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+2 PRACTICAL WORKBOOK

PHYSICS

TN 12th Physics Practical Workbook

1

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PROCEDURE ANGLE OF THE PRISM After making preliminary adjustments, the prism is placed on the prism table. The slit is illuminated by a monochromatic source of light say, sodium vapour lamp. Both the faces AB and AC receive parallel rays from the collimator. 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. The telescope is then rotated to the position T2 where 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. The difference between these two reading give twice the angle of the prism. Half of this gives the angle of the prism. ANGLE OF MINIMUM DEVIATION 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. 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. It will be found that, as the prism table is rotated in the same direction, 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. The telescope is now adjusted so that its vertical cross wire coincides with the image and reading of the verniers are noted. 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. The difference between the two readings give the angle of minimum deviation (D). The refractive idex of the material of the prism is calculated using the formula



TN 12th Physics Practical Workbook

A D 2 A sin 2

sin

2

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www.nammakalvi.weebly.com 1. SPECTROMETER - µ OF A SOLID PRISM AIM

: To determined the angle of a given prism and its angle of minimum deviation and hence calculate its refractive index.

FORMULA : Refractive index of the material of the given prism

AD 2  A sin 2 Where A is the angle of the prism D is the angle of minimum deviation sin

DIAGRAM: To find the angle of Prism

TN 12th Physics Practical Workbook

To find the angle of minimum deviation

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OBSERVATION i) To find the angle of Prism VERNIER I RAY

MSR

VC

TR = MSR+ (VCLC)

VERNIER II MSR

VC

TR = MSR+ (VCLC)

Reading of the image reflected from the one face (R1) Reading of the image reflected from other face (R2) 2A= R1 R2

2A= R1 R2

Mean 2A = A=

ii)

To find the angle of minimum deviation VERNIER I RAY

MSR

VC

TR = MSR+ (VCLC)

VERNIER II MSR

VC

TR = MSR+ (VCLC)

Reading of the image in minimum deviation position (R3) Reading of the direct image (R4) D = R3 R4

D = R3 R4

Mean D =

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Calculation: To find “A” 2A 2A

= R1 R2 =

2A

= R1 R2 =

=

2A

=

+

AVERAGE 2A =

A =

=

2

2

=

2

=

To find “D” D = R3 R4 =

D = R3 R4 =

D=

D= +

Average D = =

=

2

=

2

D= To find “ ”

𝐴+𝐷 𝑠𝑖𝑛 ( ) 𝑠𝑖𝑛 ( 2 𝜇= = 𝐴 𝑠𝑖𝑛 ( 𝑠𝑖𝑛 ( ) 2

𝜇=

𝑠𝑖𝑛 ( 𝑠𝑖𝑛 (

2 2

𝜇=

) )

=

+ 2

) )

2

) )

𝑠𝑖𝑛( 𝑠𝑖𝑛(

=

= RESULT: i) The angle of the prism

A

=

(degree)

ii) The angle of minimum deviation

D

=

(degree)

iii) Refractive index of the material of the given prism µ = TN 12th Physics Practical Workbook

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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 be 00 — 1800. Telescope is then rotated through 90° and fixed. Prism table is rotated to get the reflected image which is made to coincide with the vertical cross wire. Keeping the platform fixed, vernier table is rotated through 45° so that the light rays from the collimator fall on the .grating. Now the grating is in normal incidence position. The direct reading RI is measured. 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  is found as R1  R2. The experiment is repeated for green and yellow spectral lines also. Number of lines per unit length of the grating is N. Wavelength of the spectral line is calculated from the formula



TN 12th Physics Practical Workbook

sin  mN

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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 FORMULA : The wavelength () of a spectral line using normal incidence arrangement of 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

Adjusting the grating for normal incidence:

Determination of angle of diffraction:

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OBSERVATION VERNIER I RAY

MSR

VC

VERNIER II

TR = MSR+ (VCLC)

Diffracted ray

Direct reading

MSR

VC

TR = MSR+ (VCLC)

RD1

RD2

BLUE

RB1

RB2

GREEN

RG1

RG2

YELLOW

RY1

RY2

TO FIND THE “” RD1 – R1



RD2 – R2

BLUE

B

GREEN

G

YELLOW

Y

m =1 N = 6 105

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Calculation: RD1 – RB1 =

RD2 – RB2 =

B =

B = Average B =

+

=

2

RD1 –RG1 =

RD2 – RG2 =

G =

G = Average B =

+

=

2

RD1 – RY1 =

RD2 – RY1 =

Y =

Y = Average Y =

B 

sin B sin( )   mN 1  6  105

G 

sin G sin( )   5 mN 1  6  10

Y 

sin Y sin( )   mN 1  6  105

+

=

2

5



 10 7 m

5



 10 7 m

5



 10 7 m

6  10

6  10 6  10

RESULT : i)

wavelength of blue colour

B =

10–7m

ii)

wavelength of green colour G =

10–7m

iii)

wavelength of yellow colour Y =

10–7m

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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 length AJ = ℓR1 is measured. JC =ℓX1 is found out as (100 - ℓR1). The experiment is repeated four more times by increasing the value of R in steps of 1 ohm. Then the resistance box R and coil X are interchanged in the gaps G1 and G2. For the same values of R as in the previous part of the experiment the balancing length AJ =ℓ X2 are measured. The balancing length JC =ℓR2 are found out as (100- ℓX2). The values of ℓX and ℓR are calculated from

X 

 X1   X 2    R2  R  R1 2 2 ℓ

The resistance of the coil is found by substituting in the formula 𝑋 = 𝑅 ℓ𝑋 𝑅

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  

TN 12th Physics Practical Workbook

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 r2X l

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3. METRE BRIDGE – DETERMINATION OF RESISTANCE AND SPECIFICE RESISTANCE 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. FORMULA Resistance of the wire X  R

lX lR

Specific resistance of the material of the wire   Where

 r2X l

R is known resistance l R is the balancing length of R l X is the balancing length of X r is the radius of the wire l is the length of the wire

Circuit diagram – Before interchanging

Circuit diagram – After interchanging

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OBSERVATION (i) To determine the resistance of the given coil

S.No

R (ohm)

1

1

2

2

3

3

4

4

5

5

(ii)

Balancing length before interchanging

 x1 (cm)

 R1 (cm)

Balancing length after interchanging

 X2 (cm)

 R2 (cm)

To determine the radius of the coil LC = 0.01 10–3m ZERO ERROR = S.No PSR HSC 1 2 3 4

lX lR (ohm)

X R

Mean

X 

 X1   X 2 2 (cm)

HSR

R 

 R1   R 2 2 (cm)

ZERO CORRECTION = CR = PSR+HSRL.C

Diameter 2r r

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Calculation:

𝑙𝑥 = 𝑙𝑥 = 𝑙𝑥 = 𝑙𝑥 =

𝑙𝑥1 +𝑙𝑥2 2 𝑙𝑥1 +𝑙𝑥2 2 𝑙𝑥1 +𝑙𝑥2 2 𝑙𝑥1 +𝑙𝑥2 2

+

=

2 +

=

2 +

=

2 +

=

2

=

𝑙𝑅 =

=

𝑙𝑅 =

=

𝑙𝑅 =

=

𝑙𝑅 =

𝑙𝑅1 +𝑙𝑅2 2 𝑙𝑅1 +𝑙𝑅2 2 𝑙𝑅1 +𝑙𝑅2 2 𝑙𝑅1 +𝑙𝑅2 2

= = = =

+

=

2 +

=

2 +

=

2 +

=

2

Calculation for X

𝑋=𝑅 𝑋=𝑅 𝑋=𝑅 𝑋=𝑅 𝑋=𝑅

𝑙𝑥 𝑙𝑅 𝑙𝑥 𝑙𝑅 𝑙𝑥 𝑙𝑅 𝑙𝑥 𝑙𝑅 𝑙𝑥 𝑙𝑅

= = = = =

Mean X =

5

=

Calculation for : 𝜌=

𝜋𝑟 2 𝑋 3.14 × = 𝑙

× 10−3 ×

× 10−3 × 1

=

RESULT: Resistance of the wire X =  Specific resistance of the material of the wire  = TN 12th Physics Practical Workbook

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PROCEDURE

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

J is pressed

in the first and the last wire and the opposite side deflections in the galvanometer shows that the connections are correct. Lechlanche 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. Daniel cell is included in the circuit using the DPDT switch, and the balancing length is measured as ℓ2. The experiment is repeated for six times by moving rheostat in one direction for changing the current in the circuit. The ratio of the emf of the two cells is found from the formula

TN 12th Physics Practical Workbook

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E1 l1  E2 l2

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4.

POTENTIOMETER – COMPARISION OF EMFS OF TWO CELLS

Aim: To compare the emfs of two primary cells using a potentiometer. Formula: E1 l1  E2 l2

E1 emf of primary cell 1 (Lechlanche 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

OBSERVATION S.No

balancing length for

balancing length for

Lechlanche cell

Daniel cell

l1 cm

l 2 cm

E1 l1  E2 l2

1 2 3 4 5 6 Mean

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E1 E2

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Calculation:

𝐸1 = 𝐸2

𝐸1 = 𝐸2

𝐸1 = 𝐸2

𝐸1 = 𝐸2

𝐸1 = 𝐸2

𝐸1 = 𝐸2

Mean

𝐸1 𝐸2

=

Result : The mean ratio of emf of the two cells = TN 12th Physics Practical Workbook

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5. Tangent Galvanometer – Determination of BH Aim : To determine the value of the horizontal component of earth’s magnetic field (BH) Formula: 𝜇0 𝑛 𝐼 ( ) 2𝑎 𝑡𝑎𝑛𝜃 BH - horizontal component of earth’s magnetic field 0 – permeability of free space n – number of turns I – current a – radius of coil  - mean deflection produced in TG 𝐵𝐻 =

Circuit diagram:

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 0 0 – 00. The current I is passed through the circuit and the deflection of the needle are measured as 1 and 2 . By reversing the current, the deflection are measured as 3 and 4. The average deflection  is found out. The experiment is repeated by varying current. The average value of 𝐼 𝑡𝑎𝑛𝜃

is found out. The radius R of the coil is found out by measuring its circumference. The

number of turns “n” of the coil is noted. The Horizontal component of earth’s magnetic induction is calculated by substituting in the formula 𝐵𝐻 =

𝜇0 𝑛 𝐼 ( ) 2𝑎 𝑡𝑎𝑛𝜃

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OBSERVATION: Deflection of T.G. (degree) S.No

Current I (A)

1

2

3

4

mean 

Tan 

I tan 

1 2 3 4 mean Calculation : Circumference of the coil (2r) = Radius (r) =

10–2 m

2

10–2 m =

𝐼 = 𝑡𝑎𝑛 𝜃

𝐼 = 𝑡𝑎𝑛 𝜃

𝐼 = 𝑡𝑎𝑛 𝜃

𝐼 = 𝑡𝑎𝑛 𝜃

Mean

𝐵𝐻 =

𝐼 𝑡𝑎𝑛 𝜃

=

𝜇0 𝑛 𝐼 4𝜋 × 10−7 × 5 × ( )= 2𝑎 𝑡𝑎𝑛𝜃 2× =

Result: The horizontal component of earth’s magnetic field (BH) = TN 12th Physics Practical Workbook

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6. SONOMETER – FREQUENCY OF AC Aim:To determine the frequency of the ac main using a sonometer Formula:The frequency of the ac main 1

𝑛 =2×

√𝑇 𝑙

×

1 √𝑚

where T is the tension of the sonometer wire ℓ is the resonating length m is the linear density of the wire 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 200 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 ℓ 1. The same procedure is repeated and ℓ2 is measured. The average ℓ1 and ℓ2 is ℓ. The experiment is repeated for the loads 400gm, 600 gm and 800 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 √𝑇 1 𝑛= × × 2 𝑙 √𝑚 Observation S.No:

Load M (kg)

(ii)

1.

0.200

2.

0.400

3.

0.600

4.

0.800

Length of the vibrating segment ℓ1(cm) ℓ2(cm)

Mean

T = Mg

ℓ (cm)

(newton)

To determine the radius of the sonometer wire LC = 0.01 10–3m ZERO ERROR = S.No PSR HSC HSR 1 2 3 Mean d Radius r =

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√𝑇

√𝑇 𝑙

ZERO CORRECTION = CR = PSR+(HSRL.C)

10–3m www.uburt.in

Calcuation: 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 =

T = 0.29.8 = 1.96

√𝑇 =

ℓ=

√𝑇 = 𝑙

T = 0.49.8 =3.92

√𝑇 =

ℓ=

√𝑇 = 𝑙

T = 0.69.8 =5.88

√𝑇 =

ℓ=

√𝑇 = 𝑙

T = 0.89.8 =7.84

√𝑇 =

ℓ=

√𝑇 = 𝑙

Mean

1

𝑛=2×

√𝑇 𝑙

=

√𝑇 𝑙

×

4

1 √𝑚

=

=

Result : The frequency of the ac main n =

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Hz

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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. Formula Forward resistance of the PN junction diode 𝑅𝑓 =

∆V𝑓 ∆𝐼𝑓

∆V𝑓 is the forward voltage ∆𝐼𝑓 is the forward current Junction diode - forward bias

Zener

diode –

reverse bias Zener diode – reverse bias

Procedure: 1)Forward Characteristic Curve :The circuit is wired as in the diagram. The forward voltage Vf is increased from zero in

steps of 0.1 V upto 1V. 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. The reciprocal of the slope of this curve above the knee point is found as forward resistance of the Diode. Forward resistance 𝑟𝑖 = (

∆𝑉𝑓 ∆𝐼𝑓

)

2) Reverse breakdown characteristics of the zener diode:The circuit is wired as in the diagram. The voltage VO is increased from zero in steps of 1V upto 10V. The corresponding values of IZ are noted. A graph is drawn with VO along X-axis and IZ along Y-axis. This is called reverse characteristic curve. At particular voltage, the current increases enormously, this voltage is called zener voltage (VZ)

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Observation Junction diode forward bias s.No:

VF (V)

1

Zener diode -reverse bias IF (mA)

s.No:

Vo (V)

0.1

1

1

2

0.2

2

2

3

0.3

3

3

4

0.4

4

4

5

0.5

5

5

6

0.6

6

6

7

0.7

7

7

8

0.8

8

8

9

0.9

9

9

10

1.0

10

10

IZ (mA)

Calculation: 𝑅𝑓 =

Result: i) ii)

∆V𝑓 𝐵𝐶 = = ∆𝐼𝑓 𝐴𝐵

=

The forward resistance of the junction diode = The zener breakdown voltage =

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8. Common Emitter NPN Transistor Characteristics Aim: To study the characteristics of a common Emitter NPN transistor and to determine its input impedance, output impedance and current gain. Formula: ∆𝑉𝐵𝐸

(i)

input impedance 𝑟𝑖 = (

(ii)

output impedance 𝑟𝑜 = (

(iii)

current gain 𝛽 = (

∆𝐼𝐶

∆𝐼𝐵

)

∆𝐼𝐵 ∆𝑉𝐶𝐸 ∆𝐼𝐶

)

)

∆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

𝑟𝑖 = (

∆𝑉𝐵𝐸 𝐵𝐶 )= ∆𝐼𝐵 𝐴𝐵

∆𝑉𝐶𝐸 𝐵𝐶 𝑟𝑜 = ( )= ∆𝐼𝐶 𝐴𝐵

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𝛽=(

∆𝐼𝐶 𝐴𝐵 )= ∆𝐼𝐵 𝐵𝐶

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INPUT CHARACTERISTICS

OUTPUT CHARACTERISTICS IB

VCE = 5V s.No:

VBE (V)

IB (mA)

=

s.No:

Vo (V)

1

0.1

1

0.5

2

0.2

2

1

3

0.3

3

1.5

4

0.4

4

2

5

0.5

5

2.5

6

0.6

6

3

7

0.7

7

3.5

8

0.8

8

4

9

0.9

9

4.5

10

1.0

10

5

20A, 40A, 60A, 80A IC (mA)

IC (mA)

IC (mA)

IC (mA)

TRANSFER CHARACTERISTIC (VCE = 5V) S.NO

IB (A)

1

25

2

50

3

75

4

100

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IC (mA)

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Calculation:

ri = (

∆VBE BC = )= ∆IB AB

ro = ( β=(

∆VCE BC = )= ∆IC AB

∆IC AB = )= ∆IB BC

PROCEDURE: The circuit is wired as in the diagram. 1.INPUT CHARACTERISTIC CURVE :The collector emitter voltage V CE is kept at a constant value. The base emitter voltage VBE is increased from zero in steps of 0.1 V upto 1V. The corresponding values of IB are noted.A graph is drawn with VBE along X-axis and IB along Y-axis. This is called input characteristic curve. The reciprocal of the slope of this curve above the knee point is found as input impedance of the transistor. Input impedance 𝑟𝑖 = (

∆𝑉𝐵𝐸 ∆𝐼𝐵

)

2. TRANSFER CHARACTERISTIC CURVE :The collector emitter voltage VCE is kept at a constant value (5V). IB is increased in steps of 25 µA from 25 µA to 100µA. The corresponding values of IC are noted. A graph is drawn with IB along Xaxis and Ic along Y-axis. This is called transfer characteristic curve. The slope of this curve gives the current gain of the transistor. ∆𝐼

Current gain 𝛽 = (∆𝐼 𝐶 ) 𝐵

3. OUTPUT CHARACTERISTIC CURVE :The base current IB is kept at a constant value. VCE is increased in steps of 0.5 V from Zero. The corresponding values of IC are noted. A graph is drawn with VCE along X-axis and IC along Y-axis. This is called output characteristic curve. The reciprocal of the slope of the output characteristic curve near horizontal part gives the output impedance (r0). Output impedance 𝑟𝑜 = (

∆𝑉𝐶𝐸 ∆𝐼𝐶

)

Result: i)

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

ii)

The input impedance ri =

iii)

The output impedance r0 =

iv)

The current gain  =

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BASIC AMPLIFIERS USING OP-AMP INVERTING AMPLIFIER:The circuit is wired as shown in the diagram using OP-AMP IC 741. RS is kept as 10 KΩ and RF as 22 KΩ. The input voltage Vin is kept as 1V and output voltage Vo is measured from the digital voltmeter. Then the experiment is repeated for input values V in = 1.5 V, 2V and 2.5 V. Second Set of readings is taken by keeping Vin = 1 V and Rs = 10 KΩ and changing RF as 10 KΩ,22KΩ,33 KΩ & 𝑉

47 KΩ. Experimental gain is found as𝐴𝑉 = (𝑉 𝑂 ) 𝑖𝑛

𝑅

Theoretical gain is found from 𝐴𝑉 = − ( 𝑅𝑓 ) 𝑠

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

NON-INVERTING AMPLIFIER:The circuit is wired as shown in the diagram using OP-AMP IC 741. RS is kept as 10 KΩ and RF as 22 KΩ. The input voltage Vin is kept as 1V and output voltage Vo is measured from the digital voltmeter. Then the experiment is repeated for input values V in = 1.5 V, 2V and 2.5V. Second Set of readings is taken by keeping Vin = 1 V and Rs = 10 KΩ and changing RF as 10 KΩ,22KΩ,33 KΩ & 𝑉

47 KΩ. Experimental gain is found as𝐴𝑉 = (𝑉 𝑂 ) 𝑖𝑛

𝑅𝑓

Theoretical gain is found from 𝐴𝑉 = 1 + ( 𝑅 ) 𝑠

Both the AV values are compared and found to be equal. SUMMING AMPLIFIER:The circuit is wired as shown in the diagram using OP AMP IC 741, The values of R1, R2 and RF are kept as 10 K Ω. The input voltages are kept as VI = 1V and V2 = 0.5V and the output voltage Vo is measured using the digital voltmeter 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). Since this is equal to experimental output voltage the summing action of the amplifier is verified.

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9. OPERATIONAL AMPLIFIER Aim : To construct the following basic amplifiers using OP-AMP IC741. i) Inverting amplifier ii) Non inverting amplifier iii) Summing amplifier Formula : 𝑅

𝑉

i)

Voltage gain of the inverting amplifier, 𝐴𝑉 = (𝑉 𝑂 ) = − ( 𝑅𝑓 )

ii)

Voltage gain of the non inverting amplifier, 𝐴𝑉 = (𝑉 𝑂 ) = 1 + ( 𝑅𝑓 )

iii)

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

𝑖𝑛

𝑠

𝑉

𝑠

𝑖𝑛

Where

𝑅

V0 output voltage Vin, V1 and V2 are the input voltages Rf and Rs are the external resistances

INVERTING AMPLIFIER

SET

Vout(V)

Experimental Gain

Rs ()

Rf ()

Vin(V)

1

10K

22K

1

-2.2

2

10K

22K

1.5

-2.2

3

10K

22K

2

-2.2

4

10K

22K

2.5

-2.2

1

10K

10K

1

-1.0

2

10K

22K

1

-2.2

3

10K

33K

1

-3.3

4

10K

47K

1

-4.7

𝑉𝑂

𝐴𝑉 = (𝑉 ) 𝑖𝑛

I

II

Theoretical Gain

S.NO

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𝑅

𝐴𝑉 = − ( 𝑅𝑓 ) 𝑠

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NON-INVERTING AMPLIFIER

SET

S.NO

Rs ()

Rf ()

Vin(V)

Vout(V)

Experimental Gain

Theoretical Gain 𝑅

𝑉𝑂

𝐴𝑉 = 1 + ( 𝑅𝑓 )

𝐴𝑉 = (𝑉 )

𝑠

𝑖𝑛

I

II

1

10K

22K

1.0

3.2

2

10K

22K

1.5

3.2

3

10K

22K

2.0

3.2

4

10K

22K

2.5

3.2

1

10K

10K

1.0

2.0

2

10K

22K

1.0

3.2

3

10K

33K

1.0

4.3

4

10K

47K

1.0

5.7

SUMMING AMPLIFIER

R1 = R2 = Rf = 10K Experimental Output voltage V0 (Volt)

S.NO

V1 (Volt)

V2 (Volt)

1

1.0

0.5

1.5

2

1.0

1.0

2.0

3

1.0

1.5

2.5

4

1.0

2.0

3.0

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Theoretical output voltage V0 = - (V1 + V2) (Volt)

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Calculation: Inverting amplifier

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝐴𝑉 = − (

𝑅𝑓 ) 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝐴𝑉 = − (

𝑅𝑓 )= 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝐴𝑉 = − (

𝑅𝑓 )= 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝐴𝑉 = − (

𝑅𝑓 )= 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝐴𝑉 = − (

𝑅𝑓 )= 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝐴𝑉 = − (

𝑅𝑓 )= 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝐴𝑉 = − (

𝑅𝑓 )= 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝐴𝑉 = − (

𝑅𝑓 )= 𝑅𝑠

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Non inverting amplifier 𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝑅𝑓 𝐴𝑉 = 1 + ( ) 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝑅𝑓 𝐴𝑉 = 1 + ( ) = 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝑅𝑓 𝐴𝑉 = 1 + ( ) = 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝑅𝑓 𝐴𝑉 = 1 + ( ) = 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝑅𝑓 𝐴𝑉 = 1 + ( ) = 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝑅𝑓 𝐴𝑉 = 1 + ( ) = 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝑅𝑓 𝐴𝑉 = 1 + ( ) = 𝑅𝑠

𝑉𝑂 𝐴𝑉 = ( ) = 𝑉𝑖𝑛

𝑅𝑓 𝐴𝑉 = 1 + ( ) = 𝑅𝑠

Summing amplifier 1) Vo = –(V1 + V2) = 2) Vo = –(V1 + V2) = 3) Vo = –(V1 + V2) = 4) Vo = –(V1 + V2) = Result : i) ii)

The inverting amplifier and non-inverting amplifier are constructed using OP-AMP and gain is determined. The inverting summing amplifier is constructed and the output voltage is found to be the sum of the applied input voltages

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10 INTEGRATED LOGIC GATE CIRCUITS Aim: To study the Truth Table of integrated Logic Gates IC 7400(NAND), 7408(AND), 7402 (NOR), 7432 (OR), 7404 (NOT) and 7486 (EXOR) 1)For IC’s 7400 (NAND), 7408(AND), 7432(OR) & 7486(EX-OR)

2) For IC 7402(NOR) - Quad 2 input

Hex inverter NOT (7404)

POSITIVE LOGIC SYSTEM :Logic 1 represents TRUE or high voltage 5V or LED ON Logic 0 represents FALSE or low voltage 0V or LED OFF OR function

When any one input or all inputs are true, output-is-true Y =A + B

AND function

Only when all inputs are true, output is true Y = AB

NOT function

Output is the complement of input ̅ Y=A

NOR function

Only when all inputs are false, output is true Y = ̅̅̅̅̅̅̅ A+B

NAND function

When any one of the inputs is false, output is true ̅̅̅̅̅̅ Y=A ∙B

EXOR function

Only when the inputs are different, output is true ̅B ̅+A Y = A⨁B = AB

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NAND gate:Power supply +5V is connected to pin 14 and gnd to pin 7 of the IC. Inputs A & B are connected to pins 1 & 2 of the IC. Output pin 3 of the IC is connected to logic level indicator. Both inputs A & B are kept at logic 0 and output LED is observed, Then the inputs are chan ged as logic 0 & logic 1, logic 1 & logic 0 and logic 1 & logic 1 and the outputs are observed each time. The inputs and outputs are tabulated in the truth table. AND, OR and EXOR gates:ICs 7408 (AND), 7432 (OR) and 7486 (EXOR) are placed on the board arid the same procedure is followed as for NAND gate and outputs are tabulated in the truth table. NOR gate :IC 7402 is placed on the board. Power supply and gnd are connected as before. 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 :IC 7404 is placed on the board. One input A is connected to pin 1 and the output to pin 2 of IC. Input is kept at logic 1 and then at logic 0 and the outputs are found and tabulated in the truth table. IC 7432(OR) Truth Table (OR) A 0 0 1 1 IC 7408 (AND)

B 0 1 0 1

Truth Table (AND) A B 0 0 0 1 1 0 1 1

IC 7404 (NOT)

Y = AB

Truth Table (NOT) 𝑌 = 𝐴̅

A 0 1 IC 7402(NOR)

Truth Table (NOR) A 0 0 1 1

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B 0 1 0 1

Y = ̅̅̅̅̅̅̅̅ 𝐴+𝐵

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IC 7400 (NAND) Truth Table (NAND)

A 0 0 1 1

IC 7486( EX-OR) Truth Table (EX-OR)

B 0 1 0 1

A 0 0 1 1

B 0 1 0 1

Y = ̅̅̅̅̅̅ 𝐴∙𝐵

Y = AB

Calculation: OR

AND

NOT

NOR

NAND

EX-OR

Result: The performance of digital gates OR, AND, NOT, NAND, NOR and EX-OR are verified using IC chips.

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