In this article, we will find out around one of the significant circuits in a simple analogue circuit plan, a Differential Amplifier. It is basically an electronic amplifier, which has two sources of inputs and amplifies the contrast between those two input voltages.

The Operational Amplifier is inside a Differential Amplifier with highlights like High Input Impedance, Low Output Impedance, and so on.

## What is Differential Amplifier?

As a standard op-amp has two input sources, i.e inverting and non-inverting. We can likewise interface signals to both of these inputs simultaneously creating another basic type of op-amp circuit called a Difference Amplifier.

A difference amplifier amplifies the difference between the two input signals. An operational amplifier is a difference amplifier. However, the open-loop voltage gain of an operational amplifier is excessively high (preferably infinite), to be utilized without a feedback connection.

Along these lines, a useful difference amplifier utilizes a negative feedback connection with control the voltage gain increase of the amplifier.

## Differential Amplifier

It appeared in the above circuit, is a mix of both inverting and non-inverting amplifiers. In the event that the non-inverting terminal is associated with ground, the circuit works as an inverting amplifier, and the input signal V_{1} is enhanced by {– (R_{3}/R_{1})}.

Additionally, if the inverting input terminal is associated with ground, the circuit acts as a non-inverting amplifier. With the inverting input terminal grounded, R3 and R1 work as the feedback components of a non-inverting amplifier.

V_{2} is conceivably isolated across resistors R_{2} and R_{4} to give V_{R4}, and afterwards, V_{R4} is amplified by (R_{3} + R_{1})/R_{1}.

V_{2}=0,

{V}_{O1}=\quad -({R}_{3}/{R}_{1})\times {V}_{1}

With V_{1}=0,

{V}_{R4}={ \left\{ { {R}_{4}/({R}_{2}+{R}_{4}) } \right\} }\times {V}_{2}\\ and\\ {V}_{O2}={ \left\{ { ({R}_{1}+{R}_{3})/{R}_{1} } \right\} }\times {V}_{R4}

Therefore,

{V}_{O2}=\left\{ ({R}_{1}+{R}_{3})/{R}_{1} \right\} \times \left\{ {R}_{4}/({R}_{2}+{R}_{4}) \right\} \times {V}_{2}

If the input resistances are chosen such that, R_{2} = R_{1} and R_{4} = R_{3}, then

{V}_{O2}={ ({R}_{3}/{R}_{1} })\times {V}_{2}

Now, according to the superposition theorem, if both the input signals V_{1} and V_{2} are present, then the output voltage is

{V}_{O}={V}_{O1}+{V}_{O2}

={ \{ -({R}_{3}/{R}_{1})\times {V}_{1}\} }+{ \{ ({R}_{3}/{R}_{1} })\times {V}_{2}\}

When the resistors R_{3} and R_{1} are of the same value, the output is the direct difference of the input voltages applied. By selecting R_{3} greater than R_{1}, the output can be made an amplified version of the difference of the input voltages.

**Input Resistance**

**What is the input resistance of a differential amplifier?**

One issue with choosing the differential amplifier resistors as R_{2} = R_{1} and R_{3} = R_{4} is that the input resistance for both inverting and non-inverting amplifiers are inconsistent.

**How do you calculate differential input resistance?**The input resistance for voltage V_{1} is R_{1} as on account of an inverting amplifier. For the non-inverting input, for example for input voltage V_{2}, the input resistance is (R_{2}+R_{4}).

This difference in the input resistance makes one of the input signals be more amplified than the others.

The output condition of the differential amplifier Vout can be gotten by making the ratio R_{4}/R_{2} equivalent to R_{3}/R_{1}, rather than making R_{2} = R_{1} and R_{4} = R_{3}.

The input resistance contrast won’t cause an issue if the signal source resistance is a lot littler than the input resistance. Likewise, it is typically attractive to have R_{2} = R_{1} and R_{4} = R_{3}, so as to limit the input offset voltages.

**Differential Gain**

**What is the differential gain of a difference amplifier?**

The differential gain of a differential amplifier is defined as the gain acquired at the output signal as for the difference in the input signals applied.

The output voltage of a difference amplifier is given as,

{V}_{O}={ A }_{ D }({V}_{1}-{V}_{2})

where, A_{D}= -(R_{3}/R_{1}) is the differential gain of the amplifier

## Common Mode Input

A differential amplifier amplifies the contrast between the two input voltages. In a short word, a common mode input Vcm would make the input sources (V_{1} + V_{cm}) and (V_{2} + V_{cm}), which will result in Vcm being cancelled when the difference of the two input voltages is amplified.

Since the output of a practical differential amplifier relies on the ratio of the input resistance, in the event that these resistors’ ratio is not actually equivalent, at that point one input voltage is amplified by a more amount than the other input.

Thusly, the common mode voltage Vcm won’t be totally dropped. Since it is practically difficult to coordinate resistor ratios perfectly, there is probably going to be some common mode output voltage.

With the presence of common mode input voltage, the output voltage of the differential amplifier is given as,

{V}_{O}={ A }_{ d }{ V }_{ d }+{ A }_{ c }{ V }_{ c }

where, Vd= Difference Voltage V_{1}-V_{2}

Vc= Common Mode Voltage (V_{1}+V_{2})/2

## Common Mode Rejection Ratio (CMRR)

The capacity of a differential amplifier to dismiss common mode input signals is expressed regarding the common mode dismissal ratio (CMRR). The common mode dismissal ratio of a differential amplifier is numerically given as the ratio of the differential voltage gain of the differential amplifier to its common mode gain.

CMRR=\left| { A }_{ d }/{ A }_{ C } \right|

Preferably, the common mode voltage gain of a differential amplifier is zero. Consequently, the CMRR is preferably infinite.

## Characteristics of a Differential Amplifier

- High differential voltage gain
- Low common mode gain
- High input impedance
- Low output impedance
- High common mode rejection ratio
- Large bandwidth
- Low offset voltages and currents

## Differential Amplifier as Comparator

A differential amplifier circuit is a valuable operation amp circuit since it tends to be designed to either “add” or “substract” the input voltages, by reasonably including more resistors in parallel with the input resistors.

A Wheatstone bridge differential amplifier circuit configuration is as appeared in the above figure. This circuit acts like a differential voltage comparator.

By associating one input to a fixed voltage and the other to a thermistor (or LDR), the circuit identifies high or low degrees of temperature as the output voltage turns into a direct linear function of the adjustments in the active part of the resistive bridge.

A Wheatstone bridge differential amplifier can likewise be utilized to locate the obscure resistance in the resistive bridge, by contrasting the input voltages over the resistors.

## Example

Decide the output voltage of a differential amplifier for the input voltages of 400µV and 280µV. The differential gain of the amplifier is 4500 and the estimation of CMRR is (I) 100 and (ii) 10^{5}

**i)** **CMRR = A _{d }/ A_{c}**

100 =4500 / A_{c}

A_{c} = 45

V_{d} = V_{1} – V_{2} = 400µV – 280µV = 120µV

V_{c} = (V_{1} + V_{2})/2 = 680µV / 2 = 340 µV

V_{O} = (A_{d}V_{d}) + (A_{c}V_{c})

= [(4500 x 120) + (45 x 340)] µV

V_{O} = 555300 µV = 555.3 mV

**ii) CMRR = 10 ^{5}**

A_{c} = A_{d} / CMRR = 4500 / 10^{5} = 0.045

V_{O} = A_{d}V_{d} + A_{c} V_{c} = [4500×120 + 0.045×340] µV

V_{O} = 540015.3 µV

Note: Ideally, Ac is zero. So the output is only A

_{d}V_{d}, which results in V_{O}= 4500 x 120 µV = 540mV.

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