voltage = Dynamic forward resistance =Static forward resistance =

QUESTIONS:

1. How does the avalanche breakdown differ from Zener breakdown?2. What are the applications of Zener diode?3. What is stabilizer?4. For the reference voltage, which is above 6V, What is the temperature coefficient?

Experiment No : 4

TRANSISTOR CB CONFIGURATION (INPUT AND OUTPUT)

AIM: 1. To obtain the input and output characteristics of a given transistor in CB configuration. 2. To measure the h-parameters of the given transistor in CB configuration.

APPARATUS:1. Power supplies 0-30V 2Nos.2. Ammeters 0-10mA 2Nos.3. Voltmeters 0-1V 1 No.

0-30V 1 No.



COMPONENTS:

1. Transistor – BC 547 1No.2. Resistor -1K 1No.

THEORY:Bipolar Junction Transistor (BJT)

The bipolar junction transistor (BJT) is a three element device formed from two junctions which share a common semiconductor layer. There are two types of transistors. One is npn transistor in which p type is sandwiched between two n type layers. The second one is pnp in which n region is common to two p type layers. The transistor is mainly used for two applications, Amplification and Switching.

The two types of transistors are represented in figures below.

B BE C E C

Emitter Base Collector Emitter Base Collector

The symbols for two transistors are shown below

pnp transistor npn transistor

The arrow in the emitter leads specifies the direction of current when emitter base junction is forward biased. The three doped regions of transistor are 1) Emitter, which is heavily doped, to emit or supply the carriers.

2) Base ,which is lightly doped and very thin. The job of the base is to pass emitter injected carriers on to the collector. The light doping of the base means that the free carriers have a long life time in the base region. Very thin base means that the free carriers have only a short distance to go to reach the collector.

For these two reasons, almost all the emitter-injected carriers pass through the base to collector.

3) Collector, whose function is to collect or gather most of the carriers from the base. The collector area is considerably larger than the emitter area. This difference is due to the fact that in most uses of BJT, the collector region must handle more power than the emitter. Hence more surface area is required for heat dissipation.

Modes of transistor operation:Each junction of the BJT can be forward or reverse biased independently. Thus four modes of

operations (given in the table below) exist. These modes of operation are defined independent of transistor configuration.

Mode of operationJunction bias conditionEmitter – base Collector – Base

Forward activeCutoff

ForwardReverse

ReverseReverse


p n p n p n


SaturationReverse-active

ForwardReverse

ForwardForward

In forward active region, the transistor behaves as a current controlled current source. It does so because the output (collector) current is controlled by the input (base) current and is independent of load resistance or the output voltage ( Ic = IB ,if ICO is neglected). Thus it can be used as an amplifier in active region.

Both junctions are reverse biased in the cutoff region. Both IE and IC are in the order of reverse saturation current and transistor behavior approximates an open switch.

Both junctions are forward biased in the saturation region. The collector current may be large, but small voltage exists across the collector region. This condition is nearly that of closed switch.

The operation of a BJT between cutoff and saturation corresponds to action of switch.

In the common base configuration, the base is common to both input and output circuit. The output (collector) current IC is completely determined by the input (emitter) current IE and the output (collector to base ) voltage VCB.

IC = f(VCB , IE)The input voltage VEB is completely determined from the emitter current and collector to base

voltage.

The early effect or base width modulation:

The width W of the depletion region of the diode increases with the magnitude of the reverse voltage. The collector junction is reverse biased in the active region. As the voltage applied at the junction increases, transition region penetrates deeper into the collector and base. As the neutrality of the charge must be maintained, the number of uncovered charges on each side remains equal. Since the doping in the base is smaller than that of the collector, the penetration of the transition region into the base is much larger than into the collector. Hence collector depletion region is neglected.

This modulation of effective base width by collector voltage is known as early effect or base width modulation. These effects have three consequences. 1) increases with increase in VCB

2) IE increases with increase in VCB

3) For extremely large voltage the base width may be reduced to zero, causing voltage breakdown in the transistor.

Input characteristics:

The input characteristics represent the forward characteristics of the emitter to base junction for various collector voltages.


VC

B=1V

VC

B=2V

IE

VEB

VC

B=0V


For fixed values of collector voltage VCB as the base to emitter voltage (VEB) is below the cut-in voltage (V) the emitter current is very small. When VEB exceeds this V the emitter current increases rapidly. The increase in the magnitude of collector voltage will, by the early effect, cause the emitter current to increase with the VEB held constant.

Output characteristics

The output characteristics will relate output current IC to an output voltage VCB for various values of IE.

The output characteristics have three basic regions.1) Active Region 2) Cut-Off Region 3) Saturation Region

Active Region

In the active region the collector current is essentially independent of collector voltage and depends only upon emitter current.

However, because of the early effect there is a small increase in IC with VCB. Because is less than 1, but almost equal to unity, the magnitude of collector current is (slightly) less than that of the emitter current.

Saturation Region

The region to the left of the ordinate, VCB = 0, and above the characteristic IE = 0, in which both emitter and collector junction are forward biased, is called the saturation region. There is a large change in the collector current with small change in collector voltage.

Cut-off Region

The region below the IE = 0 characteristic, for which emitter and collector junctions are both reverse biased, is referred to as a cut-off region.


SATU

RAT

ION

REG

ION

IC

IE = 3mA

IE = 2mA

IE = 1mA

IE = 0mA

VCB

0

ACTIVE REGION

CUTOFF REGION


h parameters:

The terminal behavior of a two port device is specified by two voltages and two currents. i1 i2

+ +input V1 V2 output port - - port

We may select two of the four quantity as independent variables and express the remainingtwo in terms of the chosen independent variables. If the current i1 and the voltage V2 are independent variables and if the two ports is linear, we may write

V1 = h11 i1 +h12V2 . i1 = h21 i1 +h22V2 .

The quantities h11, h12, h21, & h22 are called the ‘h’ or hybrid parameters because they are not all alike dimensionally. The h parameters are defined as follows.

hi = h11 = V1 = input resistance with output short circuit(ohms) i1 V2 = 0

hr = h12 = V1 = reverse open circuit voltage amplification V2 i2 = 0 (dimensionless)

hf = h21 = i2 = short circuit current gain ( dimensionless) i1 V2 = 0

ho = h22 = i1 = output conductance with open circuit (mhos) V2 i1 = 0

The model: By using four h-parameters we can construct mathematical model of the two port active device.

Transistor Hybrid Model:

For small signals, the transistor operates with reasonable linearity, so that the transistor can be replaced by hybrid equivalent circuit.


Two-port active device

+ +

+

-

hi

hrV2

- -hfi1

i1 i2

ho

+

-

V2V1

+ +

+

-

hib

hrbVcb

- -hfbie

Ie Ic

hob

+

-

VcbVeb


The hybrid small signal model for C.B. configuration

E C

B B

The h-parameter equations are

Veb = hibie+hrbvcb

ic = hfb ie + hobVcb.

and h-parameter given by

CIRCUIT DIAGRAM:

PROCEDURE:

Input Characteristics:

1. Connect the circuit as shown in the diagram. Keep the VCB 0V(constant). Increase the VEB and observe the IE for different values of VEB.

2. Tabulate the results. Plot the graph VEB Vs IE. Repeat the experiment for VCB = 1V & VCB = 1.5V.3. Calculate h-parameters, Using the above formulas.

Output Characteristics:

1. Starting with VEB = 0, increase it to get IE = 1mA. Then increase the VCB in steps and note down the values of Ic, without exceeding the rated values.2. Tabulate the results. Repeat for different values of IE , say 2mA & 3mA and draw the family of characteristics.3. Calculate h-parameters, Using the above formulas.


BC 547

V+

0-30V

A +

0-10mA

0-30V

1K

0-30V

A +

0-10mA

V+

0-1V

Fig: 4.3 INPUT AND OUTPUT CHARACTERISTICS OF CB TRANSISTOR


OBSERVATIONS:

RESULTS: The h-parameters in CB configuration are

hib, = hfb = hrb = hob =

QUESTIONS

1. What is base width modulation (The early effect)?2. The common base amplifier is also current follower, why?3. What is meant by active region?4. Because of early effect, there ia a small increase in IC, with VCB, explain.5. How do the h-parameter vary with quiescent point and temperature?

Experiment No : 5

TRANSISTOR CE CONFIGURATION (INPUT AND OUTPUT)


S.No VCB VCB VCB

VEB IE VEB IE VEB IE

S.No IE IE IE

IC VCB IC VCB IC VCB


AIM: 1. To obtain the input and output characteristics of a given transistor in CE configuration. 2. To measure the h-parameters of the given transistor in CE configuration.

APPARATUS:1. Power supplies 0-30V 2Nos.2. Ammeters 0-1mA, 1 No.

0- 10 mA, 1 No.0-100mA 1 No.

3. Voltmeters 0-1V 1 No.0-30V 1 No.

COMPONENTS:1. Transistor BC 547 1No.2. Resistor 1K 1No.

THEORY:Input characteristics

The input characteristic is the plot of the input current IB Vs input voltage VBE for a range of values of output voltage VCE.

With VCE = 0 and emitter junction forward biased, the input characteristic is essentially that of a forward bias. As VCE increases with constant VBE, it causes a decrease in base width and results in a decreasing recombination base current.

Output CharacteristicsThe output characteristics are a plot of the output current IC versus output voltage VCE for a range of

values of input current IB.


SAT

UR

AT

ION

RE

GIO

N

IC IB=600A

IB=550A

IB=500A

IB=0A

VCE

ICEO = ICBO

CUTOFF REGION

ACTIVE REGION

VC

E=1V

VC

E=0VIB

VBE

VC

E=2V


Active Region

In the active region the collector junction is reversed biased and emitter junction is forward biased. In this region the transistor output current IC responds most sensitively to an input signal.In the active region the output characteristic IC is given by

IC = IB + (1+) ICO.Note that IB >>ICO, and hence IC IB in the active region.The curves of the output characteristics are not as horizontal as those of the output characteristics in the common base configuration.

Assume that, because of the early effect , increases by only one half of 1% from 0.98 to 0.985, as VCE increases from a few volts to 10 volts, then the value of Beta increases from (0.98/1-0.98) = 49 to (0.985/1-0985) = 66, or about 34%. This numerical example illustrates that a very small change in reflects a very large change in the value of , and hence upon the common emitter curves.

Cut-off regionThe cut-off region is defined as the condition where the collector current is equal to the reverse

saturation current ICO and the emitter current is zero. For the transistor to be in cut-off , IE=0, IC=ICO,IB = -IC = -ICO, and VBE is a reverse voltage whose magnitude is of the order of 0.1V for germanium and 0 V for silicon transistor.

Saturation regionIn the saturation region the collector junction and emitter junction are forward biased by at least the

cut-in voltage (V). Since the voltages VBE and VBC across a forward biased emitter junction and collector junction have a magnitude of only a few tenths of a volt (V), VCE = VBE - VBC = zero, (ideal) or VCE = few tenths of the volt (practical ) at saturation. Hence, in output characteristics the saturation region is very close to zero voltage axis.In saturation region the collector current is approximately equal to base current, for given values of V CC and RC.

h parameters:

The terminal behavior of a two port device is specified by two voltages and two currents.

i1 i2

+ +input V1 V2 outputport - - port

We may select two of the four quantity as independent variables and express the remainingtwo in terms of the chosen independent variables. If the current i1 and the voltage V2 are independent variables and if the two ports is linear, we may write

V1 = h11 i1 +h12V2 . i1 = h21 i1 +h22V2 .

The quantities h11, h12, h21, & h22 are called the ‘h’ or hybrid parameters because they are not all alike dimensionally. The h parameters are defined as follows.

hi = h11 = V1 = input resistance with output short circuit(ohms) i1 V2 = 0

hr = h12 = V1 = reverse open circuit voltage amplification V2 i2 = 0 (dimensionless)

hf = h21 = i2 = short circuit current gain ( dimensionless)


Two-port active device


i1 V2 = 0

ho = h22 = i1 = output conductance with open circuit (mhos) V2 i1 = 0

The model: By using four h-parameters we can construct mathematical model of the two port active device.

Transistor Hybrid Model:

For small signals, the transistor operates with reasonable linearity, so that the transistor can be replaced by hybrid equivalent circuit.

The hybrid small signal model for common emitter configuration.

For common emitter configuration, the input voltage and input current are V BE and IB. the output voltage and current are VCE and IC.

The h-parameter equations are VBE = hieib+hrevce

Ic = hfe Ib + hoeVce

B C

E E

The h-parameters are


+ +

+

-

hi

hrV2

- -hfi1

i1 i2

ho

+

-

V2V1

+ ++

-

hie

hreVce

- -hfeib

Ib Ic

hoe

+

-

VceVbe


CIRCUIT DIAGRAM

PROCEDURE:

Input Characteristics:

1. Connect the circuit as shown in the diagram. Keep the VCE 0V(constant). Increase the VBE and observe the IB for different values of VBE.

2. Tabulate the results. Plot the graph VBE Vs IB. Repeat the experiment for VCE = 1V & VCE = 2V.

3. Calculate h-parameters, Using the above formulas.

Output Characteristics:

1. Starting with Vbb = 0, increase it to get IB = 500uA. Then increase the VCC in steps and note down the values of IC, without exceeding the rated values.

2. Tabulate the results. Repeat for different values of IB , say 550A & 600A and draw the family of characteristics.

3. Calculate h-parameters, Using the above formulas.


0-30V

V+

0-1V

A+

0-10 mA

V+

0-30V

A +

0-100mA

1KΩ

0-30V

BC547

INPUT CHARACTERISTICS

-

-

-

-+

-

+

-

V+

0-1V

A0-1 mA

V+

0-30V

A +

0-100mA

0-30V

1KΩ

0-30V

BC547

OUTPUT CHARACTERISTICS

++ -

-

-

-

-

+

-


OBSERVATIONS:

Input characteristics output characteristics

RESULTS: The h-parameters in CE configuration are

hie, = hfe = hre, = hoe =

QUESTIONS

1. What is ICBO, and how does it varies with temperature.2. What should be the value of the VBE to keep the transistor in the cut-off. (both Ge and Si)3. What is the order of magnitude of the temperature co-efficient of VBE(SAT), VBC(SAT), and VCE(SAT)?4. Is VBE(SAT), greater or less than VCE(SAT),? Explain.5. Why do the O/P characteristics of CE configuration are not as horizontal as those of C.B configuration O/P characteristics?


S.No VCE VCE VCE

VBE IB VBE IB VBE IB

S.No IB IB IB

IC VCE IC VCE IC VCE


Experiment No : 6

RECTIFIERS WITHOUT FILTERS(FULL WAVE AND HALF WAVE )

AIM: To study the half wave and full wave rectifier circuits without filters.

APPARATUS:1. Ammeters 0-100mA 1No.2. Digital multi-meter 1No.3. Decade resistance box 1No.

COMPONENTS:1. Diode 1N4007 2 Nos.2. Transformer 230V / 16-0-16 V, 0.5A, 1No.

THEORY:Rectifier:

A device, such as the semiconductor diode, which is capable of converting a sinusoidal input waveform (whose average value is zero) into a unidirectional (though not constant) waveform with a non zero average component is called a rectifier.

All most all electronic circuits requires DC source of power. For portable low-power system batteries may be used. More frequently, however electronic equipment is energized by power supply, a circuit which converts the AC waveform of the power lines to direct voltage of constant amplitude.

The block diagram of dc supply is shown below.

The functions of the various circuits are as listed 1. Transformer: adjust the ac level such that the approximate dc amplitude is achieved2. Rectifier: A device, such as the semiconductor diode, which is capable of converting a sinusoidal input waveform (whose average value is zero) into a unidirectional (though not constant) waveform with a non zero average component.3. Filter: ‘Smoothes’ the waveform by eliminating the ac component from the rectifier output4. Regulator: Maintains a constant voltage level independent of load conditions or variation in the amplitude of the ac supply. The basic circuit for half wave rectifier is as shown below.

Let Vi = VmsintVm is the peak value.The load current is given byi = Imsin t if 0<t < Πi = 0 if Π<t < 2Π


Transformer Rectifier Filter RegulatorAC Input To load

0 2 t

0 2 t


Im = Vm / (Rf +RL)

During the interval, t = 0 Π, the Vi is positive and the diode will be forward biased and the equivalent circuit is shown below (the diode is replaced by forward resistance Rf )

i = Vi / (Rf +RL) = Vm sint / (Rf +RL) = (Vm / (Rf +RL) ) sintFor the interval t = п 2 п, the vi is negative and the diode becomes reverse biased. The equivalent circuit is shown below. (the diode is replaced by open circuit).

Thus the output current is unidirectional and it will have non-zero average value.Its average value = area of one cycle divided by the base

[ because = ωt, base = 2Π, area of one cycle = ]

Idc =

Idc = =

The dc output voltage is

Vdc = Idc.RL=

Irms = = = im / 2


1N4007

RL

AC InputV

i

V+ -

Vi

+ V -

Rf

RL

+

-

i

Vi

+ V -

RL

+

-

ii=0v=vi


PIV: The peak –inverse –voltage (PIV) or [ PRV (peak reverse voltage)] rating of the diode is primary importance in the design of rectification system. It is the voltage rating that must not be exceeded in the reverse bias region of the diode.

In the half wave rectifier, when the diode is reverse biased, the voltage appears across the diode is ‘v i’ and its maximum value is vm. This vm should not exceed the PIV of the diode. Otherwise the diode will enter the Zener avalanche region.

For half wave rectifier PIV rating > vm

Full wave rectifier:The circuit of the full wave rectifier is shown below. This circuit is seen to comprise two half-wave

circuits connected so that conduction takes place through one diode during one half cycle and thorough other diode during the second half of the cycle.

During the positive cycle of the input signal D1 is forward biased and i1 current is flown through D1

and RL. During negative half cycle of the input signal, D2 is reverse biased and i2 current is flowing through D2

and RL. The current to the load, which is the sum of these currents, i = i 1+i2. The dc and rms values of the load current and load voltages are.

Idc = Average value = Area of one cycle divided by the base [because here base (time) = Π ]

Idc =

= = 2Im / Π

Where Im = ,

Vm is the peak transformer secondary voltage from one end to the center tap. The dc output voltage of the fullwave connection is twice that for the half wave circuit. Because the area above the axis for one full AC input cycle is twice that obtained for a half-wave system.

I2rms = =

= = Im2 / 2 Irms =


AC

mai

ns

A B

C D


PIV:When any one of the two diodes is reverse biased, (let that is replaced by open circuit) the maximum

voltage appears across that open circuit. This can be solved by following equivalent circuits

vm vm

RL RL

vm vm

when D1 open circuited(reverse biased) when D1 short circuited(forward biased) D2 short circuited( forward biased) D2 open circuited(reverse biased)

By applying the KVL we can find out , VAB or VCD

VAB = VCD = 2vm

Thus, when any diode is reverse biased, then a maximum of 2vm appears across that diode. For safe operation, the PIV rating of that diode should be greater that or equal to 2vm

PIV > 2vm for center tap transformer full-wave rectifier

HALF WAVE RECTIFIER:

CIRCUIT DIAGRAM:

PROCEDURE:

1. Connect the circuit as shown in the diagram.2. Give input from AC mains and measure the secondary voltage Vs of the transformer.3. Measure the no load DC voltage using DMM. Let this be VNL. 4. Now connect the DRB. Vary the DRB and nte the values of Idc, in steps of 10mA until the current reaches 100mA.5. At each step measure the VDC and VAC. Calculate ripple factor as the ratio of VAC VDC.

OBSERVATIONS:


1N4007

V+

DMM

A+0-100mA

DRB

AC

mai

ns

Half wave rectifier Full wave rectifier


VNL=S.No IDC VDC VAC RL = VDC / IDC = VAC/VDC %regulation

VNL-VDC × 100 VDC

Verify the following theoretical calculations:1. Expected DC output voltage.2. Ripple factor of half wave.3. Observe the output waveforms on CRO.

FULL WAVE RECTIFIER:

CIRCUIT DIAGRAM:

PROCEDURE:

Repeat the experiment for a full wave rectifier with the connections as shown in the above figure and tabulate the result

OBSERVATIONS: VNL=S.No IDC VDC VAC RL = VDC / IDC = VAC/VDC %regulation

VNL-VDC × 100 VDC

Plot the graph of VDC Vs IDC . Vs IDC and calculate the theoretical values of expected DC output voltage and observe the output on CRO. Model wave forms:


1N4007

V+

DMMDRB

1N4007

A+0-100mA

AC

mai

ns


RESULTS:

VDC = Vac = V =

QUESTIONS

1. Why does ripple factor is good for full wave rectifier than the half wave rectifier?2. The frequency of the full wave rectifier is twice that of a input signal frequency and the frequency of the half wave rectifier is same as that of the I/P signal frequency, Explain why?3. Why should be the ‘PIV’ rating of the diode in full wave rectifier is twice that of diode in half wave rectifier?


AC

mai

ns


Experiment No : 7

RECTIFIERS WITH FILTERS (FULL WAVE AND HALF WAVE )

AIM: To determine the following parameters of half wave and full wave circuits with -section filters. 1. ripple factor 2. Variation in % regulation 3. to observe the o/p on CRO.

APPARATUS:

1. Ammeters 0-100mA 1No.2. Digital multi-meter 1No.3. Decade resistance box 1No. 4. Decade inductance box 1No

COMPONENTS:

1. Diode 1N4007 2 No.2. Capacitors 1000µF / 63V 2 No.3. Transformer 230V / 16-0-16 V, 0.5A, 1No.

THEORY:

Rectifiers with filters:A rectifier circuit is necessary to convert a signal having zero average value into one that has a non-

zero average. A filter circuit is necessary to provide a more steady DC voltage.A most popular filter circuit is a capacitor filter circuit is shown below.

AC input

The action of this system depends upon the fact that the capacitors stores energy during the conduction period and delivers this energy to the load during the non-conducting period. In this way the time during which the current passes through the load is prolonged, and the ripple is considerably decreased.

The half wave capacitor filter is shown below.


Rectifier circuit

Capacitor filter

DC load RL


i

+ + v - iL iC C vo vi RL

-

The diode will be forward biased (short circuited) when the transformer voltage v i exceeds the capacitor voltage, then the capacitor starts charging in stepping with applied voltage.The diode will be reverse biased (open circuited) when the transformer voltage v i falls below the capacitor voltage. Then the capacitor starts discharging through the load resistor.The following waveforms describes the effect of the capacitor on the circuit response. Let the capacitor is initially charged during the first quarter cycle the diode conducts and the capacitor charges with input voltage up to vo = vm . When vi is falls below vm the diode is not conducting the capacitor discharges at slower rate than input voltage. If the time constant RLC is large as compared period of input waveform the discharge is slow. Thus only a small decrease in vo occurs between t1 and t2. At time t = t2, the vi

equals the capacitor voltage. The diode again conducts between t2 and t3 does not conduct between t3 and t4.

The process is repeated.

Ripple factor:The filtered output has a DC value and some ac variations (ripple). These smaller the ac variation with

respect to the DC level the better the filter circuit operation.The filter voltage waveform with DC and ripple voltages is shown below. Ripple factor(r) = rms value of ac component of a signal 100% = Vr (rms) 100%

Average value of signal Vdc

Voltage regulation:The variation of dc output as a function of dc load current is called regulation. The percentage

regulation is given as % regulation = V no load – V load 100% V load

V no load dc output voltage under zero load current (RL = )V load dc output voltage at normal load current at which regulation is determined. For ideal power supply the output voltage is independent of the load current or the load resistance and the percentage regulation is zero

CIRCUIT DIAGRAM:

PROCEDURE:


N2

N3

1N4007

1000F63V

DIB

V DMM

A+0-100mA

DRBAC

mai

ns

1000F63V

+

-+

-

+

-


1. Connect the circuit as shown in the diagram.2. Give input from AC mains and measure the secondary voltage Vs of the transformer.3. Measure the no load DC voltage using DMM. Let this be VNL. 4. Now connect the DRB. Vary the DRB and nte the values of Idc, in steps of 10mA until the current reaches 100mA.5. At each step measure the VDC and VAC. Calculate ripple factor as the ratio of VAC VDC.

OBSERVATIONS: S.No

IDC VDC VAC RL = VDC / IDC = VAC VDC %regulation VNL-VDC × 100 VDC

Plot the graphs of VDC Vs IDC, Vs IDC and % regulation Vs IDC

FULL WAVE RECTIFIER:CIRCUIT DIAGRAM:

Model Graphs

Ripples in half wave rectifier Ripples in full wave rectifier

PROCEDURE:

Repeat the experiment for a full wave rectifier with the connections as shown in the above figure and tabulate the results.

OBSERVATIONS:

S.No IDC VDC VAC RL = VDC / IDC = VAC VDC %regulation


N2

N3

1N4007

1000F63V

DIB

V DMM

A+0-100mA

DRBAC

mai

ns

1000F63V

+

-+

-

+

-

1N4007


VNL-VDC × 100 VDC

RESULTS:

VDC = Vac = V =

QUESTIONS:

1. How the capacitor reduces the ripples?2. Explain how DC voltage is finally obtained, starting from AC input voltage to the rectifier circuit.3. The capacitor rectifier circuit also acts as a envelope demodulator, Explain?

Experiment No : 8

FET CHARACTERISTICS

AIM: To obtain the static characteristics (VDS vs ID.). To obtain the transfer characteristics (VGS vs ID )of a given FET and to calculate the parameters of FET (rd, gm, µ)

APPARATUS:

1. Power supplies 0-30V 2Nos.2. ammeter 0- 10mA 1 No3. Voltmeters 0-10V 1 No

0-30V 1 No

COMPONENTS:

1. FET BFW 11 1No2. Diode 1N4007 1 No

Theory:

FET Characteristics

The field effect transistor is a three terminal semiconductor device. The three terminal are Gate, Drain and Source. The FET is a majority carrier (Unipolar) device and it is a voltage controlled current source. It is a voltage controlled device since the output drain current is controlled by Gate to source voltage. It can also acts as a current source since the output drain current independent on the drain resistance or drain to source voltageThe high input impedance and low noise are two of the advantages of FET. The main disadvantage of FET is relatively small gain bandwidth product as compared to transistor.An important feature of the FET is that, it is often simpler to fabricate and occupies less space on the chip then does the BJT. To the resultant component density can be extremely high, of a exceeding of 105 MOSFET per



chip. MOS devices can be connected as resistor and capacitors. Exploitation of this features makes the MOSFET the dominate device in VLSI.There are two types of field effect transistors.The junction field effect transistor ( JFET or simply FET).And the Insulated Gate field effect transistor (IGFET) more commonly called the Metal Oxide semiconductor (MOS) transistor. (MOST or MOSFET).

MOSFET are of two types .Depletion type MOSFET and Enhanced type MOSFET. The JFET and depletion type MOSFET can be used in amplifier s. The enhancement type MOSFET is more popular in digital circuits.

JFET Volt- Ampere CharacteristicsThere are two types of JFET one is n-Channel and the other is p- Channel FET. The Characteristic of

n-channel FET are shown below. For n-channel FET the gate to source voltage is negative and for the p-channel FET gate to source voltage is positive.

Static characteristicStatic characteristic are the plot of ID versus VDS for a range of values of VGS. With VGS = 0, and by

applying VDS, this VDS along the n-type bar reverse biases the gate-junction. For small values of V DS , ID

increases linearly with VDS. A voltage VDS is reached at which the channel is pinch-off, ID begins to level off and approaches constant values (IDSS).IDSS is the maximum drain current for JFET is defined by the condition.VGS = 0V and VDS is greater than VP.

The Ohmic RegionThe region to the left of the pinch off locus is referred as a ohmic Region or Voltage controlled

resistance Region. In this region the JFET can actually be employed as a variable resistor. (Resistance = V D / ID ). Is controlled by the VGS.

Pinch-off Region or Saturation Region

In this region ID responds very slightly to VDS, it depends upon VGS and is given by ID = IDSS 1 – VGS

2

VP

Where VP = Pinch – off Voltage.

The pinch – off Voltage is defined as, It is the gate to source voltage at which the drain current becomes zero. The FET acts as voltage controlled current source and can be used as an amplifier by properly biased.

Transfer Characteristics:

The transfer Characteristics is a plot of ID Versus VGS at a constant value of VDS.

When VGS = 0, ID = IDSS.When VGS = VP, ID = 0.

The transfer characteristics are defined by the equation ID = IDSS 1 – VGS 2

VP

SYMBOLS

N- Channel JFET P – Channel JFET


G

D

S

G

D

S

Source

DrainGate 1

Gate 2


Lead configuration

The direction of the arrow indicates the directions in which the gate current would flow of the gate junction were forward biased.

CIRCUIT DIAGRAM:

PROCEDURE:

A) Static Characteristics:1. Connect the circuit as shown in the figure. The diode is only for protection. Keep the gate voltage V GS = 0V and increase the drain voltage VD from zero to 10 V in convenient steps, noting the drain current ID at each step.2. plot the graph between VD and ID .3. Repeat the experiment for different values of gate voltage.

B) Transfer Characteristics:Keep the VDS constant say 10 V and record ID for different (-ve ) values of VGS starting from zero.

Tabulate the result and draw the graphs.

STATIC CHARACTERISTICS:

S.No VGS = 0 VGS = -1V VGS = -2VVDS (volts) ID(mA) VDS (volts) ID(mA) VDS (volts) ID(mA)

TRANSFER CHARACTERISTICS


0-30V

BFW 11

1N4007

V+

0-10V0-30V V+

0-30V

A+0-10mA

+

-

+

-


S.No VDS = 10v VDS = 2VVGS (volts) ID(mA) VGS (volts) ID(mA)

MODEL GRAPHS:

RESULT:From these values of above graphs determine

rd = VDS /ID at constant VGS gm = ID / VGS at constant VDS

and µ as the product of gm and rd.

QUESTIONS:

1. Define IDSS , Vp, VGS (off) gm, rd, µ2. What is biasing voltage for gate to source junction.3. What is the input impedance if a Junction FET.


VDS

I D

VDS = 0V

VDS = -1V

VDS = -2V

VDS = 5V VDS = 2V


Experiment No : 9SCR CHARACTERISTICS

AIM: To draw the forward and reverse characteristics of SCR.

APPARATUS:

1. Power supply (dual channel) 0-30V 1 No.2. Voltmeters 0-30V 2 Nos3. Ammeters 0-10mA 1 No

0-100mA 1 No

COMPONENTS:

1. Resistors – (1K -2 No.)2. SCR TYN – 64

THEORY:The structure of silicon controlled rectifier (SCR) consists of four alternate p and n type layers. The

name “Silicon Controlled Rectifier” indicates that the SCR is a rectifier constructed of silicon material with a third terminal for control purposes. This third terminal in the SCR, called a gate, determines when the rectifier switches from the open circuit to short circuit state. The graphic symbol for SCR is shown below.

The volt-ampere characteristics


VF

Gate

Cathode Anode

IA IGT

+ -

Cathode Gate Anode

Silicon Controlled Rectifier TYN 604

TYN 604


The volt–ampere characteristic of an SCR is shown for various gate currents. We can observe from the characteristics that the firing voltage is decreasing with increasing gate current. In fact the ‘’ increases with increasing gate current, thus the breakover (firing) voltage decreases. The anode current after breakdown may be larger by a factor of 1000 than the current before breakdown. When the gate current is very large, breakover may occur at low voltage, the characteristic has the appearance of a simple p-n diode.

Forward breakover voltage (VBO), is that voltage above which the SCR enters the conduction region (on state). Holding current(IH) , is that value of current below which the SCR switches from the conduction state to the forward blocking region( off state).Holding voltage(VH), is that voltage required to turn off the SCR.

SCR is used in relay control, motor control, phase control, heater control, battery chargers, inverters, regulated power supplies and static switches.

CIRCUIT DIAGRAM:


VH

IA

IH

VA

IG2> IG1>0

IG2 IG1 IG = 0

VFBO

IG Large

TYN 604

1KA

+0-100mA

V+

0-30VV1 A +

0-10mA

0-30V

1K-

-

-

+

-

SCR FORWARD CHARACTERISCTICS


PROCEDURE:Forward characteristics:1. Connect the circuit as shown below.2. Keep the gate voltage to Zero and increase the input voltage from zero to 15, not down the corresponding ammeter readings IA and VA

3. Set the gate voltage to 10V and increase the anode voltage from 0 to say 15Volts, and observe that at a particular anode voltage the anode current shoots-off, resulting the anode voltage falls towards zero (SCR firing).4. If it does not happened, repeat the above step till you observe the firing. 5. Note down the value of IG at which the Firing happens.6. Now slightly increase(decrease) by 0.1V of gate voltage and note down the values of IA for different values of VA stating from zero to VBO and beyond. 7. Plot the graph VA VS IA for different values of IG.

Reverse Characteristics:1. Connect the circuit as shown in the figure.2. Increase the anode voltage V1 from zero to 20V in steps of 1V.3. Note down the anode current IA for every step of VA.4. Plot the graph VA VS IA

Tabulation for Forward characteristics:

S.NoIG= IG= IG=

VA IA VA IA VA IA


TYN 604

1KA+

0-500uA

V+

0-30VV1-

-

+

-

SCR REVERSE CHARACTERISCTICS


Tabulation for Reverse characteristics:

S. No VA IA

RESULTS :

VH =IH =

QUESTIONS:1. What are the applications of SCR.2. Does SCR works as a switch?

Experiment No : 10

UJT CHARACTERISTICS

AIM: To obtain the static characteristics of the given UJT 2N2646 and hence obtain the –ve resistance of the device. APPARATUS:

1. Power supplies 0-30V 2Nos.2. Ammeter 0-100mA 1No.3. Voltmeter 0-10V 1No.

COMPONENTS:

1. UJT – 2N2646 – 1No.2. Resistors - 220 - 2No.

CIRCUIT DIAGRAM:

Maharaj Vijayaram Gajapathi Raj College of Engineering, Vizianagaram 480-30V

2N2646

0-30VV+

0-10V

220ΩA+

0-100mA

220Ω

-

-

+

-


THEORY: The UJT is a three terminal device, is used in oscillators, trigger circuits, saw-tooth generators and

voltage-or current- regulated supplies. The single P-N junction accounts for the terminology uni-junction. It was originally called a duo(double) base diode due to the presence of two base contacts.

The symbol for the uni-junction transistor is provided in fig below. Note that the emitter leg is drawn at an angle to the vertical line representing the slab of n-type material. The arrow head is pointing in the direction of conventional current (hole) flow when the device is in the forward –biased active, or conducting state.

Symbolic representation Lead Configuration

UJT Characteristics:

The characteristics of a UJT are shown. For the left of the emitter potentials to the peak point, the I E is IEO (in microampere).This region is called the Cut-off region. Once conduction is established at VE =VP, decreases with increases in IE, this device, therefore ,has a negative resistance region. The negative resistance region is due to the holes injected into the n-type slab from the aluminum p-type rod when conduction is established. The increased holes content in an n-type material will result in an increase in the number of free electrons in the slab, producing an increase in conductivity (G) and a corresponding drop in resistance ‘R’.

When the VE approaches the valley point, any further increase in IE will place the device in the saturation region. In this region the characteristics approach that of the semiconductor diode.

PROCEDURE:

1. Connect the circuit as shown in the diagram. Keep VBB at 10V, vary VEE and measure VE and IE. Until VEE reaches a particular value emitter base1 diode presents and open circuit. The voltmeter reads the source voltage VEE since the current is zero.2. If VEE is further increased UJT fires. The voltage at which the UJT fires is known as the peak voltage V P

and the resultant current is known as peak current IP 3. Further in of IE results in decrease of VE .4. This process will continue until the valley point is reached, with co-ordinates VV and IV. For values of emitter voltage VE greater than IV, the emitter portion of the device behaves as a conventional forward-biased PN junction diode.

OBSERVATIONS:


E +

VE

B2

B1

VBB

-

+

-

Emitter

Base 1

Base -2

VP

-ve resistance region

VV

IPIV IE


Serial number VE (Volts) IE (mA)

MODEL GRAPH:

RESULT: negative resistance = VE / IE

QUESTIONS:1. What are the applications of UJT?2. What is intrinsic standoff ratio of the UJT?3. What is the function of UJT in relation oscillator?

Experiment No : 11COMMON EMITTER AMPLIFIER

AIM: 1. To design a common emitter amplifier and obtain its voltage gain, current gain, input impedance output impedance.

2. To plot the frequency response and to determine the bandwidth of the amplifier.

APPARATUS:

1. Power supply 0-30V 1 No.2. CRO 1 No.3. Digital multimeter. 1 No.4. Signal generator 1 No.

COMPONENTS:

1. Resistors 220 1No.



other resistors of designed values.2. Capacitors 10 µF 2Nos

100µF 1No.3. Transistors BC 547 1No.

THEORY Amplifiers:

An amplifier is a circuit where output signal is at a higher level of energy than input and faithfully reproduce the input waveform.

The device which can be used as an amplifier should have Ideal controlled source characteristicsThe BJT and FET have controlled source characteristics in the active region and pinch of regions

respectively. These are biased in their appropriate regions by means of externally applied direct voltages or current (biasing). i.e., these controlled sources establish operating point (Quiescent point). And time varying input signal is superimposed on the quiescent level to produce time varying output signal.

Multistage AmplifiersA number of stages can be used in cascade to amplify a signal from source, the microphone, to a level

which is suitable for the operation of another transducers, such as loud speakers. These cascaded amplifiers are called multistage amplifiers. The overall gain of the multistage amplifier is equal to product of the individual gain.

Quiescent point (operating point or Q point)The co-ordinates of IC, VCE with no signal is applied is called as operating point. The operating point is a fixed point as the characteristics. So it is called Quiescent point (Q-point).

When no signal is applied, the power supply (Vcc) establishes a fixed level of current and voltages (I C

and VCE). For transistor amplifiers this resulting DC current and voltages establishes an operating point on the characteristics that define the region that will be employed for the amplification of the signal.

For amplification, the Q-point should be selected in the middle of the active region. When time varying input signal is applied, the Q point moves towards saturation and cut-off regions. We have to choose the operating point in such way that, for the entire range of input signal the Q-point should not be entered into the saturation and cutoff regions for faithful amplification.

Biasing: The process of establishing the operating point in the active region by means of a fixed power supply

is called biasing. The biasing is necessary to operate the transistor in the active region for the entire range of input signal, so that a faithful amplification results. If no biasing or no proper biasing is applied, the transistor may be saturated or cutoff and the output will be clipped off. Frequency response.

It is the plot which gives the relation between the gain of the amplifier and frequency.The frequency distortion occurs when the signal components of different frequencies amplified

differently. This distortion may be caused by the internal device capacitances, coupling components or the load (of reactive). Under these conditions, the gain A is a complex number, whose magnitude and phase angle depend upon the frequency of the input signal.

If the frequency response characteristics is not a horizontal straight line over the range of frequencies under consideration, the amplifier is said to have frequency distortion over these range.

Bandwidth For an amplifier stage the frequency characteristic may be divided into three regions.

1. Mid band : This is a range, over which the amplification is reasonable constant and equal to Ao.2. Low frequency region: In this region, below mid band, an amplifier stage may behave like a high pass circuit.



3. High frequency region: In this region, above mid band, an amplifier stag may behave like a simple low pass circuit.

Lower cutoff or lower 3dB frequency ( FL )It is the frequency (in low frequency region) at which the gain becomes 1/ 2 times its mid band

value. It can also be defined as (if gain is expressed in decibels)the frequency at which the gain is reduced by 3 decibels from mid band value.

Upper cutoff frequency or upper 3 dB frequency (FH) It is the frequency (in high frequency region) at which the gain is reduced to 1/2 times its mid band

value.The frequency range from FL to FL is called the bandwidth of the amplifier stage. We may anticipate

in general way that a signal whose frequency range lie within the range of FL to FH will pass through the stage without distortion.Decibels:

The power gain in decibels is obtained by AdB = 10log10(Po / Pi ) where Po = output power Pi = input power

VdB = 20log10(Vo / Vi )where Vo = output voltage Vi = input voltageCE amplifier:

The CE amplifier is having a voltage and current gain greater than unity. The values of input resistance Ri and output resistance Ro lie between those for CB and CC amplifiers. The CE amplifier in this laboratory uses self biasing or emitter biasing configuration. The power supply Vcc, resistors R1, R2, RC

provide the biasing. The emitter resistor RE is meant for stabilization. The coupling capacitor blocks the DC voltages but freely passes the signal voltages. If the input coupling capacitor C1 is not used, the dc or average value of the signal source may offer the biasing of the amplifier. The output signal voltage may be applied to the input of another amplifier without effecting its bias, because of the blocking capacitor C2 . The emitter resistor RE provides stabilization as follows. If Ic tends to increase.(let ICO has risen due to increase in temperature). The current in RE is IE = IC + IB also increases. The voltage drop across RE also increases and the drop is in the direction to reverse bias the emitter junction so that the base current is decreased. Finally the collector current, IC decreases. Hence IC will increase less than it would have, had there been no self biasing resistor RE.

The Bypass capacitor CE is mainly used to prevent the loss of amplification due to negative feedback provided by the RE . This bypass capacitor will effect the low frequency response of the amplifier.

CIRCUIT DIAGRAM:


BC 547

RCR1

R2 RE CE 100F

RS 220Ω C1 10F

+

Signal Generator

C2 = 10F

+

CRODRB

+VCC = 12V

+ -

+ -

+-


DESIGN:

Design 1

Determine the values of RC, RE, R1, R2 for the network for the operating points indicated VCC = 12 V, ICQ = 5mA, VCEQ = 6V.VEN = VCC / 10RE = VEN / IE = VEN / IC (IE IC )RC = VRC / IC = (VCC – VCEQ - VEN) / IC

VBN = VBE + VE = R2 (1 / 10 ) RE find R2 , VBN = (R2 /(R1+R2)) VCC Find R2.

Alternate DesignVcc=12v, Choose operating point as (IcQ= 5mA,VCEQ= 6v)Rc = 1K Choose stability factor S=12Measure ‘’ value of the transistor using digital multimeter Calculate IB = IC/ =Re = ( Vcc- Vce- Ic Rc)/ (IB +Ic)S = (1+) [1+ (Rb/Re)]/ [1++ (Rb/Re)]Calculate Rb/Re, and then calculate Rb.VBN =VBE + IE Re , VBE =0.7vV = VBN +IBRb (where V is thevinen’s voltage considering Vcc, R1, R2)(V/Rb) =Vcc/R1, then calculate R1Calculate IR1 =(Vcc - VBN)/R1Calculate IR2 = IR1 – IB

R2 = ( VBN/ IR2)[Rb =R1R2/ (R1+R2),V=VccR2/( R1+R2) in the thevinen’s circuit]

PROCEDURE:

1. Connect the circuit with component values calculated.2. Verify the co-ordinates of operating points and note down any deviations from the designed values (V CEQ, ICQ and VBE). 3. Connect the signal generator with a sine wave of 1KHz frequency to the input and increase the input to such a level that the output waveform of the signal as observed on CRO is not distorted.

4. Measure the input and output voltages and calculate the gain of the amplifier. Av = (VO/P / VI/P) .5. To measure the input impedance, find the voltage drop across the known resistance R S. The input current therefore is measured as the voltage across Rs / Rs value. Input impedance Zi = Vi /Ii 6. To measure the input impedance, measure the output signal voltage VO/P without any load. Connect a resistive load and then adjust the load until the new output signal VO/P equal to the one half of the original signal. Remove the ROUT from the circuit and measure its value. The measured value is the output impedance of the circuit.



7. To measure the current gain AI, note down the output signal voltage when Ro is connected and divide it by Ro to get the output current. Now current gain = output current / input current. The power gain is the product of voltage gain and current gain. 8. Vary the frequency of the input signal from 50Hz to 1MHz in suitable steps and calculate gain at each step. Plot the graph between gain in dB Vs frequency. Note down the half power points and find the bandwidth of the amplifier. 9. Observe the phase relation between input and output signals at different frequencies.

MODEL GRAPHS:

AV Max 3dB point

Bandwidth

Frequency

TABULATION:

S.No Frequency Input Voltage Output Voltage Gain Gain in Decibels (dB)

12.....

50 Hz100 Hz200 Hz500 Hz800 Hz1k Hz2k Hz...1MHz

RESULTS:Voltage gain (AV) =Current gain (AI) =Input impedance (Zi) =



Output impedance (ZO) =Bandwidth. =

QUESTIONS:

1. Why should we use CE amplifier in the intermediate stage of multistage amplifiers?2. In active region the transistor acts as a current controlled current source and so that it can be used as an amplifier. Explain?3. Why is a capacitor coupling is used to connect a signal source to an amplifier?4. Why does the gain of the amplifier vary with the frequency?5. How the CE amplifier does produce 1800 phase shift between output and input?

Experiment No : 12



COMMON COLLECTOR AMPLIFIER

(EMITTER FOLLOWER)

AIM: To study the performance of a common collector (emitter) amplifier. And obtain its voltage gain, current gain , input impedance, output impedance and bandwidth.

APPARATUS:


COMPONENTS:

1. Resistors 220 1 No.33K 1 No.8.2K, 1 No.10K 1 No.

2. Capacitors 10 µF 2Nos.3. Transistors BC 547 1No.

THEORY

When the output is taken from the emitter terminal of the transistor as shown in the fig. The network is referred to as Common collector transistor amplifier. This configuration is also called emitter follower, because its voltage gain is close to unity, and hence a change across the load at the emitter. In other words the emitter follows the input signal.

In a.c analysis the collector is grounded, so, actually a common collector configuration the input resistance Ri of emitter follower is very high (Hundreds of Kilo ohms) and output resistance RO is very low (tens of ohms). Hence the common collector circuit can be used as a buffer stage which performs the function of resistance transformation.(From high to low resistance) over a wide range of frequencies, with voltage gain close to unity. The emitter follower increases the power level of the signal, i.e., it provides power gain.

For the common collector amplifier, the current gain AI is high (approximately equal to common emitter stage) AV is less than unity (but close to unity), RI is the highest and RO is the lowest of the three (CE, CB & CC) configurations. This circuit is widely applied as a buffer stage between a high impedance source and low impedance load.

The emitter follower is frequently used for impedance matching purposes; the emitter follower (due to its low output resistance) is often used to drive capacitive loads.



CIRCUIT DIAGRAM:

PROCEDURE: 1. Connect the circuit as shown in the figure 2. The operating points VCEQ , IEQ and VBE are measured.3. Connect the signal generator with a sine wave of 1KHz frequency to the input and increase the input to such a level that the output waveform of the signal as observed on CRO is not distorted.4. Measure the input and output voltages and calculate the gain of the amplifier. Av = (VO/P / VI/P) .5. To measure the input impedance, find the voltage drop across the known resistance RS, which gives drop in RMS Value. Convert this RMS value to peak to peak value for further calculations. The input current therefore is measured as the voltage across Rs(VRS(p_p)), VRS(p_p) / Rs value. Input impedance Zi = Vi /Ii 6. To measure the input impedance, measure the output signal voltage VO/P without any load. Connect a resistive load and then adjust the load until the new output signal VO/P equal to the one half of the original signal. Remove the ROUT from the circuit and measure its value. The measured value is the output impedance of the circuit.7. To measure the current gain AI, note down the output signal voltage when Ro is connected and divide it by R0 to get the output current. Now current gain = output current / input current. The power gain is the product of voltage gain and current gain. 8. Vary the frequency of the input signal from 50Hz to 1MHz in suitable steps and calculate gain at each step. Plot the graph between voltage gain Vs frequency. Note down the half power points and find the bandwidth of the amplifier.


Q1 BC 547

33 KΩ

8.2 KΩ10 KΩ

C2 10F

RS 220Ω C1 10F

+Signal Generator +

CRO

+Vcc = 12V


TABULATION:


12.....

50 Hz100 Hz200 Hz500 Hz1k Hz2k Hz...1MHz

RESULTS:

Voltage gain (AV) =Current gain (AI) =Input impedance (Zi) =Output impedance (ZO) =Bandwidth =

QUESTIONS:1. Why the common collector amplifier is also called emitter follower?2. Which of the amplifiers (CE, CB or CC) has a highest input resistance (Zi)?3. How will the size of the biasing resistor Ri effect the input impedance of the circuit?



Experiment No : 13

RC PHASE SHIFT OSCILLATOR

AIM: To design and to find the frequency of the RC phase shift Oscillator.To measure the phase shift at each stage.

APPARATUS:


COMPONENTS:

1. Resistors As per designed values2. Capacitors As per designed values

100µF 1 No.3. Potentiometer 1K 1 No. 4. Transistors BC 547 1 No.

THEORY:

Sinusoidal Oscillator

Sinusoidal Oscillator is a feedback amplifier designed to have a closed looped poles on the j –axis at the frequency of the desired output.

An Oscillator is a feedback amplifier (the gain with feedback A f = A / (1+A) ) with the closed loop gain , A = -1.

When A = -1, Af becomes infinite thus, an infinitesimal signal (noise voltage) can be provided measurable output voltage, at this circuit acts as oscillator even without an input signal.Barkhausen Criteria

The two conditions must be satisfied by a circuit to sustain Oscillation.1. The Phase shift through the amplifier and feedback network must be 3600.2. The magnitude of the gain of the amplifier and feedback network must be unity.These two conditions indicates that -A = 1. This is called the barkhausen Criteria. The RC Phase shift Oscillator, Wein bridge Oscillator, Colpitts Oscillator, Hartley oscillator are the examples of sinusoidal oscillator. All these oscillators should satisfy the Barkhausen criteria to produce Oscillations. The RC Phase shift Oscillator consists of transistor amplifier and three cascaded (series) RC sections. The output of the last RC combination being returned to the base of the transistor. If the loading of the phase shift network on the amplifier can be neglected. The amplifier introduces 1800 Phase shift to the voltage which appears on the base. The Three RC sections will give additional phase shift, at some frequency the phase shift



introduced by the RC network will be equal to 1800, at this frequency the total phase shift from the base around the circuit and back to the base will be exactly zero. This particular frequency will be the one at which the circuit will oscillates, provided the magnitude of amplification is sufficiently large. The frequency of oscillation is given by

F =(1 / 2RC) 1 / (6 +4K).Where K = RC / R.

The requirement is that the magnitude of A must exceed unity. In order for oscillations to start, the circuit needs the condition

hFE > 4k+23+(29/K)The phase shift oscillator is particularly suited to the range of audio frequencies. At frequencies in the

range of mega hertz, has no marked advantage over circuits employing tuned LC networks. The frequency of oscillations can be varied by changing any one of the impedance elements ( R or C ) in the phase shifting networks.

CIRCUIT DIAGRAM

DESIGN: K = 2.7 = Rc / R, find out the value of R.

find out the value of C Find out the value of C for the frequency of 1 KHz.

PROCEDURE:

1. Connect the circuit as shown in the figure.2. First check the amplifier section (i.e., find out the operating points)3. Adjust the potentiometer such that oscillations are sustained. Monitor the output waveform using CRO.4. Measure the voltages at each RC section. 5. Measure the phase shift introduced by each section forming Lissajous figures. Use the sine wave at reference voltage.6. Measure the waveform frequency and compare it with theoretical value.


Q1 BC 547

RCR1

R2 RE

C

R

VCC = 12V

CE 100uF

C

R

C

1K POT

A B C D


Measurement of phase shift using Lissajous figures:

= sin-1 y1 / y2 = sin-1 x1 / x2.

Y Major Axis.

y1 y2

X x1

x2

RESULTS:

Designed Theoretical Frequency = Practical Waveform Frequency = Phase Shift across Node A and B = Phase Shift across Node B and C = Phase Shift across Node C and D = Over All Phase Shift (Practical) =Theoretical Phase Shift =

QUESTIONS:

1. How the Barkhausen Criteria is satisfied in the RC phase-shift oscillator?2. How do you vary the frequency of the RC phase-shift oscillator?3. What is the maximum phase-shift that can be obtained by a single RC section?4. Under what conditions the amplifier behaves as an oscillator?5. What is the frequency range that can be obtained by the RC phase shift oscillator?



Experiment No: 14CLASS ‘A’ POWER AMPLIFIER

AIM: To observe and calculate the conduction angle and efficiency of a class A power amplifier.

APPARATUS: 1. Power supply 0-30V 1 No2. Signal generator 1Hz – 1MHz 1 No3. CRO 20MHz 1 No

COMPONENTS: 1. Resistors 220Ω, 5.6KΩ, each one,2. 47KΩ, 1KΩ3. Capacitors 4.7µF 2 No,

47µF 1 No4. Transistors 2N3055 1 No.

THEORY:

Class A Amplifier

Common emitter voltage amplifiers are the most commonly used type of amplifier as they have a large voltage gain. They are designed to produce a large output voltage swing from a relatively small input signal voltage of only a few millivolt's and are used mainly as "Small Signal Amplifiers". However, sometimes an amplifier is required to drive large resistive loads such as a loudspeaker and for these types of applications where high switching currents are needed Power Amplifiers are required.

The main function of the Power amplifier, which are also known as a "Large Signal Amplifier" is to deliver power, which is the product of voltage and current to the load. Basically a power amplifier is also a voltage amplifier the difference being that the load resistance connected to the output is relatively low, for example a loudspeaker of 4 or 8Ωs resulting in high currents flowing through the Collector of the transistor. Because of these high load currents the output transistor(s) used for power amplifier output stages need to have higher voltage and power ratings than the general ones used for small signal stages.



Since we are interested in delivering maximum AC power to the load, while consuming the minimum DC power possible from the supply we are mostly concerned with the "Conversion Efficiency" of the amplifier. However, one of the main disadvantage of power amplifiers and especially Class A type amplifiers is that their overall conversion efficiency is very low. Percentage efficiency in amplifiers is defined as the r.m.s. output power dissipated in the load divided by the total DC power taken from the supply source as shown below.

OUTPUT POWER: is given by POUT =Vrms2/RL

Vrms =1/(2.√2) VPP

POUT =VOUT2/8RL

POUT(max) = VPP2/8RL

Power Amplifier Efficiency

Where:

η% - is the efficiency of the amplifier.

Pout - is the amplifiers output power delivered to the load.

Pdc - is the DC power taken from the supply.

DC Power=



For a power amplifier it is very important that the amplifiers power supply is well designed to provide the maximum available continuous power to the output signal.

Angle of flow or conduction angle

Power amplifier circuits (output stages) are classified as A, B, AB and C for analog designs, and class D and E for switching designs based upon the conduction angle or angle of flow, Θ, of the input signal through the (or each) output amplifying device, that is, the portion of the input signal cycle during which the amplifying device conducts. The image of the conduction angle is derived from amplifying a sinusoidal signal. (If the device is always on, Θ = 360°.) The angle of flow is closely related to the amplifier power efficiency.

Class A100% of the input signal is used (conduction angle Θ = 360° or 2π); i.e., the active element

remains conducting (works in its "linear" range) all of the time. Where efficiency is not a consideration, most small signal linear amplifiers are designed as Class A. Class A amplifiers are typically more linear and less complex than other types, but are very inefficient. This type of amplifier is most commonly used in small-signal stages or for low-power applications (such as driving headphones).

CIRCUIT DIAGRAM

PROCEDURE:1. Connect the circuit as shown in the figure.2. Provide the appropriate input signal and observe the corresponding output on CRO.3. Adjust the amplitude of the signal generator so as to get undistorted output.4. Note down the values of input and output voltages.5. Now increase the amplitude of the input signal and note down the amplitude above which distorted signal is observed.

Model waveforms:


C E 47 u F

2N3055

C2 4.7uF

R

1 33

K

C1 4.7uFRS

VS

VCC = 12V

V0

R

2 5.

6K

RC

1 K

R

E 22

0

CLASS-A POWER AMPLIFIER

+ -

+ -+

-

http://en.wikipedia.org/wiki/Electrical_efficiency

http://en.wikipedia.org/wiki/Analog_circuit


RESULT:

Input sinusoidal voltage applied = Standard input Frequency applied = Output voltage obtained = Voltage Gain = DC Power=

Conduction angle has been observed and distortions too.

QUESTION: 1. What is the conduction angle of a class A power amplifier? And why?2. What kinds of distortions have been observed? State the reason.3. State the reason for using 2N3055 transistor. State how is it different from normal transistor.4. State the applications of class A power amplifiers.


InputVoltage

OutputVoltage


Experiment No : 15

WEIN BRIDGE OSCILLATOR

AIM: To design a Wein Bridge oscillator and study its performance.

APPARATUS:1. Power supply (dual channel) 0-30V 1 No.2. CRO 1 No.3. Digital multimeter. 1 No.

COMPONENTS:1. Resistors (R1=R2=R= 47K ) 47K 2 Nos. (R3) 10K 1 No.

DRB 1 No.2. Capacitors 2 No.

(C1=C2=C= 0.1F CC1=CC2=10uf,CE=100uf)3. Transistors BC 547 2 No.s4. Resistors designed values

(R11,RC1=RC2, R12,R22, RE2)

THEORY:

An oscillator, in which a balanced bridge is used as the feedback nework is known as Wien bridge oscillator, is shown below.The active element is an operational amplifier which has a very large positive voltage gain, If the bridge balance is desired, in order to get oscillation, then R1and R2 must be chosen so that , vi=0 ( R4 / (R3 + R4) = 1/3 or R3 = 2 R4.)Maharaj Vijayaram Gajapathi Raj College of Engineering, Vizianagaram 66


The frequency of oscillation (f0 = 1/2RC) is precisely the null frequency of the balanced bridge continuous variation of frequency is accomplished by varying simultaneously the two capacitors

CIRCUIT DIAGRAM:

output

Wien Bridge Oscillator

Design : Calculate the biasing resistors R11,R12,R22 RC1,RC2,RE2 using design procedure as mentioned in the CE amplifier experiment.

Calculate the frequency of oscillations f0 = 1/2RC

PROCEDURE: 1. Connect the circuit as shown in the figure above.2. In place of R4 connect a DRB.3. Connect CRO at the output terminal.4. Vary the DRB until the oscillations are sustained.5. Find the frequency of oscillations, and verify the observed output with theoretical values.6. repeat the above steps 5 and 6 for various values of C i.e., 0.01 and .001f. OBSERVATIONS

S.No Value of C fO theoretical fO practical

RESULTS:

Theoretical Frequency =


10f 10f

100f

Q1 Q2

Rc1 Rc2R12R11

R22 RE2 CE

CC1 CC2

Vcc = 12V

R1 R3

R2

C1

C2R4 DRB


Practical Frequency =

QUESTIONS:

1. How the Barkhausen criterion is satisfied in the Wien Bridge oscillator?2. What determines the frequency of oscillations?3. How do you vary the frequency of oscillations?4. What is the frequency range of oscillations?

Experiment No : 16

FET AMPLIFIER(COMMON SOURCE)

AIM: 1. To design a FET common source amplifier for a voltage gain of 82. To obtain experimentally the voltage gain of the amplifier.3. To obtain the frequency response characteristics of the amplifier

APPARATUS:1. Power supply 0-30V 1No.2. CRO – 1No.3. Digital multi-meter. 1 No.4. Signal generator 1 No.

COMPONENTS:1. Resistors 47k 1 No.

RS and RD as per design2. Capacitors 10 µF 3Nos.3. FET BFW 11 1No.

CIRCUIT DIAGRAM:


BFW 11

RD

RS C3 10uFRG

C1 10uF

+

INPUT

+VDD 12 V

C2 10uF

CRO


DESIGN:Select VGSQ near the max. gm value say VP / 2, IDSS = 10 mA, VP = -4V, Yos = 2S, rd = 1 / Yos

IDQ = IDSS

VDD = ID(RD+RS) + VDSQ

At bias point gm = gmo =

Av = gm (RD// rd) , VGSQ = - ID.RS

PROCEDURE:

1. Connect the circuit as shown in the figure.2. note the operating points VDSQ, VGSQ, IDQ

3. set the output of the signal generator just below the point of distortion, so that the maximum undistorted output appears on the CRO. 4. note this input voltage and keep it constant through out the experiment. 5. Vary the input frequency from 50Hz to 1MHz, and note down the output signal voltage at every setting of frequency. 6. Plot the frequency response for the amplifier and calculate the bandwidth.

TABULATION:


12....

50 Hz100 Hz200 Hz500 Hz1k Hz2k Hz



. ...1MHz

RESULTS:

Voltage gain =Bandwidth =

QUESTIONS:

1. How do you explain the wedge shape of the depletion regions?2. What is the mechanism responsible for constant resistance operation of FET?3. Compare the characteristics of BJT and FET.4. Is it possible to operate FET with gate forward biased?

Experiment No : 17

FET AMPLIFIER(COMMON DRAIN)

AIM: 1. To obtain experimentally the voltage gain of the FET amplifier in Common Drain Configuration.

2. To obtain the frequency response characteristics of the amplifier.

APPARATUS:1. Power supply 0-30V 1No.2. CRO 1No.3. Digital multi-meter. 1 No.4. Signal generator – 1 No.

COMPONENTS:

1. Resistors 47k (RG) 1 No.



800(RS) 1 No.2. Capacitors 10 µF 2No.3. FET BFW 11 1No.

CIRCUIT DIAGRAM:

PROCEDURE:1. Connect the circuit as shown in the figure.2. set the output of the signal generator just below the point of distortion, so that the maximum undistorted output appears on the CRO. 3. note this input voltage and keep it constant through out the experiment. 4. Vary the input frequency from 50Hz to 1MHz, and note down the output signal voltage at every setting of frequency. 5. Plot the frequency response for the amplifier and calculate the bandwidth.

TABULATION:


12.....


RESULTS:Voltage gain =Maharaj Vijayaram Gajapathi Raj College of Engineering, Vizianagaram 71

BFW 11

RS

C2 10F

RG

C1 10F

+

INPUT

+ VDD 12V

+

CRO

D

G

S


Bandwidth =

QUESTIONS :

1. Why the common drain amplifier is also called source follower?2. What are the applications of common drain amplifier?

Experiment No : 18

CURRENT SERIES FEEDBACK AMPLIFIER

AIM: To design current series feedback amplifier and obtain the voltage gain , current gain, input impedance, output impedance and bandwidth of the amplifier with and without feedback.

APPARATUS:


COMPONENTS:


Transresistanceamplifier

Transconductanceamplifier


1. Resistors 220 1No. other resistors of designed values

2. Capacitors 10 µF 2Nos. 100µF 1No.

3. Transistors BC 547 1No.

THEORY:

Feedback amplifiers:

Feedback is the process whereby a portion of the output is returned to the input to form part of the system excitation. The block diagram of the feedback amplifier is shown below.

VS + Vi VO

+

Vf

A = Vo /Vi, = Vf / Vo and Af = Vo / VsDepending on the relative polarity of the signal being feedback into the input negative feedback or

positive feedback results. If the feedback signal is of same polarity to the input signal, positive feedback results. This feedback drives the amplifier into oscillations. The gain of the amplifier with positive feedback is, Af = A / (1-A).

If the feedback signal is of opposite polarity to the input signal, negative feedback results. The gain of the amplifier with negative feedback is Af = A / (1+A). It has many advantages, among them being 1. Better stabilized gain, that is gain is insensitive to variations in individual transistor parameters. 2. Improves the frequency response (increases the bandwidth).3. Reduces the noise, frequency distortion and non-linear distortion

The four basic amplifier types:A circuit used for electronic amplification can be designed to respond to either voltage or current as

its primary input signal. Similarly, the circuit can be designed to supply either a voltage or a current as its primary output signal. Depending on its input and output signals, an amplifier can be classified into one of the four basic types summarized by figure below. Voltage amplifier Vin Vout

Iin Current amplifier Iout


Basic amplifier (A)

Feedback network

Basic amplifier (A)

Feedback network

Av

Ai

RM gm

Inpu

t sig

nals

Output signals


A voltage amplifier with gain Av accepts a voltage as the input signal and provides a voltage as its output signal. A current amplifier with gain Ai has input and output signals that are both currents.

A circuit in which the input signal is a voltage and output signal a current is called a transconductance amplifier, or some times a voltage-to-current converter. A transresistance amplifier with gain ‘R M’ accepts a current as its input signal and provides a voltage as its output signal. Some times, the transresistance amplifier is called current to voltage converter.

The amplification factor gm for transconductance amplifier is the ratio of Iout/ Iin , has units of amperes / volt or conductance.

The amplification factor RM for transresistance amplifier is defined as the ratio of Vout / Iin and has units of volts / amp or resistance.

The four feedback topologies: Each of the four basic amplifier types- voltage, current, transresistance and transconductance has its

own appropriate feedback topology determined entirely by the amplifiers input and output signal types.If the basic amplifier is having a voltage at input then the feedback voltage is mixed in series with

input voltage, if the input signal is current, then the feedback current should be mixed in parallel (shunt) with the input current.

If the basic amplifier is having a voltage at its input, then the output voltage is sampled by connecting the feedback network in shunt across the output. If the output is current, then the output current is sampled by connecting the feedback network in series with the output.

A voltage amplifier, for which the input and output signals are both voltages, requires voltage, or series, mixing at the output and voltage, or or shunt sampling, at the output. This feedback topology is called the series-mixing/ shunt-sampling (series/shunt) connection.

The resultant feedback is called voltage series feedback amplifier.The shunt mixing / series sampling (shunt / series) feedback topology appropriate for current

amplifier , and resultant feedback amplifier is called current shunt feedback amplifier. The shunt-mixing / shunt-sampling (shunt/shunt) feedback topology appropriate for a transresistance amplifier.

The series-mixing / series sampling (series/series) feedback topology appropriate for a transconductance amplifier. The resultant feedback amplifier is called current series feedback amplifier.

The effects of feedback topology on a) input resistance: If the feedback signal is returned to the input in series (shunt) with the applied voltage, regardless of whether the feedback is obtained by sampling the output current or voltage , it increases (decreases) the input resistance. b) Output resistance: The feedback which samples the output voltage(current), regardless of how this signal is returned to the input, tends to decrease (increase) the output resistance.

The common emitter transistor amplifier with a resistance in the emitter and FET common source amplifier with a resistor in the source lead are examples of current series – topology. A current series feedback amplifier improves the characteristics of transconducatnace amplifier, that is both input resistance and output resistances are increased. This topology stabilizes the transconductance gm.

In current series feedback amplifier, as shown in figure, the feedback signal is the voltage Vf across emitter resistor and sampled signal is the load current through the collector resistor. Hence this is the case of current-series feedback. To remove the feedback the emitter resistor must be removed or bypassed by a capacitor

CIRCUIT DIAGRAM:


Q1 BC 547

RCR1

R2 RE CE 100F

RS 220Ω C1 10F

+Signal Generator40mV/1KHz

C2 = 10F

+

CRODRB

+VCC = 12V


PROCEDURE:

1. Connect the circuit with component values calculated.2. Verify the co-ordinates of operating points and note down any deviations from the designed values (VCEQ, ICQ and VBE). 3. Connect the signal generator with a sine wave of 1KHz frequency to the input and increase the input to such a level that the output waveform of the signal as observed on CRO is not distorted.4. Measure the input and output voltages and calculate the gain of the amplifier. Av = (VO/P / VI/P) .5. To measure the input impedance, find the voltage drop across the known resistance RS. The input current therefore is measured as the voltage across Rs / Rs value. Input impedance Zi = Vi /Ii 6. To measure the input impedance, measure the output signal voltage VO/P without any load. Connect a resistive load and then adjust the load until the new output signal VO/P equal to the one half of the original signal. Remove the ROUT from the circuit and measure its value. The measured value is the output impedance of the circuit.7. To measure the current gain AI, note down the output signal voltage when Ro is connected and divide it by Ro to get the output current. Now current gain = output current / input current. The power gain is the product of voltage gain and current gain. 8. Vary the frequency of the input signal from 50Hz to 1MHz in suitable steps and calculate gain at each step. Plot the graph between voltage gain Vs frequency. Note down the half power points and find the bandwidth of the amplifier. 9. Observe the phase relation between input and output signals at different frequencies. 10. Repeat the above steps by connecting (disconnecting) the emitter bypass capacitor CE. the readings with CE gives the response of the amplifier without out feedback. The readings without the CE gives the performance of the amplifier in current series feedback mode.

MODEL GRAPHS

Without feedback - 3dB line - 3dB line


Gai

n in

dB


With feedback

Frequency

TABULATION:


With Feedback

12.....


RESULTS:

Voltage gain with /without feedback =Current gain with /without feedback =Input impedance with /without feedback =Output impedance with /without feedback =Bandwidth with /without feedback. =

QUESTIONS:1. What is meant by Feedback?2. Distinguish between negative and positive feedback.3. Distinguish between current and voltage feedback.4. Discuss the effect of negative feedback on 1) gain 2) Input impedance 3) output impedance 4) distortion.

Experiment No : 19

VOLTAGE SERIES FEEDBACK AMPLIFIER



AIM: To study the performance of a voltage series feedback amplifier. And obtain its voltage gain, current gain, input impedance, output impedance and bandwidth.

APPARATUS:


COMPONENTS:

1. Resistors 220 1 No.33K 1 No.8.2K 1 No.10K 1 No.

2. Capacitors 10 µF 3Nos. 3. Transistors BC 547 1No. THEORY:

The FET common drain amplifier (source follower) and the BJT common collector amplifier (emitter follower) are the examples of voltage series topology.

Voltage series feedback amplifier improves the characteristics of voltage amplifier. i.e., Input resistance increases and output resistance decreases.

Rif = Ri (1+A) Rof = Ro / (1+A)

Af = A / (1+A) FLF= FL / (1+A) FHF = FH (1+A)

In this experiment, the negative feedback is introduced by connecting the emitter resistor in the emitter lead. The feedback signal is the voltage across emitter resistor, which is in series with the input voltage and the output is the voltage, thus it is the voltage series feedback amplifier.

CIRCUIT DIAGRAM:


Q1 BC547

R1 33KΩ

R2 8.2KΩ RE 10KΩ

C2 10

RS 220Ω

C1 10F

+

Signal Generator

+

CRO

VCC +12V


PROCEDURE:

1. Connect the circuit as shown in the figure 2. The operating points VCEQ , IEQ and VBE are measured.3. Connect the signal generator with a sine wave of 1KHz frequency to the input and increase the input to such a level that the output waveform of the signal as observed on CRO is not distorted.4. Measure the input and output voltages and calculate the gain of the amplifier. Av = (VO/P / VI/P) .5. To measure the input impedance, find the voltage drop across the known resistance R S. The input current therefore is measured as the voltage across Rs / Rs value. Input impedance Zi = Vi /Ii 6. To measure the input impedance, measure the output signal voltage VO/P without any load. Connect a resistive load and then adjust the load until the new output signal VO/P equal to the one half of the original signal. Remove the ROUT from the circuit and measure its value. The measured value is the output impedance of the circuit.7. To measure the current gain AI, note down the output signal voltage when Ro is connected and divide it by Ro to get the output current. Now current gain = output current / input current. The power gain is the product of voltage gain and current gain. 8. Vary the frequency of the input signal from 50Hz to 1MHz in suitable steps and calculate gain at each step. Plot the graph between voltage gain Vs frequency. Note down the half power points and find the bandwidth of the amplifier. 9. Repeat the above steps by connecting (disconnecting) the emitter bypass capacitor C E. The readings with CE give the response of the amplifier without out feedback. The readings without the CE give the performance of the amplifier in current series feedback mode.



MODEL GRAPHS:

AV Max 3dB point

Bandwidth

Frequency

TABULATION:


With Feedback

12.....


RESULTS:Voltage gain =Current gain =Input impedance =Output impedance =Bandwidth. =



Experiment No : 20

COLPITTS OSCILLATOR

AIM: To design a colpitts oscillator and study its performance.

APPARATUS:1. Power supply 0-30V 1 No.2. CRO 1 No.3. Digital multimeter. 1 No.

COMPONENTS:1. Resistors Resistors as per designed values2. Capacitors C1 and C2 are as per designed values

10F 2 Nos.100µF 1 No.

3. Inductor as per designed value.4. Transistors BC 547 1 No. THEORY

Colpitts OscillatorThe RC phase shift and wien bridge oscillator ar eTC tunable oscillators. That I, the frequency at

oscillator is determined by the resistance and capacitance values used. These oscillators are particularly suited to one range of frequencies from several Hertz to several hundred Kilohertz and so include the range of audio frequencies. At frequencies in the Megahertz range , these have no advantages over circuits employing tuned LC networks.

Tuned LC oscillators are used in many applications, including radio transmitters, AM and FM receivers, and sinusoidal function generators. One popular LC oscillator configuration, shown in fig, is called the COLPITTS oscillator. The resonant combination of C1,C2 and L is called the tank circuit. The oscillation frequencies changed via the adjustable inductor L.

CIRCUIT DIAGRAM:


Q1 BC 547

RCR1

R2 RE

CC2 = 10uF

VCC +12V

CC1 10uF

C1 C2

L


Design L = 1mHC = C1 in series with C2 , find out Chfe = > C2 / C1 find out C1 and C2

find out the values of frequency for different values of L values by using the formula

fo =

PROCEDURE:

1. Connect the circuit as shown in the figure.2. The output oscillations are to be observed by CRO.3. Vary the values of C1 for sustained oscillations and compare these values with theoretical values. 4. Now vary the L and note down the frequency of oscillations and compare them with theoretical ones.

S.No Value of C1 Value of C2 Value of L fO theoritical fO practical

RESULTS:Theoretical Frequency = Practical Frequency =

QUESTIONS:

1. How the Barkhausen criterion is satisfied in Colpitts oscillator?2. What is the feedback network in the oscillator?3. What is the frequency range that can be obtained by the Colpitts oscillator?4. What are applications of LC tuned oscillators?



Experiment No : 21

HARTLEY OSCILLATOR

AIM: To design a Hartley oscillator and study its performance.

APPARATUS:1. Power supply 0-30V 1 No.2. CRO 1 No.3. Digital multimeter. 1 No.4. Bread Board 1 No.

COMPONENTS:1. Resistors Resistors as per designed values2. Capacitors C as per designed values

10F 2 Nos.100µF 1 No.

3. Inductor L1 and L2 are as per designed value.4. Transistors BC 547 1 No.

THEORY:

The Hartley oscillator uses the LC network as a feedback network. This Hartley oscillator is formed by changing the capacitors in the Colpitts tank circuit to inductors and by changing the tunable inductor to an adjustable capacitor. Both tha Colpitts and Hartley oscillator configuration requires an amplifier with inverting( 1800 phase shift) gain in order to sustain oscillations.

CIRCUIT DIAGRAM:


Q1 BC 547

RCR1

R2 RE

C2 = 10u

VCC +12V

C1 10uF

L1L2

C

CE 100uF


PROCEDURE:

1. Connect the circuit as shown in the figure.2. The output oscillations are to be observed by CRO.3. Vary the values of C for sustained oscillations and compare these values with theoretical values. 4. Now vary the L1 AND L2 and note down the frequency of oscillations and compare them with theoretical ones.

OBSERVATIONS:

S.No Value of L1 Value of L2 Value of C fO theoretical fO practical

RESULTS:

Theoretical Frequency = Practical Frequency =

QUESTIONS:

1. How the Barkhausen criterion is satisfied in the Hartley oscillator?2. What is the feedback network in the Hartley oscillator?3. What is the frequency range that can be obtained by Hartley oscillator?4. What is the frequency stability criterion for the sinusoidal oscillator?


Transistor BC 547 / BC 557

1.collector 2. Base 3. Emitter


Lead Configuration


Source

DrainGate 1

Gate 2

JFET BFW-11

Emitter

Base 1

Base -2

Uni-Junction Transistor 2N2646

Anode Cathode

Diode – 1N 4007 / Zener Diode

OP-AMP LM 741

Offset Null

INV – I/P

Non- INV-I/P

-VEE

NC

+VCC

Output

Offset Null

1

2

3

4

8

7

6

5Cathode Gate Anode

Silicon Controlled RectifierTYN 604

TYN 604

jawaharlal nehru technological … · web viewa rectifier circuit is necessary to convert a signal...

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