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    EXP. NO : 1

    DATE : LINEAR OPERATIONAL AMPLIFIER CIRCUITS

    ----------------------------------------------------------------------------------------------

    Aim:

    To design, construct and test the following linear operational amplifier circuits

    (1) Inverting amplifier (2) Non-inverting amplifier

    (3) Voltage follower (4) Summing Amplifier

    (5) Integrator (6) Differentiator

    (7) Subtractor

    Components Required:

    S. No Component Name Range Type Quantity

    1 Op-amp

    2 Power Supply

    3 Resistor

    4 Capacitor

    5 Voltmeter

    6 Breadboard

    7 Signal generator

    Theory:

    Inverting Amplifier:

    The inverting amplifier is the most widely used in all the op-amp

    circuits. The output voltage V O is fed back to inverting input terminal through the R f - R 1

    network where R f is the feedback resistor. The input signal V i is applied to the inverting

    input terminal through R 1 and non-inverting input terminal of op-amp is grounded.

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    VO = (R f / R 1) V i

    ACL = Vo / V i = - R f / R 1

    The Negative sign indicates a phase shift of 180 between input (V i) and Output (Vo).

    Non-Inverting Amplifier:

    The non inverting amplifier circuit amplifies without inverting the input

    signal. In this circuit, the input is applied to the non inverting input terminal and inverting

    input terminal is grounded such a circuit is called non inverting amplifier. It is also having a

    negative feedback system as output is fed back to the inverting input terminal.

    VO = (1+ (R f / R 1) V i

    ACL = V o / V i = 1+ (R f / R 1)

    Circuit Diagram of Inverting Amplifier:

    Design Procedure:

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    ACL = 2; R 1 = 1K

    ACL = V o / V i = 1+ (R f / R 1) = 2

    Rf = (A CL-1) R 1 = 1K

    Tabulation:

    S. NoInput Voltage (V in )

    in Volts

    Output Voltage V O in Volts

    Theoretical Value Practical Value

    1

    2

    3

    4

    5

    6

    Voltage Follower:

    In this circuit, the output voltage is equal to the input voltage both in

    magnitude and phase i.e. output follows the input. So, the circuit is called voltage follower.

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    Input is applied to non inverting input and the output is directly connected to inverting

    input.

    VO = V i

    Procedure:

    1. Make the connections as per the circuit diagram.

    2. Vary the input voltage using regulated DC power supply then measure and

    tabulate the corresponding output voltage.

    3. Compare theoretical Output with the actual output obtained from the circuit.

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    Circuit Diagram for Voltage Follower:

    Design Procedure:

    VO = V i

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    Tabulation:

    S. NoInput Voltage (V in )

    in Volts

    Output Voltage V O in Volts

    Theoretical Value Practical Value

    1

    2

    3

    4

    5

    6

    Summing Amplifier (summer):

    The operational amplifier is used to design a circuit whose output is

    equal to sum of the several input signals. Such a circuit is known as summer. It has two

    configurations as (1) Inverting Summer amplifier (2) Non inverting summing amplifier. (1)Inverting summing amplifier, the input is given to the inverting input terminal of op-amp.

    The circuit output is the inverted sum.

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    VO = [(R f / R 1) V 1 + (R f / R 2) V 2

    (2) Non inverting summing amplifier, the input is given to the non inverting input terminal

    of operational amplifier. The output of the circuit is non inverted sum.

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    Circuit Diagram for Summing Amplifier:

    Design Procedure:

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    VO = [-(R f / R 1 ) V 1 + (R f / R 2) V 2]

    R1= R 2 = R f

    R1= R 2 = R f = 1K

    VO = - [V 1 + V 2]

    Tabulation:

    S. No

    Input Voltage (V in )

    in VoltsOutput Voltage V O in Volts

    V1 V2 Theoretical Value Practical Value

    1

    2

    3

    4

    5

    6

    Difference Amplifier (Subtractor):

    A circuit that amplifies the difference between the two signals is called

    difference amplifier. This type of amplifiers is mostly used in instrumentation circuit.

    VO = (R f / R 1) (V 1-V 2)

    The main purpose of the difference amplifier stage is to provide high gain to the difference

    mode signal and cancel the common mode signal i.e., i t should have high CMRR.

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    Differentiator:

    One of the simplest of the op-amp circuits that contain capacitor is the

    differentiating amplifier or differentiator. As the name Differentiator suggests, the circuit

    performs the mathematical operation of differentiation. That is, the output waveform is the

    derivative of input waveform. But by using the differentiator at high frequencies, it may

    becomes unstable and break into oscillation. The impedance at input also decreases with

    increase in frequency; thereby making the circuit sensitive to high frequency noise.

    Analysis of Practical Differentiator:

    As the input current of op-amp is zero, there is no current input at node

    B. Hence it is at the ground potential. From the concept of virtual ground, node A is also atthe ground potential and hence V B = V A = 0V.

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    Circuit Diagram for Difference Amplifier:

    Design Procedure:

    R1= R 2 = 1K

    R3 = R f = 4.7 K

    VO = (R f / R 1) [V 1 - V 2]

    Tabulation:

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    S. No

    Input Voltage (V in )

    in VoltsOutput Voltage V O in Volts

    V1 V2 Theoretical Value Practical Value

    1

    2

    3

    4

    5

    6

    For the current I, we can write,

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    in

    V - Vin AI = - - - - - - - - - - - - - - - - - V = 0 AZ1VinI = Where Z = R in series with C1 1 1Z1

    So in Laplace domain we can write,

    1+ sR C1 1 1Z = R + =1 1 sC sC1 1sC V (s)1I =1+ sR C1 1

    Now the current I is, 1V - V VA O OI = = -1 R Rf f

    V (s)OIn Laplace, I = -1 Rf d (V - V ) dVA O Oand I = C = - C2 f f dt dt

    Taling Laplace Transform we get,

    I = - sC V (s)o2 f Applying KCL

    at node A,

    I = I + I1 2

    V (s)sC Vin (s) O1 = - - sC V (s)of 1+sR C R1 1 f

    -sR C V (s)1 inf V (s) =O [1+ s C R ] [1+ s C R ]1 1f f

    If C R = C R then1 1f f

    -sR C V (s)1 inf V (s) =O 21+ s C R1 1

    The time constant R C is much greater than R C or R C and hence1 1 1f f f

    the equation becomes

    V (s) =O - sR C V (s)1 inf

    The output voltage is the R C times the differentiation of the input.1f

    dV (t) dinV (t) = - R C ------------ as s=O f 1 dt dt

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    Circuit Diagram for Differentiator:

    Design:

    dViV = - R C1O f dt

    R =1.6K f

    R =1K 1

    1

    f =1KHz =2 R C 1f

    1C = = 0.16 f 1 2 f R f

    ;R R1 f R = = 615 eq R +R1 f

    Model Graph:

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    Integrator:

    One of the simple op-amp circuits that also contain the capacitor is known as

    integrator. As the name integrator suggests, the circuit performs the mathematical

    operation of integration. That is, the output waveform is the integration of input waveform.

    Analysis of Practical Integrator:

    As the input of op-amp is zero, the node B still at ground potential. Hence the node A

    is also at the ground potential from the concept of virtual ground. So, V A = 0.

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    (V - V )in AI = - - - - - - - - - - - - - - - - - V = 0 AR1VinI =R1

    d (V - V )A OSimilarly I = - C1 f dtdV

    A= - C f dtV - VA OAnd I =2 Rf

    VO= -Rf

    Tabular Column: Differentiator

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    S.NOInput Voltage (V i) involts

    Output Voltage (V o)in volts

    Time period in ms

    Tabular Column: Integrator

    S.NOInput Voltage (V i) involts

    Output Voltage (V o)in volts

    Time period in ms

    Procedure for Differentiator and Integrator:

    1. Make the Connections as per the circuit diagram.

    2. Set the 1 KHz square wave input using function generator and obtain the output

    waveform on the CRO.

    3. Determine and tabulate the amplitude, time period of the output waveform. 4. Draw the graph for output.

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    At node A, applying KCL

    I = I + I1 2dV VV O Oin = - C -

    f R dt R1 f Taking Laplace transform of above Equation,

    V (s)OV (s) / R = - s C V (s) -in 1 Of Rf

    .

    1= - V (s) [s C + ]O f Rf

    [1+ s C R ]f f = - V (s)ORf

    Rf V (s) = - V (s)inO R [1+ s R C ]1 f f 1

    V (s) = - V (s)inO R1sR C +1 f Rf

    When R is very large than R / R can be neglected and hence circuit behaves like1f f

    an ideal integrator as1

    V (s) = - V (s)inO sR C1 f 1V (t) = - V (t) dt --------- as 1/s = dtO insR C1 f

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    Circuit Diagram for Integrator:

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    Design: R1 = 1K, C=1F and f = 1KHZ

    T = 1/f

    f=1/2 R 1Cf

    VO = V in T / (R 1C 1) = 1.6K

    R eq = R 1*R f / (R 1 +R f ) = 1.610 6 /2.610 3 = 615

    Model Graph:

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    Result:

    Thus the linear operational amplifier circuits were designed, constructed and its

    performance was tested using op-amp IC 741.

    DescriptionMax.

    Marks

    Marks

    Secured

    Preparation 30

    Performance 40

    Viva Voce 10

    Record 20

    Total 100

    Staff Signature


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