The timestamps for different topics covered in this video are as follows, 1) 00:07 - Introduction, 2) 00:40 - Common mode signals Vs Differential mode signals, 3) 02:01 - Noise in Single ended amplifier Vs Differential amplifier, 4) 02:31 - OP-Amp Differential mode configuration, 5) 03:23 - OP-Amp Common mode configuration, 6) 04:03 - Voltage gains of OP-Amp, 7) 04:30 - Practical illustrations, 8) 05:48 - CMRR formula Explanation, 9) 06:41 - Conversion table of CMRR in dB and Data sheet, 10) 07:00 - Importance of CMRR, 11) 08:35 - Examples and Solutions (Example:1), 12) 09:50 - Example :2, 13) 11:49 - CMRR Vs Frequency, 14) 12:44 - Thanks for watching, 15) 12:57 - Next video.
I think , anyone can't explaine this much simply this CMRR topic. For last 2days I have been watching so many videos having CMRR. Finally I got correct idea. Thank you. Very well explained.
2:29 why is there still an output of the differential amplifier when the two incoming signals are in phase? I understand that the noise gets to zero, but the output also has to become zero right? Help me :)
It is due to imperfections within an actual op-amp, some common-mode voltage will appear at the output, It is also influenced by imbalanced resistor values at the input terminals, It also depends on the frequency of the common mode input signal. Watch these videos: 1) ruclips.net/video/GvAlSuXSt5o/видео.html (Op-Amp Input and Output offset voltage), 2) ruclips.net/video/WBp-EF3PCwc/видео.html (Op-Amp as Differential amplifier).
Due to imperfections within an actual op-amp, some common-mode voltage will appear at the output, Watch these videos: 1) ruclips.net/video/GvAlSuXSt5o/видео.html (Op-Amp Input and Output offset voltage), 2) ruclips.net/video/WBp-EF3PCwc/видео.html (Op-Amp as Differential amplifier).
Sir i have doubts in a question pls help me the question is If the ratio of differential gain to common mode gain is 10000, in certain differential amplifier, then CMRR is
CMRR is the ratio of differential gain(ADM) to common mode gain(ACM), In your case its 10,000, So, CMRR dB = 20 log10 ADM/ACM, = 20 log 10 *10000 = 80dB. Refer Example :1 in this video,
Hello Sir, how was the conversion of 90dB. When I make mine, I get 31623 for the 90dB ~ 90, then if I replace it back to subscription, I get the 31623/20 then finally 150/And ( 150/(31623/20) ). Finally my Acm = 0.095V
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The timestamps for different topics covered in this video are as follows,
1) 00:07 - Introduction,
2) 00:40 - Common mode signals Vs Differential mode signals,
3) 02:01 - Noise in Single ended amplifier Vs Differential amplifier,
4) 02:31 - OP-Amp Differential mode configuration,
5) 03:23 - OP-Amp Common mode configuration,
6) 04:03 - Voltage gains of OP-Amp,
7) 04:30 - Practical illustrations,
8) 05:48 - CMRR formula Explanation,
9) 06:41 - Conversion table of CMRR in dB and Data sheet,
10) 07:00 - Importance of CMRR,
11) 08:35 - Examples and Solutions (Example:1),
12) 09:50 - Example :2,
13) 11:49 - CMRR Vs Frequency,
14) 12:44 - Thanks for watching,
15) 12:57 - Next video.
I think , anyone can't explaine this much simply this CMRR topic. For last 2days I have been watching so many videos having CMRR. Finally I got correct idea. Thank you. Very well explained.
You are welcome Dear Xavi.
Thank you for explanation
You are welcome.
2:29 why is there still an output of the differential amplifier when the two incoming signals are in phase? I understand that the noise gets to zero, but the output also has to become zero right? Help me :)
It is due to imperfections within an actual op-amp, some common-mode voltage will appear at the output,
It is also influenced by imbalanced resistor values at the input terminals,
It also depends on the frequency of the common mode input signal.
Watch these videos:
1) ruclips.net/video/GvAlSuXSt5o/видео.html (Op-Amp Input and Output offset voltage),
2) ruclips.net/video/WBp-EF3PCwc/видео.html (Op-Amp as Differential amplifier).
Ex2- Noise on each input is 1mv . So as it is a common mode voltage , it should get rejected? Where I am lagging .
Due to imperfections within an actual op-amp, some common-mode voltage will appear at the output,
Watch these videos:
1) ruclips.net/video/GvAlSuXSt5o/видео.html (Op-Amp Input and Output offset voltage),
2) ruclips.net/video/WBp-EF3PCwc/видео.html (Op-Amp as Differential amplifier).
Love u sirji ❤❤❤
Thank you.
Awesome explanation sir
Thank you
Where did the 31623 come from
Expressing the 90 dB common-mode rejection ratio as a numerical value gives CMRR = 31623. (Antilog)
This Clarified my doubt bro :)
Nice words.
Amazing explanation, thank you so much!
You are welcome.
Thanks dude! Great explanation
You are welcome.
Nicely explained ☺️
Thank you.
Nice information
Thank you mam.
Simply superb
Thank you
Sir i have doubts in a question pls help me the question is
If the ratio of differential gain to common mode gain is 10000, in certain differential amplifier, then CMRR is
CMRR is the ratio of differential gain(ADM) to common mode gain(ACM),
In your case its 10,000,
So, CMRR dB = 20 log10 ADM/ACM,
= 20 log 10 *10000 = 80dB.
Refer Example :1 in this video,
@@RobertRobert-bn6fr tysm sir
@@BeerusOPPYou are welcome
SUPER .. THANK YOU
You are welcome.
I have a few problems can you help me solve it? In the instrumentation! I would really appreciate your help!
Yes sure.
Super explanation, Thank you sir
You are welcome.
Great tutorial, thank you.
You are welcome.
excellent
Thank you
Superb explanation. simple correction ACM doesn't have unit because it's gain
best explanation
Thank you
Sir electronics and communication ka 5th and 6th semester ka course banao.
Ok, Dear Sudhir.
@@ElectronicsAD thank you sir, much appreciated.
how 31623 coming?
CMRR = antilog 90dB/20 = 31623,
I hope this cleared your doubt.
Hello Sir, how was the conversion of 90dB. When I make mine, I get 31623 for the 90dB ~ 90, then if I replace it back to subscription, I get the 31623/20 then finally 150/And ( 150/(31623/20) ). Finally my Acm = 0.095V
CMRR = antilog 90dB/20 = 31623,
I hope this cleared your doubt.
1V/10mV is actually 1000 and not 100 at 9:04
New sub
Thank you.