EXPERIMENT 1 Date: AIM: To study the frequency response...

PDDC SEM 3 Advanced Electronics Electronics & Communication Engineering Department

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EXPERIMENT – 1 Date:

AIM: To study the frequency response of single stage RC couple amplifier & to find it’s…

1) Cut off Frequency

2) Bandwidth

3) Midband Gain

Appratus: Connecting wires, function generator, CRO, voltage supply and as follows…

Component Value

Resistor 2KΩ, 4 KΩ, 22KΩ, 100KΩ, 220KΩ

Capacitor 1µF, 47µF

Transistor BC177

Theory:

Frequency response of an amplifier

The frequency response of any amplifier is helpful for analysis the behavior of the amplifier of

different frequencies of input signal. It is a graph of amplifier output voltage frequency of input

signal. Ideally the frequency response should be flat over entire frequency range.

Different region in frequency response :-

1) Low frequency region.

2) Mid frequency region.

3) High frequency region.


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Bandwidth of an Amplifier :-

The mid frequency region of frequency response is useful for amplification as amplifier gain is

constant at a value called midband voltage gain Av. This range of frequency will be of practical

use as you can use amplifier effectively over this frequency range.

Using the frequency response graph we come to know that there is a band of frequency in

which magnitude of output voltage of gain is either equal or relatively close to their mid

frequency band value. To fix frequency range in which the amplitude gain is relatively range a

cutoff level of 0.707 A is chosen the corresponding frequency f1 & f2 are known as cutoff

frequency 3db of frequency

BW=f2-f1 Hz

PROCEDURE:

To study frequency response of single stage amplifier at frequency 1000Hz provide input at few

milli volts at base of circuit, check the output. Calculate voltage gain this is midband voltage

gain.

Now start taking readings for f=1,2,3,4……….30,40 continue till the maximum possible value of

frequency in the patterns.

Plot these values of gain in db v/s frequency

Calculations:

From the Graph:

Fl = 22.90

Fh + 14.45

Bandwidth B.W. = Fh-Fl

= 22.90-19.45

= 8.45

Av (Mid) = 10.890 & in db =20.741 db

For db = 20 log (Av (mid))

= 20 log (10.890)

= 20.741 db

CONCLUSION:

From the above experiment we can find out frequency bandwidth & voltage gain of single step amplifier

& we conclude has following points….

1) In the low frequency, the gain or output voltage decrease due to the increased reactance by

coupling bypass capacitor.

2) In high frequency region output voltage & gain will decrease due to the transistor inlend

capacitor.


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EXPERIMENT –2

AIM: To study the frequency response of multi stage RC couple amplifier & to find it’s…

4) Cut off Frequency

5) Bandwidth

6) Midband Gain

Apparatus: Connecting wires, function generator, CRO, voltage supply and as follows…

Component Value

Resistor 2KΩ, 4 KΩ, 22KΩ, 100KΩ, 220KΩ

Capacitor 1µF, 2µF, 91µF

Transistor BC177

Theory:

The most applications a single stage transistor will not be able to meet the specification such as

voltage, current gain.

Transistors are then connected as shown in circuit fig. behind.

In such cases, if more than one amplifier are cascaded then the input takes are of the input

impedence which the output stage takes care of the impedence matching requirements and the

middle stage will fulfill the high voltage gain requirement.


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Thus the cascading of amplifier is to be done when,

1) Amplification provided by a single stage amplifier is not sufficient large.

2) The input & output impedance are not of correct value.

Output voltage gain of double stage are

Av = Vo = Vo Vol = Av2 × Av1

Vin VDi Vin

The bandwidth of the multi amplifier is always less than the bandwidth of single stage amplifier.

Bandwidth of double stage amplifier = fH(a) – fL(a)

PROCEDURE:

To draw the frequency response of a double stage amplifier give input at few mV at input side &

check output.

Now take reading for F=1, 2, 3, 4, 5…..100,200….MHz till the minimum possible value of

frequency in this pattern plot the value of gain in graph of db v/s frequency.

CALCULATION:

We can from the graph

fL = 30 Hz ; Fw = 900 KHz

B.W. = FH – FL

= 900 × 103 – 30 ≈ 900 KHz

Av = 126.284

Av in db = 20 log10 (126.284) = 42.027db

CONCLUSION:

We can conclude that with increase in the no. of stage ‘n’ the gain increase but the bandwidth goes on

decreasing.


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Experiment:- 3

AIM: To Study the Inverting & Non-Inverting Operational Amplifier.

Apparatus: 1) IC 741

2) Resistor 1.5KΩ, 2.2KΩ

3) Connecting wires

4) Function Generator

5) CRO

Inverting Amplifier

The signal to be amplified has been connected to the inverting terminal via resistor R1 and other

resistor(Rf) is connected between output and inverting terminal in called feedback resistance.

It introduces a negative feedback.

Non-Inverting terminal is connected to ground.

Output is an amplified version of input signal.

Expression voltage gain Af is

Af=-(Rf/R1)*Vin

Here –ve sign is due to opposite phase of output voltage that of Input Voltage.


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Non-Inverting Amplifier

In Non-Inverting amplifier using operational amplifier signal which is to be amplified is

connected to the non-inverting terminal of amplifier.

Input & Output voltage are in phase with each other.

Negative feedback is in corporate in this circuit via feedback resistor Rf which is

connected between output & Inverting terminal of operational amplifier.

Expression voltage gain Af is

Af= (1+Rf/R1)*Vin

Voltage Follower:

When a Non-Inverting amplifier is configured for unity gain, it is called a voltage follower.

The output voltage is equal to and in phase with the input.

It is preferred because it has much higher resistance and the output amplitude is exactly equal

to the input.

In the voltage follower simply open R1 & short Rf.

Output voltage is feedback into inverting terminal of op-amp, the gain of feedback circuit is

called one.


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Experiment – 4 Date:

AIM: To study summing inverting and non-inverting amplifier using op-amp. Appartus:

Function generator CRO Power supply Multimeter Resistors Bread board Connecting wires

Theory:

The input impedance of the op-amp is very high so it can be used as summing circuit because

signal can be applied simultaneously. Output of the summing amplifier gives the sum of the signal applied to the input terminals. These are two types of summing amplifier circuits as shown in figure-1 & figure-2.

Inverting summing amplifier:

In inverting summng amplifier, input V1, V2 is applied to the inverting input

through R1, R2 respectively. THe non-inverting input is connected to ground so, bias current is almost zero & input current flows through feedback resistor Rf.


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I = I1 + I2 = V1 / R1 + V2 / R2

V0 = -Rfi = -Rfi [ V1/R1 + V2/R2] If R1 = R2 = Rin then, V0 = -Rf / Rin

Thus, the output voltage is proportional to sum of individual input voltage. If Rf = - Rin then,

Average -Inverting amplifier

In average inverting amplifier,

I = I1 + I2 = V1 / R1 + V2 / R2 V0 = - Rf I2 = -Rf (V1 / R1 + V2/R2) If R1 =R2 = Rin then, V0 = -Rf / Rin (V1 + V2)

Non - inverting summing amplifier

In non-inverting summing amplifier terminal in ground through resistor R2. The signal to the

added are V1 & V2 are applied to the non-inverting terminal through resistor R1 & R2 feedback resistor is Rf.

Vout = (1 + Rf / Rin) Vin

Vout = (1 + Rf + R1) (V1 + V2)/ 2

If Rf = R1 then V0 = 2 * (V1+V2)/2

If R0 = 0, R1 = ∞ then V0 = (1+0) [(V1 + V2)/2]

V0 = (V1+V2) / 2


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Procedure:

For summing inverting amplifier, connect the circuit as shown in figure.

Give supply to the circuit, using regulated power supply

Adjust the V1, V2 and take reading of output voltage respectively.

For summing inverting amplifier, connect the ckt as shown figure.

Give supply to the ckt with the help of regulated power supply.

Calculation:

For summing amplifier, Vout = -Rf / Rin (V1 + V2)

Rf = 2.2 KΩ, V1 = 5.08 V and Rin = 2.2 KΩ, V2 = 7.72 V

Vout = -2.2 / 2.2 (5.08 + 7.72) = -12.80 V

Average inverting amplifier,

Vout = -6.40 V

Summing Non – inverting amplifier, Vout = (1+ Rf / R1) (V1 + V2) = 10.016 V

Rf = 2.2KΩ, R1 =2.2 KΩ and V1 = 5.08 V, V2 = 5.08 V

Average non-inverting amplifier, Vout = (1+Rf/R1) [(V1+V2)/2]

Here, Rf = 0, V1 = 5.08 V and R1 = ∞, V2 = 5.08 V

So, Vout = 10.16 / 2 = 5.8 V

Conclusion:

From this experiment, we can conclude that the inverting and non-inverting amplifier

satisfied the condition.


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Experiment-5 Date:

Aim:-To Study Op-Amp as integrator & differentiator

Apparatus: - Component Value

Resistor 2.2kohm

Capacitor 0.1uF

CRO, Function generator, connecting wire

Figure:

Figure 1 Differentiator circuit

Figure 2 Integrator circuit


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Theory:-

1) Differentiator:-

Op-Amp can be used as differentiator by connecting a capacitor of 0.01uF at the place of input

resistance.

When square wave is applied to the Op-Amp due to presence of capacitor the value of output

voltage will not increase suddenly but was take some time.

Here, transistor in form of obtained in the output on CRO screen.

When sine wave is applied to the Op-Amp cosine wave is obtained.

2) Integrator:-

Op-Amp can be used as integrator by connecting a capacitor of 0.1uF at the place of feedback

resistor.

Here, when square wave is applied to Op-Amp due to presence of capacitor. Here output will

decrease or increase as seen at the output. Output is triangular wave form.

Now, when sine wave is applied the output obtained is cosine.

Procedure:-

For differentiator, input voltage is given to the inverting terminal.

Here, we apply square & sine wave alternately & increase output on CRO screen.

For Integrator, Input is given at non-inverting terminal

Here we apply square & sine wave alternately &measure o/p on CRO screen.

Conclusion:-

From this practical of differentiator & integrator we come to learn their application of

Op-Amp as differentiator & integrator.


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Experiment-6 Date:

Aim: To Study about differential Amplifier.

Apparatus: Power Supply, Function generator, CRO, Op-Amp., Resistor, Connecting wires.

Theory:

A differential Amplifier is used to amplify the difference of the signals applied to the inverting & non-

inverting input terminal. Two input terminal Vin1 and V in2 are applied to the non-inverting and inverting

inputs none of the signal is grounded.

The circuit is the combination of inverting and non-inverting amplifier. The figure shows below is the

differential Amplifier.


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In the figure,

I1 =

,

I2 =

,

If =

Vout Vout (a) Vout (b)………………. (I)

Vout(a) =

, ………………. (II)

Vout (b) =(

) (

) ………………. (III)

Now with equation (I) and (II) and (III)

Vout =

(

) (

)

Vout (a) =

Procedure:

(1). Connect the circuit as shown in figure

(2). Prepare C.R.O. for the measurement and connect.

(3). C.R.O. Probes to the input & output.

(4). Give the supply from the power supply.

(5). Observe the output and take the observation.

(6). Switch off the supply.

Conclusion:

From the practical, we can conclude that output voltage is amplified signal of two input signal

(Difference of two input signal).


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Experiment-7 Date:

AIM: To verify the effect of output offset voltage.

Apparatus: Power supply, CRO, Op-amp, resistor, veriable resistor, connecting wire

Theory:

This o/p voltage produced due to input offset voltage is called as output offset

voltage.

VO = Vio (1+Rf/R1)

Equation shows that VO not only depends on Vio but also depends on Rf &

R1.

Due to input offset voltage, there is an output voltage produced when

the signal input is zero. The compensation n/w would be used in order to

amplify this output voltage.

The total output offset voltage result mainly due to the input offset

voltage.

Ios & it is given by the following equation

VO = Vio (1+Rf/R1) + Rf IOS

The total output voltage without bias current compensating resistance

VO = Vio (1+Rf/R1) + Rf Ib

Ramp is given by

VO = - [Vref/2n * Rf/R]*R

= - [(Vref* Rf)/R]*[D1/2 +D2/22 +D2/23+- - - - - - - - - - - +Dn/2n]

Procedure:

Connect the circuit as shown in figure.

Switch on the power supply.

Measure the output voltage of pin no-6 of op-amp with respect to ground for

given input signal of the pin no-2 of op-amp.

Switch off the power supply.


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Calculations:

VO = - [Vref/2n * Rf/R]

Vo= - [5/22 * 1/1]

Vo= -5/4

Vo=-1.20V

Conclusion:

By this experiment we can conclude that by using R-2R ladder circuit to covert digital

signal into the analog signal.


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Experiment 8

Aim: To Study Binary Weighted DAC using Op- amp.

Apparatus: Power Supply, Op-amp, Resistor, Connecting Wires, Bread Board.

Theory:

Digital to Analogue converter is basically consisting of resistive network digitally controlled

electronics switches, a voltage reference and current to voltage converter.

Fig shows the binary weighted resistor type digital to analogue converter this circuit uses a

network of binary weighted resistor and a running amplifier the resistor are from the network of binary

weighted resistor. There are two electronics switches used one per digital bit and the position of the

moving arm of every switch is controlled by the binary input word thus, n – bit digital word will decide

the position of switches and connect their corresponding binary weighted resistor to either – Vr.

Depending upon the positions of switches the current I1, I2 will start following these resistor

respectively.

Resolution / Sample Space δ V = Va (max) – Va (min)

2n – 1

When n = no of input V0 = Vn1 [ Rf * b1 + Rf * b0] R0 / I’ R0 / I’ Vn1 * R1 (2’b1 + 2b0) R2 Procedure: i) Connect the circuit as shown in figure beside. ii) Switch on the power supply. iii) Give the appropriate input signal. iv) Take the output at the pin 6 respects to ground. v) Switch off the supply.

Calculations if we want analogue signal, from 0 – 15 V then Vmax = 15V, Vmin = 0V

δ V = V (max) – V (min)

2n – 1

Now, δ V = sample space

Vmax = 15 V; n= v

Vmin = 0 V


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δ V = 15 – 0

2 - 1

δ V = 15

3

= 5V

δ V = Vh1 – R1

R0

Rf = δ v R0

Vh1

= 5 (2.2) KΩ

5

= 2.2 KΩ

Conclusion:

By using, Binary Weighted DAC using Op- amp we can convert analog to digital.


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Experiment:- 9 Date:

Aim: To Study 2 bit R-2R Ladder digital to Analogue converter.

Apparatus: Power Supply, Op-Amp., Resistor, Switch, Connecting wires, Bread Board.

Theory:

A fig (a) shows the basic R-2R ladder network. It consists of resistor of only two values, R and 2R. R-2R network is the ladder type DAC and the one shown in figure (a) is a 2 bit. b0, by is a 4 bit digital input. R-2R ladder network, consist of resistor of only values R and 2R. This will simplify the network

and also the selection of network component V0 is the analogue O/P voltage which is proportional to the digital i/p while analysis. V0 = Resolution x D Where D = Binary Data

Procedure:

Connect the circuit as shown in figure beside

ii) Prepare the CRO for measurement & connect the input & output to CRO.

iii) Measure the observation

iv) Switch off the power – supply

Observation:

Initial Input Voltage = 1.6mV

Input Voltage V1 =V2=0V

Output Voltage with potentiometer is connected 0.00V.

Resistance: 1.05KΩ

V1 = V2 =0V → V005 = -1.6mV

Rv = 10KΩ → V005 =0 V, K = 1.05KΩ

Conclusion:

From this practical, we determines the valve of resistance at which the output voltage of Op-

amp to be zero.


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Experiment:- 10

AIM: - To generate frequency using RC Phase shift & Operational Amplifier.

Apparatus:-

1) Power Supply

2) Operational Amplifier

3) Resistance

4) Capacitor

5) Connecting wires

Theory:-

The oscillating frequency can be changed by changing either resistance or capacitor in the RC

phase shift networkas the frequency is dependent on R and C. All the resistence and all the capacitor

should be varied simultaneous and equally because we want the phase shift introduce by each R-C

section to remain unchanged.

The RC phase shift oscillation using operational amplifier is shown in the fig (a) the feedback

network used the resistor connected between output and inverting terminal. This network will

introduce a phase shift of 180 these feedback network attneuates the signal at its input and feedback

it to the amplifier input the level of attenuation is decided by the feedback factor B.

The gain of the inverting amplifier is decided by the values of Rf and R1. This gain is adjusted in

such a way that the product |AB| is slightly gretor than 1.

The expression for frequency of oscillation of an RC phase shift using op- amp is given by

Fo = 1/2 * 3.14 *RC*/--- 6

Procedure:-

Connect the ckt as shown in the fig (a)

Connect the probs to the power supply and switch on the supply.

Measure the resistance and capacitance respectively and take the observation.

Switch off the supply.


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Calculation:-

Here R1 = R2 = R3 = R

C1 = C2 = C3 = C

1) R = 10 Kohm

C = 470 pF

Fo = 1/2 * 3.14 *RC*/--- 6

F = 13.82 Khz

2) R = 10 Kohm

C = 270 pF

Fo = 1/2 * 3.14 *RC*/--- 6

F= 24.07 Khz

Conclusion:-

By performing this practical we conclude the we can generate frequency using

RC phase shift & operational amplifier.

EXPERIMENT 1 Date: AIM: To study the frequency response...

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