ECE 315 Homework 7 Solution
Spring 2009
1.
(Simple CS amplifier design)
An NMOS amplifier is to be designed to provide an output voltage
centered at 1.5V with 0.5V peak-to-peak variation across a
R
D
=50k
Ω
load in the circuit below.
(a)
If a small-signal gain of 5V/V is needed, what
g
m
is required if
r
o
of the NMOSFET can be
ignored?
(2 pts)
If we ignore
r
o
of the NMOS, then the small signal
A
V
is
–g
m
R
D
for the CS amplifier.
We will not
deal with the negative sign here explicitly, since in small-signal circuits that is just a constant phase
shift.
Therefore
g
m
= 5/50k
Ω
= 0.10mS.
(b) For
V
DD
=3V, what values of
I
D
and
V
OV
will you choose for the output center point?
(4 pts)
The center point has the output at 1.5V, and therefore the current will be (3.0 – 1.5)/50k
Ω
=
0.030mA.
For the given gm in (a), we will need a
V
OV
of 0.6V.
(c)
What
W/L
ratio is required if
k
n
’
=
μ
n
C
ox
= 0.1mA/V
2
?
(2 pts)
For the large-signal current in saturation ignoring the Early effect, we have
2
'
2
1
OV
n
D
V
k
L
W
I
=
, which gives 1.67 or
(W/L)
= 5/3.
Additional info: The
W/L
is often given in scalable numbers in technology whose minimal definable
dimension is
2
λ
. This
2
is
called the gate pitch, which is the naming convention of the technology,
i.e., 65nm CMOS technology means the distance between two closely packed gate is 65nm (see
below).
In digital circuits, we usually work with
L
=2
λ
, and a wide transistor denoted as 200
λ
/2
λ
or
simply 200/2.
In analog circuits, due to the concerns of a larger
V
A
, we often have the minimal
transistor length at 3
λ
.
The answer 5/3 above is a typical minimal-sized transistor in analog circuits
(while in digital circuits it will be 3/2).
For some analog design to generate a small voltage or
current in current mirrors or steering, devices of 3/1,000 are not strange either.
To drive an external
circuis, a very strong transistor of 10,000/2 is not uncommon.
Notice that for a square layout design,
even for 10,000/2
W/L
, the foot print can still be smaller than 100
μ
m by 100
μ
m in 0.35
μ
m
technology.
(d) If
V
th
=0.8V, what is the large-signal
V
GS
and
V
DS
as the
Q
point?
(2 pts)
V
OV
=0.6V, therefore,
V
GS
= 1.4V.
V
DS
is at 1.5V as previously presented.
This is the Q point.
2
1