# Sensor Systems

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theory:sensor_technology:stb_common_mode_rejection_ratio_cmrr

# Common Mode Rejection Ratio (CMRR)

The Common Mode Rejection Ratio is is the ratio between the output of an amplifier due to a common signal with repect to a differential signal. Consider a certain amplifier with two inputs $V_{A}$ and $V_{B}$. We define

• the gain of the amplifier as $A$ - meaning that differential voltages are amplified $A$ times

$$V_{O} = A (V_{A}-V_{B})$$

• an amplification of common signals of $B$ - meaning that the average of $V_{A}$ and $V_{B}$ is amplified $B$ times

$$V_{O} = B (V_{A}+V_{B})/2$$

The Common Mode Rejection Ratio is defined as $$CMRR = \frac{A}{\left |B \right |} \label{eq:CMRR}$$

How does this work with the simple OpAmp amplifier of figure 1?

Fig. 1: Simple OpAmp based amplifier

We can start with superposition of the output $V_{O}$ due to voltage $V_{B}$ and $V_{A}$. To simplify the problem, we start with the output $V_{O}$ due to voltage $V_{A}$ and $V^{+}$: \begin{aligned} V_{O}&=V^{+} \cdot \frac{R_{4}+R_{3}}{R_{3}}-V_{B} \cdot \frac{R_{4}}{R_{3}} \\ &= V^{+} \cdot \frac{m \cdot R+R}{R}-V_{B} \cdot \frac{m \cdot R}{R} \\ &= V^{+} \cdot \frac{m +1}{R}-V_{B} \cdot m \end{aligned} \label{eq:Vout2}

where $V^{+}$ can be calculated as

\begin{aligned} V^{+} &= \frac{R_{2}}{R_{1}+R_{2}} \cdot V_{A} \\ &= \frac{m \cdot R}{m \cdot R + R} \cdot V_{A} \\ &=\frac{m}{m+1} \cdot V_{A}. \end{aligned} \label{eq:Vplus} Which yields

\begin{aligned} V_{O}&= \frac{m}{m+1} \cdot V_{A} \cdot \frac{m +1}{R}-V_{B} \cdot m \\ &= m \cdot \left ( V_{A} - V_{B} \right ) \end{aligned} \label{eq:Vouttotal}

This means that in the ideal situation, the gain $A$ is equal to $m$, and there is no effect of a common signal. However, we made an assumption that the factor $m$ in $R_{4}$ is equal to the factor $m$ in $R_{2}$. What would happen if we distinguish between $m_{4}$ and $m_{2}$?

Start again with \begin{aligned} V_{O}&=V^{+} \cdot \frac{R_{4}+R_{3}}{R_{3}}-V_{B} \cdot \frac{R_{4}}{R_{3}} \\ &= V^{+} \cdot \frac{m_{4} \cdot R+R}{R}-V_{B} \cdot \frac{m_{4} \cdot R}{R} \\ &= V^{+} \cdot \frac{m_{4} +1}{R}-V_{B} \cdot m_{4} \end{aligned} \label{eq:Vout2_imbalance}

where $V^{+}$ can be calculated as

\begin{aligned} V^{+} &= \frac{R_{2}}{R_{1}+R_{2}} \cdot V_{A} \\ &= \frac{m_{2} \cdot R}{m_{2} \cdot R + R} \cdot V_{A} \\ &=\frac{m_{2}}{m_{2}+1} \cdot V_{A}. \end{aligned} \label{eq:Vplus_imbalance} Now we find $$V_{O}= \frac{m_{2}}{m_{2}+1} \cdot V_{A} \cdot \frac{m_{4} +1}{R}-V_{B} \cdot m_{4} \label{eq:Vouttotal_imbalance}$$ which can no longer be simplifed. However, we can bring it to the form of $$V_{O}= A \left ( V_{A} - V_{B} \right )+ \frac{B}{2}\left ( V_{A} + V_{B} \right ) \label{eq:Vout_as_sum}$$ The factors $A$ and $B$ (the differential gain and common gain respectively) are

\begin{aligned} A &= \frac{m_{2}+2m_{2}m_{4}+m_{4}}{2 \left ( m_{2}+1 \right )} \\ B &= \frac{m_{2}-m_{4}}{ m_{2}+1 } \end{aligned}

Here is $A$ the desired differential gain and $B$ the residu due to $m_{2} \neq m_{4}$. This means $$CMRR = \frac{A}{\left |B \right |} = \frac{m_{2}+2m_{2}m_{4}+m_{4}}{\left | m_{2}-m_{4} \right |} \label{eq:CMRR_OpAmp}$$ where we can see that if $m_{2}$ becomes equal to $m_{4}$, the CMRR becomes infinite.

# Sensor Technology TOC

These are the chapters for the Sensor Technology course: