Homework 8 Solns
1. KCL implies that the same current,
I
out
, must flow through both transistors. Assuming they
are all operating in saturation, it follows from the EKV model that
,
V
V
V
c
in

=
κ
which
implies that
(
29
in
c
V
V
V

=
regardless of whether the transistors are operating in weak,
moderate, or strong inversion.
2. To keep the bottom transistor saturated, we must have that
DSsat
V
V
≥
. Substituting how
V
depends on
c
V
and
in
V
into this inequality, we have that
/
DSsat
in
c
V
V
V
+
≤
. The
minimum allowable
c
V
would thus be
k
V
V
DSsat
in
/
+
. If
c
V
were set at this value, then we
must have that
DSsat
DSsat
DSsat
out
V
V
V
V
2
=
+
≥
. If
0
T
in
V
V
, we have that
(
29
0
T
in
DSsat
V
V
V

=
. If
0
T
in
V
V
<
, we have that
T
DSsat
U
V
5
=
.
3. When
V
in
is at its smallest value,
I
out
is at its smallest value, and
V
will adjust itself to its
highest value, which will be just far enough below
V
c
so that the cascode transistor just
passes the
I
out
set up by the bottom transistor. Initially, the bottom transistor will be
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 Spring '07
 SPENCER
 Microelectronics, Transistor, Vin, cascode transistor

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