On
page 525, Eq. (18.3), you assume M1
is in saturation when the Schmitt trigger switches. However,
for
M1 to be in saturation
Vx
>= Vin
– VTHN1 and
thus Vx
>= Vx
+ VTHN2 – VTHN1
Since
VTHN2
increases due to the body effect we can write VTHN2
> VTHN1
so for M1 to be in
saturation
0
>= to a positive number (so M1 is
NOT operating in the saturation region).
The
connection of M1 and M2 is a
split-length device; see Problem 6.14 on page 160. So, M1 operates
in
either the cutoff or triode regions.
Your conclusion appears to be correct. However, as discussed
in
section 6.3.2 starting on page 143,
on the border between the saturation and triode regions (when VDS
= VDS,sat
= VGS
– VTHN) the
triode/saturation equations for the MOSFET’s
drain current are equal. So,
what
we need to show is that M1 is
operating with Vx
= Vin – VTHN1 (VDS,sat
for
M1) for it to be okay
to
use the saturation equation as seen
in Eq. (18.3). As pointed out above and in the book this is indeed
the
case since VTHN1
= VTHN2
as M1/M2 turn on (M1/M2 behave as a single device, again, see problem
6.14).