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,
PACHAIYAPPAS HIGHER SECONDARY SCHOOL,
KANCHIPURAM - 631501.
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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.
APPARATUS REQUIRED :
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.
APPARATUS REQUIRED :
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.
APPARATUS REQUIRED :
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.
APPARATUS REQUIRED :
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.
APPARATUS REQUIRED :
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.
APPARATUS REQUIRED :
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.
APPARATUS REQUIRED :
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.
APPARATUS REQUIRED :
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
APPARATUS REQUIRED :
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:-
i) The circuit connections are made as shown in the diagram.
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
Experimental output Vo = - 4.05 volt
Theoretical output
Vo = (V1 + V2) = - (1.5 + 2.5) = - 4.00 V
S.No: 3
Experimental output Vo = - 5.09 volt
Theoretical output
Vo = (V1 + V2) = - (2 + 3) = - 5.00 V
S.No: 4
Experimental output Vo = - 6.06 volt
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
APPARATUS REQUIRED :
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:-
i) The circuit connections are made as shown in the diagram.
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
Experimental output Vo = - 3.08 volt
Theoretical output
Vo = (V1 + V2) = - ( 1 + 2 ) = - 3.00 V
S.No: 2
Experimental output Vo = - 4.05 volt
Theoretical output
Vo = (V1 + V2) = - (1.5 + 2.5) = - 4.00 V
S.No: 3
Experimental output Vo = - 5.09 volt
Theoretical output
Vo = (V1 + V2) = - (2 + 3) = - 5.00 V
S.No: 4
Experimental output Vo = - 6.06 volt
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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TABULAR COLUMN AND OBSERVATIONS :
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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,
PACHAIYAPPAS HIGHER SECONDARY SCHOOL,
KANCHIPURAM - 631501.
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