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What's wrong with this bias circuit?

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Hi all: I am a fresh man to design OPA, After I reading this two paper
1."Design Procedure for Two-Stage CMOS Operational Amplifiers Employing Current Buffer" Jirayuth Mahattanakul, IEEE TRANSACTIONS ON CIRCUITS AND SYSTEMS 2005, November, 2005
2."A Compensation Strategy for Two-Stage CMOS Opamps Based on Current Buffer" IEEE Trans. Circuits and Systs. I,vol.44, pp.257-262, Mar.1997,

I design a OPA in the file opa.pdf which can be found in the attachment, at the same, I design a bias circuit in the file opa_bias_improve.pdf which can be found in the attachment too, but afer I simulating the circuit use Hspice, I found the CurrentBuffer in the opa.pdf i.e. the transistor M6 is in the cutoff region, not in the saturation region. The netlist is in the file use_opa_bias_improve.sp.

At the same time, I found that the Vth of M6 is almost equal to the Vgs of M6, even larger than Vgs of M6, no matter what I change the w/l of the MB4 and MB6 which lie in the file opa_bias_improve.pdf.

After I use the other bias circuit in the file opa_bias.pdf which can be found in the attachment too, afer I simulating the circuit use Hspice, I found the CurrentBuffer in the opa.pdf i.e. the transistor M6 can be in the saturation region. The netlist is in the file use_opa_bias.sp

I do not know why, so I beg help. Help me please, thanx!!!!
 

Hi all: may be I don't show my doubt, my meaning is if I use the bias circuit in the file opa_bias.pdf, the transistor M6 can work normally, but if I ues the bias circuit in the file opa_bias_improve.pdf, M6 is in the cutoff region, so I want to know why!

Help me please, thanx!!!
 

One thing that could be causing the trouble is that the gate length of M6 is only 0.6u while the gate length of M11 is 4u. It would be a good idea to have these two transistors having the same Vgs-Vth voltage. Since they will conduct the same current they should be identical in size.

Right now since W/L of M6 is so large compared to M11, the circuit is pulling it into cutoff or triode instead of saturation.
 

vnu=vgsmb10-vdsmb8+vgsmb8
other trace vnu=vgsmb10+vdsmb6
you must satify: vdsmb6=-vdsmb8+vgsmb8
so the w/L of mb6 should be smaller . mb4 will be cut off. so vnu=0
if the w/L of mb6 is large, vdsmb8 is large, mb10 is at triode region.
so vnu= vgsmb8 or so. it is too low.
 

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