Arithmetic Circuits

# Arithmetic Circuits

In the previous chapter, we discussed about the basic applications of op-amp. Note that they come under the linear operations of an op-amp. In this chapter, let us discuss about arithmetic circuits, which are also linear applications of op-amp.

The electronic circuits, which perform arithmetic operations are called asÂ arithmetic circuits. Using op-amps, you can build basic arithmetic circuits such as anÂ adderÂ and aÂ subtractor. In this chapter, you will learn about each of them in detail.

An adder is an electronic circuit that produces an output, which is equal to the sum of the applied inputs. This section discusses about the op-amp based adder circuit.

An op-amp based adder produces an output equal to the sum of the input voltages applied at its inverting terminal. It is also called as aÂ summing amplifier, since the output is an amplified one.

TheÂ circuit diagramÂ of an op-amp based adder is shown in the following figure âˆ’

In the above circuit, the non-inverting input terminal of the op-amp is connected to ground. That means zero volts is applied at its non-inverting input terminal.

According to theÂ virtual short concept, the voltage at the inverting input terminal of an op-amp is same as that of the voltage at its non-inverting input terminal. So, the voltage at the inverting input terminal of the op-amp will be zero volts.

TheÂ nodal equationÂ at the inverting input terminal’s node is

0âˆ’V1R1+0âˆ’V2R2+0âˆ’V0Rf=00âˆ’V1R1+0âˆ’V2R2+0âˆ’V0Rf=0
=>V1R1âˆ’V2R2=V0Rf=>V1R1âˆ’V2R2=V0Rf
=>V0=Rf(V1R1+V2R2)=>V0=Rf(V1R1+V2R2)

IfÂ Rf=R1=R2=RRf=R1=R2=R, then the output voltageÂ V0V0Â will be âˆ’

V0=âˆ’R(V1R+V2R)V0=âˆ’R(V1R+V2R)
=>V0=âˆ’(V1+V2)=>V0=âˆ’(V1+V2)

Therefore, the op-amp based adder circuit discussed above will produce the sum of the two input voltagesÂ v1v1Â andÂ v1v1, as the output, when all the resistors present in the circuit are of same value. Note that the output voltageÂ V0V0Â of an adder circuit is having aÂ negative sign, which indicates that there exists a 1800Â phase difference between the input and the output.

## Subtractor

A subtractor is an electronic circuit that produces an output, which is equal to the difference of the applied inputs. This section discusses about the op-amp based subtractor circuit.

An op-amp based subtractor produces an output equal to the difference of the input voltages applied at its inverting and non-inverting terminals. It is also called as aÂ difference amplifier, since the output is an amplified one.

TheÂ circuit diagramÂ of an op-amp based subtractor is shown in the following figure âˆ’

Now, let us find the expression for output voltageÂ V0V0Â of the above circuit usingÂ superposition theoremÂ using the following steps âˆ’

### Step 1

Firstly, let us calculate the output voltageÂ V01V01Â by considering onlyÂ V1V1.

For this, eliminateÂ V2V2Â by making it short circuit. Then we obtain theÂ modified circuit diagramÂ as shown in the following figure âˆ’

Now, using theÂ voltage division principle, calculate the voltage at the non-inverting input terminal of the op-amp.

=>Vp=V1(R3R2+R3)=>Vp=V1(R3R2+R3)

Now, the above circuit looks like a non-inverting amplifier having input voltageÂ VpVp. Therefore, the output voltageÂ V01V01Â of above circuit will be

V01=Vp(1+RfR1)V01=Vp(1+RfR1)

Substitute, the value ofÂ VpVpÂ in above equation, we obtain the output voltageÂ V01V01Â by considering onlyÂ V1V1, as âˆ’

V01=V1(R3R2+R3)(1+RfR1)V01=V1(R3R2+R3)(1+RfR1)

### Step 2

In this step, let us find the output voltage,Â V02V02Â by considering onlyÂ V2V2. Similar to that in the above step, eliminateÂ V1V1Â by making it short circuit. TheÂ modified circuit diagramÂ is shown in the following figure.

You can observe that the voltage at the non-inverting input terminal of the op-amp will be zero volts. It means, the above circuit is simply anÂ inverting op-amp. Therefore, the output voltageÂ V02V02Â of above circuit will be âˆ’

V02=(âˆ’RfR1)V2V02=(âˆ’RfR1)V2

### Step 3

In this step, we will obtain the output voltageÂ V0V0Â of the subtractor circuit byÂ adding the output voltagesÂ obtained in Step1 and Step2. Mathematically, it can be written as

V0=V01+V02V0=V01+V02

Substituting the values ofÂ V01V01Â andÂ V02V02Â in the above equation, we get âˆ’

V0=V1(R3R2+R3)(1+RfR1)+(âˆ’RfR1)V2V0=V1(R3R2+R3)(1+RfR1)+(âˆ’RfR1)V2
=>V0=V1(R3R2+R3)(1+RfR1)âˆ’(RfR1)V2=>V0=V1(R3R2+R3)(1+RfR1)âˆ’(RfR1)V2

IfÂ Rf=R1=R2=R3=RRf=R1=R2=R3=R, then the output voltageÂ V0V0Â will be

V0=V1(RR+R)(1+RR)âˆ’(RR)V2V0=V1(RR+R)(1+RR)âˆ’(RR)V2
=>V0=V1(R2R)(2)âˆ’(1)V2=>V0=V1(R2R)(2)âˆ’(1)V2
V0=V1âˆ’V2V0=V1âˆ’V2

Thus, the op-amp based subtractor circuit discussed above will produce an output, which is the difference of two input voltagesÂ V1V1Â andÂ V2V2, when all the resistors present in the circuit are of same value.