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# 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 as **arithmetic circuits**. Using op-amps, you can build basic arithmetic circuits such as an **adder** and a **subtractor**. In this chapter, you will learn about each of them in detail.

## Adder

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 a **summing amplifier**, since the output is an amplified one.

The **circuit diagram** of an op-amp based adder is shown in the following figure −

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 the **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.

The **nodal equation** at the inverting input terminal's node is

$$\frac{0-V_1}{R_1}+\frac{0-V_2}{R_2}+\frac{0-V_0}{R_f}=0$$

$$=>\frac{V_1}{R_1}-\frac{V_2}{R_2}=\frac{V_0}{R_f}$$

$$=>V_{0}=R_{f}\left(\frac{V_1}{R_1}+\frac{V_2}{R_2}\right)$$

If $R_{f}=R_{1}=R_{2}=R$, then the output voltage $V_{0}$ will be −

$$V_{0}=-R{}\left(\frac{V_1}{R}+\frac{V_2}{R}\right)$$

$$=>V_{0}=-(V_{1}+V_{2})$$

Therefore, the op-amp based adder circuit discussed above will produce the sum of the two input voltages $v_{1}$ and $v_{1}$, as the output, when all the resistors present in the circuit are of same value. Note that the output voltage $V_{0}$ of an adder circuit is having a **negative sign**, which indicates that there exists a 180^{0} 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 a **difference amplifier**, since the output is an amplified one.

The **circuit diagram** of an op-amp based subtractor is shown in the following figure −

Now, let us find the expression for output voltage $V_{0}$ of the above circuit using **superposition theorem** using the following steps −

### Step 1

Firstly, let us calculate the output voltage $V_{01}$ by considering only $V_{1}$.

For this, eliminate $V_{2}$ by making it short circuit. Then we obtain the **modified circuit diagram** as shown in the following figure −

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

$$=>V_{p}=V_{1}\left(\frac{R_3}{R_2+R_3}\right)$$

Now, the above circuit looks like a non-inverting amplifier having input voltage $V_{p}$. Therefore, the output voltage $V_{01}$ of above circuit will be

$$V_{01}=V_{p}\left(1+\frac{R_f}{R_1}\right)$$

Substitute, the value of $V_{p}$ in above equation, we obtain the output voltage $V_{01}$ by considering only $V_{1}$, as −

$$V_{01}=V_{1}\left(\frac{R_3}{R_2+R_3}\right)\left(1+\frac{R_f}{R_1}\right)$$

### Step 2

In this step, let us find the output voltage, $V_{02}$ by considering only $V_{2}$. Similar to that in the above step, eliminate $V_{1}$ by making it short circuit. The **modified circuit diagram** 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 an **inverting op-amp**. Therefore, the output voltage $V_{02}$ of above circuit will be −

$$V_{02}=\left(-\frac{R_f}{R_1}\right)V_{2}$$

### Step 3

In this step, we will obtain the output voltage $V_{0}$ of the subtractor circuit by **adding the output voltages** obtained in Step1 and Step2. Mathematically, it can be written as

$$V_{0}=V_{01}+V_{02}$$

Substituting the values of $V_{01}$ and $V_{02}$ in the above equation, we get −

$$V_{0}=V_{1}\left(\frac{R_3}{R_2+R_3}\right)\left(1+\frac{R_f}{R_1}\right)+\left(-\frac{R_f}{R_1}\right)V_{2}$$

$$=>V_{0}=V_{1}\left(\frac{R_3}{R_2+R_3}\right)\left(1+\frac{R_f}{R_1}\right)-\left(\frac{R_f}{R_1}\right)V_{2}$$

If $R_{f}=R_{1}=R_{2}=R_{3}=R$, then the output voltage $V_{0}$ will be

$$V_{0}=V_{1}\left(\frac{R}{R+R}\right)\left(1+\frac{R}{R}\right)-\left(\frac{R}{R}\right)V_{2}$$

$$=>V_{0}=V_{1}\left(\frac{R}{2R}\right)(2)-(1)V_{2}$$

$$V_{0}=V_{1}-V_{2}$$

Thus, the op-amp based subtractor circuit discussed above will produce an output, which is the difference of two input voltages $V_{1}$ and $V_{2}$, when all the resistors present in the circuit are of same value.