1.EM +2 Physics practical guide contents

PACHAIYAPPAS HIGHER SECONDARY SCHOOL,

KANCHIPURAM - 631501.

PHYSICS PRACTICAL HANDBOOK

HIGHER SECONDARY SECOND YEAR

Prepared by

B.ELANGOVAN. M.Sc., M.Ed., M.Phil.,

(Tamilnadu Dr.Radhakrishnan Best Teacher Award recipient - 2011)

P.G.Teacher in Physics,



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PHYSICS PRACTICAL

HIGHER SECONDARY - SECOND YEAR

S.NO DATE NAME OF THE EXPERIMENT PAGE

1 Spectrometer - Prism

2 Spectrometer - Grating

3 Metre Bridge

4 Potentiometer

5 Tangent Galvanometer

6 Sonometer

7 PN - junction diode and Zener Diode

8 NPN Transistor - Part-1

9 NPN Transistor - Part-2

10 Operational Amplifier - Inverting

11 Operational Amplifier - Non-inverting

12 Integrated Circuits- Logic gates

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1. SPECTROMETER - OF A SOLID PRISM

FORMULA REQUIRED: Refractive index of the material of the given prism is

2

Asin

2

DAsin

Where, A is the angle of the prism

D is the angle of minimum deviation

Unit of refractive index =no unit.

DIAGRAM:

To find the angle of Prism To find the angle of minimum deviation

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1. SPECTROMETER - OF A SOLID PRISM

AIM :

To determine the angle of a given prism and its angle of minimum deviation and

hence to calculate its refractive index.

APPARATUS REQUIRED :

Spectrometer, solid Prism, sodium vapour lamp and reading lens.

PROCEDURE :

1) ANGLE OF THE PRISM

i) After making initial adjustments, the prism is placed on the prism table.

ii) The slit is illuminated by a sodium vapour lamp.

iii) The telescope is rotated until the image of the slit formed by reflection at the

face AB is made to coincide with the vertical cross wire of the telescope in the

position T1. The reading of the verniers are noted.

iv) The telescope is then rotated to the position T2 .The image of the slit formed by

reflection at the face AC coincides with the vertical cross wire. The readings

corresponding to the verniers are again noted.

v) The difference between these two reading gives twice the angle of the prism.

Half of this gives the angle of the prism.

2) ANGLE OF MINIMUM DEVIATION

i) The prism is placed on the prism table so that light from the collimator falls on

one refracting face. The refracted image is observed through the telescope.

ii) The prism table is now rotated so that the refracted image moves towards the

direct ray. If necessary the telescope is rotated so as to follow the image.

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iii) It will be found that, the image moves towards the direct ray upto a point and

then turns back. The position of the image where it turns back is the minimum

deviation position and the prism table is fixed in this position.

iv) The telescope is now adjusted so that its vertical cross wire coincides with the

image and reading of the verniers are noted.

v) Now the prism is removed and the telescope is turned to receive the direct ray

and vertical cross wire is adjusted to coincide with the image. The reading of the

verniers are noted.

vi) The difference between the two readings give the angle of minimum deviation (D).

vii) The refractive idex of the material of the prism is calculated using the formula

2sin

2sin

A

DA

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CALCULATIONS:

To find A

2A = R1 R2 = 227o20 - 127o16 2A = R1 R2 = 407

o20 - 287o16 2A = 120o4 2A = 120o4

A = 60o2 A = 60o2

Average A = 60o2

To find D

D = R3 R4 = 39o44 - 0o0 D = R3 R4 = 219

o44 - 180o0

D = 39o44 D = 39o44

Average D = 39o44

To find

=

+ 2

2

=

602 + 3944 2

602

2

=

9946 2

602

2 =

4953

301

= 0.7647

0.5003

= 1.528 ( no unit )

RESULT: i) The angle of the prism A = 60o 2

ii) The angle of minimum deviation D = 39o44

iii) Refractive index of the material of the given prism = 1.528 (no unit)

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2. SPECTROMETER GRATING WAVELENGTH OF COMPOSITE LIGHT

FORMULA REQUIRED :

The wavelength () of a spectral line using the grating is given by

Nm

sin

Where,

is the angle of diffraction

m is the order

N is the number of lines per unit length drawn on the grating

Determination of angle of diffraction :

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2. SPECTROMETER GRATING WAVELENGTH OF COMPOSITE LIGHT

AIM:

To determine the wavelength of the composite light using a diffraction grating and a spectrometer.


Spectrometer, solid Prism, sodium vapour lamp and reading lens.

PROCEDURE:

The preliminary adjustments of the spectrometer are made. The slit is illuminated

by white light from mercury vapour lamp. The grating is mounted on the prism table. The

direct image (white) of the slit is adjusted to coincide with the vertical cross wire. The

direct reading RI is measured using the verniers.

Now the telescope is released to get the first order (n= 1) diffracted image of the

slit in the left side. It is adjusted so that the vertical cross wire coincides with violet

spectral line. Readings corresponding to both the verniers are taken as R2. The angle of

diffraction for violet is found as R1 R2. The experiment is repeated for green and

yellow spectral lines also.

The number of lines per unit length of the grating is N. Wavelength of the

spectral line is calculated from the formula

Nm

sin

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CALCULATIONS :

TO FIND THE

RAY RD1 R1 RD2 R2

BLUE 13o 34- 0o0 = 13o34 193o34- 180o0 = 13o34 B = 13o34

GREEN 15o 45- 0o0 = 15o45 195o45- 180o0 = 15o45 G = 15o45

YELLOW 16o 42- 0o0 = 16o42 196o42- 180o0 = 16o42 Y = 16o42

m = 1 and N = 5 105 lines/m

The wavelength of the Blue spectral line :

mNm

o

BB

5

551004692.0

105

2346.0

1051

)'3413sin(sin

B = 4692 Ao

The wavelength of the Green spectral line :

mNm

o

GG

5

551005429.0

105

2714.0

1051

)'4515sin(sin

G = 5429 Ao

The wavelength of the Yellow spectral line :

mNm

o

YY

5

551005748.0

105

2874.0

1051

)'4216sin(sin

Y = 5748 Ao

RESULT :

i) Wavelength of blue colour B = 4692 Ao

ii) Wavelength of green colour G = 5429 Ao

iii) Wavelength of yellow colour Y = 5748 Ao

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3. METRE BRIDGE

FORMULA REQUIRED :

1) When the known resistance Q is in the right gap,

Resistance of the wire 2

1

l

lQP ohm

2) When the known resistance Q is in the left gap,

Resistance of the wire 1

2

l

lQP ohm

Specific resistance of the material of the wire ml

Pr

2

Where, Q is known resistance

1l is the balancing length of P

2l is the balancing length of Q

r is the radius of the wire

l is the length of the wire

CIRCUIT DIAGRAM :

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3. METRE BRIDGE

AIM :

To determine the resistance of the given coil of wire using a meter bridge and to

calculate the specific resistance of the material of the wire.


The metre bridge, battery, key, galvanometer, known and unknown resistances,

high resistance and connecting wires.

PROCEDURE :

The connections are made as in the circuit diagram. The jockey J is pressed near

the ends A and C and if the deflections in the galvanometer are in the opposite directions,

then the circuit is correct. Now the jockey is moved over the wire and its position J is

found when there is no deflection in the galvanometer. The balancing lengths AJ = 1

and JC =2 are measured. The experiment is repeated four more times by increasing the

value of Q in steps of 1 ohm.

When the known resistance Q is in the right gap G2, the resistance of the wire

unknown resistance 2

1

l

lQP .

Then the resistances Q and P are interchanged in the gaps G1 and G2. The

unknown resistance P is calculated from the formula 1

2

l

lQP .

The length () of the coil is measured using scale and radius(r) of the coil is

measured using screw gauge. The specific resistance of the coil is calculated using the

formula

ml

Pr

2

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CALCULATIONS :

RESULT : 1) Resistance of the wire P = 4.569

2) Specific resistance of the material of the wire = 1.39 x 10 -6 m

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4. POTENTIOMETER COMPARISON OF EMFS OF TWO CELLS

FORMULA REQUIRED :

The ratio of the emfs of the two cells is

( NO UNIT )

1E emf of primary cell 1 (Leclanche cell)

2E emf of primary cell 2 (Daniel cell)

1l is the balancing length for cell 1

2l is the balancing length for cell 2

CIRCUIT DIAGRAM:

Here,

Bt = Battery, K = Key, Rh = Rheostat, G = Galvanometer, HR = High resistance

J = Jockey

2

1

2

1

l

l

E

E

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4. POTENTIOMETER COMPARISON OF EMFS OF TWO CELLS

AIM:

To compare the emfs of two primary cells using a potentiometer.


Potentiometer, Battery, Key, Rheostat, Galvanometer, High resistance,

the two given cells, Jockey and connecting wires.

PROCEDURE :

i) The connections are made according to the circuit diagram. The jockey

J i s p r e s s e d in the first and the last wire and the opposite side

deflections in the galvanometer shows that the connections are

correct.

ii) Leclanche cell is included in the circuit using the DPDT switch.

The jockey is moved over the potentiometer wire to get zero deflection in

the galvanometer. The balancing length AJ is measured as 1.

iii) Daniel cell is included in the circuit using the DPDT switch, and

the balancing length is measured as 2.

iv) The experiment is repeated for six times by moving rheostat in one

direction for changing the current in the circuit.

v) The ratio of the emf of the two cells is found from the formula

2

1

2

1

l

l

E

E

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CALCULATIONS:

RESULT : The mean ratio of emf of the two cells using the Potentiometer = 1.344 (no unit)

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5. TANGENT GALVANOMETER Determination of BH

AIM : To determine the value of the horizontal component of earths magnetic field (BH)

using the Tangent Galvanometer.


Tangent galvanometer, key, Rheostat, ammeter, commutator

and connecting wires.

FORMULA REQUIRED :

Horizontal component of earths magnetic field =0

2

tesla

0 permeability of free space

n number of turns

I current

a radius of coil

mean deflection produced in TG

CIRCUIT DIAGRAM:

Here,

Bt = battery, K = key

A = ammeter, C = commutator

TG = Tangent galvanometer, Rh = rheostat

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5. TANGENT GALVANOMETER Determination of BH

AIM : To determine the value of the horizontal component of earths magnetic field (BH)

using the Tangent Galvanometer

FORMULA :

Horizontal component of earths magnetic field =0

2

tesla

0 permeability of free space n number of turns I current a radius of coil

mean deflection produced in TG

PROCEDURE:

The battery, rheostat, ammeter and tangent galvanometer are connected as in

the circuit diagram. The coil in the tangent galvanometer is adjusted to be along the

magnetic meridian. Then the compass box alone is rotated so that the aluminum pointer

read 00 00.

The current I is passed through the circuit and the deflections of the needle are

noted as 1 and 2 . By reversing the current, the deflection are noted as 3 and 4.

The average deflection is found out.

The experiment is repeated by varying the current.

The average value of

is found out. The radius a of the coil is found out by

measuring its circumference. The number of turn n of the coil is noted.

The Horizontal component of earths magnetic induction is calculated by the

formula

=0

2

tesla.

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CALCULATIONS :

Radius (r) = 7.5 102 m

Mean

=

0.7961+0.8052+0.7725+0.7555

4 = 0.7823

To calculate the horizontal component of earths magnetic field (BH)

=0

2

= 41075 0.7823

2 7.5 102

= 49.12844

15 105 = 3.28 105

RESULT :

The horizontal component of earths magnetic field (BH) = 3.28 X 10

- 5 Tesla

.: 1

tan=

0.6

37 =

0.6

0.7536

= 0.7961

.: 2

tan=

0.7

41 =

0.6

0.7536

= 0.8052

.: 3

tan=

0.8

46 =

0.6

1.0355

= 0.7725

.: 4

tan=

0.9

50 =

0.9

1.1918

= 0.7555

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6. SONOMETER FREQUENCY OF AC

AIM: To determine the frequency of the ac main using a sonometer.

FORMULA REQUIRED:

The frequency of the ac main

=1

2

1

where, T is the tension of the sonometer wire

is the resonating length

m is the linear density of the wire

DIAGRAM :

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6. SONOMETER FREQUENCY OF AC

AIM: To determine the frequency of the ac main using a sonometer.


The sonometer, 6V AC power supply, Different loads, bar magnets, knife

edges and connecting wires.

PROCEDURE:

The ac mains voltage is brought down to 6 V by means of step down

transformer. The secondary of the transformer is connected to the ends of the

sonometer wire. A bar magnet is held below the sonometer wire at the centre. The

magnetic field is horizontal and at right angles to the length of the wire.

With 250 gms (M) added to the weight hanger, the a.c. current is passed through

the wire. Now the wire is set into forced vibrations. The length between the two

knife edges is adjusted so that it vibrates in one segment. The length between the

knife edges is measured as 1. The same procedure is repeated and 2 is measured. The

average 1 and 2 is . The experiment is repeated for the loads 500gm, 750 gm and

1000 gm.

The radius of the wire r is measured using screw gauge. The linear density of the wire

is m = r2, where is its density.

The frequency of the a.c. mains is calculated from the formula

=1

2

1

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CALCUATIONS :

Diameter of the wire d =

Radius of the wire r =

2 =

Density of the steel wire () = 7800kgm3

Linear density m = 2 =

= ( 1.72 X10-3 ) = 4.147 X 10 -2

S.No: 1

T = 0.250 9.8 = 2.45 = 1.565 = 0.332 m

=

1.565

0.332= 4.714

S.No: 2

T = 0.500 9.8 =4.90 = 2.214 = 0.455 m

=

2.214

0.455= 4.863

S.No: 3

T = 0.750 9.8 =7.35 = 2.711 = 0.559

=

2.711

0.559= 4.849

S.No: 4

T = 1.000 9.8 =9.80 = 3.130 = 0.644

=

3.13

0.644= 4.860

Mean

=

4.714+4.863+4.849+4.860

4= 4.8215

=1

2

1

=

1 4.8215

2 4.147 102

= 58.13 Hz

RESULT : The frequency of the ac main n = 58.13 Hz

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7. JUNCTION DIODE AND ZENER DIODE

FORMULA REQUIRED :

Forward resistance of the PN junction diode =V

Here,

V is the forward voltage

is the forward current.

CIRCUIT DIAGRAM:

PN - JUNCTION DIODE - forward bias

CIRCUIT DIAGRAM:

ZENER DIODE reverse bias

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7. JUNCTION DIODE AND ZENER DIODE

AIM :

a) To study the forward bias characteristics of a PN junction diode and to determine

the forward resistance of the diode.

b) To study the reverse breakdown characteristics of the zener diode.


PN-junction diode, zener diode, variable voltage source, milliammeter, voltmeter

and connecting wires.

PROCEDURE :

1) Forward Characteristic Curve of a PN junction diode:-

i) The circuit connections are made as in the diagram.

ii) The forward voltage Vf is increased from zero in steps of 0.1 V upto 1V.

iii) The corresponding values of If are noted. A graph is drawn with Vf along X-

axis and If along Y-axis. This is called forward characteristic curve.

iv) The reciprocal of the slope of this curve above the knee point is found as

forward resistance of the Diode.

v) Forward resistance =

2) Reverse breakdown characteristics of the zener diode:-

i) The circuit is wired as in the diagram.

ii) The voltage VO is increased from zero in steps of 1V upto 8V. The

corresponding values of IZ are noted.

iii) A graph is drawn with VO along X-axis and IZ along Y-axis. This is called

reverse characteristic curve.

iv) At particular voltage, the current increases enormously, this voltage is called

zener voltage (VZ)

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CALCULATIONS :

RESULT : i) The forward resistance of the PN-junction diode = 33.33 ohm.

ii) The zener breakdown voltage = 7.3 volt.

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8. COMMOMN EMITTER NPN TRANSISTOR - PART-I

AIM :

To study the characteristics of a common Emitter NPN transistor and to determine

its input impedance and output impedance.

FORMULA REQURIED :

(i) Input impedance =

(ii) Output impedance =

Here,

is the change in base emitter voltage

is the change in base current

is the change in collector emitter voltage

is the change in collector current

Input characteristics curve Output characteristics curve

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8. COMMOMN EMITTER NPN TRANSISTOR - PART-I

AIM :

To study the characteristics of a common Emitter NPN transistor and to determine

its input impedance and output impedance.


NPN transistor, milliammeter, microammeter, voltmeters, variable voltage sources

and connecting wires

PROCEDURE :

The circuit connections are made as in the diagram.

1.INPUT CHARACTERISTIC CURVE :-

i) The collector emitter voltage VCE is kept at a constant value(2 V).

ii) The base emitter voltage VBE is increased from zero in steps of 0.1 V

upto 1V. The corresponding values of IB are noted.

iii) A graph is drawn with VBE along X-axis and IB along Y-axis. This is called

input characteristic curve.

iv) The reciprocal of the slope of this curve above the knee point is found as

input impedance of the transistor.

v) The Input impedance =

2. OUTPUT CHARACTERISTIC CURVE :-

i) The base current IB is kept at a constant value.

ii) VCE is increased in steps of 0.5 V from Zero. The corresponding values of IC

are noted.

iii) A graph is drawn with VCE along X-axis and IC along Y-axis. This is called

output characteristic curve.

iv) The reciprocal of the slope of the output characteristic curve near

horizontal part gives the output impedance (r0).

Output impedance =

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RESULT :

i) The static characteristic curves of the transistor in CE configuration are drawn.

ii) The input impedance ri = 2 k

iii) The output impedance r0 = 700

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9. COMMOMN EMITTER NPN TRANSISTOR - PART-II

FORMULA REQUIRED :

i) Output impedance =

ii) Current gain =

(No unit)

Here,

is the change in base emitter voltage

is the change in base current

is the change in collector emitter voltage

is the change in collector current.

CIRCUIT DIAGRAM:

Output characteristics curve Transfer characteristics curve

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9. COMMOMN EMITTER NPN TRANSISTOR - PART-II

AIM :

To study the characteristics of a common Emitter NPN transistor and to

determine its output impedance and the current gain.


NPN transistor, milliammeter, microammeter, voltmeters, variable voltage

sources and connecting wires.

PROCEDURE:

1. OUTPUT CHARACTERISTIC CURVE :-

i) The base current IB is kept at a constant value.

ii) VCE is increased in steps of 0.5 V from Zero. The corresponding values of IC

are noted.

iii) A graph is drawn with VCE along X-axis and IC along Y-axis. This is called

output characteristic curve.

iv) The reciprocal of the slope of the output characteristic curve near

horizontal part gives the output impedance (r0).

Output impedance =

2. TRANSFER CHARACTERISTIC CURVE :-

i) The collector emitter voltage VCE is kept at a constant value (2V).

ii) IB is increased in steps of 25 A from 25 A to 100A. The corresponding

values of IC are noted.

iii) A graph is drawn with IB along X-axis and Ic along Y-axis. This is called

transfer characteristic curve.

iv) The slope of this curve gives the current gain of the transistor.

Current gain =

(no unit)

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CALCULATIONS:

RESULT :

I) The static characteristic curves of the transistor in CE configuration are drawn.

II) The output impedance r0 = 700

III) The current gain = 100 ( no unit )

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10. OPERATIONAL AMPLIFIER - Inverting amplifier

FORMULA REQUIRED :

i) Voltage gain of the inverting amplifier, =

=

(no unit)

ii) The output voltage of the inverting summing amplifier, V0 = (V1 +V2) volt

Here, V0 output voltage

Vin, V1 and V2 are the input voltages

Rf and Rs are the external resistances

CIRCUIT DIAGRAMS :

INVERTING AMPLIFIER :

SUMMING AMPLIFIER :

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10. OPERATIONAL AMPLIFIER - Inverting amplifier

AIM : To construct the following basic amplifiers using OP-AMP IC741.

i) Inverting amplifier ii) Summing amplifier


Operational amplifier(IC-741), dual power supply, 10K, 22K, 33K resistors, digital

voltmeter and connecting wires.

PROCEDURE :

INVERTING AMPLIFIER:-

i) The circuit connections are made as shown in the diagram.

ii) RS is kept as 10 K and RF as 22 K.

iii) The input voltage Vin is kept as 1V and output voltage Vo is measured

from the digital voltmeter.

iv) Then the experiment is repeated for input values Vin = 1.5 V, 2V and

2.5 V.

v) Experimental gain is found as =

vi) Theoretical gain is found from =

vii) Both the AV values are compared and found to be equal.

SUMMING AMPLIFIER:-


ii) The values of R1, R2 and RF are kept as 10 K . The input voltages are kept

as VI = 1V and V2 =2.0V and the output voltage Vo is measured using the

digital voltmeter

iii) Then the experiment is repeated for different sets of values for V1 and V2.

Theoretical output v o l t a g e i s found from V0 = - (V1 + V2).

iv) The theoretical and experimental output values are compared.

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CALCULATIONS :

1. Inverting amplifier

EXPERIMENTAL GAIN THEORETICAL GAIN

S.No : 1

=

= 2.26

1.0= 2.26

=

= 22

10= 2.20

S.No: 2

=

= 3.42

1.5= 2.28

=

= 22

10= 2.20

S.No: 3

=

= 4.54

2.0= 2.27

=

= 22

10= 2.20

S.No: 4

=

= 5.73

2.5= 2.29

=

= 22

10= 2.20

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CALCULATIONS :

Summing amplifier

S.No : 1

Experimental output Vo = - 3.08 volt

Theoretical output

Vo = (V1 + V2) = - ( 1 + 2 ) = - 3.00 V

S.No: 2


Theoretical output

Vo = (V1 + V2) = - (1.5 + 2.5) = - 4.00 V

S.No: 3


Theoretical output

Vo = (V1 + V2) = - (2 + 3) = - 5.00 V

S.No: 4


Theoretical output

Vo = (V1 + V2) = - (2.5 + 3.5) = - 6.00 V

RESULT :

i) The inverting amplifier and summing amplifier are constructed using OP-AMP and

the experimental and the theoretical outputs are compared.

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11. OPERATIONAL AMPLIFIER - Non-Inverting amplifier

FORMULA REQUIRED :

i) Voltage gain of the non-inverting amplifier, =

= 1 +

(no unit)

ii) The output voltage of the inverting summing amplifier, V0 = (V1 +V2) volt

Here, V0 output voltage

Vin, V1 and V2 are the input voltages

Rf and Rs are the external resistances

CIRCUIT DIAGRAMS :

NON-INVERTING AMPLIFIER :

SUMMING AMPLIFIER :

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11. OPERATIONAL AMPLIFIER - Non -inverting amplifier

AIM : To construct the following basic amplifiers using OP-AMP IC741.

i) Non-inverting amplifier ii) Summing amplifier


Operational amplifier(IC-741), dual power supply, 10K, 22K, 33K resistors, digital

voltmeter and connecting wires.

PROCEDURE :

1. NON-INVERTING AMPLIFIER:-

I) The circuit connections are made as shown in the diagram.

II) RS is kept as 10 K and RF as 22 K.

III) The input voltage Vin is kept as 1V and output voltage Vo is measured

from the digital voltmeter.

IV) Then the experiment is repeated for input values V in = 1.5 V, 2V and

2.5 V.

V) Experimental gain is found as =

VI) Theoretical gain is found from =

= 1 +

VII) Both the AV values are compared and found to be equal.

2. SUMMING AMPLIFIER:-


ii) The values of R1, R2 and RF are kept as 10 K . The input voltages are kept

as VI = 1V and V2 =2.0V and the output voltage Vo is measured using the

digital voltmeter

iii) Then the experiment is repeated for different sets of values for V1 and V2.

Theoretical output v o l t a g e i s found from V0 = - (V1 + V2).

iv) The theoretical and experimental output values are compared.

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CALCULATIONS : Non- Inverting amplifier

EXPERIMENTAL GAIN THEORETICAL GAIN

S.No : 1

=

=

3.26

1.0= 3.26

= 1 +

= 1 +

22

10= 3.20

S.No: 2

=

=4.86

1.5= 3.24

= 1 +

= 1 +

22

10= 3.20

S.No: 3

=

=6.56

2.0= 3.28

= 1 +

= 1 +

22

10= 3.20

S.No: 4

=

=8.05

2.5= 3.22

= 1 +

= 1 +

22

10= 3.20

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CALCULATIONS :

Summing amplifier

S.No : 1


Theoretical output

Vo = (V1 + V2) = - ( 1 + 2 ) = - 3.00 V

S.No: 2


Theoretical output

Vo = (V1 + V2) = - (1.5 + 2.5) = - 4.00 V

S.No: 3


Theoretical output

Vo = (V1 + V2) = - (2 + 3) = - 5.00 V

S.No: 4


Theoretical output

Vo = (V1 + V2) = - (2.5 + 3.5) = - 6.00 V

RESULT :

The non-inverting amplifier and summing amplifier are constructed using OP-AMP

and the experimental and the theoretical outputs are compared.

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12. INTEGRATED LOGIC GATE CIRCUITS AIM:

To study the Truth Table of integrated Logic Gates IC 7400(NAND), IC 7402 (NOR), IC 7404 (NOT), IC 7408(AND), IC 7432 (OR), and IC 7486 (EXOR)

1) For ICs 7400 (NAND), 7408(AND), 7432(OR) & 7486(EX-OR)

2) For IC 7402(NOR) 3) For NOT (7404)

FORMULA REQUIRED :

OR gate: Boolean equation

Y =A + B

AND gate: Boolean equation

Y = AB

NOT gate: Boolean equation

Y = A

NOR gate: Boolean equation Here, A, B = inputs

Y = A + B and Y = output

NAND gate: Boolean equation

Y=A B

EX-OR gate: Boolean equation

Y = AB = AB + A B

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12. INTEGRATED LOGIC GATE CIRCUITS

AIM : To study the Truth Table of integrated Logic Gates IC 7400(NAND), IC 7402

(NOR), IC 7404 (NOT), IC 7408(AND), IC 7432 (OR), and IC 7486 (EXOR)

APPARATUS REQUIRED : Logic Gates IC 7400(NAND), IC 7402 (NOR), IC 7404 (NOT), IC 7408(AND), IC 7432

(OR), and IC 7486 (EXOR)

PROCEDURE :

For NAND gate, AND gate, OR gate and EXOR gate:-

i) Power supply +5V is connected to pin 14 and ground to pin 7 of the IC.

ii) Inputs A & B are connected to pins 1 & 2 of the IC.

iii) Output pin 3 of the IC is connected to logic level indicator.

iv) Inputs A & B are kept at 0 & 0 and output LED is observed. Then the

inputs are changed as 0 & 1, 1 & 0 and 1 & 1 and the outputs are

observed each time. The inputs and outputs are tabulated in the truth

table.

v) Similarly, ICs 7408 (AND), 7432 (OR) and 7486 (EXOR) are placed on the

board and the same procedure is followed as for NAND gate and outputs

are tabulated in the truth table.

NOR gate :-

i) IC 7402 is placed on the board. Power supply and ground are connected

as before.

ii) The inputs are connected to pins 2 & 3 and the output to pin 1 of IC.

Then the same procedure is repeated and tabulation is done in the truth

table.

NOT gate :-

i) IC 7404 is placed on the board. One input A is connected to pin 1 and

the output to pin 2 of IC. I

ii) Input is kept at logic 1 and then at logic 0 and the outputs are

found and tabulated in the truth table.

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TABULAR COLUMN AND OBSERVATIONS :

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CALCULATIONS :

RESULT :

The performance of digital gates OR, AND, NOT, NAND, NOR and EX-OR gates

and their truth tables are verified using IC chips.

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Best wishes to get centum (200/200) in Physics

L L

Prepared by

B.ELANGOVAN. M.Sc., M.Ed., M.Phil.,

(Tamilnadu Dr.Radhakrishnan Best Teacher Award recipient - 2011)

P.G.Teacher in Physics,



y

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1.EM +2 Physics practical guide contents