CMFB for 2stage OTA - one or two?

[ee484]

Hi, I have a question about common-mode feedback circuits for two-stage OTA.

In page 324 of Razavi's analog book, "It is important to note that fully-differential two stage opamps such as that in fig. 9.22 require two CMFB networks, one for the output of each stage. An example is described in [10-Rabii, JSSC]"

However, I don't see why we need to have two cmfb circuit for fig. 9.22 case.
If one cmfb circuit is adjusted to cm desired level, it feedback to the first stage tail current so that certain Vgs voltage at M9 and M10. Thus, the common-mode output voltage in the first stage is the same a Vdd-Vgs9,10.

Could somebody argue my statement?

Please, justfy why we need two cmfb for two-stage ota. Is two-stage ota have to have two cmfb in any case?

Thanks in advance!

 

[rfsystem]

Your argument is right. Only if M9 and M10 are connected as diffamp and biased by a current source you need two CMFB stages.

You can look back from the output to the input for a comon mode signal. If the output should be at some common mode level the current trough M9 and M10 should be equal to the ouput stage bias. That define the gate voltage of M9+M10. This gate voltage define the potential of two counteracting current sources. The first is the PMOS bias current sources from top. The second if the bias current source from the NMOS input diffamp below. So if either the PMOS bias is regulated or the NMOS bias the regulation loop gets closed with full determined biasing.

I think you write a program which analyses the toplogy of circuit and determine the rank deficiency of bias stability.

 

[safwatonline]

helo,
i am really a newbie in analog ic design so dont take my words as facts, anyway the idea is that why we need CMFB in first place , to make sure that any error in the current is eliminated (i.e. to prevent forcing any MOS in triode),which isnt insured by saving the CM on a certain level but by simultainiously adjusting the bias voltage of the MOS loads.
so taking first stage first u must have CMFB for Vb3 for instance , then taking 2nd stage , u need to do another CMFB for Vb4 , i dont see any relation between the current of the first with the second (i.e. if u set Vb3, what will make Vb4 correct and vice-versa)
anyway this is just my opinion.
regards,
a.safwat

 

[nxing]

I agree with safwatonline' opinion and you can refer back the the paper razavi's reference paper.

 

[rfsystem]

safwatonline,

the schematic shown above is w/o the CMFB path. It is ask how many CMFB are needed to make CMD stabilisation.

Otherwise if you have a simple two-stage, symmetric input, asymmetric output, opamp the output node of the mirror driving the output stage gate input is also not defined. It is defined by the external feedback!

 

[avlsi]

only one CMFB is enough,as I used only one in folded cascode OTA.I donot think,we need two.just refer paul gray book chapter 16,as he has given quantitative analysis.