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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