Instrumentation Amplifier Explained (with Derivation)
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- Опубліковано 21 кві 2018
- In this video, the instrumentation amplifier has been explained with the derivation of the output voltage.
And also in this video, it has been explained that how this instrumentation amplifier is superior to the normal differential amplifier (Difference amplifier) and in certain industrial applications why instrumentation amplifier is prefered (Because of high gain, high CMRR and high input impedance) over the differential amplifier.
What is Instrumentation Amplifier:
The instrumentation amplifier is one kind of differential amplifier which provides very high gain, high CMRR and high input impedance and it is designed for very specific applications.
It is used in certain industrial applications (to amplify the output of the transducers) and in test and measurement equipment.
Because of its high CMRR and high input impedance, it is prefered in the harsh environmental conditions where it is possible to have very large common mode noise or interference signals and very low input differential signals.
The timestamps for the different topics covered in the video is given below:
0:19 What is Instrumentation Amplifier?
0:55 Why Instrumentation amplifier is prefered over the differential amplifier in certain applications
11:20 Instrumentation amplifier circuit explained with the derivation
This video will be helpful to all the students of science and engineering to learn about the instrumentation amplifier.
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Music Credit:
www.bensound.com/ - Наука та технологія
The timestamps for the different topics covered in the video is given below:
0:19 What is Instrumentation Amplifier?
0:55 Why Instrumentation amplifier is prefered over the differential amplifier in certain applications
11:20 Instrumentation amplifier circuit explained with the derivation
sir,it could be better if a make a playlist of this amplifier which are the application of op amp.because i find ur schmitt trigger after a long.Thanks for this fab video.
Any thumb rule is there for achieve maximum gain?
How many of you have exam tomorrow
Literally mine!
Today 😅
@@Shravankumar_888 in half hourrrr☠️☠️☠️☠️☠️
Tomorrow 😅
I do have!
This whole video lecture playlist is very helpful one night before exam.. 😂😂
S
Today is that night for me
@@ryucoder5427 same here😂
@@ryucoder5427 same here
I have my exam in just 3 hrs 🙃
I think you should compose a book, your lectures are very powerful and should be documented as a reference.
Great video, simple and easy to understand graphics, no time wasted, straight and to the point
I was going to use instrumentation amplifier for my final project. In proteus simulation i get the values that i want but in application, i got way more different output voltages from instrumentation amplifier. So anyone has an explanation for that situation?
have seen all these circuits 11:20 in Gate ques. Today I get to know their practical significance. Thanks
This explanation is very understandable. Thank you 👍
Instrumentation amplifiers well explained. Thank you !
someone give this man a medal
Or Bhai
how will it impact if we keep r5 and r6 different ? will this still work
Thank you for such a wonderful lectures,Sir...❣️
Most brilliant vedio explanation I have ever seen in my life thankyou
This whole video lecture is very useful.
What parameters must be considered while selecting an opamp to construct instrumentation amplifier?
could someone tell me why would the common mode input signal will be amplified?(at 11:00)
would you please give us any example of monolithic IC for Instrumentation amplifier ?
4:39 sir how do we get the equations on the va and vb
Thanks for a very good explanation! Appreciate your efforts! Keep up the good work! :)
how did we put va and vb in terms of vcm and vd?
7:15 Bottom right corner,
Vo = 10.009Va MINUS 10Vb, not PLUS
Confused me for a good minute please edit
yeah, thanks for the comment. I thought I was making a mistake somewhere.
yeah bro .... that got me confused for 1 min but later solved perfectly
Yeah Thanx brooo...
LOL I left this comment while cramming for an exam two years ago, glad it's been helpful to some!
Please someone tell me why diferential voltage is vd/2 in each terminal 4:44
You doing excellent job keep doing
at 4.36 minute how can you tell that Vcm is the average of Va and Vb??
you have the best electronic lcthurs ever
Thank you very much man, really great explanation!
Beautifully explained 👏
My exam is after 4 hours and currently I am watching this playlist...🤣😂 Very helpful👍
kaisa raha exam?
@@snehasishdhali9925 Achha gaya..... This video was very helpful
@@funfactff3167 yes bhaiya☺
Very good lecture sir.... Thanks a lot
Sir videos are really good please make videos on EMTL and Antennas
Best channel for electronics
Great video for starters to learn about instrumentation amplifiers. The content is well- arranged and the explanations are detailed, though It is not very easy to follow since English is my second language.
Aoa bhai mny 0.3 mv signal ko amplify krna hai...ic bta skty ho
Thanks for making this video. It was helpful.
The best video there is ..HATS off!!
9:38
Hello, can i ask why is it that the input impedance at the inverting terminal is equal to R1? Shouldn't the input impedance of an op amp be very high? Similar question for non inverting terminal.
If you watch my previous videos on op-amp, I have calculated the input impedance for both inverting and non-inverting configuration.
Please refer those videos.
Here is the link:
ua-cam.com/video/uyOfonR_rEw/v-deo.html
Also please use the timestamps I have provided in the pinned comment to go to a specific topic.
Mira macho, se que no te vas a enterar de na, pero eres la polla!!!!!
Muchísimas gracias por tus vídeos, éste cuatri me lo saco gracias a ti y además aprendiendo.
Keep going!!
At 7:07 there is a confusion, if Va= Vcm+Vd/2 & Vb=Vcm-Vd/2 then Vo should be 0.009Vd+20.009Vcm
same thing
Same doubt @@hydrodynamics6038
Awesome video, .... Helped a lot
Why we used non inverting op amp
Best explanatory video ever
Well explained. Keep going..
Looks so difficult but you made it so easyy
Another excellent presentation. Are you saying in some cases if you use unity gain op amps on the inputs of another op amp with matched resistors you can probably not have to purchase a IN amp which can be expensive ?
Yes true. But moreover that, Instrumentation amplifier has high CMRR and low noise than regular op-amps.
Thnk you very much beaultiful explanation
How buffer circuit solve the problem of low input impedance problem at input terminal??
The input impedance of the buffer circuit is very high. So, basically, it isolates the input signal from the op-amp.
thanks for a good video!!
Hello sir. CAn you please tell me, how to design a Instrumentation amplifier and find the resistance values only with Input and output values?
At the later part of the video, I have already given the expression. Using that equation, you can set the gain.
What is the use of R5 and R6 resistors in instrumentation amplifier ? can we just wire them?
These resistors provide the gain. As I mentioned at 10:51, they provide the initial gain.
You made it clear!!!
Another doubt bro. In Duration 14:52 how did you arrive at the expression (1+2R5/RG)
When i do it I'm getting (1+2R5) and the RG is getting cancelled out.
Ig is equal to (Vb - Va)/ Rg. If you multiply this term with (2R5 +Rg) then you will get (Vb- Va) (2R5/Rg + 1)
I think you must have made some minor mistake during the calculation.
Please do it again. You will get it.
so we modified the Subtractor design or differential amplifier design by using pre-stage which acts as voltage buffer for common mode voltage & Differential amplifier for differential voltages and eliminating the issue of Resistor mismatch which results in finite CMRR
Great Video Sir
Can you Please make a video on transducer also
Excellent
I am watching your playlist just before my BARC interview
good luck for the interview !!
@@ALLABOUTELECTRONICS thankyou so much
@@ALLABOUTELECTRONICS I want to inform you that I have been selected as Scientist at BARC
@@harshitsrivastava7700 congrats 👏
@@Joe-el7yk Thankyou 🙏
Such an indianised definitions... Hand collar indian pro
bussin work uncle
wonder full explanation!!!!
Thank you so much~
Really good!
Excellent !
very useful video!!
I hope everybody don't mind me ask this.
at 3:36, how to arrange from first into second expression because I tried several times, still couldn't get it.
Just multiply and divide the left term by ( R3 + R4) / R3. I hope, you will get it now.
@@ALLABOUTELECTRONICS That's clear my confusion, thanks for your help.
Sir, at 4:45 can you please explain how did you write the equations for Va and Vb?
Here, VCM is (Va + Vb)/2. The average of the Va and Vb. While Vd is the difference between Va and Vb.
So, if we write Va and Vb in terms of VCM and Vd then it can be written like that.
I hope it will clear your doubt.
@@ALLABOUTELECTRONICS Thank you so much, sir!
great explanation bro.
thanks!
It's all clear 😍😍
on 7:04 how did value come in terms of vcm and vd?
Va= Vcm + (Vd/2) and Vb= Vcm - (Vd/2)
Just put these values and solve it. You will get the output in terms of Vcm and Vd.
Hi, how did you get this equation Va= Vcm + (Vd/2) and Vb= Vcm - (Vd/2)? is it because you assume that the op-amp is ideal and by this equation Vcm will get cancel away and left with Vd?
@@ALLABOUTELECTRONICS how? How is 10.009(1+5mV/2)= .009?
@@ALLABOUTELECTRONICS 10.009vcm +10vcm = 20.009vcm .. Where d u get ur numbers? Also with vd
Hello! This is awesome! helped me a lot on my technologist certification. Would you be able to show us though how we could derive transfer functions of different op-amp circuitries? I hope you really can.. Thanks again!!
In some of the earlier videos on active filters, I have derived the transfer function. (Butterworth filter design using op-amp). In that video, I have provided the link for the detailed derivation. It might be helpful to you. Will also try to cover more such examples on the second channel soon.
Pls explain how Vout=R2/R1*(VA'-VB')
that is the condition for differential amplifier. this is also a kind of differential amplifier with a different arrangement of op amps.
At 14:23, It should be R5 + RG + R6.
Keep going....
God bless you
thank you
Thank you sir
YOU ARE THE BEST😘😘😘🥰
Tomorrow is my exams and I'm studying this now
nice !
I have a question.why we use differential amplifier in the instrumentation amplifier if we get differential voltage from buffer amplifier which is used in instrumentation amplifier?
Pls answer soon
To improve the signal to noise ratio. Please watch the video from 10:53 onwards.
Sir at 9:53 why the differential input will be lesser?
Because the effective voltage which will be available at the inputs of the differential amplifier is Zin*Vd/ (Zin + Rs).
e.g if at the inverting end input impedance is R1, and let's say the resistance of the bridge circuit is also R1.
Then the effective voltage available at the inverting terminal is Vd1*R1 / (R1 + R1) = Vd1/2
Normally we assume that the source has zero source resistance, which means entire input appear across R1.
But because of the source resistance, the effective voltage will reduce.
Here the resistance of the bridge will also act as a source resistance. That's why differential input will reduce if the source resistance is comparable to the input impedance.
I hope it will clear your doubt.
For more info, you can check the video on input and output impedance:
Here is the link:
ua-cam.com/video/7jw2_x8dyQ8/v-deo.html
@@ALLABOUTELECTRONICS yes sir, thanks for the help😄
nice one
very nice explanation, but why do you speak with so many pauses at random moments?
He is a mathematician! 5☆
i think noone can explain electronic as you do
10.009(vcm+vd2)+10(vcm-vd/2)= 20.009vcm + 0.0045vd what am i missing here?? 7:27
He has done a previous step wrong: it should be V●=10.009Va-10Vb, refer the original formula, his final answer is actually correct
@@aritrasarkar8578 thanks you, i will check later!
o my guru, which book r u referring
You may refer Op-amp by ramakant gayakwad. But you won't get all the details. But still, for reference, it is a good book.
Do you provide notes too?
Some of notes are available on the website.
Please check the website.
www.allaboutelectronics.org
You will find the link in the about section of the channel.
Pronunciation needs some work (not that I can speak your language). Other than that, good video. +1
I think the critical intuition about instrumentation amps is that the first two op-amps totally isolate the input away from the gain resistor network. This prevents the amplifier from loading down the transducer.
There's subtitle though?
7:15 vo =0.009vcm+10.0045vd how??
Just put the Va = Vcm + Vd/2 and Vb = Vcm - Vd/2. But there is a sign mistake in the video. I mean Vo = 10.009 Va - 10 Vb. I hope, now you will get it.
bhai time ni hy
If you are here for instrumentation amp start from 11:25 😊
Time 9:43 Shouldn't it be the input impedance of the non-inverting terminal is R3 parallel R4 and not R3+R4?
To find input impedance, we apply the input and find current. The ratio of input voltage to the input current gives the input impedance looking from that end. So, considering ideal opamp, no current is flowing into opamp terminal. Hence, input current is Vin / R3 + R4. Or input impedance is R3+ R4. I hope it will clear your doubt. For more information, please check the video on non inverting opamp. I have derived the expression of input impedance.
Thank you..
ye itna tough kyu hai😔
16:21
Ig current will be Va-Vb not Vb-Va
IIT will be vb - va only
this shit is Too complicated, I'll just memorise it
why do indeans speak like this ? can someone explain LOL
Indians speak dozens of other languages and usually never speak English