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Applications of Operational Amplifier - Subtractor using Operational Amplifier | Linear Integrated Circuits : Applications of Operational Amplifier

Chapter: Linear Integrated Circuits : Applications of Operational Amplifier

Subtractor using Operational Amplifier

If all resistors are equal in value, then the output voltage can be derived by using superposition principle.

Subtractor:



A basic differential amplifier can be used as a subtractor as shown in the above figure. If all resistors are equal in value, then the output voltage can be derived by using superposition principle.

 

To find the output V01 due to  V1 alone, make V2 = 0.

Then the circuit of figure as shown in the above becomes a non-inverting amplifier having input voltage  V1/2 at the non-inverting input terminal and the output becomes

V01 =  V1/2(1+R/R) =  V1 when all resistances are R in the circuit.

Similarly the output V02 due to V2 alone (with  V1 grounded) can be written simply for an inverting amplifier as

V02 = -V2

Thus the output voltage Vo due to both the inputs can be written as

V0 =V01 - V02 = V1 - V2

 

Adder/Subtractor:



It is possible to perform addition and subtraction simultaneously with a single op-amp using the circuit shown in figure 2.16.

The output voltage Vo can be obtained by using superposition theorem. To find output voltage V01 due to  V1 alone, make all other input voltages V2, V3 and V4 equal to zero.

The simplified circuit is shown in figure 2.17. This is the circuit of an inverting amplifier and its output voltage is, V01= -R/(R/2) * V1/2= -  V1 by Thevenin‘s equivalent circuit at inverting input terminal).

Similarly, the output voltage V02 due to V2 alone is,

V02= - V2

Now, the output voltage V03 due to the input voltage signal V3 alone applied at the (+) input terminal can be found by setting  V1, V2 and V4 equal to zero.

V03=V3

The circuit now becomes a non-inverting amplifier as shown in fig.(c).

So, the output voltage V03 due to V3 alone is

V03 = V3

Similarly, it can be shown that the output voltage V04 due to V4 alone is

V04 = V4

Thus, the output voltage Vo due to all four input voltages is given by

 Vo = V01 = V02 = V03 = V04

Vo = - V1 -V2 +V3+ V4

Vo = (V3 +V4) – (V1 +V2)

So, the circuit is an adder-subtractor.

 

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