diode connected CS amp

[zxcv2201]

I am studying the picture above.

It makes sense that as I1 approaches 0, VDD-Vth becomes. However, I don't know why it can't go up to VDD.
I would like to know how to understand it correctly.

 

[FvM]

Drain current is > 0 in subtreshold range.

 

[zxcv2201]

I am studying diode connected cs amp while reading Razavi's book.

At this time, I thought that finding the swing would be as follows.

1) Vout_max = When M1 is cut-off and does not operate,
* Vout = VDD-Vth2

2) Vout_min = When M1 reaches saturation,
* Vout = Vds1-Vth1

In summary, Vds1-Vth1 < Vout < Vdd-Vth2


Therefore, I thought that the factors affecting swing were M1's Vov and Vth2.

However, the book says that when the gain increases (ex. Av=5), the output swing is severely limited.

I don't know the correlation between Av=Vov2/Vov2 and output swing.
I wonder where I didn't understand correctly.

 

[D.A.(Tony)Stewart]

2. * Vout = Vds1
Av=Vov2/Vov2 ??

I think this.
Voh = Vdd-Vth2 max where Voh=Vdd-Vds2 and Rds2=Vds2/Id2
Vol = RdsOn1 * Ith2 min (Ith2 = current at Vgs2=Vth2=Vds2)
Thus Aol = gain is related Voh max - Vol min or Rds2/Rds1
Rds is related to beta and Vgs

In the saturation region, Ids = beta * (Vgs - Vt)^2 / 2.

Test

If |Vth| = 1.5 for both thus Voh max = 3.5V
beta1 = 20m like CD4xxx series
beta2 = 80m like 74ALCxx series so what is Ids max?


Vdd =5V Vgs bias = 1.75V what is Vol min
What is Aol max?
Razavi says.. Aol= |Vgs2-Vth2)| / (Vgs1-Vth1)

Using my example, Vgs2= |5-1.75-1.5|=3.25 and Vgs2-Vth2=1.75
Vgs1=1.75, so Vgs1-Vth1= 0.25

Aol =1.75/0.25 = 7
with 500 mVpp in and 3Vpp out from 0.5 to 3.5V out.
Avg=2.0V out, 1.75V in

hmm my simulation http://tinyurl.com/yp732lj4
gives me Aol=6
...close but no cigar


Anyone else?

 

[D.A.(Tony)Stewart]

An easier formula for gain of this config is as follows;

sqrt ( beta1 / beta2 )= Av max