Calculation of ICMR and OCMR of two different architectures

[Winny_Puuh]

Hi guys, I just tried to calculate the input and output common mode range of 2 architectures… unfortunately, most of the results are totally unrealistic and I cannot find my mistakes. Hope you guys see more than me. Btw, the numbering of the transistors goes from bottom to top like: M0, M12, M34, M56

telescopic architecture:


Vb1 = 1.1 V
Vth0 = 595.9 mV, Vth12 = 752.2 mV, Vth34 = 776.9 mV, Vth56 = 661.5 mV
Vdd = 1.8 V
Vin = 900 mV
Vcm = set by CMFB = 1.094 V for smallest current  Vgs0 = 550 mV

ICMR:
Vin,max = Vdd - Vds56 - Vds34 - Vgd12 = Vdd - (Vgs5-Vth5) - (Vgs34-Vth34) - (Vds12-Vgs12)
= Vdd - (Vdd-Vcm-Vth5) - (Vb1-Vds12-Vds0-Vth34) - (Vds12-Vgs12)
= Vdd- (Vdd-Vcm-Vth5) - (Vb1 -Vds0-Vth34) - (-Vgs12)
= - (-Vcm-Vth5) - (Vb1 -Vds0-Vth34) - (-Vgs12)
= -(-Vcm-Vth5) - (Vb1 - Vds0 -Vth3) - (-Vin +Vds0)
= -(-Vcm-Vth5) - (Vb1 -Vth3) - (-Vin)
= Vcm + Vth5 - Vb1 + Vth3 + Vin = 1.094 + 0.6615 - 1.1 + 0.7769 + 0.9 = 2.3324 V……… wrong
Vin,min = Vds0 + Vgs12 = Vds0 + (Vth12) = Vgs0 - Vth0 + Vth12 = 550mV – 0.5959 + 752.2 mV = 706.3 mV  maybe right

OCMR:
Vout,max = Vdd – Vds56 = Vdd – (Vgs56-Vth56) = Vdd- (Vdd-Vcm-Vth56) = Vcm+Vth56 = 1.094 + 0.6615 = 1.7555 V  unrealistic
Vout,min = Vds34 + Vds12 + Vds0 = (Vgs34-Vth34) + (Vgs12 – Vth12) + (Vgs0 – Vth0)
= (Vb1-(Vds12+Vds0)-Vth34) + (Vin-Vds0 – Vth12) + Vds0
= (Vb1-(Vds12+Vds0)-Vth34) + (Vin– Vth12)
= (Vb1-(Vgs12-Vth12+Vds0)-Vth34) + (Vin– Vth12)
= (Vb1-(Vin-Vds0-Vth12+Vds0)-Vth34) + (Vin– Vth12)
= (Vb1-(Vin -Vth12)-Vth34) + (Vin– Vth12)= Vb1 –Vin+ Vth12-Vth34+Vin-Vth12
= Vb1 - Vth34 = 1.1 – 0.7769 = 0.3231 V  finally sth. realistic =)

2nd architecture


The current sources at top and bottom are just M0 and M5
Vg0 = 800 mV
Vg5 = 800 mV
Vth0 = 581.1 mV , Vth12 = 567.7 mV, Vth34 = 761 mV, Vth5 = 652.8 mV
Vdd = 1.8
ICMR:
Vin,max = Vdd- Vds5 – Vds34 – Vdg12 = Vdd – Vds5 – (Vdd - Vds5 – Vds12- Vds0) – ((Vds12 + Vds0) – (Vgs12 + Vds0)) = – (– Vds12- Vds0) – ((Vds12 + Vds0) – (Vgs12 + Vds0)) = – (– Vds12- Vds0) – (Vds12 – Vgs12) = Vds0 + Vgs12 = Vgs0 – Vth0 + Vth12 = 0.8 – 0.5811 + 0.5677 = 0.7866

Vin,min = Vds0 + Vth12 = Vgs0 –Vth0 + Vth12 = 0.7866 --> fantastic ICMR… 0V

OCMR:
Vout,max = Vdd – Vds5 – Vds34 = Vdd – Vds5 – Vgs34 = Vdd – (Vgs5 - Vth5) – Vgs34 = Vdd – Vds5 – Vgs34 = Vdd – (Vgs5 - Vth5) – (Vdd - Vds5 - Vout) = – (Vgs5 - Vth5) – (- Vgs5-Vth5 - Vout) = 2*Vth5 + Vout = 0.4056 V

Vout,min = Vds0 + Vds12 = Vds0 + Vgs12 – Vth12 = Vgs0 – Vth0 + Vgs12 –Vth12= Vgs0 – Vth0 = 800m – 581.1m = 218.9 mV

Oh, and another question, architecture 1 is used in a closed-loop operation. Does that change anything in the output common mode range calculation?
Thanks for any help!
Winny