In this lecture, we are going to learn the concept of the Ideal Differential Amplifier, how it works, and some important terms like differential gain, common mode gain, and common mode rejection ratio (CMRR) of the Ideal Differential Amplifier.

## Basic Concept of Ideal Differential Amplifier

The function of an Ideal Differential Amplifier is to amplify the difference between two signals.

The differential amplifier is used in many applications where the response from dc to many, megahertz is required.

It is also the basic stage of an integrated operational amplifier with differential input.

The block diagram of an Ideal Differential Amplifier is shown in the below figure. It has two input terminals and one output terminal.

Two input signals V_{1} and V_{2} are connected to the input terminals. The voltages V_{1} and V_{2} are measured with respect to ground.

In an Ideal Differential Amplifier, the output voltage V_{o} is proportional to the difference between the two input signals.

Hence the expression for the output voltage is given by,

**V _{o} ∝ ( V_{1} – V_{2} )**

**Differential Input Signal**

- The difference between the input signals V
_{1}and V_{2}is called the differential signal V_{d}.

**∴ Differential Signal V _{d} = V_{1} – V_{2}**

- From the above equation, it is clear that the amplifier output will be non-zero if and only if the difference signal ( V
_{1}– V_{2}) has a non-zero value.

- So this amplifier will amplify only the difference ( V
_{1}– V_{2}) between the input signals. That is why the name differential amplifier.

**Common Mode signal**

- A common signal to both the inputs (i.e. V
_{1}= V_{2}= V) is called a common mode signal. The output voltage produced by an ideal differential amplifier is zero for the common mode signal.

## Differential Gain of Differential Amplifier

- The expression for the output voltage is given by,

**V _{o} ∝ ( V_{1} – V_{2} )**

- The above equation can be modified by removing the signal of proportionality as,

**V _{o} = A_{d} ( V_{1} – V_{2} )**

Where A_{d} is called the differential gain.

- The differential gain of a differential amplifier can be defined as the gain with which the differential amplifier amplifies the difference signal V
_{d}.

- As V
_{d}= V_{1}– V_{2}, we can get the above equation,

**V _{o} = A_{d} V_{d}**

- The expression for the differential gain is,

\boxed{\mathbf{Differential\;Gain,\, A_d = \frac{V_o}{V_d}}}

- Instead of defining it as a ratio, the differential gain is sometimes expressed in decibels (dB) as,

\boxed{\mathbf{A_d(dB) = 20 \log_{10} [V_o/V_d]}}

## Common Mode Gain of Differential Amplifier

- The output voltage of an ideal differential amplifier will be zero if V
_{1}= V_{2}= V as per the below equation.

**V _{o} = A_{d} ( V_{1} – V_{2} )**

- However, the above equation does not describe a practical differential amplifier. In practice, the output voltage V
_{o}of a differential amplifier depends not only on the difference signal “V_{d}” but also depends on an average voltage level called “Common Mode Signal V_{c}“.

- The expression for common mode signal V
_{c}is,

\mathbf{V_c = \frac{V_1 + V_2}{2}}

- The gain with which a practical differential amplifier amplifies the common mode signal (V
_{c}) is called the “Common Mode Gain A_{c}“.

- The output voltage produced due to the common mode signal is given by,

**V _{o} = A_{c} V_{c}**

- The total voltage of a differential amplifier is given by,

\boxed{\mathbf{V_0 = A_d V_d + A_c V_c}}

- For an ideal differential amplifier, the differential gain (A
_{d}) should be infinite and the common mode gain (A_{c}) should be zero so that the output voltage is proportional only to the differential input signal.

- However, practically it is not possible to achieve. To know how far the amplifier is successful in rejecting the common mode signal, let us define a very important parameter called the Common Mode Rejection Ratio (CMRR).

You can also read CMRR in detail:What is CMRR? | Common Mode Rejection Ratio

## CMRR of Differential Amplifier

- The signal which is present at both the input terminals of a differential amplifier is called the common mode signal.

- The best example of a common mode signal in practice is noise.

- The differential amplifier should produce a very small output voltage corresponding to the common mode signal. In other words, it should be capable of rejecting the common mode signal.

- Common mode rejection ratio (CMRR) is the ability of a differential amplifier to reject the common mode signal successfully. It is called the figure of merit of the differential amplifier.

- CMRR is defined as the ratio of differential gain A
_{d}and Common mode gain A_{c}. It is denoted by the letter “ρ”.

\boxed{\mathbf{CMRR = \rho = \left | \frac{A_d}{A_c} \right |}}

**Ideally, the CMRR should be infinite and practically it should be as high as possible. **

- Sometimes CMRR is expressed in decibels (dB).

\boxed{\mathbf{\therefore CMRR (dB)= 20 \log_{10}\left | \frac{A_d}{A_c} \right |}}

Also Read:What are the different methods to improve CMRR in Differential Amplifiers?

## Features of Differential Amplifier

The important features of a differential amplifier are as follows:

- High differential gain and low common mode gain
- High common mode rejection ratio (CMRR)
- High input impedance
- Lower output impedance
- High gain
- Large bandwidth

## Example of Differential Amplifier

**1. For a differential amplifier the two input signals are V _{1} = 100 μV and V_{2} = -100 μV. Calculate the values of the differential input signal and common mode input signal. If the values of differential and common mode gains are 2500 and 0.5 respectively, calculate V_{o} and CMRR in dB.**

**Solution:**

1. Differential Input V_{d} = V_{1} – V_{2} = 100 – (-100) = 200 μV

2. Common Mode input signal V_c = \frac{V_1 + V_2}{2} = \frac{100 + (-100)}{2} = 0

3. Output voltage V_{o} = A_{d} V_{d} + A_{c}V_{c} = 2500 x 200 μV = 0.5 Volt.

4. CMRR (dB) = 20 log_{10}[A_{d}/A_{c}] = 20 log_{10}[2500/0.5] = 73.97 ≈ 74 dB

Also Read:Block diagram of op-amp