[SOLVED] [Moved]: OTA-C balanced cascode IC design

FvM said:

V5 should be removed. Presently the amplifier output is shorted.

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however if I removed v5, when I remove the feedback wire after setting the dc operating points and my gm value wouldnt the circuit be incomplete as the differential pair are not identical ?

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a question concerning the gm value, if the value of gm is increasing through my bandwidth as follows, how can I make it constant over my required bandwidth ?

however if I removed v5, when I remove the feedback wire after setting the dc operating points and my gm value wouldnt the circuit be incomplete as the differential pair are not identical ?

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I only wanted to point out the posted circuit doesn't implement negative feedback as frankrose suggested but a shorted amplifier output. It's a different question how the complete circuit will be connected. Surely neither with voltage sources at both inputs.

if the value of gm is increasing through my bandwidth as follows, how can I make it constant over my required bandwidth?

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You don't show the measurement setup generating the magnitude curve but I doubt that you see gm in the plot. It looks more like capacitive feedforward with zero Id (= zero gm).

FvM said:

I only wanted to point out the posted circuit doesn't implement negative feedback as frankrose suggested but a shorted amplifier output. It's a different question how the complete circuit will be connected. Surely neither with voltage sources at both inputs.


You don't show the measurement setup generating the magnitude curve but I doubt that you see gm in the plot. It looks more like capacitive feedforward with zero Id (= zero gm).

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Well I have just changed the capacitor at the output with a resistance of 0 Ohms and changed the vdc and NM3 with a AC voltage of 0.6 DC magnitude and +100mv AC and plotted the current passing through this wire divided by the input AC voltage to get the gm value through my required bandwidth 1 MHz to 3 MHz, shouldn't that be the gm value output of the this ota through my bandwidth ?

Use this test setup to simulate the Gm:



Set input Vcm to 0.6V, the series AC source to 1V AC magnitude. Choose Rbias to 10MOhm, C1 and C2 to 1uF.
Calculate the Gm with this expression: 1/(mag(IF("/C2/PLUS")))
I think you have problems with basics of electronics, you cannot short the output to ground with a 0 Ohm resistor or 0V voltage source. Read more about biasing, operating point, etc. Forum cannot replace a school.

frankrose said:

Use this test setup to simulate the Gm:
Set input Vcm to 0.6V, the series AC source to 1V AC magnitude. Choose Rbias to 10MOhm, C1 and C2 to 1uF.
Calculate the Gm with this expression: 1/(mag(IF("/C2/PLUS")))
I think you have problems with basics of electronics, you cannot short the output to ground with a 0 Ohm resistor or 0V voltage source. Read more about biasing, operating point, etc. Forum cannot replace a school.

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Well I think you are being a little harsh on me in the first post I explicitly wrote in it that I did not get the hang of the basics yet, I am still learning, I tried to learn the basics for a long time however I always forget them so I decided to learn by practice and start implementing, I know I am really bad at it, but at least I am trying to learn that is why I am proud of trying even if I fail a lot and still cannot understand everything you tell me to do, and yes forum cannot replace a school, but forum is helping me to learn and I think if I look at me now and me from a month ago I think I have made a great progress for myself thanks to you and FVM he has been helping me alot in the past few weeks in different posts, well if you do not mind I've decided to take the circuit a level down and implement the circuit attached, balanced without cascodes to solve the problem of the assymetrical stages and it would be much easier for me in biasing especially, i've biased the circuit on cadence virtuoso and was able to achieve the gm value of my first ota from the ideal circuit at the output wire, however I was advised as you said that I need an extra common mode feedback circuit, however I do not understand what do you both mean by an extra common mode circuit if you do not mind elaborating of course , I've attached my new circuit and the value of the output gm at the output node from this circuit of my first ota which is ideal value should be 4.7087u, I achieved a value of 4.71u would that be close enough ? Thank you for your great help even if I make you feel I am a burden for you.

Totally good the 4.71u. And the point is not to disappoint you, to help you. I have read hundreds of pages then I got experience, not I invented the solutions above, everything is written in good books, free online materials. If you read them you can find out solutions by yourself. It is not easier to get the basics from this forum, but if something is not written or hard to find, then feel yourself free to ask anything. This is just my advice, to spend more time with the mentioned topics by reading books instead of learning from any forum. This is not guaranteed here that you get correct answer from me for example. I am sure wikipedia is better many times. So that was just an advice.

frankrose said:

Totally good the 4.71u. And the point is not to disappoint you, to help you. I have read hundreds of pages then I got experience, not I invented the solutions above, everything is written in good books, free online materials. If you read them you can find out solutions by yourself. It is not easier to get the basics from this forum, but if something is not written or hard to find, then feel yourself free to ask anything. This is just my advice, to spend more time with the mentioned topics by reading books instead of learning from any forum. This is not guaranteed here that you get correct answer from me for example. I am sure wikipedia is better many times. So that was just an advice.

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Yes I know I need time to learn the basics however the problem is that I signed in a group project we are designing a low energy bluetooth receiver which would affect my grades, I thought that would be a good way to get to learn the basics, I've improved alot but still I am not good enough, so the problem is I do not have the time to learn all the basics now because I need to deliver my part within a week and half so right now I am looking for quick solutions and I will try to understand the basics along the way until I deliver the project, after it I will have all the time I want to read books and learn the basics to understand them well before signing in another project, if you do not mind would you help me understand what is the meaning of implementing an extra common mode feedback circuit to bias the output at my desired value ? I was advised to do so in my full circuit however there are a couple of things I do not understand, first would be the previous thing about the common mode, the second would be if you may took a look at the following circuit, and my non ideal circuit from the previous post, in the previous post that circuit corresponds to the first ota in this post, in order to correspond to the grounded input Vi1 should I adjust the AC voltage of V5 to be equal zero ? or what should I do?

I didn't advise you common mode feedback, this is not a fully differential circuit. I just mentioned negative feedback, and common mode voltage, don't mix these! In your circuit negative feedback is used to set the output voltage close to the common mode voltage, where your non-ideal circuit can operates similar as the ideal equivalent on your last figure.

After you have put together your designed Gm cells, use this setup:
Connect Vi1 and Vi3 to 0.6V DC voltage without AC magnitude, and connect Vi2 to 0.6V DC voltage with 1V AC magnitude.

If you simulate the whole 3 stage circuit in DC you should see that every node voltage on top level is close to 0.6V because of the negative feedback.
If you simulate the whole 3 stage circuit in AC you should get a transfer function which corresponds to 2nd input, beacuse only at the Vi2 you have set AC magnitude to 1V.

frankrose said:

I didn't advise you common mode feedback, this is not a fully differential circuit. I just mentioned negative feedback, and common mode voltage, don't mix these! In your circuit negative feedback is used to set the output voltage close to the common mode voltage, where your non-ideal circuit can operates similar as the ideal equivalent on your last figure.

After you have put together your designed Gm cells, use this setup:
Connect Vi1 and Vi3 to 0.6V DC voltage without AC magnitude, and connect Vi2 to 0.6V DC voltage with 1V AC magnitude.

If you simulate the whole 3 stage circuit in DC you should see that every node voltage on top level is close to 0.6V because of the negative feedback.
If you simulate the whole 3 stage circuit in AC you should get a transfer function which corresponds to 2nd input, beacuse only at the Vi2 you have set AC magnitude to 1V.

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just as an update in the new balanced circuit without cascodes I've went fully differential as biasing is much easier now, will I still go with common mode voltage or should I go with common mode feedback - which still I do not know what it means exactly- ?

Your last circuit is not fully differential! Fully differential means the input is differential and the output is differential. On your last circuit only the input is differential, the output is asymmetric or non-differential.

Here is a fully differential OTA:
https://file.scirp.org/Html/10-7600157/58c58d2b-f15a-4e50-b424-614ab0dc0ce8.jpg
As you can see it has got 5 terminal, 2 inputs and 2 outputs and common-mode feedback input, CMFB. You have got 2 inputs and 1 outputs.

Check my previous post, I updated it. But I won't explain more about it, please go forward with your original design. I can't teach you everything.

frankrose said:

Your last circuit is not fully differential! Fully differential means the input is differential and the output is differential. On your last circuit only the input is differential, the output is asymmetric or non-differential.

Here is a fully differential OTA:
https://file.scirp.org/Html/10-7600157/58c58d2b-f15a-4e50-b424-614ab0dc0ce8.jpg
As you can see it has got 5 terminal, 2 inputs and 2 outputs and common-mode feedback input, CMFB. You have got 2 inputs and 1 outputs.

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I understand the part of the fully differential outputs yes I know the screenshot above is not fully differential, that is why i told you in the previous post as an update I have went fully differential, but I still do not understand whats is the cmfb input part, I understand you can't teach me everything but would you recommend a reference or somewhere I can learn what is the CMFB node is?