*A differential amplifier is an op amp circuit which is designed to amplify the difference input available and reject the common-mode voltage.* It is used for suppressing the effect of noise at the output. Since the noise present will be having the same amplitude at the two terminals of the op-amp. So when the difference between terminals is taken, the noise will cancel each other. The differential amplifier output is proportional to the difference of the input terminals. Hence the output is free from noise.

**Contents**show

## Differential amplifier

Figure 1 shows the basic differential amplifier. The two input signal V_{1 }and V_{2} are applied to the op amp.

**Working**

Apply superposition theorem to find out the output voltage. First of all, deactivate V_{2} and connect it to ground as shown in figure 2.

Applying voltage division method

(1)

Since the op-amp is ideal and negative feedback is present, the voltage of the inverting terminal (V_{−}) is equal to the voltage of the non-inverting terminal (V_{+}), according to the **virtual short concept**.

V_{− }= V_{+}

The currents entering both terminals of the op-amp are zero since the op-amp is ideal.

(2)

(3)

Apply KCL at node **P**

(4)

From equations (2), (3) and (4), we have

From equation 1, we have

(5)

Now deactivate V_{1} and connect it to ground as shown in figure 3.

As we can see that the voltage across R_{4} is zero. Therefore V_{+} = 0 V. Since the op-amp is ideal and negative feedback is present, the voltage of the inverting terminal (V_{−}) is equal to the voltage of the non-inverting terminal (V_{+} = 0), according to the **virtual ground concept**.

V_{− }= V_{+} = 0 Volt

The currents entering both terminals of the op-amp are zero since the op-amp is ideal.

(6)

(7)

Apply KCL at node **P**

(8)

From equations (6), (7) and (8), we have

(9)

Adding equations (5) and (9), we get the output voltage V_{o}

(10)

(11)

**Case-1: when **

then

The output V_{o} is

Now

and

The output V_{o} is given by

where A_{d} = differential gain and A_{c} = common mode gain

As

Therefore, we have

and

so

### Common mode rejection ratio (CMRR)

using above value of A_{d} and A_{c}, we have

**Note:** Ideally CMRR is infinite. So CMRR value for this circuit to be infinite, .

**Case2: when **

since

and

then we can write it as

and

Therefore equation (10) becomes

(12)

(13)

Comparing equation (12) and (13), we have

and

Then CMRR is given as

**Note:**

**Note:** For a better differential amplifier, CMRR should be as high as possible.

**Note:** CMRR depends upon the circuit and not depend upon the applied input.