Op-Amp: CMRR (Common Mode Rejection Ratio) Explained (with example)
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- Опубліковано 6 лют 2025
- In this video, what is Common Mode Rejection Ratio (CMRR) in op-amp and what is the importance of CMRR has been explained with the example.
What is CMRR?
CMRR is the ratio of differential gain and the common mode gain.
For ideal op-amp, the value of CMRR is infinite, but for practical op-amp's the value of CMRR used to be in the range of 80 to 100 dB.
Common mode gain is the gain of op-amp when same input is applied or same input is present at both input terminals.
Op-Amp in open loop condition acts as a differential amplifier and amplifies the difference between the two input terminals.
So, if both inputs are equal then the output of the op-amp in ideal condition should be zero.
But actually, some output used to be present at the output terminal.
And the ratio of this common output to the input voltage is known as the common mode gain.
For ideal op-amp, the value of common mode gain should be zero. But practical op-amp has some finite value (less than 1) of common mode gain.
Noise is the main source of common mode input signal. And op-amp should be able to suppress this noise as much as possible.
And how well it is able to suppress this noise is represented by this CMRR.
The timestamps for the different topics covered in the video.
0:20 What is CMRR and what is the importance of CMRR.
4:58 Example
This video will be helpful to all the students of science and engineering in understanding the concept of CMRR in op-amp.
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The timestamps for the different topics covered in the video.
0:20 What is CMRR and what is the importance of CMRR.
4:58 Example
good explanation. thanks
I have seen many tutors chanting CMMR like a parrot, but here I got the applicability of the term in a vivid voice. Thanks for your efforts.
Seriously!! Professors kept chanting even before even introducing CMRR.
after nptel/mit ocw/behzad razavi i found your lectures most clear and to the point.i am watching this because some topics were not covered in those lecture series....really loved your teaching...wish you a very very good fortune
This series is the best explanation of op amp on the internet!Thank you sir for making this.
Hai sir I m from visakhapatnam
Today itself i watched ur videos
Very clear and understanding manner
I came here to cram, and i understand it all the better, appreciate the work !
Yall should be grateful, dont see what all the complaining is for....Thanks ALL ABOUT ELECTRONICS ....you remind me of another teacher we have on here by the name Neso
They are the same I guess
another student of neso academy
From there :)
Best explanation of CMMR on the internet! Thank you! Subscribed :)
Nice explanation sir. From Africa Ethiopia
Best Channel for OP-AMPS tutorials! thank you
Best content
Excellent video. Very good delivery of speech.
i think this guys explains best but don't tell us 100% of any part and thinks we know it already but we don't
It's great, better explained than in my school.
A very good explanation of CMRR.
Thank you for your lectures ....ur all the lectures are very good clearing all the concepts and doubts....PLZZ create a quiz of some doubtful questions at the end so it become easier for us to clear our doubts more easily.....
Amazing thing is that these stuffs are totally free!!
Remember.. If you are not paying for the product,then you are the product.
Totally understand CMRR after watching the video. Thank you.
Thanks for the applied problem.
But please correct formula at arround 7:50.
Acm = (1/Ad) * 10^(90dB/20)
not understand.. why?
Made it so easy for self learning, thanks buddy!
All videos are very good
Awsome explaination on youtube
Thanks a lot for explaining such a confusing topic in this short video 🙏
Thanks a lot sir ,i was struggling to understand cmrr
Amazing style of teaching sir.....
Thank you very much sir very good explanation love from kerala
CORRECTION :- If input frequency increases then CMRR decreases. You have mentioned incorrectly. Kindly update if I am wrong.
The best explanation!
Sir you are going great excellent videos
Sir at 9:18 you said that as the frequency increases the ability to suppress the common mode signal will also decrease. But sir as the freq increases then the common mode gain also decreases so the common mode signal would be attenuated more at higher freq just as the differential input signal. So how the ability decreases?
Very well explained 👍🏻
The information is there, buts whats with "THE" ?
I like your video very much. It's really great. I'll keep an eye on your channel. I am your fan and I will support you.
Sir, at 6:11,i personally think that there should be a '---' in the front of the R2/R1.
It would be there if it is written as R2/R1 (V1- V2)
But its (V2 -V1). So, the negative sign is already considered. I hope it will clear your doubt.
@@ALLABOUTELECTRONICS But actually i think the Vinput is defined by the difference between non-inverting terminal and inverting terminal, so it should just be (V2--V1), if together with '--', it turns out to (V1--V2).
If you closely observe the input at the non-inverting is V2.
thank you very much! You helped me a lot.
The setup box we use .Is a opamp?
absolute god
Sir, could you please reply, how to calculate (formula) the Common mode gain, if CMRR is not given in a Numerical where all other parameters are same as you explained in above example....
I think it you can find out cmrr in data sheet of that particular IC
Please confirm ,
* so basically common mode gain depends on the open/closed loop gain through CMRR .. ie A_cm it's not a fixed value
* CMRR is calculated at what frequency ? in specs sheet in the begining of the video no frequency is mentioned.
10 ^ 4.5 where did it come from??
I have the same problem
90=20log->. Log(ad/acm)=90/20=4.5->10^4.5
You are the best! Thanks!
that means since we try to reduse commen mode signal is it not a sinal that is not be provided by us
Nice Sir
Thanks A lot Sir
sir is common mode gain is somthing associated with op amp
At 7:37 how he wrote 10^4.5,
Anyone please explain
90 = 20 log (Ad/Acm).
That means 4.5 = log (Ad/Acm)
Here the base of the log is 10. Hence, 10^4.5 = Ad/Acm.
I hope, it will clear your doubt.
Hi i have a comparator supplied with +5v/-5v and when I measure the inverting or non inverting pins I read 2.5v what's the reason of that???
Have you applied any input at the inverting or non-inverting terminal. Try to connect any one of them to the specific voltage level and then again check. Are you getting that applied voltage ?
@ALLABOUTELECTRONICS it was common mode voltage of that comparator thank u👍
Sir,if we interchange the pins of op-amp ic pin-2 (inverting terminal) and pin-3 (non-inverting terminal),then what will be happen?
I didn't get it. You mean the external connections ??
well explained
is coomen mode signal is only for differntial amplifers
In the numerical the output voltage should be -50mv as it is a inverting amplifier .... Please clarify my doubt sir Thank you
First of all, the op-amp is used as a differential amplifier. The output would be -50 mV, if the differetial input V1 - V2 is 5mV. Here, we have assumed that the V2 - V1 is the differential input. And hence, the output is positive. I hope, it will clear your doubt.
The entire video is a rhetorical question
Best sir 👏👏
Sir ,may i know,what is the reason behind the amplification of common mode signal,why op-amp can't suppress that even if their difference is zero?
For more info, please check my couple of videos on BJT - Differential amplifier. I have explained it using the small-signal analysis. Once you will go through it, it will get clear to you. The basic reason is, the first stage of the op-amp is differential pair. Since the pair is not ideal, we have amplification of common-mode signal.
is common mode input apper becouse of noise signals
sir why the common mode input appers at input terminalds
sir why we try to reduse cm output voltage
HELLO SIR
can you make one video for PSRR (power supply rejection ratio) of OP-AMP
I too need the same
How we can say CMRR Depend on frequency??
Very helpful, thank you!
What is common mode source in fact?please
Sir what will be the next videos on this opamp??
I will let you know very soon.
Next couple of videos will be on DC offsets/ DC imperfections (input offset voltage, input bias current etc.) in op-amp.
ALL ABOUT ELECTRONICS thanks for this notification.... By the way sir, can you say about the completion of the opamp course, how many lectures still left? This would aid in my preparation.
it will be even more helpful if by drawing given input and outputs.
Sir, in the decibal scale op amp gain is given by A -> 10log(A). Then the CMRR should also be given as 10log(Ad/Acm). Please clarify my doubt
The voltage gain of the op-amp is defined as 20 log (A). When we are representing the power gain in dB then it is defined as 10 log (Ap).
For more info please go through this video:
ua-cam.com/video/ta1sUTiJNkY/v-deo.html
Awesome!!
Sir make some detailed video on sensors and how it is used in IC engine for daily uses like in car etc
where is above the head button
Bro suggest me book name to study electronics
You can start with Electronic Principles by Albert Malvino
Don't throw the wards in 100 100 velocity
I have a question can u solve it please....
THANKS BROTHER...
which software is this you use to draw the circuits?
Good one
Subtitles covering most of the equations written.
You can turn it off manually in the video settings.
In numerical calculation of gains ,Is sign of voltage considered?
No, no need to consider the sign of the voltage. Because the gain defines the ratio of output voltage to the input voltage in absolute terms and it is independent of the sign of the voltage. Gain = | Vout/Vin|
But suppose if the output voltage is inverted with respect to the input voltage (in case of inverting Op-amp) then we can say that there is a 180-degree phase shift between the output and the input.
E.g in case of inverting op-amp, if the input is 1V and output is -5V, then we can say that the gain of the op-amp is 5, and there is a 180-degree phase shift between the output and the input.
I hope it will clear your doubt.
@3.06 Why the term Acm*Vcm is included when we apply differential voltage instead of common mode voltage?
Because Common mode signal is always present.
Sir make some video on VLSI,MICROPROCESSOR AND RESISTORS
it is 0.063 microvolts, not 0.63microvolts
in the end
It will be 0.63 uV. Please check it again, you will get it .
Sir at 6:10 vo negative or not I mean it's input applied In inverting input plz sir reply
As I mentioned the overall output voltage is R2 / R1 * (V2 - V1)
If V2 > V1 then Vo will be positive else it will be negative.
@@ALLABOUTELECTRONICS thanks
Nice explanation thank you!!
Why we want to compress the common signal
The noise, for example, is a common-mode signal. Because it is present at both terminal. And to improve the signal to noise ratio, it is required to remove or compress this common-mode signal.
Can you replace my lecturer at uni?
Why we use 20 log here
I believe it's because the formula refers to voltage gain. If it was power gain it would be 10 log.
Great!
Thank you!
Excellent!!
Thank you very much
Thank you
What if noise signal frequency is same as that of information signal in that case how lpf differentiate noise and information signal
If information signal and noise signal frequency are same then, of course, LPF will not be able to differentiate between the two signal.
But usually, the output signals from the sensors are not very high-frequency signals. And at those frequencies, differential amplifier adequately suppresses the common mode noise signals. (In the case when noise signal frequency is same as signal frequency).
But still by chance in your application, if information signal and the noise signal are high-frequency signals, in that case, you need select op-amp which has high CMRR at that operating frequency.
I hope it will clear your doubt.
why Ad/Acm = 10^4.5 ?
oh no it's okay..got it hahaa
90/20= 4.5 and if you take the antilog then it will be 10^4.5.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS the anti log of what.. how? im very confused w this .... u have an unknown inside the log.. not sure how to do this
i just remember how its done. I was confused with the 10, and it obviously comes from the log base 10.. thanx for the videos!!
Thank you sir
Please upload video to technics of increasing cmrr
we know, CMRR=(change in input)/output
here you have used CMRR is 90db,,,
it is wrong...
here it must be CMR is 90db...
i think i have not learnt at undergrade but now get the clear idea ..thanks. can u provide the pdf file or slide of this videoes
I am hearing that more and more. You are lucky to have you tube and the internet. When I graduated there was no internet, and no youtube. It was more of a struggle to learn.
Nice
bro please make bjt tutorials using pspice
I will include the simulation in the upcoming videos.
Please describe it more details
awesome
what language is this? It says English but the person isn't speaking English
bruh, it is english lmao. what are you on?
You should be thanking him for not speaking in his mother tongue you ingrate
thoda dhire samjhaya kijie ....and important chij repeat kiya kro...
Great...
please talk your words slower.. hard to understand your english
Volleyoghurt use x0.25 speed😎
@@kannavsharma8453 or lower his baudrate 😎😋
Volleyoghurt 😃
yess exactly..it is too hard to understand and making notes at the same time
I watch his videos on 2x and have no problem, what r u saying man?