www.Padasalai.Net
PACHAIYAPPA’S 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,
PACHAIYAPPA’S HIGHER SECONDARY SCHOOL,
KANCHIPURAM - 631501.
www.Padasalai.Net
www.Padasalai.Net
PHYSICS PRACTICAL
HIGHER SECONDARY - SECOND YEAR
S.NO
DATE
NAME OF THE EXPERIMENT
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
PAGE
www.Padasalai.Net
1. SPECTROMETER - µ OF A SOLID PRISM
FORMULA REQUIRED:
Refractive index of the material of the given prism is

Where,
AD
2
A
sin
2
sin
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
www.Padasalai.Net
1. SPECTROMETER - µ 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 T 2 .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.
www.Padasalai.Net
www.Padasalai.Net
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

A D
2
A
sin
2
sin
www.Padasalai.Net
www.Padasalai.Net
CALCULATIONS:
To find “A”
2A
= R1 R2 = 227o20’ - 127o16’
2A
= 120o4’
2A = R1 R2 = 407o20’ - 287o16’
2A
A = 60o2’
= 120o4’
A = 60o2’
Average A = 60o2’
To find “D”
D = R3 R4 = 39o44’ - 0o0’
D = R3 R4 = 219o44’ - 180o0’
D = 39o44’
D = 39o44’
Average D = 39o44’
To find “”
𝜇=
𝐴+𝐷
60°2’ +
39°44’
𝑠𝑖𝑛
2
2
=
60°2’
𝐴
𝑠𝑖𝑛
𝑠𝑖𝑛
2
2
𝑠𝑖𝑛
μ=
𝑠𝑖𝑛
𝑠𝑖𝑛
μ=
=
99°46′
2
60°2′
2
=
𝑠𝑖𝑛 49°53′
𝑠𝑖𝑛 30°1′
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 µ = 1.528 (no unit)
www.Padasalai.Net
2. SPECTROMETER – GRATING – WAVELENGTH OF COMPOSITE LIGHT
FORMULA REQUIRED :
The wavelength () of a spectral line using the grating is given by

sin
mN
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 :
www.Padasalai.Net
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 R 2. 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

sin
mN
www.Padasalai.Net
www.Padasalai.Net
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 :
sin B sin( 13o34' )
0.2346
B 


mN
1 5 105
5 105

0.04692 105 m
λB = 4692 Ao
The wavelength of the Green spectral line :
sinG sin(15o 45' )
0.2714
G 


 0.05429 105 m
5
5
mN
1 5 10
5 10
λG = 5429 Ao
The wavelength of the Yellow spectral line :
sinY sin(16o 42' )
0.2874
Y 


 0.05748  10 5 m
5
5
mN
1  5  10
5  10
λY = 5748 Ao
RESULT :
B = 4692 Ao
i)
Wavelength of blue colour
ii)
Wavelength of green colour G = 5429 Ao
iii)
Wavelength of yellow colour Y = 5748 Ao
www.Padasalai.Net
3. METRE BRIDGE
FORMULA REQUIRED :
1) When the known resistance Q is in the right gap,
Resistance of the wire P  Q
l1
ohm
l2
2) When the known resistance Q is in the left gap,
Resistance of the wire P  Q
l2
ohm
l1
Specific resistance of the material of the wire  
Where,
 r2X
R is known resistance
l1 is the balancing length of P
l 2 is the balancing length of Q
r is the radius of the wire
l is the length of the wire
CIRCUIT DIAGRAM :
l
m
www.Padasalai.Net
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 = ℓ1
and JC =ℓ2 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 P  Q
l1
.
l2
Then the resistances Q and P are interchanged in the gaps G1 and G2. The
unknown resistance P is calculated from the formula P  Q
l2
.
l1
The length (ℓ) 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  
 r2X
l
m
www.Padasalai.Net
www.Padasalai.Net
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
www.Padasalai.Net
4.
POTENTIOMETER – COMPARISION OF EMFS OF TWO CELLS
FORMULA REQUIRED :
The ratio of the emfs of the two cells is
E1 l1

E2 l2
( NO UNIT )
E1 emf of primary cell 1 (Leclanche cell)
E 2 emf of primary cell 2 (Daniel cell)
l1 is the balancing length for cell 1
l 2 is the balancing length for cell 2
CIRCUIT DIAGRAM:
Here,
Bt = Battery, K = Key, Rh = Rheostat, G = Galvanometer, HR = High resistance
J = Jockey
www.Padasalai.Net
4.
POTENTIOMETER – COMPARISION 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 ℓ1.
iii)
Daniel cell is included in the circuit using the DPDT switch, and
the balancing length is measured as ℓ2.
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
E1 l1

E2 l2
www.Padasalai.Net
www.Padasalai.Net
CALCULATIONS:
RESULT :
The mean ratio of emf of the two cells using the Potentiometer =
1.344 (no unit)
www.Padasalai.Net
5. TANGENT GALVANOMETER – Determination of BH
AIM :
To determine the value of the horizontal component of earth’s magnetic field (B H)
using the Tangent Galvanometer.
APPARATUS REQUIRED :
Tangent galvanometer, key, Rheostat, ammeter, commutator
and connecting wires.
FORMULA REQUIRED :
Horizontal component of earth’s magnetic field𝐵𝐻 =
𝜇0𝑛
𝐼
2𝑎
𝑡𝑎𝑛𝜃
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
tesla
www.Padasalai.Net
5. TANGENT GALVANOMETER – Determination of BH
AIM :
To determine the value of the horizontal component of earth’s magnetic field (BH)
using the Tangent Galvanometer
FORMULA :
Horizontal component of earth’s magnetic field𝐵𝐻 =
0
n
I
a

–
–
–
–
–
𝜇0𝑛
𝐼
2𝑎
𝑡𝑎𝑛𝜃
tesla
permeability of free space
number of turns
current
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
𝐼
𝑡𝑎𝑛𝜃
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 earth’s magnetic induction is calculated by the
formula
𝐵𝐻 =
𝜇0𝑛
𝐼
2𝑎
𝑡𝑎𝑛𝜃
tesla.
www.Padasalai.Net
www.Padasalai.Net
CALCULATIONS :
Radius (r) =
7.5  10–2 m
𝑆. 𝑁𝑂: 2
𝑆. 𝑁𝑂: 1
𝐼
0.7
0.6
=
=
tan 𝜃
𝑡𝑎𝑛41°
0.7536
𝐼
0.6
0.6
=
=
tan 𝜃
𝑡𝑎𝑛37°
0.7536
= 0.8052
= 0.7961
𝑆. 𝑁𝑂: 3
𝑆. 𝑁𝑂: 4
𝐼
0.8
0.6
=
=
tan 𝜃
𝑡𝑎𝑛46°
1.0355
𝐼
0.9
0.9
=
=
tan 𝜃
𝑡𝑎𝑛50°
1.1918
= 0.7725
Mean
𝐼
𝑡𝑎𝑛 𝜃
=
= 0.7555
0.7961+0.8052+0.7725+0.7555
4
= 0.7823
To calculate the horizontal component of earth’s magnetic field (BH)
𝐵𝐻 =
=
=
𝜇 0𝑛
𝐼
2𝑎
𝑡𝑎𝑛𝜃
4𝜋 ×10 −7 ×5× 0.7823
2× 7.5 ×10 −2
49.12844
𝑋 10−5 𝑡𝑒𝑠𝑙𝑎 = 3.28𝑋 10−5 𝑡𝑒𝑠𝑙𝑎
15
RESULT :
The horizontal component of earth’s magnetic field (BH) = 3.28 X 10 - 5 Tesla
www.Padasalai.Net
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
𝑇
2
𝑙
𝑛= ×
×
1
𝑚
where, T is the tension of the sonometer wire
ℓ is the resonating length
m is the linear density of the wire
DIAGRAM :
www.Padasalai.Net
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 ℓ1. The same procedure is repeated and ℓ2 is measured. The
average ℓ1 and ℓ2 is ℓ. 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
𝑛=
1
𝑇
1
×
×
2
𝑙
𝑚
www.Padasalai.Net
www.Padasalai.Net
CALCUATIONS :
Diameter of the wire d =
Radius of the wire r =
𝑑
2
=
Density of the steel wire () = 7800kgm–3
Linear density m = 𝜋𝑟 2 𝜌 =
𝑚 = ( 1.72 X10-3 ) ½ = 4.147 X 10 -2
S.No: 1
T = 0.250 9.8 = 2.45
𝑇 = 1.565
ℓ= 0.332 m
𝑇
1.565
=
= 4.714
𝑙
0.332
𝑇 = 2.214
ℓ= 0.455 m
𝑇
2.214
=
= 4.863
𝑙
0.455
𝑇 = 2.711
ℓ= 0.559
𝑇
2.711
=
= 4.849
𝑙
0.559
𝑇 = 3.130
ℓ= 0.644
𝑇
3.13
=
= 4.860
𝑙
0.644
S.No: 2
T = 0.500 9.8 =4.90
S.No: 3
T = 0.750 9.8 =7.35
S.No: 4
T = 1.000 9.8 =9.80
𝑇
Mean
𝑙
=
4.714+4.863+4.849+4.860
4
1
𝑇
2
𝑙
𝑛= ×
×
1
𝑚
=
= 4.8215
1 𝑋 4.8215
2 𝑋 4.147 𝑋 10 −2
= 58.13 Hz
RESULT :
The frequency of the ac main n =
58.13 Hz
www.Padasalai.Net
7. JUNCTION DIODE AND ZENER DIODE
FORMULA REQUIRED :
Forward resistance of the PN junction diode 𝑅𝑓
Here,
∆V𝑓 is the forward voltage
∆𝐼𝑓 is the forward current.
CIRCUIT DIAGRAM:
PN - JUNCTION DIODE - forward bias
CIRCUIT DIAGRAM:
ZENER DIODE – reverse bias
=
∆V 𝑓
∆𝐼𝑓
Ω
www.Padasalai.Net
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 Xaxis 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 𝑟𝑖 =
∆𝑉𝑓
∆𝐼𝑓
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)
www.Padasalai.Net
www.Padasalai.Net
CALCULATIONS :
RESULT :
i)
The forward resistance of the PN-junction diode = 33.33 ohm.
ii)
The zener breakdown voltage =
7.3 volt.
www.Padasalai.Net
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 𝑟𝑖 =
(ii)
Output impedance 𝑟𝑜 =
∆𝑉𝐵𝐸
∆𝐼𝐵
∆𝑉𝐶𝐸
∆𝐼𝐶
Ω
Ω
Here,
∆𝑉𝐵𝐸 is the change in base emitter voltage
∆𝐼𝐵
is the change in base current
∆𝑉𝐶𝐸 is the change in collector emitter voltage
∆𝐼𝐶
is the change in collector current
Input characteristics curve
Output characteristics curve
www.Padasalai.Net
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 V CE 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 𝑟𝑖 =
∆𝑉𝐵𝐸
∆𝐼𝐵
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 I C
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 𝑟𝑜 =
∆𝑉𝐶𝐸
∆𝐼𝐶
www.Padasalai.Net
www.Padasalai.Net
RESULT :
i)
The static characteristic curves of the transistor in CE configuration are drawn.
ii)
The input impedance ri
iii)
The output impedance r0 =
=
2 kΩ
700 Ω
www.Padasalai.Net
9. COMMOMN EMITTER NPN TRANSISTOR - PART-II
FORMULA REQUIRED :
i)
Output impedance 𝑟𝑜 =
ii)
Current gain
𝛽 =
∆𝑉𝐶𝐸
∆𝐼𝐶
∆𝐼𝐶
∆𝐼𝐵
Ω
(No unit)
Here,
∆𝑉𝐵𝐸 is the change in base emitter voltage
∆𝐼𝐵
is the change in base current
∆𝑉𝐶𝐸 is the change in collector emitter voltage
∆𝐼𝐶
is the change in collector current.
CIRCUIT DIAGRAM:
Output characteristics curve
Transfer characteristics curve
www.Padasalai.Net
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 I C
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 𝑟𝑜 =
∆𝑉𝐶𝐸
∆𝐼𝐶
Ω
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 µA from 25 µA to 100µA. 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 𝛽 =
∆𝐼𝐶
∆𝐼𝐵
(no unit)
www.Padasalai.Net
www.Padasalai.Net
CALCULATIONS:
RESULT :
I)
The static characteristic curves of the transistor in CE configuration are drawn.
II)
The output impedance r0 =
III)
The current gain
=
700 Ω
100 ( no unit )
www.Padasalai.Net
10. OPERATIONAL AMPLIFIER - Inverting amplifier
FORMULA REQUIRED :
𝑉𝑂
Voltage gain of the inverting amplifier, 𝐴𝑉 =
ii)
The output voltage of the inverting summing amplifier, V0 = – (V1 +V2) volt
Here,
𝑉𝑖𝑛
V0 output voltage
Vin, V1 and V2 are the input voltages
Rf and Rs are the external resistances
CIRCUIT DIAGRAMS :
INVERTING AMPLIFIER :
SUMMING AMPLIFIER :
=−
𝑅𝑓
i)
𝑅𝑠
(no unit)
www.Padasalai.Net
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 KΩ and RF as 22 KΩ.
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.
𝑉𝑂
v)
Experimental gain is found as𝐴𝑉 =
vi)
Theoretical gain is found from 𝐴𝑉 = −
vii)
Both the AV values are compared and found to be equal.
𝑉𝑖𝑛
𝑅𝑓
𝑅𝑠
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 Ω. 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 V 1 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.
www.Padasalai.Net
www.Padasalai.Net
CALCULATIONS :
1. Inverting amplifier
EXPERIMENTAL GAIN
S.No : 1
𝑉𝑂
2.26
𝐴𝑉 =
= −
= −2.26
𝑉𝑖𝑛
1.0
THEORETICAL GAIN
𝐴𝑉 = −
𝑅𝑓
22
=−
= −2.20
𝑅𝑠
10
𝐴𝑉 = −
𝑅𝑓
22
=−
= −2.20
𝑅𝑠
10
𝐴𝑉 = −
𝑅𝑓
22
=−
= −2.20
𝑅𝑠
10
𝐴𝑉 = −
𝑅𝑓
22
=−
= −2.20
𝑅𝑠
10
S.No: 2
𝑉𝑂
3.42
𝐴𝑉 =
= −
= −2.28
𝑉𝑖𝑛
1.5
S.No: 3
𝑉𝑂
4.54
𝐴𝑉 =
= −
= −2.27
𝑉𝑖𝑛
2.0
S.No: 4
𝑉𝑂
5.73
𝐴𝑉 =
= −
= −2.29
𝑉𝑖𝑛
2.5
www.Padasalai.Net
www.Padasalai.Net
CALCULATIONS :
Summing amplifier
Theoretical output
S.No : 1
Experimental output Vo = - 3.08 volt
Vo = – (V1 + V2) = - ( 1 + 2 )
= - 3.00 V
Theoretical output
S.No: 2
Experimental output Vo = - 4.05 volt
Vo = – (V1 + V2) = - (1.5 + 2.5) = - 4.00 V
Theoretical output
S.No: 3
Experimental output Vo = - 5.09 volt
Vo = – (V1 + V2) = - (2 + 3)
= - 5.00 V
Theoretical output
S.No: 4
Experimental output Vo = - 6.06 volt
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.
www.Padasalai.Net
11. OPERATIONAL AMPLIFIER -
Non-Inverting amplifier
FORMULA REQUIRED :
𝑉𝑂
Voltage gain of the non-inverting amplifier, 𝐴𝑉 =
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 :
𝑉𝑖𝑛
=1+
𝑅𝑓
i)
𝑅𝑖𝑛
(no unit)
www.Padasalai.Net
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 KΩ and RF as 22 KΩ.
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.
𝑉𝑂
V)
Experimental gain is found as 𝐴𝑉 =
VI)
Theoretical gain is found from 𝐴𝑉 =
VII)
Both the AV values are compared and found to be equal.
𝑉𝑖𝑛
𝑉𝑂
𝑉𝑖𝑛
=1+
𝑅𝑓
𝑅𝑖𝑛
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 Ω. 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 V 1 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.
www.Padasalai.Net
www.Padasalai.Net
CALCULATIONS :
Non- Inverting amplifier
EXPERIMENTAL GAIN
S.No : 1
𝑉
3.26
𝐴𝑉 = 𝑉 𝑂 = 1.0 = 3.26
THEORETICAL GAIN
𝐴𝑉 = 1 +
𝑖𝑛
S.No: 2
𝑉𝑂
4.86
𝐴𝑉 =
=
= 3.24
𝑉𝑖𝑛
1.5
𝐴𝑉 = 1 +
S.No: 3
𝑉𝑂
6.56
𝐴𝑉 =
=
= 3.28
𝑉𝑖𝑛
2.0
𝐴𝑉 = 1 +
S.No: 4
𝑉𝑂
8.05
𝐴𝑉 =
=
= 3.22
𝑉𝑖𝑛
2.5
𝐴𝑉 = 1 +
𝑅𝑓
𝑅𝑠
𝑅𝑓
𝑅𝑠
𝑅𝑓
𝑅𝑠
𝑅𝑓
𝑅𝑠
22
= 1 + 10 = 3.20
22
= 1 + 10 = 3.20
22
= 1 + 10 = 3.20
22
= 1 + 10 = 3.20
www.Padasalai.Net
www.Padasalai.Net
CALCULATIONS :
Summing amplifier
Theoretical output
S.No : 1
Experimental output Vo = - 3.08 volt
Vo = – (V1 + V2) = - ( 1 + 2 ) = - 3.00 V
Theoretical output
S.No: 2
Experimental output Vo = - 4.05 volt
Vo = – (V1 + V2) = - (1.5 + 2.5) = - 4.00 V
Theoretical output
S.No: 3
Experimental output Vo = - 5.09 volt
Vo = – (V1 + V2) = - (2 + 3) = - 5.00 V
Theoretical output
S.No: 4
Experimental output Vo = - 6.06 volt
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.
www.Padasalai.Net
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 IC’s 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
Y=A+B
NAND gate:
Boolean equation
Y=A ∙ B
EX-OR gate:
Boolean equation
Y = A⨁B = AB + AB
Here, A, B = inputs
and
Y = output
www.Padasalai.Net
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.
www.Padasalai.Net
TABULAR COLUMN AND OBSERVATIONS :
www.Padasalai.Net
TABULAR COLUMN AND OBSERVATIONS :
www.Padasalai.Net
TABULAR COLUMN AND OBSERVATIONS :
www.Padasalai.Net
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.
www.Padasalai.Net
“ 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,
PACHAIYAPPA’S HIGHER SECONDARY SCHOOL,
KANCHIPURAM - 631501.
y