Differential Amplifier – Working & Applications

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.

Differential amplifier

Figure 1 shows the basic differential amplifier. The two input signal V1 and V2 are applied to the op amp.

Differential amplifier
Fig. 1 Differential amplifier circuit

Working

Apply superposition theorem to find out the output voltage. First of all, deactivate V2 and connect it to ground as shown in figure 2.

differential amplifier working
Fig. 2 differential amplifier working

Applying voltage division method

(1)   \begin{equation*}  V_+ = \frac{R_4}{R_3+R_4}V_1 \end{equation*}

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)   \begin{equation*}  I_1=\frac{0-V_+}{R_1}=-\frac{V_+}{R_1} \end{equation*}

(3)   \begin{equation*}  I_2=\frac{V_+-V_{o1}}{R_2} \end{equation*}

Apply KCL at node P

I_{1} = 0 + I_2

(4)   \begin{equation*}  I_1= I_{2} \end{equation*}

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

\Rightarrow -\frac{V_+}{R_1}=\frac{V_+-V_{o1}}{R_2}

\Rightarrow \frac{V_{o1}}{R_2}=V_+[\frac{1}{R_1}+\frac{1}{R_2}]

From equation 1, we have

(5)   \begin{equation*}  V_{o1}=(1+\frac{R_2}{R_1})(\frac{R_4}{R_3+R_4})V_1 \end{equation*}

Now deactivate V1 and connect it to ground as shown in figure 3.

differential amplifier working
Fig. 3 differential amplifier working

As we can see that the voltage across R4 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)   \begin{equation*}  I_1=\frac{V_2-0}{R_1}=\frac{V_2}{R_1} \end{equation*}

(7)   \begin{equation*}  I_2=\frac{0-V_{o2}}{R_2}=-\frac{V_{o2}}{R_2} \end{equation*}

Apply KCL at node P

I_{1} = 0 + I_2

(8)   \begin{equation*}  I_1= I_{2} \end{equation*}

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

\Rightarrow \frac{V_2}{R_1}=-\frac{V_{o2}}{R_2}

\Rightarrow V_{o2}=-\frac{R_2}{R_1}V_2

(9)   \begin{equation*}  V_{o2}=-\frac{R_2}{R_1}V_2 \end{equation*}

Adding equations (5) and (9), we get the output voltage Vo

V_o=V_{o1}+V_{o2}

(10)   \begin{equation*}  V_o=(1+\frac{R_2}{R_1})(\frac{R_4}{R_3+R_4})V_1-\frac{R_2}{R_1}V_2 \end{equation*}

(11)   \begin{equation*}  V_o=\frac{R_4}{R_3}(1+\frac{R_2}{R_1})(\frac{1}{1+\frac{R_4}{R_3}})V_1-\frac{R_2}{R_1}V_2 \end{equation*}

Case-1: when \frac{R_2}{R_1}=\frac{R_4}{R_3}

then

V_o=\frac{R_2}{R_1}(1+\frac{R_2}{R_1})(\frac{1}{1+\frac{R_2}{R_1}})V_1-\frac{R_2}{R_1}V_2

\Rightarrow V_o=\frac{R_2}{R_1}(1+\frac{R_2}{R_1})(\frac{1}{1+\frac{R_2}{R_1}})V_1-\frac{R_2}{R_1}V_2

The output Vo is

    \[ \boxed{V_o=\frac{R_2}{R_1}(V_1-V_2)} \]

Now

V_d=V_1-V_2

and

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

The output Vo is given by

V_o=A_dV_d+A_cV_c

where Ad = differential gain and Ac = common mode gain

As V_o=\frac{R_2}{R_1}(V_1-V_2)

Therefore, we have

A_d=\frac{R_2}{R_1}

and A_c=0

so V_o=A_dV_d

Common mode rejection ratio (CMRR)

    \[CMRR=\left | \frac{A_d}{A_c} \right |\]

using above value of Ad and Ac, we have

    \[CMRR=\left | \frac{\frac{R_2}{R_1}}{0} \right |=\infty\]

Note: Ideally CMRR is infinite. So CMRR value for this circuit to be infinite, \frac{R_2}{R_1}=\frac{R_4}{R_3}

.

Case2: when \frac{R_2}{R_1}\neq\frac{R_4}{R_3}

since

V_d=V_1-V_2

and

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

then we can write it as

V_1=V_c+\frac{V_d}{2}

and

V_2=V_c-\frac{V_d}{2}

Therefore equation (10) becomes

V_o=(1+\frac{R_2}{R_1})(\frac{R_4}{R_3+R_4})(V_c+\frac{V_d}{2})-\frac{R_2}{R_1}(V_c-\frac{V_d}{2})

(12)   \begin{equation*}  V_o=\frac{V_d}{2}[(1+\frac{R_2}{R_1})(\frac{R_4}{R_3+R_4})+\frac{R_2}{R_1}]+V_c[(1+\frac{R_2}{R_1})(\frac{R_4}{R_3+R_4})-\frac{R_2}{R_1}] \end{equation*}

(13)   \begin{equation*}  $V_o=A_dV_d+A_cV_c$ \end{equation*}

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

A_d=\frac{1}{2}[(1+\frac{R_2}{R_1})(\frac{R_4}{R_3+R_4})+\frac{R_2}{R_1}]

and

A_c=[(1+\frac{R_2}{R_1})(\frac{R_4}{R_3+R_4})-\frac{R_2}{R_1}]

Then CMRR is given as

CMRR=\left | \frac{A_d}{A_c} \right |

\Rightarrow CMRR=\left | \frac{\frac{1}{2}[(1+\frac{R_2}{R_1})(\frac{R_4}{R_3+R_4})+\frac{R_2}{R_1}]}{[(1+\frac{R_2}{R_1})(\frac{R_4}{R_3+R_4})-\frac{R_2}{R_1}]} \right |

    \[ \boxed{CMRR=\left | \frac{\frac{1}{2}[(1+\frac{R_2}{R_1})(\frac{R_4}{R_3+R_4})+\frac{R_2}{R_1}]}{[(1+\frac{R_2}{R_1})(\frac{R_4}{R_3+R_4})-\frac{R_2}{R_1}]} \right |} \]

Note: (CMRR)_{dB}=20\log_{10}\left | \frac{A_d}{A_c} \right |

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.

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