**What is CMRR and what is the importance of CMRR.**

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**What is CMRR and what is the importance of CMRR.**

**What is CMRR **Or Common Mode Rejection Ratio

**So, if you see the datasheet of an op-amp**, then you will find that out of the many parameters, the one parameter which is defined for the op-amp is the common-mode rejection ratio.**So, let’s understand in this article**, what do we mean by this common-mode rejection ratio and what is the importance of this common-mode rejection ratio in the op-amp.**Now, usually, this parameter is defined**for the differential amplifier. And in the earlier introductory articles on the op-amp, we have seen that the op-amp is also a differential amplifier and it amplifies the differential input signal which is available between the two terminals.

**So, let’s say if voltages V1 and V2 are applied to this op-amp**, then the output of the op-amp in open loop configuration will be open-loop gain multiplied by the difference between the two input voltages.**So, here this gain is known as the open-loop gain**or simply differential gain. Because it amplifies the difference between the two input voltages.**Now, suppose if two input voltages V1 andV2 are equal**, in that case, the output of the op-amp should be equal to zero. Or in another word, we can say that if two input terminals are tied together to the common input voltage, in that case, the output of the op-amp should be equal to zero.

**Op-Amp: CMRR Configuration**

**So, this configuration where two input voltages are equal or two input voltages**are tied together to the common input voltage, that configuration is known as the common mode configuration. And the input voltage which is applied to both the input terminal is known as the common-mode voltage.**So, in the ideal condition in this configuration, the output of the op-amp should be zero**. But actually, if you see, some voltage will be available at the output. And this output is known as the common-mode output voltage.**Now, usually, this output voltage used to be less than the applied input signal**. Or we can say that the ratio of this output divided by the input is always less than one.**So, this ratio or the gain is commonly known as the common-mode gain**of this op-amp. So, now let’s see some of the sources which act as a common-mode input signal for this op-amp.

**Op-Amp: CMRR Common source**

**So, the most common source for this common-mode input signal is the 50 or 60 Hz noise signal**. Which used to be present at both input terminals. Apart from that whenever the circuit is operated in the high electromagnetic field, that case, the electromagnetic interference which is coupled to the circuit is also common to both input terminals.**So, these are some of the sources of the common-mode input signal**. Now, suppose if we apply the differential input signal to the op-amp, in that case, the output of the op-amp ideally should be equal to this differential gain multiplied by this differential input voltage. But because of these common-mode input signals, you will see another term in the output. That is common-mode input gain multiplied by this common-mode input signal.**So, this term basically represents the common-mode output voltage**which appears at the output side. And for an op-amp, this term should be as minimum as possible.**So, in other words, we can say that op-amp should be able to suppress this common-mode input signal**, which is present at both input terminals. And how well it is able to suppress this common-mode input signal is defined by the parameter which is known as the common-mode rejection ratio.- And it is the ratio of this differential gain divided by the
**common-mode gain**. And in decibel, it can be represented as 20log (Ad/Acm). Now, usually, this parameter is defined for the open-loop configuration. and in fact, the discussion that we have carried out so far is also for the open-loop configuration.

## Op-Amp: CMRR Practical Cases

**But in practical cases**, this op-amp is used as a differential amplifier in the closed-loop configuration. So, the value of this common-mode rejection ratio which is listed in the datasheet for the open-loop configuration can be used approximately for the closed-loop configuration. And using this we can find the common-mode output voltage which is present at the output terminal.**Now, usually, this differential amplifier is used to amplify the differential input signal**which is coming from the sensors. And many times this differential input signal used to be a very small value.- S
**o, now whenever this differential input is comparable to the common-mode input signal**, in that case, this common-mode rejection ratio used to be a critical parameter. **So, in this case, it is important that this differential amplifier**should be able to suppress this common-mode input signal which is present at both input terminals.

## Op-Amp: CMRR Example

**So, based on this let’s take one example**and let’s understand how well the differential amplifier is able to suppress this common-mode input signal.**So, in this circuit**, the op-amp is used as a differential amplifier in a closed-loop configuration.**So, here the differential input that is being applied to the input terminal**is a 1 kHz sine wave with a peak value of 5mV. And here, the common-mode input signal or the common-mode noise, which is present at both terminals is a 50 Hz signal with a peak value of 2mV. And for the given circuit, the common-mode rejection ratio for the op-amp is 90 dB.**So, let’s find**out in this example, how well the circuit is able to suppress this common-mode input signal.**So, first**of all, let’s find out the output voltage whenever the differential input signal is applied between the two terminals.**So, in the earlier article on the differential amplifier have seen that for this configuration,**the output voltage can be given by the expressionR2 divided by R1, multiplied by the (V2-V1).**So, here the ratio of R2 and R1 defines the differential gain of this amplifier**. And here the difference between this voltageV2 and V1 is nothing but the differential input voltage which is applied between the two terminals.**So, we can say that the output voltage**is nothing but the differential gain multiplied by the differential input voltage that is applied between the two terminals.**So, in this example**, if you see, the value of R2 and R1 are 10-kilo ohm and 1-kilo ohm respectively. So, we can say that for this circuit the differential gain is 10. And the differential input voltage that is applied between the two terminals is the 1KHz sine wave with a peak value of 5mV.**So, for the given circuit the output voltage will be a 1KHz**sine wave with a peak value of 50mV. So, this will be the output voltage because of the differential input that is applied between the two terminals.**Now, let’s see the effect of this common-mode input signal**which is present at both terminals. So, here, the common-mode rejection ratio for the given circuit is equal to 90dB. And as we have discussed, we can approximately consider this value for the closed-loop configuration.**So, we can say that CMRR is equal**to20 log (Ad/Acm) Or we can say that the differential gain divided by the common-mode gain is equal to 10 to the power 4.5. And if you calculate this value, it will be roughly around 31623.**So, we can say that the common-mode rejection ratio for the given circuit is equal to 31623**. Now, as we know it is nothing but the differential gain divided by the common-mode gain.**So, we can say that the common-mode gain will be equal to this differential gain**divided by this common-mode rejection ratio. Now, here for the given circuit, we have found that the value of this differential gain is 10. And we have already found the value of this common-mode rejection ratio as 31623. So, this will be the common-mode gain for the given circuit.**Now, the common-mode output voltage**can be defined as the common-mode gain multiplied by the common-mode input signal. Now, here we have already found the value of this common-mode gain. That is equal to 10 divided by 31623. And here, this common-mode input signal is50 Hz signal with a peak value of 2mV.

**So, if you calculate**it, then we will find the value of this common-mode output voltage as 0.63 UV.**So, at the output**, we will get the common-mode output voltage of 50Hz with a peak value of 063 UV.- S
**o, in this way,**these differential amplifier suppresses the common-mode input signal which is present at the input terminals. And how well it is able to suppress this common-mode input signal, it depends upon the value of this common-mode rejection ratio. Now, this common-mode rejection ratio is also a function of frequency. **So, as the frequency will increase**, the ability of the circuit to suppress this common-mode noise will also reduce.**So, for the given example**, if the common-mode input signal which is present at both input terminals is high-frequency noise, in that case, the circuit will not be able to suppress that noise as efficiently as it was able to suppress the 50 Hz noise. But in that case, suppose the input signal frequency is way less than the high-frequency noise, then still input signal can be filtered out using the low pass filter.**So, these are some design aspects**which you should need to take care of while designing the circuits.

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**Last World About CMRR**

So, I hope in this article, you understood what is common-mode rejection ratio and what is the importance of this parameter in op-amp circuits.

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