Question

The CMRR is important for comparators and other difference amplifiers. If the output voltage changes by 5V for an input difference voltage $v_{\text{S}}$of $300\mu \text{V}$ , and by 1.0 V for a common mode input voltage of 300 mV , what is the CMRR of the amplifier?

Short answer

The common mod rejection ratio of the amplifier is $5.2\times {10}^3$or$74\text{dB}$.

Step-by-step solution

Step 1. Introduction

The expression for common mod rejection ratio of an operational amplifier.

$\text{CMRR}=\frac{A_d}{A_{\text{cm}}}\dots \dots \left(1\right)$

Here, $A_d$is difference gain, and $A_{\text{cm}}$is common mode gain of an operational amplifier.

Write the expression for gain of an operational amplifier.

$A=\frac{v_o}{v_i}\dots \dots \left(2\right)$

Here,$v_o$ is output voltage change, and$v_i$ input voltage.

Step 2. Explanation

$1.7\times {10}^4$Calculate the difference gain of an operational amplifier.

Substitute for , and for in equation (2).

$A_d=\frac{5\text{V}}{300\mu \text{V}}=\frac{5\text{V}}{\left(300\mu \text{V}\right)\left(\frac{{10}^{-6}\text{V}}{1\mu \text{V}}\right)}=\frac{5\text{V}}{\left(300\times {10}^{-6}\text{V}\right)}=1.7\times {10}^4$

Calculate the common mode input gain of an operational amplifier.

Substitute 1V for $v_o$, and 300 mV for $v_i$in equation (2).

$A_i=\frac{1\text{V}}{300\text{mV}}=\frac{1\text{V}}{\left(300\text{mV}\right)\left(\frac{{10}^{-3}\text{V}}{1\text{mV}}\right)}=\frac{1\text{V}}{\left(300\times {10}^{-3}\text{V}\right)}=3.3$

Calculate the common mod rejection ratio of an operational amplifier.

Substitute $1.7\times {10}^4$for $A_d$, and 3.3 for $A_i$in equation (1).

$\text{CMRR}=\frac{1.7\times {10}^4}{3.3}=5.2\times {10}^3$

CMMR convert in to dB

$\text{CMRR}=20\log \left(5.2\times {10}^3\right)=74\text{dB}$

Therefore, the common mod rejection ratio of the amplifier is $5.2\times {10}^3$or $74\text{dB}$.