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(fully)-differential circuit and CMFB

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AllenD

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(fully)-differential circuit tail current cource and CMFB

Hi Team
I have a question about differential circuit design and its common mode rejection.
Assuming a simple differential circuit: nmos tail current source(M1) with nmos diff pair(M2 M3) as the input stage and 2 diode connected pmos(M4 M5) as load. The gates of the nmos are the 2 input and drains of the pmos are the 2 output.

To my understanding so far
1.The tail current source is definitely provide common mode rejection.
2. I also need to add CMFB to increase common mode rejection.
3. If I instead , use a current mirror as the load, I don't need the CMFB because the current mirror itself is going to provide common mode rejection.

Q1. If my "statement 1" is correct, then the common mode is rejected by the tail current source, why do I need CMFB to reject common mode further?
Q2 The way I see CMFB is to define/regulate the output DC voltage. But if you consider a single input single output common source amplifier, the output DC voltage is undefined just like a fully differential common source amplifier without CMFB. Why nobody is talking about such feedback for a single input single output common source amplifier?
Q3 Is the tail current is for input common mode rejection and CMFB for output common mode rejection?

Thanks
Allen
 

CMFB is required in usual fully differential amplifiers to set the output common mode level to the intended value.

Your imagined fully differential circuit isn't a usual (neither useful) amplifier, by using diode loads, it has effectively no common mode or differential gain and thus doesn't need CMFB. A useful fully differential amplifier could e.g. use PMOS current sources as output load. Then CMFB can adjust the "tail" current source.

By inserting a current mirror load, it's no longer a fully differential amplifier. No CMFB needed either.

Regarding Q2. Most amplifiers with high (ideally infinite) gain need feedback to set the DC bias and/or the signal gain. Every analog design text book is talking about feedback.

The specific point with CMFB is that it can't be achieved by an external feedback network, it need to access internal amplifier nodes.
 

Q1 CMFB is used to set operating points, to keep the suitable transistors in active region, not for common mode rejection. it is just biasing of devices.
Q2 simple CS amplifier's operating point has to be also defined, just somehow else, still with biasing (regurarly with current mirrors and/or resistors). those as feedforward and/or feedback devices. you can implement CMFF instead of CMFB to bias differential amplifier too, but CMFB is more reliable for high gain circuits.
Q3 there is no output common-mode rejection. you reject common mode noise at the input so it can't appear at the output.
 

Hi FVM
Thanks for the reply!
CMFB is required in usual fully differential amplifiers to set the output common mode level to the intended value.
Did you mean that CMFB is setting the output point and is not intended to improve CMRR?

Your imagined fully differential circuit isn't a usual (neither useful) amplifier, by using diode loads, it has effectively no common mode or differential gain and thus doesn't need CMFB.

Regarding Q2. Most amplifiers with high (ideally infinite) gain need feedback to set the DC bias and/or the signal gain. Every analog design text book is talking about feedback.
It seems you emphasized the high gain of the amplifier and to my understanding from your reply, only the amplifier with high gain(opamp) need CMFB. That makes a lot of sense! Thanks. However, the reason I imagine the example is that I am trying to design a low gain amplifier. The input signal to my amplifier is almost 1/3 of the VDD to GND and the gain of the amplifier is -gm2/gm4=~-2. Hence I want the output common mode/DC signal to be exactly 0.5VDD so that the output signal is not distorted by lack of headroom. This is also why I care a lot about the output common mode/DC voltage to be what I want. In this case, do you agree I need the CMFB to fight the process variation and etc?

Thanks
Allen

- - - Updated - - -

Hi frankrose
Thanks for the reply!
Q1 CMFB is used to set operating points, to keep the suitable transistors in active region, not for common mode rejection. it is just biasing of devices.
Q2 simple CS amplifier's operating point has to be also defined, just somehow else, still with biasing (regurarly with current mirrors and/or resistors). those as feedforward and/or feedback devices. you can implement CMFF instead of CMFB to bias differential amplifier too, but CMFB is more reliable for high gain circuits.
Q3 there is no output common-mode rejection. you reject common mode noise at the input so it can't appear at the output.

1. I understand now that CMFB the output biasing technique and not a CMRR improvement technique! Thanks!
2. It seems when people talk about CMFB, the amplifier is usually an opamp- with large gain. I understand that with high gain, the output operating point is easy to go off without CMFB. However, what if the gain is small? In my example, I am trying to design a low gain amplifier. The input signal to my amplifier is almost 1/3 of the VDD to GND and the gain of the amplifier is -gm2/gm4=~-2. This makes the 2/3 of the headroom at the output. Hence to keep all the mosfets at satuation, I want the output common mode/DC signal to be exactly 0.5VDD so that the output signal is not distorted by lack of headroom. This is also why I care a lot about the output common mode/DC voltage to be what I want. In this case, do you think I need the CMFB to fight the process variation and etc?

Thanks
Allen
 

Gmx/Gmy amplifiers, like a diff.pair with diode loads can tolerate process variaion well, doesn't require CMFB, but for large signals the headroom and distortion will be always worse than for a differential OPAmp with CMFB, because the last has got high loop gain and feedback.
The point of using high loop gain OPAmp for small gain with feedback network to reach good headroom, high linearity and small distortion.
On the edge of saturation the OPAmp's output still can maintain some linearity because the feedback and loop gain. Gmx/Gmy amplifiers don't have that.
 

I am trying to design a low gain amplifier. The input signal to my amplifier is almost 1/3 of the VDD to GND.
You surely need a degenerated differential pair to process the large input swing linearly. It's also unlikely that the diode load can handle the respective output swing without signal distortion.
 

Hi frankrose
Thanks for the recommendation.

Is process variation the main reason CMFB is applied in the circuit? Can you please elaborate why a fully differential amplifier with large gain(opamp) need CMFB, while a fully differential amplifier with small gain(Gmx/Gmy) doesn't.

My understanding is that if the gain is large, the output common mode/DC point is easier to deviate away from the intended value. However, it's about the room of error.

For example, in the case of Gmx/Gmy amplifier, if the output common mode swings are within +/-1mV, which is not a big value. But I can only tolerate +/- 0.5mV common mode swing. Should I use CMFB fix the common mode swing? Or is there a better way?

Thanks
Allen
 

Q1 CMFB is used to set operating points, to keep the suitable transistors in active region, not for common mode rejection. it is just biasing of devices.

And after thinking about it. Can I ask you a follow up question?
The CMFB is sensing the output common mode voltage and reduce its amplitude using feedback technique. So if a circuit have CMFB, the Vout_cm should be reduced compare to a circuit without CMFB. A_cm=Vout_cm/Vin_cm is reduced which means the CMRR is reduced.

Do you agree with the analysis above? Thanks
 

Is process variation the main reason CMFB is applied in the circuit?
No, every variation like DC input level, supply, temperature variation and mismatch.
Can you please elaborate why a fully differential amplifier with large gain(opamp) need CMFB, while a fully differential amplifier with small gain(Gmx/Gmy) doesn't.
Small gain amps, like a simple CS stage with resistor load won't be sensitive for variations, the output operating point will stay close to the defined value but if you put a current source into the circuit as load instead of the resistor the output operating voltage will sit on one of the supply rail for a small difference in any parameter of the circuit. So if you have huge gain you have to apply feedback to control the operating point, and for fully differential circuits you need a common feedback to set the 2 outputs to the same operating voltage. If you don't use common feedback between the outputs there will be an offset, which is not good for the next stage. In small gain circuits this is not a big problem if the outputs have a small DC difference, because it won't "overcontrol" the next small gain differential stage.
So if a circuit have CMFB, the Vout_cm should be reduced compare to a circuit without CMFB. A_cm=Vout_cm/Vin_cm is reduced which means the CMRR is reduced.
Yes, I agree it reduces the common mode output voltage, but very important that the common mode input doesn't create differential output voltage, only creates common mode output voltage which is reduced by CMFB. In fully differential systems the input common mode variation induced differential output variation comes from asymmetries of the system., and this output component is the important, it has to be low, and CMFB doesn't help on this.
 

A_cm=Vout_cm/Vin_cm is reduced which means the CMRR is reduced.

CMRR is the ratio of Vin_cm to equivalent Vin_dm (apparent differential input voltage produced by common mode signal). A_cm has nothing to do with CMRR, except you are abusing a fully differential amplifier as BALUN converter, relying on a high CMFB gain and large CMFB bandwidth.
 

Small addition, that CMFB cannot reduce the CMRR directly due to above reasons, mostly the asymmetries, mismatch determine CMRR of fully differential system, but it reduces the common mode signal level in the cascaded system, so indirectly actually it helps on the total CMRR.
 

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