LAB MANUAL
II B.Sc. Allied Physics
List of Experiments
S.No.
Experiment
1.
Compound Pendulum
2.
Torsion Pendulum
3.
Young’s modulus- uniform bending using microscope
4.
Coefficient of Viscosity –Poiseuille’s flow method
5.
Dispersive power of a prism
6.
Newton`s rings
7.
Air wedge
8.
Zener diode characteristics
9.
Grating N and λ
10.
Sonometer- Frequency of AC mains
11.
Logic gates using discrete components
12.
Logic gates using ICs
2
1. Compound Pendulum
Aim: To determine
(i)
the acceleration due to gravity(g) and
(ii)
the radius of gyration of the compound pendulum about its centre of gravity.
Apparatus required: Compound Pendulum, stop clock, meter scale, etc…
Formula:
(i)
The acceleration due to gravity is given by
𝑔 = 4𝜋 2
𝐿
𝑇2
𝑚/𝑠 2
g = acceleration due to gravity in metre/second2
where,
L = length of the equivalent simple pendulum in metre
T= time period in seconds
(ii)
The radius of gyration is calculated to be
𝐾=
𝑃𝑄
𝑐𝑚
2
Where PQ = points corresponding to minimum periods in the graph.
Procedure:
(i)
The bar pendulum is suspended by passing the knife edge through the first hole
from end A (say).
(ii)
It is allowed to oscillate with small amplitude without wobbling.
(iii)
The time period for 20 oscillations is noted and the period of oscillation is
calculated.
(iv)
In this way, the periods are determined by suspending the pendulum through all
the holes on one side of the centre of gravity of the bar.
(v)
The bar is then inverted and suspended by passing the knife edge through different
holes from the other end. But the distance of knife edge is measured from the
same end (A) of the bar.
(vi)
The readings are tabulated (Table 1).
(vii)
A graph is drawn taking distance of knife edge along x-axis and the period of
oscillation along y-axis.
(viii)
The graph is symmetrical about the line passing through the centre of gravity (c)
parallel to y-axis. It consists of two similar curves on either side of c.
3
(ix)
A line is drawn parallel to the x-axis corresponding to a time period cutting the
curve at four points A, B, C and D.
(x)
Then the length of the equivalent simple pendulum is calculated using the relation
L =(AC+BD)/2
(xi)
A number of such horizontal lines are drawn corresponding to different periods are
found and tabulated (Table 2). The mean value of L/T 2 is substituted in theformula
and the acceleration due to gravity is determined.
(xii)
From the graph , the radius of gyration is calculated using the relation K=PQ/2
Model graph:
4
Table 1
S.No.
Distance from
marked end(A)
Time for 20 oscillations (s)
Trial 1
Trial 2
Time for one
Mean
oscillation(s)
in cm
Table 2 (from graph)
S.No.
Period (T) in
T2
L=(AC+BD)/2
sec
(S2)
(m)
L/T2
g=4π2(L/T2)
(m/s2)
Result:
(i)
Acceleration due to gravity
(ii)
Radius of gyration
=____________ m/s2
= ____________ m
5
2. Torsion Pendulum
Aim: To determine
(i)
the rigidity modulus (G) of the material of the wire and
(ii)
the Moment of inertia(I) of a given disc by torsional oscillations.
Apparatus required:
Torsion pendulum, stop clock, two equal cylindrical masses, screw gauge, meter scale,
etc…
Formula:
(i)
𝑮=
(ii)
𝑰=
𝟏𝟔𝝅𝒍𝒎(𝒅𝟐𝟐 −𝒅𝟐𝟏)
𝒂𝟒(𝑻𝟐𝟐 −𝑻𝟐𝟏 )
𝟐𝒎(𝒅𝟐𝟐 −𝒅𝟐𝟏 )𝑻𝟐𝑶
(𝑻𝟐𝟐−𝑻𝟐𝟏 )
in N/m2
𝒊𝒏 𝑲𝒈/𝒎𝟐
Where G is the rigidity modulus of the material of the wire in N/m2
l is the length of the suspension wire in meter.
m is the mass of the cylinders placed on the disc in Kg.
d1 is the distance between the chuck and the cylinder when kept closed in meter
d2 is the distance between the chuck and the cylinder when kept at the rim of the
disc in meter.
T0 is the time period when no masses are placed on the disc in second
T1 is the time period when equal masses are placed at a distance d 1 in second
T2 is the time period when equal masses are placed at a distance d2 in second
a is the radius of the wire in meter.
6
Torsion pendulum setup to find the period of oscillations T 0, T1 and T2
Procedure:
(i) The mass (m) of each of the equal cylindrical masses is found by weighing.
(ii) Then the two equal cylindrical masses are placed on the disc symmetrically on either
side, close to the suspension wire at equal distance d1(closest distance). Now the disc is
rotated through an angle and is then released. The time taken for 10 oscillations is noted
in two trials. The mean time period (T1) is calculated.
(iii)The two equal masses are now moved to the extreme ends of the disc at equal distance
d2 from the centre of the disc. The torsional oscillations are executed by twisting the
disc and the corresponding period (T2) is determined.
(iv) The two equal masses are now removed and the time period T 0 is found when the disc
alone is vibrating.
(v) The length(l) of the wire between two chucks is measured.
(vi) The radius(a) of the wire is determined accurately with a screw gauge.
(vii) By substituting all these values in the above-mentioned formula, the rigidity
modulus (G) of the wire and moment of inertia (I) of the disc are calculated.
Table 1 To find the period of oscillation
Position of the
symmetrical
Time for 10 oscillations (s)
Trial 1
Trial 2
Time for one oscillation
mean
(s)
weights
Without weights
(T0)
(d0)
With weights at
(T1)
distance d1
With weights at
(T2)
distance d2
Observation :
Length(l) of the wire =_________ cm
Mass of the cylinder(m) = _________ kg
7
Table 2 To find the radius (a) of the wire using screw gauge
L.C =0.01 mm
S.No.
PSR(mm)
Z.E = ____ div
HSC(div)
Z.C = ____ div
CHSC=HSC+Z.C Total
(div)
reading
=(PSR+CHSCXL.C)
(mm)
Mean diameter = _____ x10-3m
Radius (a)
= _____ x10-3 m
Result:
(i)
The rigidity modulus of the material of the wire, G = _____ Nm-2
(ii)
Moment of inertia of the given disc, I
= _____ Kgm2
8
3. Young’s Modulus by Uniform bending
Aim:
To determine the Young’s modulus of the material of the give beam subjecting it to
Uniform bending
Apparatus required:
A uniform rectangular beam, two knife edges, two weight hanger with slotted weights,
pin, microscope, Vernier caliper, screw gauge
Formula:
𝐸=
3𝑚𝑔𝑎𝑙2
2𝑏𝑑3 𝑦
Nm-2
Where
m - Mass at each end of the bar.
a - Distance between the point of suspension of the mass and nearer knife edge.
g - Acceleration due to gravity.
l is the length of the bar between the knife edges.
y - Elevation of the midpoint of the bar for a mass m at each end.
b is the breadth of the bar and
d is the thickness of the bar.
Procedure:
(i) The bar is placed symmetrically on two knife edges.
9
(ii) Two weight hangers are suspended at equal distance from the knife edges.
(iii)The distance l between knife edges and distance a of the weight hanger from knife
edges are measured.
(iv) A pin is fixed vertically at the midpoint of the bar with its pointed end upwards.
(v) The microscope is arranged in front of the pin and focused at the tip of the pin.
(vi) The slotted weights are added one by one on both the weight hangers and removed
one by one a number of times, so that the bar is brought into an elastic mood.
(vii) With the some "dead load" W on each weight hanger, the microscope is adjusted so
that the image of the tip of the pin coincides with the point of intersection of cross
wires.
(viii) The reading of the main scale and vernier of microscope are taken.
(ix) Weights are added one by one and corresponding reading are taken.
(x) From these readings, the mean elevation (y) of the mid-point of the bar for a given
mass is determined.
(xi) The breadth of the bar (b) is measured by using vernier calipers and thickness of the
bar (d) is measured by using screw gauge. Hence calculate the Young's modulus of
the material bar.
Table1:To find the elevation ‘y’ using microscope
S.No
Load x103
kg
Microscope reading
Loading
Mean
Elevation
Microscope yx10-2 m
Unloading
reading
x10-2m
1
W
2
W+500
3
W+1000
4
W+1500
5
W+2000
MSR VSR TR
MSR VSR TR
cm
cm
Mean y=__________
10
Table 2: Breadth of the beam using Vernier Caliper
Zero error=
S.No
Zero correction=
MSR
VSC
Least Count=0.01cm
VSR
TR=MSR+(CVSRXLC)
cm
Table 3: Thickness of the beam using Screw gauge
Zero error=
S.No
Zero correction=
PSR
HSC
Least Count=0.01mm
CHSC
TR=PSR+(HSCXLC)
mm
Result:
The Young’s modulus of the given material of the beam by Uniform bending =_______Nm-2.
11
4. Coefficient of Viscosity –Poiseuille`s flow method
Aim:
To determine the coefficient of viscosity of the given liquid by Poiseuille`s flow
method.
Apparatus required:
Burette, stop clock, given liquid, capillary tube, vernier microscope, meter scale,
beaker , etc…
Formula:
 =
𝛑𝛒𝐠𝐚𝟒 𝐡𝐭
( ) 𝐢𝐧 𝐍𝐬𝐦−𝟐
𝟖𝒍
𝐕
Where  is the coefficient of viscosity of the given liquid in Nsm-2
ρ is the density of the given liquid (1000 Kg/m3)
g is the acceleration due to gravity in m/s2
a is the radius of the capillary tube in m.
l is the length of the capillary tube in m.
h is the driving height in m.
t is the time of flow in S.
V is the volume of the given liquid in m3
Diagram:
12
PROCEDURE
(i)
The burette is fixed vertically in the stand and filled fully with the liquid for which
the viscosity is to be measured.
(ii)
At the lower end of the burette, a capillary tube is attached using a rubber tube.
(iii)
The capillary tube is placed on a table such that the tube is in horizontal position
(iv)
This arrangement allows the liquid to flow freely through the capillary tube without
the influence of gravity.
(v)
The knob in the bottom of the burette is opened and the water is allowed to drain
through the capillary tube. When the liquid level reaches zero-mark
level,
the
stop clock is started.
(vi)
The time taken to reach (say) 5, 10…50 cc is noted. Then the time interval for each
5 cc, namely 0-5, 5-10 … 45-50 is found and tabulated.
(vii)
The driving height h for every 5 cc is calculated by measuring the height from the
centre of the capillary tube to the middle of each 5 cc.
(viii)
The mean value of (ht/V) is calculated. The radius of the capillary tube is measured
using a travelling microscope.
(ix)
The length (l) of the capillary tube is measured using a meter scale.
(x)
Substituting the values in the given formula, the coefficient of viscosity can be
calculated.
Table 1: To find ht/V
S.NO.
Burette
Time
reading
taken(s)
Range
Time of
Pressure
ht/V in
flow(t) in
height(h) in
s/m2
s
10-2m
Average= ______
13
Table 2 :
RESULT
The co-efficient of viscosity of the given liquid, = ……….………. Nsm-2
14
5. Spectrometer- Dispersive power of Prism
Aim:
To determine the dispersive power of the material of the given prism for any two
marked lines of the mercury spectrum.
Apparatus: Spectrometer, prism, mercury vapour lamp, etc…
Formula:
The dispersive power of the prism is given by,
𝛚=
(𝟐 −𝟏 )
( − 𝟏)
Where,
=
𝐀+𝐃
)
𝟐
𝐀
𝐬𝐢𝐧( 𝟐 )
𝐬𝐢𝐧(
and
 =
(𝟏 +𝟐 )
𝟐
ω is the dispersive power of the material of the prism (no unit)
 is the refractive index of the material of the prism (no unit)
A is the angle of the prism in deg.
D is the angle of minimum deviation in deg.
1, 2 are the refractive indices of the particular lines chosen (no unit)
15
Procedure:
(i) Preliminary adjustments are made in the Spectrometer.
(ii) The collimator slit is illuminated by the light from mercury vapour lamp.
(iii) Light from the collimator is allowed to incident on the faces of the prism at an angle
almost equal to 90ᵒ. The image of the slit as reflected by the prism is viewed with the
naked eye.
(iv) The colours (VIBGYOR) obtained are viewed with the telescope.
(v) The crosswire is made to coincide with any one colour(say, green) and the prism
table is rotated. The reflected image tends to move towards the direct ray.
(vi) The image is followed with the telescope. At one stage, the image slow down and
turns back. This is the minimum deviation position.
16
(vii) Now the telescope is turned towards the initial position and the crosswire is made to
coincide with each colours obtained and the readings of both the verniers are taken.
(viii) The prism is removed now and the direct ray image is viewed through the telescope.
The readings are noted.
(ix) Substituting the values of  , D and A (=60ᵒ) , the dispersive power of the prism is
calculated from the given formula.
Table 1: To find the refractive index
Direct ray reading : VA = _______ VB = ___________
S.N
Colou
o
r
Spectrometer reading

Angle of
minimum
deviation
Vernier A
Vernier B
=
d1
d2
D=
M.S.
VS
T.
M.S.
V.S.
T.
de
de
(d 1
R
C
R
R
C
R
g
g
+d
deg
min de
deg
min
de
g
𝐀+𝐃
)
𝟐
𝐀
𝐬𝐢𝐧 ( )
𝟐
𝐬𝐢𝐧 (
2)/2
(deg
g
)
Table 2: To find the dispersive power(ω)
s.no.
lines
𝛚=
(𝟐 −𝟏 )
( − 𝟏)
Result:
The dispersive power of the given prism = __________.
17
6. Newton`s Rings
Aim: To determine the radius of curvature of the given Plano convex lens by Newton’s rings
method.
Apparatus: Newton’s rings apparatus, travelling microscope, sodium vapour
lamp, plano convex lens, etc…,
Formula:
R = (D2 n+m –D2n ) /4mλ in metre
Where, R = Radius of curvature of the lens in meter.
λ = Wavelength of sodium light =5893 Å
m = order of the rings
Dn+m =Diameter of the (n+m)th dark ring in metre.
Dn = Diameter of the n th dark ring in metre.
Experimental set up:
Procedure:
(i)
The apparatus set up is as shown in the figure.
(ii)
The travelling microscope is placed such that its objective is directly above the
Plano–convex lens.
(iii)
The inclined glass plate is tilted so that the light rays from the monochromatic
source are reflected on the plane glass plate and the field of view is brightly
illuminated.
(iv)
The focus of the microscope is adjusted such that the Newton’s rings are clearly
seen.
(v)
The traveling microscope is adjusted such that the point of intersection of the
cross wires coincides with the center of the ring system.
18
(vi)
In this system, the central ring will be a dark ring. A good number of clear rings
are available on either side of the central spot.
(vii)
Now the microscope is moved towards the left from centre, taking the first clear
dark ring as “n” and counting the rings upto , say n+18 th dark ring and fixed at
that position , where the vertical cross wire is tangential.
(viii)
Note the reading of the position of the microscope in horizontal scale.
(ix)
Then the microscope is moved towards right and the reading of every ring is
noted down till the 18th dark ring on the other side is reached.
The readings are entered in the tabular column and the mean value of D2n+m –D2n
(x)
is calculated.
(xi)
Knowing the wavelength of source light, the radius of curvature of the Plano
convex lens is calculated using the given formula.
Tabular column:
L.C. =0.001 cm
Orde
Microscope reading(cm)
Diamete (Diameter
r of
)2
D2n
(cm)2
(cm)2
r (cm)
Left
the
rings
Right
M.S.
V.S.
T.R=(M
M.S.
V.S.
T.R=(M.S
R
C
.S.R+V.
R
C
.R+V.S.C
(cm)
(div)
S.C
(cm)
(div)
XL.C)
XL.C)
D2n+m –
(cm)
(cm)
Result:
The radius of curvature of the given plano convex lens = __________ m.
19
7. Air Wedge
Aim:
To determine the thickness/ diameter of the thin wire by observing the interference
bands formed using an air wedge.
Apparatus required:
Two optically plane glass plates, plane glass plate, sodium light, thin wire, vernier
microscope etc.
Formula:
d= λl/2β meter, where
d= the diameter of the wire used in the air wedge
λ= wavelength of the light used
l= distance of the wire from the line of contact of the two plates of the air wedge
β= bandwidth
20
21
Result:
Thickness of the wire=_________ m.
22
8. Zener diode characteristics
Aim:
To study the current voltage characteristics of the given Zener diode.
Apparatus required:
Zener diode, D.C power supply, Voltmeter, milliammeter, micro ammeter, connecting
wires, bread board, etc…
Procedure:
(i) The circuit is connected in forward bias as shown in the fig.1.
(ii) Vary the source voltage gradually and note down the voltage across the diode and
note down the corresponding current values
(iii) Remove the circuit connections and reconnect as shown in fig. 2, which is the reverse
bias circuit.
(iv) Change the values of voltage in steps of 1 V and the corresponding changes in current
are measured and the readings are tabulated.
(v) Note down the Zener break down voltage.
(vi) Plot the curve with voltage along x-axis and current along y –axis.
Model graph:
23
Table 1: Forward bias
Vf in volts
If in mA
24
Table 2: Reverse bias
VR in volts
IR in A
Result: i) The current-voltage characteristics of the given Zener diode are studied. ii) The
break down voltage of the Zener diode = ………….. V
25
9. Grating N and λ
Aim:
(i) To determine the number of lines per meter of the given grating.
(ii) To determine the wavelength of the prominent lines of mercury spectrum.
Apparatus required:
Mercury vapour lamp, grating, spectrometer and sodium vapour lamp.
Formulae:
(i)
The number of lines per meter of the given grating,
N=
𝑠𝑖𝑛𝜃
𝑚𝜆
lines/m
where, λ – wavelength of the light from sodium vapour lamp(Ao) [λ = 5893 Ao].
(ii)
The wavelength of the prominent lines of
mercury spectrum,
𝑠𝑖𝑛𝜃
𝑚𝑁
where, λ – wavelength of the prominent lines of mercury spectrum (Ao),
𝜆=
N – number of lines per meter of grating (lines / m)
m – order of spectrum
θ – angle of diffraction (deg),
Procedure:
I) Adjustment of grating for normal incidence:
(i) Preliminary adjustments of the spectrometer are made. The grating (G) is mounted
on the grating table with its ruled surface facing the collimator.
(ii) The slit is illuminated by a source of light (mercury vapor lamp) and is made to
coincide with the vertical cross wire.
(iii)The vernier scales are adjusted to read 0o and 180o for the direct ray.
(iv) The telescope is rotated through an angle 90o and is fixed.
(v) The grating table is adjusted until the image coincides with the vertical cross wire.
Both the grating table and the telescope is fixed at this position.
(vi) Now the vernier table is rotated through 45o in the same direction in which the
telescope has been previously rotated.
26
(vii)
The light from the collimator falls normally on the grating. The telescope is
released and is brought on line with the direct image of the slit.
(viii)
Now the grating is said to be in the normal incidence position.
II) Determination of number of lines per meter (N):
(i) Light from sodium vapour lamp is diffracted by the grating.
(ii) The telescope is released to get the diffracted images of the first order on the left
side of the central direct image.
(iii)The vertical cross wire is coincided with the first order diffracted image.
(iv) The readings of two verniers VA and VB are tabulated.
(v) The same procedure is repeated on the right side of the central direct image.
(vi) The difference between the left and right-side readings for each line gives the value
of 2θ from which the angle of diffraction θ for each line is found.
(vii)
Substituting the value of θ and the value of the wavelength (5893 Ao) of the
sodium vapour lamp in formula (1), the value of number of lines per meter (N) of
the grating is calculated.
III) Determination of the wavelength ( λ ) of mercury spectrum:
(i)
Sodium vapour lamp is replaced by mercury vapour lamp.
(ii)
The vertical cross wire is coincided with prominent lines namely violet, blue,
green, yellow-I, yellow-II and red of the mercury spectrum.
(iii)
The readings of two verniers VA and VB are tabulated for each line.
(iv)
The same procedure is repeated on the right side of the central direct image.
(v)
The difference between the left and right side readings for each line gives the
value of 2θ from which the angle of diffraction θ for each line is found.
(vi)
Substituting the value of N and the angle of diffraction of each prominent
lines of mercury spectrum in formula (2), the wavelength (λ) of each line is
calculated.
27
To find the value of N:
m = 1,
λ = 5893 A0,
L.C = 1'
Total Reading (TR) = MSR +(VSC x L.C)
Spectral
2θ
Angle
N
(deg)
of
=
Spectrometer reading (deg)
line
Left
VA
Right
VB
VA
VB
M VS
T MS VS T MS VS T MS VS T
SR C
R R
C
R R
C
R R
C
V
V
A
B
diffractio (sin
θ/
n
R
(θ)
mλ
(deg)
)
Yellow
To determine the wavelength of prominent lines of mercury spectrum:
m = 1, N=
,
L.C = 1'
Total Reading (TR) = MSR +(VSC x L.C)
Spectral
2θ
Spectrometer reading (deg)
line
Left
(deg)
Right
Angle
λ=
of
(sin
diffractio θ /
VA
VB
VA
VB
MS VS T MS VS T MS VS T MS VS T
R
C
R R
C
R R
C
R R
C
R
V
V
A
B
n
mN)
(θ)
(deg)
(Ao
)
Violet
Blue
Green
Yellow
-I
Yellow
-II
Red
28
Result:
Colour
Actual range of wavelength
Observed
wavelength
Violet
Blue
Green
Yellow-I
Yellow-II
Red
The number of lines per meter of grating, N = __________ lines / m.
29
10. Sonometer – Frequency of A.C. mains
Aim:
To determine the frequency of A.C. mains supply using Sonometer.
Apparatus required:
Sonometer, slotted weights, bar magnets, screw gauge, meter scale, knife edges etc.,
Formula:
𝟏
𝐓
𝐧 = 𝟐𝒍 √𝐦 in 𝐇𝐳
Where n is the frequency of the A.C mains supply in Hz.
l is the length of the vibrating loop in m.
T = Mg is the tension applied.
M is the mass applied to the string in Kg.
g is the acceleration due to gravity in m/s2
m = πr2ρ is the linear density of the material of the wire in Kg/m
r is the radius of the wire in m
ρ is the density of the given wire in Kg/m3
Procedure:
(i)
Place the sonometer on the table.
(ii)
Attach a weight hanger at the free end of the string which passes over the pulley.
(iii)
Stretch the wire by loading a suitable maximum mass on the weight hanger.
(iv)
The sonometer wire is connected to the secondary of the step-down transformer.
(v)
The bar magnets are mounted below and above the wire at the middle of sonometer
bed so as to produce a magnetic field perpendicular to the wire.
(vi)
The opposite poles of the magnet must face each other.
(vii)
The bridges are placed on either side of the magnet at equal distance from the
magnet and are close to each other.
(viii)
A light paper rider is placed on the wire between the bridges of the sonometer.
(ix)
The A.C. supply is switched on.
(x)
The wire begins to vibrate.
(xi)
The length of the wire between the two bridges is adjusted till the wire vibrates with
maximum amplitude. At this stage, the paper rider placed on the wire is thrown off,
which shows the condition of resonance.
(xii)
The length of the wire between the two bridges is measured. This is called the
resonating length l.
30
(xiii)
Repeat the experiment in steps of 50gm and note down the corresponding
resonating length.
(xiv)
The linear mass density of the wire, m, can be calculated using the relation, m =
πr2ρ, where r is the radius of the wire which can be measured using the screw gauge
and ρ is the density of the material of the wire used.
(xv)
By knowing the linear density, m, of the wire, the frequency of A.C. mains supply
is calculated using the given formula.
Table 1
s.no.
Mass applied
Length of the loop
(M) in
(l) in10-2 m
Frequency
𝐧=
-3
10 Kg
𝟏 𝐓
√ in 𝐇𝐳
𝟐𝒍 𝐦
Table 2 To find the radius (r) of the wire using screw gauge
L.C =0.01 mm
S.No.
PSR(mm)
Z.E = ____ div
HSC(div)
Z.C = ____ div
CHSC=HSC±Z.C Total
(div)
reading
=(PSR+CHSCXL.C)
(mm)
Mean diameter = _____ x10-3m
Radius (r)
= _____ x10-3 m
Result: The frequency of A.C mains supply using Sonometer = _______ Hz.
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11. Logic gates using discrete components
Aim:
To construct the basic logic gates AND, OR and NOT using discrete components and
to verify their truth tables.
Apparatus:
Diodes, transistor (BC 107), resistors, power supply, voltmeter, etc…
Boolean equations.
Boolean Expression:
The Boolean expressions are,
(i)
Y = A.B for AND gate
(ii)
Y = A+B for OR gate
(iii)
Y=Ā for NOT gate
Where A,B are input voltages in volts
Y is the output voltage in volts.
Procedure:
A logic gates is an electronic circuit which make decisions. It has one output and one
or more inputs.
(i ) AND Gate:
The AND gate is an electronic circuit that gives a high output (1) only if all its inputs
are high. A dot (.) is used to show the AND operation. Connect the circuit using the diodes D 1
and D2 as shown in fig. The resistance R is connected to the positive terminal of the battery.
The negative terminal of the battery is grounded and it corresponds to 0 level, while the positive
terminal of the battery corresponds to level 1. The output Y is measured at the point c with
respect to ground.
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(ii) OR-gate:
The OR gate is an electronic circuit that gives a high output (1) if one or more of its
inputs are high. A plus (+) is used to show the OR operation.
(iii)NOT gate:
It is also called inverter because it inverts the input. It has one input and one output as
shown in figure below. The output is 0 when the input is 1, and the output is 1 when the input
is 0. The circuit is designed with a transistor in common emitter mode.
The input voltage is given through the base resistance and the output is obtained across
the collector terminal. The emitter terminal is grounded.
When the input voltage is 5V, the base current flows driving the transistor into
saturation. In this condition, the voltage across the collector and emitter is nearly zero. Hence
the output voltage will be at zero level. Similarly when the input voltage is 0V, the transistor
is driven into cut-0ff state and the voltage drop across the collector resistance is zero. The
output voltage will be at 5V with respect to the ground
Circuit diagram:
(I)
And gate:
V
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Truth table
Verification table
A
B
Y=A.B
0
0
0
0
1
0
1
0
0
1
1
1
A(V) B(V) Y=A.B(V)
OR gate:
V
Truth table
Verification table
A
B
Y=A+B
0
0
0
0
1
1
1
0
1
1
1
1
A(V) B(V) Y=A+B(V)
NOT gate
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Truth table
Verification table
A
Y=Ā
0
1
1
0
A(V) Y=Ā(V)
Result:
The logic gates AND, OR and NOT are constructed using discrete components and
their truth tables are verified.
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12.Logic gates – using IC’s
Aim:
To verify the truth table of logic gates using integrated circuits (IC’s)
Apparatus reequired:
IC 7408 – AND Gate, IC 7432 – OR Gate, IC 7404 – NOT Gate, IC 7400 –
NAND Gate, IC 7402 – NOR Gate, IC 7486 – EX-OR Gate, IC power
supply, voltmeter.
Formulae:
The Boolean expression for gates,
AND gate, Y = A.B
OR gate,
Y = A+B
NOT gate, Y = 𝐴̅
NAND gate, Y = ̅̅̅̅̅
𝐴. 𝐵
NOR gate, Y = ̅̅̅̅̅̅̅̅
𝐴+𝐵
EX-OR gate, Y = 𝐴𝐵̅ + 𝐴̅𝐵
where A and B are inputs and Y is output.
Procedure:
The basic logic gates are (i) AND gate, (ii) OR gate, (iii) NOT gate. These
gates can be combined to form other types of logic gates such as NAND, NOR and
EX-OR gates.
AND Gate:
The AND gate is a circuit which gives an output of 1 state only when all the
inputs are in 1state.
OR Gate:
The OR gate is circuit which gives an output of 1 state if one or more inputs
are in 1 state.
NOT Gate:
The NOT gate has only one input and one output. Its function is to invert
the input. If the input is in 0 state, then the output will be in 1 state. If the input is in
1 state, then the output will be in 0 state.
NAND Gate:
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A NAND gate is a negate of AND Gate, that is, the NOT gate followed by
AND gate is called NAND gate.
Ex-OR gate:
The Ex-OR gate is a circuit that gives an output of 1 state when only one of
the input is in 1 state.
AND Gate:
Truth table
A
B
Y
0
0
0
1
0
0
0
1
0
1
1
1
Verification of truth table:
Inputs
A (V)
B (V)
Output
Y (V)
37
OR Gate:
Truth table
A
B
Y
0
0
0
1
0
1
0
1
1
1
1
1
Verification of truth table:
Inputs
A (V)
B (V)
Output
Y (V)
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NOT Gate:
Truth table
A
Y
0
1
1
0
Verification of truth table:
Input (A) (V)
Output (Y) (V)
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NAND Gate:
Truth table
A
B
Y
0
0
1
1
0
1
0
1
1
1
1
0
Verification of truth table:
Inputs
A (V)
B (V)
Output
Y (V)
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NOR Gate:
A NOR gate is a negate of OR Gate, that is, the NOT gate followed by OR
gate is called NOR gate.
Truth table
A
B
Y
0
0
1
1
0
0
0
1
0
1
1
0
Verification of truth table:
Inputs
A (V)
B (V)
Output
Y (V)
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Exclusive- OR gate (Ex-OR Gate):
Truth table
A
B
Y
0
0
0
1
0
1
0
1
1
1
1
0
Verification of truth table:
Inputs
A (V)
B (V)
Output
Y (V)
In all the IC’s except IC’s 7402 and 7404, the pins 1, 2, 4, 5, 9, 10, 12, 13
are inputs and the pins 3, 6, 8, 11 are outputs.
In IC 7402 (NOR gate), the pins 2, 3, 5, 6, 8, 9, 11, 12 are inputs and the
pins 1, 4, 10, 13 are outputs.
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In IC 7404 (NOT gate), the pins 1, 3, 5, 9, 11, 13 are inputs and the pins 2,
4, 6, 8, 10, 12 are outputs.
In all IC’s 7th pin is always grounded and 14th pin is connected to +Vcc.
The IC’s are fixed on the bread board, connections are made as per configurations
and the outputs are noted on voltmeter.
Result:
The logic gates AND, NOT, OR, NOR, NAND and EX-OR are constructed
and their truth tables are verified.
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