lic lab manual

LINEAR INTEGRATED CIRCUITS LAB P a g e | 1

INVERTING AMPLIFIER: 5KΩ

1KΩ V0 RPS (0-10V)

833Ω

TABULAR COLUMN:

S.NO:

Vin (Volts)

Vout

(Volts)

Av = Vout/Vin

MODEL GRAPH: 0 Vi(v)

Slope = - 5

Vo(sat)

Vo(v)


1(a). INVERTING AND NON-INVERTING AMPLIFIER

EX.NO. :

DATE :

AIM: - To design and verify the experimental and theoritical loop gains of amplifiers using IC 741 in the inverting and non-inverting modes.

APPARATUS REQUIRED:

Equipments & Components Range Quantity

1.Dual Power Supply 2.Resistors 3.Regulated Power supply 4.Voltmeter 5.IC-741

(0-30)V

1KΩ

10KΩ (0-30)V (0-50)V

1 1 2 1 1 1

DESIGN:

INVERTING AMPLIFIER:

Let the desired gain be 5 and VCC =15v & -VCC =-15v.We know That gain for inverting amplifier AV = -Rf / R1

That is |AV| =|1

f

R

R|

Let R1 =1Kohm & given AV = 5

So 5 = ΩK1

R f

Rf = 5Kohm [Since 5Kohms is not a standard value, two 2.5 Kohms resistors are used in series or two 10 Kohms is used in parallel are used for Rf].

And Rcomp =f1

f1

R+R

RR= 833 ohms [Since 833ohms is not a standard value, 1

Kohms & 5Kohms resistors are used in parallel]

NON- INVERTING AMPLIFIER:

Let the desired gain be 6 and VCC =15v & -VCC =-15v.We know that gain for non

inverting amplifier AV = 1+1

f

R

R


NON-INVERTING AMPLIFIER: 5KΩ

1 KΩ

Vo

833 Ω

TABULAR COLUMN:

S.NO:

Vin

(Volts)

Vout

(Volts)

Av =Vout/Vin

VI (0 – 10 V)


Let R1 =1Kohm & given AV= 6

So 6 = 1+ΩK1

R f

Rf = 5Kohm [Since 5Kohms is not a standard value, two 2.5 Kohms resistors are used in series or two 10 Kohms is used in parallel are used for Rf]. And

Rcomp =f1

f1

R+R

RR = 833 ohms [Since 833ohms is not a standard value, 1 Kohms &

5Kohms resistors are used in parallel] TO CALCULATE VI (MAX): - Vo (sat) =VCC =15v-0.7v =14.3v VI(max) = Vo (sat) /gain For inverting amplifier VI(max) =14.3 v/-5 =-3v For non inverting amplifier VI (max) =14.3 v/6 =2.4v

THEORY:

The basic equation for the op-amp is Ae = V0 where A is open loop gain of the Op-amp at an operating frequency f and is positive. e is measured as per the arrow direction shown in the fig 1. This equation is valid for the open loop condition and closed loop condition [only for negative feedback]. Using the above formula all the gain equations for the different amplifier configuration can be derived. It is very important to note that A varies with frequency.

e = A

V0

A is of the order of 105 to 106 at 5 Hz Hence e = 0 for range of frequencies. This implies that the non-inverting terminal voltage follows the inverting terminal voltage or the inverting terminal voltage follows the non-inverting terminal voltage. In other words the potential difference between the inverting and non-inverting terminal is zero volt at a specified frequency the above condition will not be valid.

Gain equation for inverting amplifier A= -1

f

R

R

For non inverting amplifier A = 1 + 1

f

R

R

For voltage follower A = 1+A

A = 1


MODEL GRAPH: Vo(v)

Vo(sat)

Slope = 6

Vi(v)

TABULAR COLUMN:

S.NO:

Vin

(Volts)

Vout

(Volts)

Av =Vout/Vin

MODEL GRAPH: Vo(v)

Vo(sat)

Slope = 1

Vi(v)


MAX WATTAGE OF RESISTANCE: Resistances used in the design are 1Kohm and 10Kohms .The maximum voltages

across the resistance will be supply voltage. Hence wattage of resistance is R

V 2

.

R

V 2

= ΩK1

225 or

ΩK10

225 which is so much lower than 1/8W or 1/4W.So resistances with

5% tolerance, carbon film resistor with 1/8W or 1/4W is used.

PROCEDURE:

1. Connections are made as shown in circuit diagrams . 2. The input voltage is given and the output voltage is noted. The maximum input

voltage that can be given to the circuit is VI[max]. 3. In all the config when VI = 0, V0 = 0. 4. The close loop gain is calculated for each input voltage and transfer

characteristics is drawn for each configuration. 5. The slope of the characteristics between input and output voltage gives the small

signal AC closed loop gain as represented in model graph.

RESULT:

Slope of the DC characteristics between input and output voltage gives the small signal AC closed loop gain provided the condition outlined in theory regarding open loop gain with frequency is maintained.


CIRCUIT DIAGRAM:

R 2 = 100KΩ

R1=1KΩ

Vo

AFO

AFO R 1 =1KΩ

R 2 = 100K Ω

TABULAR COLUMN:

S.NO:

V1

(Volts)

V2

(Volts)

V0 (Volts)

gain

a

a


1(b).DIFFERENTIAL AMPLIFIER

EX.NO.:

DATE:

AIM: To design and test the operation of Differential amplifier.

APPARATUS REQUIRED:

Equipments & Components Range Quantity

1.Dual Power Supply 2.Resistors 3.AFO 4.CRO 5.IC-741

(0-30)V

1KΩ

100KΩ (0-1)MHz (0-20)KHz

1 2 2 1 1 1

DESIGN:

Gain = 100, & Let R1 = 1 KΩ AD = R2 / R1 So R2 = AD * R 1

R2 = 100 * 1KΩ = 100KΩ.

THEORY:

A Circuit that amplifies the difference between two signals is called a differential amplifier. This type of amplifier is very useful in instrumentation circuits. For differential amplifier, though the circuit is not symmetric, but because of the mismatch, the gain at the output with respect to positive terminal is slightly different in magnitude to that of negative terminal. So even with the same voltage applied to both the inputs, the output is not zero.


PROCEDURE:

1. Connections are given as per the circuit diagram. 2. Set the input Voltages V1 = 50mV & V2 =40mV. 3. Note down the Output Voltage 4. Vary the input Voltages and note down the output voltages. 5. Calculate the gain & Compare it with the Theoritical gain.

RESULT:

Thus the Differential amplifier is designed & tested.


CIRCUIT DIAGRAM:

DIFFERENTIATOR:

31.8 KΩ VO (CRO)

AFO INTEGRATOR:

VO (CRO)

SPECIFICATION FOR IC741: +Vcc = +15V, - Vcc = -15V Ambient Temparature : 250 C Input offset voltage : 6 mV(Max) Input offset current : 200nA(Max) Input bias current : 500nA(Max)

Input resistance : 2MΩ

Output resistance : 75Ω Total Power dissipation : 85mW.

0.1µF

Vin= 0.1Vpp

50Hz a

100KΩ

AFO

10KΩ

Vin = 1Vpp, 50Hz

a

3.1KΩ

0.1µF


2. DIFFERENTIATOR & INTEGRATOR

EX.NO.:

DATE:

AIM: To design and study the operation of 1) Differentiator 2) Integrator.

APPARATUS REQUIRED:

Equipments & Components

Range Quantity

1.Dual Power Supply 2.Resistors 3.Capacitors 4.IC741 6.AFO 7.CRO

(0-30)V

31.8KΩ,3.1KΩ,10KΩ,100KΩ

1KΩ

0.1µF (0-1)MHz (0-20)MHz

1 each 1 3 1 1 1 1

DESIGN:

DIFFERENTIATOR: The transfer function of differentiator is given by

i

0

V

V =

11

1f

RsC+1

CsR-

If sC1R1 <<1then V0 / VI = -sRfC1 The above equation can be rewritten as

AV = af

f where far =

1f CR28.6

1

Let fa =fmax=50 Hz, assume C1 =0.1 µFD

Rf = 1aCf28.6

1 = 31.8KΩ

To prevent loading Rf =10R1

So R1 =10

R f = 3.1KΩ

INTEGRATOR:

To find R1 & Rf in the lossy integrator, so that the peak gain is 20dB and the gain

is 3dB down from it’s peak when ω = 10,000rad/sec.

Assume C = 0.001µF

We know that A(dB) = 20log10 Rf / R1 / [1+(ωRfCf)2]1/2


PINDIAGRAM FOR IC741:

IC741 1 = Offset Null, 2 = Inverting input terminal, 3 = Non inverting input terminal, 4 = -Vcc, 5 = Offset Null, 6 = Output, 7 = +Vcc, 8 = No connection

MODEL GRAPH:

DIFFERENTIATOR:

Vi(v) 0V

t (msec)

Vo(v) 0V

t (msec)

1

2

3

4

8

7

6

5


To find the peak value in decibel

Put ω = 0. So A(dB) = 20. So R1 = Rf / 10.

At ω = 10,000 rad/sec, gain in dB is down by 3dB from it’s peak of 20dB & thus is 17dB. So 20 log10 10/[1+(104 * 10-9 * Rf )

2]1/2+ = 17dB

By solving 1+(10-5 * Rf )2 = 103/10

(10-5 * Rf )2 = 1

Rf = 100KΩ, So R1 = 10KΩ.

VERIFICATION: From the design C1 = 0.1MFD & R1 =3.1kΩ So SR1C1 =0.097 Consider an input with

Vmax =1v and f=50Hz VI =Vmax sinwt =sinwt We know that

Vo = -RfC1 dt

dVI = -31.8 * 103 *0.1 * 10-6 dt

d (sin wt)

= (-0.00318)(6.28f)cos(6.28ft) The transfer function of integrator is given by

I

O

V

V =

)sC

1+R(R

sC

1*R

-

f

f1

f

f

= )1+sCR(R

R-

ff1

f

If sRfCf >>1 then

I

O

V

V =

)sCR(R

R-

ff1

f

= f1CsR

1-

Vo = f1CR

1- ∫ VI dt

Let Cf = 0.1MFD & R1 =10kohm Vo = Vm sin wt =0.1 sin wt

Then Vo = - 6_3 10*1.0*10*10

1.0 ∫ sin wt = -100 cos wt/w

If f =50 Hz, Vo = -0.3cos wt


DIFFERENTIATOR:

Vi(v) 0V t (msec)

Vo(V)

0V

t (msec)

TABULAR COLUMN:

Amplitude(V) Time period(ms)

Input

Output


INTEGRATOR:

Vi(v)

0V t (msec)

Vo (V) 0V t(msec)


INTEGRATOR:

Vi (V)

0V t(msec) Vo(V) 0V t(msec)

TABULAR COLUMN:

Amplitude(V) Timeperiod(ms)

Input

Output


THEORY:

DIFFERENTIATOR:

The circuit which produces the differentiation of the input voltage at it’s output is called differentiator. The Differentiator circuit can be obtained without using active devices like op-amp, transistors etc. In such a case Differentiator is called passive

Differentiator. Differentiator using an active device like op-amp is called active differentiator. This can be obtained by exchanging the positions of R & C in the basic integrator circuit. Vo = -C1 Rf dVin / dt

The negative sign indicates that there is a phase shift of 1800 between input & output. The main advantage of differentiator is the small time constant required for differentiation.

INTEGRATOR:

In an integrator circuit , the output voltage is the integration of the input voltage. The integrator circuit can be obtained without using active devices like op-amp, transistors etc. In such a case an integrator is called passive integrator. While an integrator using an active devices like op-amp is called active integrator.

Vo = - 1/R1Cf ∫ Vin dt + Vo (0) 0 where Vo (0) is the constant of integration indicating the initial output voltage. The negative sign indicates that there is a phase shift of 1800 between input & output.The main advantage of such an active integrator is the large time constant.

PROCEDURE:

DIFFERENTIATOR: 1. Circuit connections are given as shown in fig 2. A sine wave with 1VPP and 50Hz frequency is given as input. 3. The output is obtained in the CRO and is verified with the theoritical verification 4. Graph is drawn as shown in model graph.

INTEGRATOR: 1. Circuit connections are given as shown in fig 2. A sine wave with 0.1VPP and 50Hz frequency is given as input. 3. The output is obtained in the CRO and is verified with the theoritical verification. 4. Graph is drawn as shown in model graph.

RESULT:

Thus the operation of Inverting Summer, integrator and differentiator was studied and the output was verified with the theoritical calculation.


CIRCUIT DIAGRAM:

Vin

10k

10k

100k

100k

10k

10k

120k

120k

10k

+

IC741

+

IC741

+

IC741

+

IC741

TABULAR COLUMN: Input Voltage (Volts) Output Voltage(Volts)


3.INSTRUMENTATION AMPLIFIER

EX.NO :

DATE :

AIM: To construct the Instrumentation amplifier using IC-741 for the gain 250.

APPARATUS REQUIRED:


Range Quantity

1.Dual Power Supply 2.Resistors 4.IC741 6.RPS 7.CRO

(0-30)V

10KΩ,

100KΩ

120KΩ (0-15)V (0-20)MHz

4 5 2 2 3 1 1

THEORY: The output of the transducer has to be amplified to drive the indicating or driving

system. This function is performed by instrumentation amplifier. The important features are: 1) High gain accuracy, 2) High Common mode rejection ratio, 3) High gain stability with low temparature coefficient,4) Low DC offset, 5) Low output impedance.

PROCEDURE:

1. Connections are given as per the circuit diagram. 2. Set the DC input voltage as 1 mV. 3. Note down the output voltage. 4. Calculate the gain and compare it with theoretical gain. 5. Repeat it for different input values.

RESULT: Thus the Instrumentation amplifier was constructed & Verified.


CIRCUIT DIAGRAM: Rf = 4.7KΩ + 1.2KΩ

R1 = 10KΩ

Vo (CRO)

AFO R=1.6KΩ R=1.6KΩ

Vin = 0.1µF 0.1µF

1Vpp

PIN DIAGRAM FOR IC741:





a

1

2

3

4

8

7

6

5


4(a). SECOND ORDER LOW PASS FILTER

EX.NO :

DATE :

AIM: To design and study the frequency response of second order butter worth filter

with cutoff frequency of 1000Hz.

APPARATUS REQUIRED:


Range Quantity

1. Dual Power Supply 2.Resistors 3.Capacitors 4.IC741 5.CRO 6.AFO

(0-30)V

4.7KΩ,1.2KΩ,10KΩ

1.5KΩ,100Ω

0.1µF (0-20)MHz (0-1)MHz

1 each 1 each 2 1 1 1 1

DESIGN: The general equation for transfer function is

I

O

V

V =

)]A-(1YY+)Y+ Y + (YY+ Y[Y

Y Y A

O32321421

21o

In a LPF Y4 = Y3 = sC and Y1 = Y2 =1/R

H(s) =

]) R

sC( +

R

sCAo) - (3 +

R

1[

R

A

2

2

2

O

Let 1/RC = w1

Then H(s) =

])w

s( +

w

s)A - (3 + [1

wA

2

11

O

2

1O

w1 / s is the normalized frequency.

Let (3 - Ao) =α For maximally flat response (3 - Ao) should be equal to 1.414.


TABULAR COLUMN: Vin =

Frequency (Hertz)

Vout

(Volts)

Gain Av(db) = 20 log

[Vout/Vin]

MODEL GRAPH:

Av(db)

AV (Max)

0.707AV (Max)

Pass band Stop band

Frequency(Hz)


When s=w1, H(s) = Ao / 1.414 which gives the half power frequency. If the desired cutoff frequency is 1KHz Then

1/RC = w1

f1 = RCπ2

1

Let C = 0.1µF f1 = 1 kHz

1KHz = R* 10 * 0.1*6.28

16-

R = 1.6Kohms. From 3-Ao = 1.414 we can find the theoritical pass band gain.

3-Ao = 1.414 Ao = 1.586

Also the op-amp is used in the non-inverting mode so, Ao = 1+1

f

R

R = 1.586

1

f

R

R = 0.586 (or) Rf =0.586 R1

Let R1 = 10Kohm Rf = 5.86[use 4.7k in series with 1.2K]

THEORY: A Filter is a circuit that is designed to pass a specified band of frequencies while attenuating all the signals outside the band. It is a frequency selective circuit. The filters are basically classified as Active filters & Passive filters. The Passive filter networks use only passive elements such as resistors, inductors and capacitors. Active filter circuits use the active elements such as op-amps, transistors along with the resistors, inductors & capacitors. A Low pass filter has a constant gain from 0 Hz to a high cutoff frequency fH. The circuit allows the range of frequencies from 0 to fH . This range is known as Passband. The range of frequencies beyond fH is completely attenuated & hence called stopband.

For a second order Butterworth active filter the roll-off rate should be -40db/decade or -12db/octave.


PROCEDURE: 1. Connections are made as shown in fig . 2. Set a sinusoidal input with peak 1v. Change the frequency of AFO in steps of

100Hz and note down the output amplitude from the CRO. 3. Find the gain in db for each input. 4. Plot the gain Vs frequency in semi log graph. Graph is drawn as shown in model

graph. 5. Verify if the roll-off rate is -40db/decade or -12db/octave.

RESULT: Thus the Second order low pass filter was designed and frequency response plot was drawn. Theoritical cut-off frequency = Practical cut-off frequency =


CIRCUIT DIAGRAM:

C2 = 0.01µF

R3 = 100KΩ

R1 = 4.7KΩ C1 = 0.01µF

Vin Vo (CRO)

R2 = 6.2KΩ

RL = 10KΩ



SPECIFICATION FOR IC741: +Vcc = +15V, - Vcc = -15V Input bias current : 500nA(Max)

Input resistance : 2MΩ Output resistance : 75Ω Total Power dissipation : 85mW. Ambient Temparature : 250 C Input offset voltage : 6 mV(Max) Input offset current : 200nA(Max)

1

2

3

4

8

7

6

5

a


4(b). SECOND ORDER BAND PASS FILTER

EX.NO :

DATE :

AIM: To design and study the frequency response of second order Band pass filter with

Central frequency 1KHz, Q = 3 & gain = 10.

APPARATUS REQUIRED:

DESIGN: f C = 1KHz, AF = 10 & Q = 3.

Let C 1 = C 2 = 0.01µF,

R 1 = Q / 2Πf C C AF

= 3 / (2Π* 1000* 0.01*10-6 * 10)

= 4.77KΩ

R 2 = Q / 2Πf C C (2Q2 – AF )

= 3 / [2Π* 1000*0.01*10-6 (2*32* - 10)]

= 5.97KΩ.

R 3 = Q / Πf C C

= 3 / Π*103*0.01*10-6

= 95.5KΩ.

So Choose R 1 = 4.7KΩ, R 2 = 6.2KΩ , R3 = 100KΩ.


Range Quantity


(0-30)V

4.7KΩ,6.2KΩ,100KΩ

0.01µF (0-20)MHz (0-1)MHz

1 each 1 2 1 1 1


MODEL GRAPH:

AF

0.707AF

Gain(dB) fL fC fH Frequency(Hz) TABULAR COLUMN: Vin =

Frequency

(Hertz)

Vout

(Volts)

Gain Av(db) = 20 log [Vout/Vin]


THEORY: There are Two types of Bandpass filters which are classified as per figure of merit or Quality factor. 1) Narrow bandpass filter (Q>10) 2) Wide bandpass filter(Q <10). This Filter is unique in following respects:

- It has two feedback paths, hence the name multiple feedback filter. - The Op-amp is used in inverting mode.

Generally the narrow band pass filter is designed for specific values of Center frequency & Q.

PROCEDURE :

1. Connections are given as shown in fig . 2. Set a sinusoidal input with peak 1v. Change the frequency of AFO in steps of

10Hz and note down the output amplitude from the CRO. 3. Find the gain in db for each input. 4. Plot the gain Vs frequency in semilog graph. Graph is drawn as shown in model

graph.

RESULT: Thus the Second order Band pass filter was designed and frequency response plot

was drawn. Practical central frequency = Bandwidth =


ASTABLE MULTIVIBRATOR USING OP AMP

MODEL GRAPH:

Vo

t t

TABULAR COLUMN:


Output

0.1µF

4.55 KΩ

10 KΩ

10 KΩ

VO(CRO)


5(a). ASTABLE MULTIVIBRATOR USING OP-AMP EX.NO :

DATE :

AIM: To design and construct an astable multivibrator using IC Operational amplifier 741.

APPARATUS REQUIRED:

THEORY: An astable multivibrator is a square waveform generator. Square wave is generated by forcing the op-amp to operate in the saturation region. The astable multivibrator is a free running symmetrical multivibrator because it does not require any external trigger.

DESIGN: Feedback factor = R2/(R1+R2)

Time period of the square wave T =2RC ln[(1+β)/(1-β)]

Let R1 =R2 10KΩ then β=0.5 Assume C = 0.1µF For a time period of 1ms T= 2RC ln 3 Rf = 4.55KΩ Component values: R1 =10KΩ R2 = 10KΩ Rf = 4.55KΩ C = 0.1µF


Range Quantity


(0-30)V

10KΩ,

4.5KΩ,1KΩ,27KΩ

0.1µF (0-20)MHz (0-1)MHz

1 2 each 1 1 1 1


PROCEDURE:

1.Circuit connections are made with the components of designed values. 2.The square wave output is obtained at the output pin of the op amp. 3.The amplitude and time period of the output waveform is noted and it is plotted on the graph.

RESULT: An astable multivibrator is designed and constructed and the square wave output is obtained. Time period of the square waveform (theoritical): Time period of the square waveform (practical):


MONOSTABLE MULTIVIBRATOR USING OP AMP 1KΩ V0 OA79

1µF

10KΩ VIN

1KΩ

0.01µF OA79

22KΩ


5(b). MONOSTABLE MULTIVIBRATOR USING OP-AMP EX.NO :

DATE :

AIM: To design monostable multivibrator circuits using op-amp.

APPARATUS REQUIRED:

THEORY: Monostable multivibrator has one stable state and one quasi-stable state. One shots

are used to set the timing of an event or to control a sequence of events in a digital system. A triggering pulse initiates the one shot action and generates a pulse of desired width. If additional triggers are applied to one - shot, when it is in the quasi-stable state, they are ineffective. The one shot needs time to recover after if returns to the stable state subsequent to a triggering event. Therefore, triggering pulses should not be applied so often to cause the ON time of the one-shot to exceed the duty cycle specified. If the duty cycle of a one shot exceeds the maximum specified value, there is jitter in the output pulse. That is the width of each pulse will not be constant.


Range Quantity

1. Dual Power Supply 2.Resistors 3.Capacitors 4.IC741 5.CRO 6.AFO 7.Diode

(0-30)V

1KΩ,

10KΩ,22KΩ

0.1µF,0.01µF (0-20)MHz (0-1)MHz 0A79

1 2 each 1 1 1 1 1 2


MODEL GRAPH:

VIN

T (msec) Vo +Vsat t (msec)

TABULAR COLUMN:


Output

-Vsat


PROCEDURE: Monostable multivibrator using op-amp:

1. Connections are made as per the figure. 2. Negative triggering pulse with 1 KHz repetition rate are applied. 3. The input and output waveforms are observed on a dual trace CRO and plotted. 4. The value of the pulse width is obtained theoretically from the following

relationship. T = RfC ln[ 1+R2/R1].

RESULT: - An Mono stable multivibrator is designed and constructed and the square wave output is obtained. Time period of the output waveform (theoritical)= Time period of the output waveform (practical) =


SCHMITT TRIGGER USING OP AMP:

AFO

MODEL GRAPH: Vi VUT T (msec) VLT V0 +VSAT T (msec)

a

1 KΩ

27 KΩ

1 KΩ

VO(CRO)


-VSAT

5(c). SCHMITT TRIGGER

EX.NO :

DATE :

AIM: To design Schmitt trigger circuit using op-amp.

APPARATUS REQUIRED:

DESIGN:

Vut =21

2

R+R

RVsat

Vlt =21

2

R+R

R -Vsat

Let Vut = +0.5v, Vlt =-0.5v

For IC 741, with supply voltages 615v, Vsat = 14v & -Vsat = -14v

Then 0.5 = 21

2

R+R

R 14v

R1= 27R2

Let R1=1kohm,then R2=27kohm

THEORY: The schmitt trigger is a circuit, which converts a slow changing waveform into a

fast changing waveform. In an schmitt trigger the o/p is in one of the two levels namely signal voltage Vsat or -Vsat. When the i/p voltage is rising, the level of the output changes when the input passes through a specific voltage Vut known as upper threshold voltage. Similarly when a falling input voltage passes through a specific voltage Vlt known as


Range Quantity

1. Dual Power Supply 2.Resistors 3.Capacitors 4.IC741 5.CRO 6.AFO 7.Diode

(0-30)V

1.5KΩ,

5.6KΩ,22KΩ,760Ω,1KΩ

0.1µF,0.01µF (0-20)MHz (0-1)MHz 0A79

1 2 each 1 1 1 1 1 1


lower threshold voltage, the level of the output changes. Vut is greater than Vlt. The difference between the two voltages is known as HYSTERESIS.

MODEL GRAPH: +VSAT VUT t(msec) -VSAT +VSAT VLT t(msec) -VSAT +VSAT VLT VUT t(msec) -VSAT


PROCEDURE:

Schmitt trigger using op-amp: 1. The circuit is connected as shown in figure . 2. The i/p dc voltage is increased from 0 to 5v and the o/p voltage is measured. The

output curve is drawn and Vut, Vlt and hysteresis are found out. 3. A sinusoidal input of peak 5v and 1khz is given. The input and output waveforms

are observed simultaneously on a dual trace CRO and plotted.

TABULAR COLUMN:

AMPLITUDE(V) TIMEPERIOD(ms)

INPUT(Sine Wave)

OUTPUT(Square wave) VUT

VLT

RESULT: A Schmitt trigger designed and constructed and the square wave output is obtained. Upper threshold voltage = Lower threshold voltage = Square output:


Amplitude = Time period =

RC PHASE SHIFT OSCILLATOR:

0.1µF 0.1µF 0.1µF

1.5KΩ 1.5KΩ 1.5KΩ

MODEL GRAPH:

V0(V) t(msec)

390KΩ

15KΩ

13KΩ

VO


6(a). RC PHASE SHIFT OSCILLATOR

EX.NO :

DATE :

AIM: To design and construct the RC phase shift oscillator of frequency of 500Hz and to plot the sinusoidal waveform.

DESIGN: The frequency of oscillation of RC phase shift oscillator is given by

fO =6RCπ2

1

Assume C =0.1 microfarad. Desired frequency of oscillator, f0 = 500Hz

Then 500 =6*10*1.0*R*π2

16_

∴R =1.3K ohm [use 1.2K in series with 100ohm] To avoid loading effect R1 = 10 R So R1 = 10 * 1.3 K =13 KΩ

For the loop gain AVβ to be greater than 1,Rf should be equal to 29R1

So Rf =29R1 =390kohms and Rcomp=f1

f1

R+R

RR =15kohms.


Range Quantity


(0-30)V

1.5 KΩ,

13KΩ,390KΩ,15KΩ

0.1µF, (0-20)MHz (0-1)MHz

1 3 1 3 1 1 1 1


Only polyester condenser should be used. Ceramic condenser should not be used. The condenser value is specified by voltage rating, tolerance & capacitor value. Standard values are 65v, 100v, 125v, 250v, and 400v


MAX WATTAGE OF RESISTANCE: Resistances used in the design are 1Kohm and 10Kohms .The maximum voltages

across the resistance will be supply voltage. Hence wattage of resistance is V2 / R. V2 /R=225 / 1.3K or 225 / 13K or225 / 390K or225 / 15K which is so much lower than 1/8W or 1/4W.So resistances with 5% tolerance, carbon film resistor with ⅛w or ¼w is used.

THEORY: The closed loop circuit of the RC phase shift oscillator is opened at point A. The

input of the op-amp is sullied from AFO under open loop conditions. The loop gain must be equal to 1 or slightly greater than 1 at the oscillating frequency and the phase shift must be equal to zero. The oscillator does not produce oscillation. This is the way of checking the oscillator working. The gain of the op-amp and attenuation of the phase shifting circuit must be separately determined and then multiplied to get the open loop gain at fo and this must be greater than 1.To adjust the gain a 470kohms carbon POT is chosen for 390k.The gain of the op-amp should be -29and the attenuation should be-1/29.The i/p voltage can be of the order of 0.1vpeak so that an o/p voltage of 2.9v peak

will be obtained.

PROCEDURE: 1. Construct the circuit as shown in circuit diagram in fig 2. Observe the output waveform on the CRO. 3. Adjust the feedback resistor Rf to get a perfect sine waveform. 4. Measure the amplitude and frequency and plot the waveform. Graph is drawn as

shown in model graph.

TABULAR COLUMN:


Output

RESULT: RC phase shift oscillator is designed to oscillate at 500Hz and the frequency of the output waveform is =


CIRCUIT DIAGRAM:

31.8 KΩ VO(CRO)

0.01µF






1

2

3

4

8

7

6

5

1.59 KΩ

1.59 KΩ

15.9 KΩ

0.01µF


6(b). WIEN BRIDGE OSCILLATOR

EX.NO :

DATE :

AIM: To design and construct the Wien Bridge oscillator of frequency of 10kHz and to

plot the sinusoidal waveform.

APPARATUS REQUIRED:


Range Quantity

1.Dual Power Supply 2.Resistors 3.Capacitors 4.IC741 5.CRO

(0-30)V

31.8KΩ,15.9KΩ

1.59KΩ

0.01µF (0-20)MHz

1 each 1 2 2 1 1

DESIGN:

The frequency of oscillation of Wien Bridge oscillator is given by

fO =RCπ2

1

Assume C =0.01 microfarad. Desired frequency of oscillator, f0 = 10KHz

Then 10000 = 6_10*01.0*R*π2

1

R =1.59K ohm To avoid loading effect

R1 = 10 R So R1 = 10 * 1.59 K =15.9 KΩ For the loop gain Av to be greater than 1,Rf should be equal to 2R1 So Rf =2R1 =31.8kohms. Only polyester condenser should be used. Ceramic condenser should not be used. The condenser value is specified by voltage rating ,tolerance &capacitor value. Standard values are 65v, 100v, 125v, 250v,and 400v.


MODEL GRAPH:

V0(V) t(msec) TABULAR COLUMN:


Output


THEORY: This is also RC Oscillator, which uses RC type of feedback network. The closed loop circuit of the wein bridge oscillator gives the negative feedback. To compensate the change ,an adaptive negative feedback is used. Since the op-amp works as a non inverting amplifier, the feedback network need not provide any phaseshift. The circuit can be viewed as weinbridge with a series RC network in one arm & a parallel RC network in the adjoining arm. Resistors R1 & Rf are connected in remaining two arms. The condition of zero phase shift is obtained by balancing the bridge.

PROCEDURE:

1. Construct the circuit as shown in circuit diagram . 2. Observe the output waveform on the CRO. 3. Adjust the feedback resistor Rf to get a perfect sine waveform.

4. Measure the amplitude and frequency and plot the waveform. RESULT: Thus the Wien Bridge oscillator circuit is designed output waveform is obtained . Theoritical frequency : 10KHz

Practical frequency :


CIRCUIT DIAGRAM :

RA

OA79

RB Vo

0.1µF

0.01µF

PIN DIAGRAM FOR IC555:

IC555 1 = Ground, 2 = Trigger, 3 = output, 4 = Reset, 5 = Control voltage,

6 = Threshold, 7 = Discharge, 8 = +Vcc

+Vcc =+5v

4 8 7

IC555 6 3 2

5 1

7.25KΩ

3.625K

Ω

1

2

3

4

8

7

6

5


7(a). ASTABLE MULTIVIBRATOR USING TIMER

EX.NO :

DATE :

AIM: To Construct astable multivibrator using IC555 timer & to generate a 1KHz square waveform.

APPARATUS REQUIRED: Equipments & Components

Range Quantity

1. Power Supply 2.Resistors 3.Capacitors 4.IC555 5.CRO 6.Diode 0A79

(0-30)V

3.625KΩ,7.25KΩ

0.01µF

0.1µF (0-20)MHz

1 each 1 1 1 1 1 1

DESIGN:

Case (I) Given f = 1KHz and D =0.5

Frequency of astable multivibrator, f = C)R+R(

45.1

BA

Then C =f)R+R(

45.1

BA

D = )R+R(

R

BA

B = 0.5

0.5RA +0.5RB = RB

RA = RB

Let C = 0.1µF, RA =RB =R

f = C)R+R(

45.1

BA

f =RC2

45.1 =>1KHz =

R*10*0.1*2

45.16-

R = 7.2Kohm


SPECIFICATION FOR IC555: Supply Voltage : +5V to +18V. Maximum current : 200mA.

FORMULA FOR DUTY CYCLE: Duty Cycle % = Ton / Ton + Toff *100.

MODEL GRAPH:

Vo(v)

Vcc

0

t(ms)

TABULAR COLUMN:


Output

1/3 Vcc

2/3 Vcc


Case (ii) Given f = 1KHz and D =0.25

Frequency of astable multivibrator, f = C)R2+R(

45.1

BA

Then C = f)R2+R(

45.1

BA

D = BA

B

R2+R

R = 0.25

RA +2RB = 4RB

RA = 2RB

Let C = 0.1µF, RA = 2RB

f = C)R2+R(

45.1

BA

f = CR4

45.1

B

=>1KHz = B

6- R*10*0.1*4

45.1

RB = 3.625Kohm Then RA =7.25Kohms

THEORY: The 555 timer is a highly stable device for generating accurate time delay or oscillation. A single 555 timer can provide time delay ranging from microseconds to hours whereas counter timer can have a maximum timing range of days. An astable multi vibrator is a square wave form generator. Square wave form is generated by forcing the Op-amp to operate in the saturation region. It is a free running symmetrical multivibrator because it does not require any external trigger.

PROCEDURE:

1. The connections are given as shown in the circuit diagram . 2. The square wave form is obtained at output pin of Op-amp. 3. Note the amplitude & Time period of the of the waveform & Plot it in

The graph. 4. Duty cycle is calculated using the formula given.

RESULT: Thus IC555 timer was operated in astable mode to generate square wave. Theoritical Duty cycle : 25% Practical Duty cycle : -----------.


CIRCUIT DIAGRAM:

+VCC = +5V

10KΩ

OA79 10KΩ

0.001µ F VIN

Vo

0.1µF

0.01µF PINDIAGRAM FOR IC555:

IC555 1 = Ground, 2 = Trigger, 3 = output, 4 = Reset, 5 = Control voltage, 6 = Threshold, 7 = Discharge, 8 = +Vcc

a

4 8 2 6 7

IC 555 3

5 1

1

2

3

4

8

7

6

5


7(b).MONOSTABLE MULTIVIBRATOR USING TIMER

EX.NO :

DATE :

AIM: To Operate the IC555 timer in Monostable mode to generate a 1KHz square waveform.


Range Quantity

1. Power Supply 2.Resistors 3.Capacitors 4.IC555 5.CRO 6.Diode0A79 7.AFO

(0-30)V

10KΩ

0.01µF

0.1µF

0.001µF (0-20)MHz (0-1)MHz

1 2 1 1 1 1 1 1 1

DESIGN:

Time period of monostable multivibrator = 1.1RC. T = 1msec.

Assume C = 0.1µF, Then R = T/1.1C R = 1*10-3 / 1.1*0.1*10-6

R ≅ 10KΩ.

THEORY: Monostable multivibrator has one stable state & one quasistable state.One shots are used to set the timing of an event or to control a sequence of events in a digital system. A triggering pulse initiates the oneshot action & generates a pulse of desired width .The oneshot needs time to recover after it returns to the stable state subsequent to a triggering event. Therefore triggering pulses should not be applied so often to cause the on time of the oneshot to exceed the duty cycle specified. If the duty cycle of a oneshot exceeds the maximum specified value, there is a jitter in the output pulse. That is the width of each pulse will not be constant.


SPECIFICATION FOR IC555: Supply Voltage : +5V to +18V. Maximum current : 200mA. MODEL GRAPH:

VIN (V)

t(ms)

Vo

t(ms)

TABULAR COLUMN:


Output


PROCEDURE: 1. The connections are given as shown in the circuit diagram (Fig 17.2). 2. The negative trigger pulses of frequency 1KHZ are applied at the trigger input. 3. The output waveform is observed and pulse duration is measured. 4. Theoritically the pulse duration is calculated as

thigh =1.1RAC

RESULT: Thus IC555 timer was operated in Monostable mode to generate square waveform. Theoritical pulse duration = 1msec. Practical pulse duration = --------.


CIRCUIT DIAGRAM: 723 VOLTAGE REGULATOR 10 V DC (0-100mA)

10Ω + -

680Ω

1kΩ 10kΩ

1kΩ 3.3kΩ

33kΩ

2.2kΩ

0.001µF

MODEL GRAPH:

Load regulation: VL(V)

RL(mA)

Line regulation: VL (V)

6 12 11

10

IC 723 2

5

3

4

7 13

a


8(a). DC POWER SUPPLY USING LM723

AIM:

To study the operation of 723-voltage regulator IC and to obtain Load regulation & Line regulation.

APPARATUS REQUIRED:


Range Quantity

1. Power Supply 2.Resistors 3.Capacitor 4.IC723 5.Ammeter 6.Voltmeter 7.Decade resistor box

(0-30) V

620Ω,

2.2KΩ,

10Ω,

33KΩ,

3.3KΩ,

0.001µF

(0-50) mA (0-10) V

1 1 1 1 1 1 1 1 1 1

THEORY:

IC723:

IC 723 general-purpose voltage regulator is inherently low current device, but can be boosted to provide 5 amps or more current by connecting external components. It has two separate sections. The Zener diode, a constant current source and reference amplifier produce a fixed voltage of about 7 volts at the terminal Vref. The constant current source forces the Zener diode to operate at a fixed point so that the Zener outputs a fixed voltage.

The other section of the IC consists of an error amplifier, a series pass current limit transistor Q2.the error amplifier compares a sample of the output voltage applied at the inverting input terminal to the reference voltage Vref applied at the NI input terminal. The error signal controls the conduction of Q1.


PROCEDURE:

1. Connections are made as shown in the circuit diagram 2. The power supply voltage Vin is adjusted to 10 V and Vref is

measured at pin 6. 3. The load current IL and load voltage VL are measured and recorded for various values of RL. 4.To measure line regulation, load resistance is kept constant and load voltage VL is measured for various values of Vin.

5.Graphs are plotted with IL versus VL and Vin versus VL.

RESULT:

Thus the DC Power supply using IC 723 is performed and graphs are plotted.


CIRCUIT DIAGRAM:

R1=240Ω

Vin

C2 =1µF Multimeter

R2 =3KΩ

C1 = 10µF

MODELGRAPH:

VL (v)

Vin (V)

LM317

Vin Vout

ADJ


8(b). DC POWER SUPPLY USING LM317

EX.NO :

DATE :

AIM: (i) To design an adjustable voltage regulator using LM317 for the following

specifications.

APPARATUS REQUIRED:


Range Quantity

1. Power Supply 2.Resistors 3.Capacitors 4.LM317 6.Voltmeter 7.Decade resistor box

(0-30)V

240Ω

10µF,1µF (0-20) V

1 1 1 1 1 1

THEORY: LM317 is a adjustable voltage regulator. They have the following performance and reliability advantages over the fixed types. *Improved system performance by having line and load regulation of a factor of 10 or better.

*Improved overload protection allows greater output current over operating temparature range.

*Improved system reliability with each device being subjected to 100% thermal limit burn- in. Thus the adjustable voltage regulators have become more popular because of versatility, performance, and reliability.

DESIGN: Designing an adjustable voltage regulator LM317 to satisfy the following specifications: Output voltage Vo = 5 to 12V. Output current Io = 1.0A IAdj = 100 micro amps maximum. If we use R1 = 240ohm., then for Vo = 5V the value of R2 = from equation Vo =VREF ( 1+ R2 / R1) + IAdjR2. Where VREF = 1.25V(Reference voltage between the output and adjustment terminals. 5 = 1.25(1+R2/240) + (10-4)R2 R2 = 3.75/(5.3)(10-3) = 0.71kohm.


similarly for Vo = 12V, the value of R2 is 12 = 1.25(1+R2/240) + (10-4)R2

R2 = 10.75/(5.3)(10-3) = 2.01kohm.

PROCEDURE:

1.Connections are made as shown in the circuit diagram 2.The power supply voltage Vin is adjusted to be greater than or equal to 15V.

3. The load voltage VL are measured and recorded for various values of R2.

4.To measure line regulation, load resistance is kept constant and load voltage VL is measured for various values of Vin.

5.Graph is plotted with Vin versus VL

RESULT: Thus the DC Power supply using LM317 is designed and graph is plotted..


Block diagram of NE 565 PLL +VCC

Input Input Demodulated output Phase Reference Comparator output VCO Output CT

+VCC RT CT

-VCC CIRCUIT DIAGRAM FOR PLL:

+6V

RT 6.8K Ω C 1µF C1 0.001µF Demodulated

output VIN Reference Output VCO output (fo)

0.001µµµµF -6V

10

2 3.6KΩ 7 3

5 6

4

8 9 1

Phase Detector

VCO

Amp.

10 8 2 7 6

NE 565 5 3 4

9 1

a


9. PLL CHARACTERISTICS AND FREQUENCYMULTIPLIER USING

NE565.

EX.NO :

DATE :

AIM: To study the operation of NE565 PLL and to use it as frequency multiplier


Range Quantity

1. Power Supply 2.Resistors 3.Capacitors 4.IC565 5.Transistor 6.IC7490 7.AFO 8.CRO

(0-30)V

6.8KΩ,20KΩ,2KΩ,10KΩ,

4.7KΩ,

0.001µF,1µF,10µF,0.01µF 2N2222 (0-1)MHz (0-20)MHz

1 each 1 1 1 1 1 1 1 1

THEORY: IC NE565 phase locked loop is available as a 14-pin DIP package. The block diagram is shown in figure1. The output frequency of the VCO (both inputs 2,3 grounded) is given by f0 = 0.25/RtCt Hz where Rt and Ct are the external resistor and capacitor connected to pin8 and pin9 .The VCO free running is adjusted with Rt and Ct to be at the center of the input frequency range. It may be seen that phase locked loop is internally broken between the VCO output and the phase comparator input. A short circuit between pins 4 and 5 connects the VCO output to the phase comparator so as to compare f0 with input signal fs. A capacitor c is connected between pin7 and pin 10(supply terminal) to make a low pass filter with the internal resistance of 3.6k.


PLL AS FREQUENCY MULTIPLIER:

+6V

20 kΩ

C=10µF

RT 2 K Ω

C1=0.001µF VIN

10kΩ fin

0.01µF -6V +6V

4.7kΩ 2N2222

11 8 2 7

NE 565 4 3 5

9 1

a

11 5

7490

(÷5)

1


PROCEDURE:

Study of PLL: 1. Connections are made as shown in fig 2. The free running frequency of Vco at pin4 is measured with the input signal Vin

set equal to zero. It is compared with the calculated value = 0.25/RtCt 3. The input signal of 1Vpp square wave at a 1khz frequency is applied to pin2. The

signal is displayed on the scope. 4. The input frequency is increased gradually till the PLL is locked to the input

frequency. This frequency f1is the lower end of the capture range. The input frequency is further increased to a frequency f2 till PLL tracks the input signal. This frequency f2 gives the upper end of the lock range. If input frequency is increased further the loop will get unlocked.

5. The input frequency is decreased gradually till the PLL is again locked. This is the frequency f3, the upper end of the capture range. The input frequency is decreased further until the loop is unlocked. This frequency f4 gives the lower end of the lock range.

6. The lock range ∆fL = f2-f4. It is compared with the calculated value of ±

7.8f0/12.also the capture range is ∆fc = f3-f1. It is compared with the calculated

value of capture range ∆fc = ± [∆fL/2Πx3.6x103xc] ½

PLL as frequency multiplier: 1. Connections are made as shown in fig. The circuit uses a 4 bit binary counter 7490

as a divide by 5 circuit. 2. The input signal is set at 1vpp square wave at 500Hz. 3. gThe VCO frequency is varied by adjusting the 20k potentiometer till the PLL is

locked. The output frequency is measured. It should be 5 times the input frequency.

4. The step 2 and 3 are repeated for input frequency of 1KHZ and 1.5KHZ RESULT:

PLL is studied and used as frequency multiplier

LINEAR INTEGRATED CIRCUITS LAB

Schematic Diagram: Instrumentation Amplifier

Waveform: Instrumentation Amplifier


Schematic Diagram: Instrumentation Amplifier

: Instrumentation Amplifier

P a g e | 73


11.SIMULATION USING PSPICE

(i) INSTRUMENTATION AMPLIFIER

EX.NO :

DATE :

Aim

To simulate instrumentation amplifier circuit using PSPICE circuit simulator and to verify the

corresponding graphs plotted.

Software required

PSPICE students’ version 9.1

Procedure

1. Draw the schematic diagram in pspice schematic editor

2. Go choose the icon “set up -> analysis”, for choosing proper analysis options.

3. Now select the option “DC sweep”.

4. Choose “voltage source” and complete the remaining options like start value and end

value.

5. Now choose the icon “set up -> Examine netlist”, and if the netlist has no errors, choose

the “simulate” option which is under “set up”.

6. The waveform will pop up after the simulation is done.

* Schematics Netlist *

R_R8 $N_0002 $N_0001 10k

R_R9 $N_0004 $N_0003 10k

R_R3 $N_0006 $N_0005 10k

R_R10 $N_0008 $N_0007 10k

R_R5 $N_0008 $N_0009 10k

R_R6 $N_0010 $N_0003 100k

R_R7 $N_0001 0 100k

R_R2 $N_0005 $N_0004 120k

R_R1 $N_0006 $N_0002 120k

E_U9 $N_0004 0 VALUE LIMIT(V($N_0009,$N_0005)*1E6,-15V,+15V)

E_U10 $N_0007 0 VALUE LIMIT(V(0,$N_0008)*1E6,-15V,+15V)



V_Vin $N_0009 0 25mV

Result

Thus the instrumentation amplifier circuit is simulated and the required graphs are plotted.


Schematic Diagram: Low pass-

Waveform: Low pass- second o


- second order filter

second order filter

P a g e | 75


(ii) LOW PASS – SECOND ORDER FILTER

EX.NO :

DATE :

Aim

To simulate low pass – second order filter circuit using PSPICE circuit simulator and to verify its

frequency response graph.

Software required


Procedure



3. Now select the option “AC sweep”.

4. Choose “Decade” for graph type and complete the remaining options like start frequency

and end frequency .





R_R4 0 $N_0001 10k

R_R3 $N_0003 $N_0002 2.5k

R_R2 $N_0002 $N_0004 2.5k

C_C1 $N_0005 $N_0002 0.1u

R_R1 $N_0001 $N_0005 6.9k

V_Vin $N_0003 0 DC 0V AC 5V


C_C2 0 $N_0004 0.1u

Result

Thus the low pass – second order filter circuit is simulated and the required frequency

response graphs are plotted.


Schematic Diagram: High pass

Waveform: High pass- second order filter


pass- second order filter

second order filter

P a g e | 77


(iii) HIGH PASS – SECOND ORDER FILTER

EX.NO :

DATE :

Aim

To simulate high pass – second order filter circuit using PSPICE circuit simulator and to verify

its frequency response graph.

Software required


Procedure





and end frequency .





V_V1 $N_0001 0 DC 0V AC 5V


C_C2 $N_0001 $N_0005 0.1u

C_C1 $N_0005 $N_0002 0.1u

R_R1 $N_0003 $N_0004 5.86k

R_R2 0 $N_0003 10k

R_R3 0 $N_0002 1.6k

R_R4 $N_0004 $N_0005 1.6k

Result

Thus the high pass – second order filter circuit is simulated and the required frequency



Schematic Diagram: Band pass

Waveform: Band pass – second order filter


Schematic Diagram: Band pass – second order filter

second order filter

P a g e | 79


(iv) BAND PASS – SECOND ORDER FILTER

EX.NO :

DATE :

Aim

To simulate band pass – second order filter circuit using PSPICE circuit simulator and to verify

its frequency response graph.

Software required


Procedure





and end frequency .





E_U1 $N_0002 0 VALUE LIMIT(V(0,$N_0001)*1E6,-15V,+15V)

R_R2 $N_0004 $N_0003 4.7k

R_R3 $N_0002 $N_0001 100k

C_C2 $N_0001 $N_0002 0.01u

C_C1 $N_0004 $N_0001 0.01u

V_Vin $N_0003 0 DC 0V AC 5V

R_R4 0 $N_0002 10k

R_R1 0 $N_0004 6.2k

Result

Thus the band pass – second order filter circuit is simulated and the required frequency



Schematic Diagram: Astable multivibrator

Waveform: Astable multivibrator


multivibrator using opamp

multivibrator

P a g e | 81


(v) ASTABLE MULTIVIBRATOR USING OPAMP

EX.NO :

DATE :

Aim

To simulate astable multivibrator circuit (opamp based) using PSPICE circuit simulator and to

verify the waveform

Software required


Procedure



3. Now select the option “transient”.

4. Choose appropriate print step (eg:10 ns) and final time.



6. The waveform window will pop up after the simulation.


R_R3 $N_0002 $N_0001 4.55k


R_R2 0 $N_0003 10k

R_R1 $N_0003 $N_0001 10k

C_C1 0 $N_0002 0.1u IC=0.001V

Result

Thus the astable multivibrator circuit using opamp is simulated and the required

waveforms are obtained.


Schematic Diagram: Monostable

Waveform: Monostable multivibrator


stable multivibrator using opamp

multivibrator using opamp

P a g e | 83


(vi). MONOSTABLE MULTIVIBRATOR USING OPAMP

EX.NO :

DATE :

Aim

To simulate monostable multivibrator circuit (opamp based) using PSPICE circuit simulator and

to verify the waveform.

Software required


Procedure









R_R1 $N_0002 $N_0001 1k

D_D2 $N_0003 $N_0004 D1N4002

R_R3 0 $N_0003 1k

D_D3 $N_0002 0 D1N4002

R_R2 $N_0003 $N_0001 10k

R_R5 $N_0006 $N_0005 4.55k

R_R6 0 $N_0007 10k

R_R7 $N_0007 $N_0005 10k

C_C2 0 $N_0002 1u

R_R4 0 $N_0004 22k



C_C3 0 $N_0006 0.05u IC=0.001V

C_C1 $N_0005 $N_0004 10u

Result

Thus the astable multivibrator circuit using opamp is simulated and the required



Schematic Diagram: SCHMITT TRIGGER

Waveform: SCHMITT TRIGGER


agram: SCHMITT TRIGGER

SCHMITT TRIGGER

P a g e | 85


(vii). SCHMITT TRIGGER USING OPAMP

EX.NO :

DATE :

Aim

To simulate Schmitt Trigger circuit (opamp based) using PSPICE circuit simulator and to verify

the waveform.

Software required


Procedure









R_R2 0 $N_0001 1k

R_R1 $N_0001 $N_0002 27k


R_R3 $N_0004 $N_0003 1k

V_V1 $N_0004 0 STIMULUS=sine

Result

Thus the Schmitt Trigger circuit using opamp is simulated and the required waveforms

are obtained.


Schematic Diagram: RC phase

Waveform: RC phase shift oscillator


Schematic Diagram: RC phase shift oscillator

: RC phase shift oscillator

P a g e | 87


(viii). RC PHASE SHIFT OSCILLATOR

EX.NO :

DATE :

Aim

To simulate the RC phase shift oscillator using PSPICE circuit simulator and to verify the

waveform

Software required


Procedure









C_C3 $N_0001 $N_0002 0.1u

R_R1 $N_0003 $N_0002 390k

R_R2 $N_0004 $N_0003 13k

R_R6 0 $N_0004 1.5k

R_R5 0 $N_0005 1.5k

R_R4 0 $N_0001 1.5k

C_C1 $N_0005 $N_0001 0.1u

R_R3 0 $N_0006 15k

C_C2 $N_0004 $N_0005 0.1u IC=.001

X_U2A $N_0003 $N_0006 v+ v- $N_0002 LM324

V_V2 0 v- 15V

V_V1 v+ 0 15V

Result

Thus the RC phase shift oscillator using opamp is simulated and the required waveforms

are obtained.


Schematic Diagram : Wien bridge oscillator

Waveform : Wien bridge oscillator


bridge oscillator

: Wien bridge oscillator

P a g e | 89


(ix). WIEN BRIDGE OSCILLATOR

EX.NO :

DATE :

Aim

To simulate the wien bridge oscillator using PSPICE circuit simulator and to verify the

waveform

Software required


Procedure









C_C2 $N_0001 $N_0002 0.01u

C_C1 0 $N_0003 0.01u IC=.1


R_R4 $N_0004 $N_0002 31.8k

R_R3 0 $N_0004 15.9k

R_R2 $N_0003 $N_0001 1.59k

R_R1 0 $N_0003 1.59k

Result

Thus the wien bridge oscillator using opamp is simulated and the required waveforms are

obtained.


Schematic Diagram: Astable multivibrator



multivibrator using 555 timer


P a g e | 91


(x). ASTABLE MULTIVIBRATOR USING 555 TIMER

EX.NO :

DATE :

Aim

To simulate the astable multivibrator (555 timer based) using PSPICE circuit simulator and to

verify the waveform

Software required


Procedure









D_D2 $N_0001 $N_0002 D1N4002

V_V1 $N_0003 0 5V

C_C2 0 $N_0004 .01u

C_C1 0 $N_0002 .01u

R_R1 $N_0001 $N_0002 3.63k

R_R2 $N_0001 $N_0003 7.3k

X_X1 0 $N_0002 $N_0005 $N_0003 $N_0004 $N_0002 $N_0001 $N_0003 555D

+ PARAMS: MAXFREQ=3E6

R_R3 0 $N_0005 100k

Result

Thus the astable multivibrator circuit using 555 timer is simulated and the required



HALF WAVE RECTIFIER

VI(v)

t(msec) Vo(v) t(msec)

a

10K

10K

10KΩ

1N4007

1N4007 AFO

VO(CRO)


12.ADDITIONAL EXPERIMENTS

(a). PRECISION RECTIFIER

EX.NO :

DATE :

AIM: To construct and study the working of half wave and full wave precision rectifiers.

APPARATUS REQUIRED:

1. IC 741 -2Nos 2. Diode(1N4007) -2 Nos 3. Resistor 10Kohm -6 Nos 4. Dual power supply (0-30) v-1 No 5. AFO (0 to 1 MHZ) -1 No 6. CRO (0 to 20 MHZ) -1 No

DESIGN: All the resistances are chosen as 10 Kohms and this condition make output voltage

is equal to the input voltage.

THEORY: Matched diodes are used .To get matched diodes CA3046 transistor array is used. This CA3046 in an IC with 5 transistors in a DIP package. If matched are not used for FWR the positive halves of the rectified wave will not be equal. Precision rectifier rectifies voltages of the order of millivolts much lower than the cut in voltage of diodes. All the resistances are chosen in kilo ohm range so that the AFO is not loaded and much greater than the o/p resistance of AFO (50 ohm).

MAX WATTAGE OF RESISTANCE: Resistances used in the design are 1Kohm and 10Kohms .The maximum voltages across the resistance will be supply voltage. Hence wattage of resistance is V2 / R. V2 /R= 225 / 10K which is so much lower than 1/8W or 1/4W.So resistances with 5% tolerance, carbon film resistor with 1/8W or 1/4W is used.

PROCEDURE: 1. Connections are made as shown in fig 2. The transfer characteristic of the FWR is done for direct voltages of positive and

negative values. 3. For this a variable supply with course knob of the supply fully anticlockwise and

fine knob is adjusted to get voltage in the range of 1-50mv in both polarities. 4. Various I/p voltages are applied to FWR and o/p voltages are noted. For the

measurement of input and output voltage DMM in dc range is used.


FULL WAVE RECTIFIER:

Vin t(msec) t(msec)

AFO

VO(CRO)

VO(v)

10KΩ

a

10KΩ 10KΩ 10KΩ

10KΩ

10KΩ 1N4007

1N4007


5. Experiment is to done for both polarities of input voltage. Note that slope=1 and

small signal gain is equal to slope. The transfer characteristics is shown in fig. 6. Alternatively an ac input can also be given from an AFO, with amplitude in

millivolts range. The input and output waveforms are viewed on the CRO & plotted as shown in model graph.

TABULAR COLUMN:

Voltage Time Frequency

Half wave Input.

Output

Full wave

Input

Output

RESULT: The input and output waveforms of the HWR & FWR are plotted.


a

1 KΩ

1 KΩ

VO(CRO)

COMPARATOR WITH POSITIVE Vref _

AFO Vref = 0.5V

MODEL GRAPH:

Vi

VREF T (msec) Vo +VSAT T (msec) -VSAT


(b). COMPARATOR

EX.NO :

DATE :

AIM:

To design the comparator in non-inverting mode using op-amp.

APPARATUS REQUIRED: -

1. IC741 -1No

2. Resistor 1Kohm -2 Nos

3. AFO -1 No

4. CRO -1 No

5. RPS -1 No

6. Linear IC trainer -1 No

7. Dual power supply -1 No

THEORY: -

A comparator is a circuit, which compares a signal voltage applied at one input of

an op-amp with a known reference voltage at the other input. It is basically an open loop

op-amp with output I V (=Vcc). There are two types of comparators. (a) Non-inverting

comparator (b) inverting comparator

NON- INVERTING COMPARATOR: -

In this circuit, a fixed reference voltage V reference is applied to inverting input

and a time varying signal Vi is applied to non- inverting input. The output voltage is at -

Vsat for Vi < Vref and V0 goes to t Vsat for Vi > Vref. The Vref may be positive or

negative voltage.


COMPARATOR WITH NEGATIVE Vref

AFO -- Vref = -o.5V

VI

T (msec) -VREF V0 +VSAT T (msec) -VSAT

a

1 KΩ

1 KΩ

VO(CRO)


PROCEDURE: - 1. The circuit is connected as shown in figure and reference=+0.5V is applied. 2. The signal generator is adjusted so that VI = 2 Vpp sine wave at 1KHz 3. Using a CRO the input and output waveforms are observed simultaneously. The

output waveform is plotted. 4. The circuit is connected as shown in fig and Vref =-0.5V is applied steps 2 and 3

are repeated.

RESULT: The operation of the comparator is studied.


TRIANGULAR WAVE GENERATOR

MODEL GRAPH:

Vo

T(msec)

VO

0.01µF


(c).TRIANGULAR WAVEFORM GENERATOR

EX.NO :

DATE :

AIM: To construct and study the operation of a triangular waveform generator using IC 741.

APPARATUS REQUIRED:

1. IC741 -2Nos 2. Resistor 10Kohm -2 Nos 3. 1Mohm -1 No 4. 100kohm -1 No 5. Capacitor 0.01MFD -1 No 6. 0.05MFD -1 No 7. Pot 100kohm -1 No 8. CRO -1 No 9. RPS -1 No 10. Linear IC trainer. -1 No

THEORY:

A triangular is generated by integrating a square waveform. The circuit uses a square waveform generator and integrator. The frequencies of the two waveforms are equal while the amplitude of triangular waveform is smaller. The decrease in amplitude is because the reactance of the feedback capacitor C2 decreases with increase in frequency. The resistor R2 is connected in parallel with C2 to avoid saturation problems at low frequencies.

PROCEDURE:

1. Connections are made as shown in circuit diagram . 2. The output voltages VO1 and VO2 are noted. 3. The amplitude and time period of the square wave VO1 and triangular wave

VO2 are recorded.

RESULT: A triangular wave generator is constructed using IC741 and studied.

lic lab manual

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