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Who can help me to analysis this AMP?

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leonken

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The architecture of amplifier is from
Op-amps and startup circuits for CMOS bandgap references with near 1-V supply
Boni, A.;
Solid-State Circuits, IEEE Journal of
Volume 37, Issue 10, Oct. 2002 Page(s):1339 - 1343
Digital Object Identifier 10.1109/JSSC.2002.803055

Summary: The design of bandgap-based voltage references in digital CMOS raises several design difficulties, as the supply voltage is lower than the silicon bandgap in electron volts, i.e., 1.2 V. A current-mode architecture is used in order to address the mai.....

The questions are:
1. How to realize the common mode feedback by using m1, m2, m0a and m0b in the input stage.
2. The m1, m2 m0a and m0b operate in weak inversion. How to achieve 60 dB DC gain of such amplifier?
3. Does m5 and m6 operate in weak inversion too, or in saturation region?
Thank you!!
 

derekqiao

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1. This is typical CMFB circuit. When Vin-com increases, Vgs(m0a) and Vgs(m0b) decrease, let the Vgs(m1) and Vgs(m2) stable.
2 & 3. M1, M2, M3, M4, M5, M6 are in saturation, they determine Gain.

By the way, Do you want to use this circuit?
 

leonken

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derekqiao said:
1. This is typical CMFB circuit. When Vin-com increases, Vgs(m0a) and Vgs(m0b) decrease, let the Vgs(m1) and Vgs(m2) stable.
2 & 3. M1, M2, M3, M4, M5, M6 are in saturation, they determine Gain.

By the way, Do you want to use this circuit?
Yes, I want to use this circuit for some low voltage application. Did you use it before?
However, I can not get a high DC gain by using such circuit.
In the attached paper, the author said the input stage M1, M2 and M0a and M0b operate in weak inversion. The transconductance of weak inversion is much smaller than the transconductance in strong inversion. I wonder that how they got a 60 dB DC gain in their paper?

And I think the M0A and M0B operate in strong inversion too?
 

zajbanlik

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Hi
For M5 I'm sure that it's not in inversion, but in saturation mode. Because it' connected like a diode.
For MOA and MOB i'm also sure that schould be in deep inversion mode. But for input differential transistors i quite strange to stay in inverse mode. Try to "play" with dimension of transistors to achieve saturation mode. Do the same for output stage. I think that also M6 schould be in saturation.

Regards
 

leonken

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What are the merits of common mode feedback?
why the common mode feedback is used?
 

zajbanlik

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Hi

I think that both transistors M0A and M0B are in fact resistors which are used for differential stage polarisation. So if would you use only one than voltage changes on drain of differential transistors would infect seriously your DC point. This way it is balanced.
My question is how good this resistor in your compensation path is? I mean tolerance, temperature dependencies, linearity etc.

Regards
 

leonken

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zajbanlik said:
Hi

I think that both transistors M0A and M0B are in fact resistors which are used for differential stage polarisation. So if would you use only one than voltage changes on drain of differential transistors would infect seriously your DC point. This way it is balanced.
My question is how good this resistor in your compensation path is? I mean tolerance, temperature dependencies, linearity etc.

Regards
Thank you zajbanlik and derekqiao!

You mean the input CMFB in the differential pair stage is used to stablize the DC operationg point of the differential pair.
Am I right?
 

zajbanlik

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Yes

I'm quite sure. The question is why not normal resistor. That's why I asked for resistor parameters.

Regards
 

leonken

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zajbanlik said:
Yes

I'm quite sure. The question is why not normal resistor. That's why I asked for resistor parameters.

Regards
If the M0A and M0B are replaed by resistors, how to realize CMFB loop?
And I already send a message to you.
Thanks.
 

zajbanlik

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Hi

Sorry I just cannot read this e-mail now. If it's not late I can replay toomorow.
Well concerning this amplifier, I simply don't think that you need common mode feedback here. This is single ended amplifier. They just used this mosfets like a resistor, because of small suply voltage they cannot use normal current source. My question was why not normal resistor, because you can achieve better stability if the resistor is with enough googd characteristics (temperature, parasitics and so on). The possible answer is that for this technology there is no qualty resistor. But then I saw it in compensation path. Because with normal resisto you are going to have polarisation current enough stabile concerning voltage of input transistors (of course if they are in saturation)
Concerning CMFB. One characteristic is preservation of stabile output signal when is no diferential exitation (this one you don`t need because you use single-ended output).The other one is more side effect it schould be made so that doesn' t afect polarisation of amplifier. I think that designer is using here this second one. Because if gate of M0A goes high because of aplied diferential voltage than other scould go down for the same amount of voltage. That means that first transistor's resistivity is smaler and second one is biger for some amount (which schouldn't be big concerning linearity)
I hope that this is frome some help


Regards
 

leonken

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Thank you zajbanlik.
I already read the Allen's book: CMOS analog circuit design.
In the chapter 7.3 Differential-output op amps, the CMFB is used for differential in differential out opamp topologies to keep the output common-mode voltage midway.
From the attached paper, I think that the main merit of such CMFB in input differential pair stage is low-voltage application. Second is to stablize the DC operationg point.
And I already understood the principle of CMFB loop.
Thanks again!
 

willyboy19

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1. Typically for this kind of single-ended amplifier, you don't need a CMFB. M0A and M0B are there to realize self-biasing.

2. Depends on the input common-mode voltage, M0A and M0B can be biased in either saturation or linear/weak inversion region. Since this is a 1V design, my bet is they are biased in linear or weak inversion region.

3. It is possible to achieve 60dB DC gain here, IF M5 is not connected as a diode as they draw in the schematic. I think this is definitely a typo error, M5 should have its gate tied to gate of M3 and M4, otherwise the compensation cap and resistor would not be needed. In this way, you should get 60dB DC gain.

Try it and let me know what you got.
 

leonken

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Is this circuit suitable for low power application?
For example, the bias current for M1 and M2 is 200nA, so, in order to force M1 and M2 operate in saturation region. The condition L>>W must be satisfied.
However, the DC gain of such circuit is proportional to sqrt{(W/L)1}.
and is reverse proportional to Id.
What can I do?
 

derekqiao

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willyboy19 write:

1. Typically for this kind of single-ended amplifier, you don't need a CMFB. M0A and M0B are there to realize self-biasing.

2. Depends on the input common-mode voltage, M0A and M0B can be biased in either saturation or linear/weak inversion region. Since this is a 1V design, my bet is they are biased in linear or weak inversion region.

3. It is possible to achieve 60dB DC gain here, IF M5 is not connected as a diode as they draw in the schematic. I think this is definitely a typo error, M5 should have its gate tied to gate of M3 and M4, otherwise the compensation cap and resistor would not be needed. In this way, you should get 60dB DC gain.


1.&2. The function of CMFB in this circuit is to get a rail current and make the operation voltage stable. Because M3 and M4 are current source loads. And M0A and M0B will operate in linear region.

3. I agree with you.

leonken:
Could you please change connection of the gate of M5 to the gate of M3? But getting a stable DC operation voltage of OUT will be difficult. You can try it.

And please pay attention to the start up circuit, it has issue in low voltage.
More questions, you can email to me.
 

leonken

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derekqiao said:
willyboy19 write:

1. Typically for this kind of single-ended amplifier, you don't need a CMFB. M0A and M0B are there to realize self-biasing.

2. Depends on the input common-mode voltage, M0A and M0B can be biased in either saturation or linear/weak inversion region. Since this is a 1V design, my bet is they are biased in linear or weak inversion region.

3. It is possible to achieve 60dB DC gain here, IF M5 is not connected as a diode as they draw in the schematic. I think this is definitely a typo error, M5 should have its gate tied to gate of M3 and M4, otherwise the compensation cap and resistor would not be needed. In this way, you should get 60dB DC gain.


1.&2. The function of CMFB in this circuit is to get a rail current and make the operation voltage stable. Because M3 and M4 are current source loads. And M0A and M0B will operate in linear region.

3. I agree with you.

leonken:
Could you please change connection of the gate of M5 to the gate of M3? But getting a stable DC operation voltage of OUT will be difficult. You can try it.

And please pay attention to the start up circuit, it has issue in low voltage.
More questions, you can email to me.

In the attached papaer, the author said the m1,m2 m0a and m0b operate in weak inversion. I'm confused by him. Anyway, I will try it.
 

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