1
ECE 315 Homework 9 Solution
Fall 2007
1.
(Design of cascode differential amplifier)
(a)
Sketch the circuit diagram of an activeload MOS differential amplifier where the input
transistors are cascoded, and a cascode current mirror is used for the active load. (8 pts)
(b)
If all of your NMOS transistors are operated at an overdrive voltage of
V
OV
and all NMOS and
PMOS transistors have the same early voltage of 
V
A
, derive the gain expression as:
A
d
=
2(V
A
/V
OV
)
2
.
Evaluate the gain for
V
OV
=0.25V and
V
A
=20V. (8 pts)
We found in the last homework the expression for a differential amplifier with active load using the
small signal model. We can apply the same results here noting now that the output resistance is the
parallel combination of the pMOS and nMOS cascode stacks. The transconductance
G
M
will be the
same transconductance of a cascode amplifier, which is just
g
m,n
G
M
= g
m,n
=
OV
D
V
I
2
g
m,p
= g
m,n
= g
m
R
o
= [(2r
o,p
+ g
m,p
r
o,p
2
)  (2r
o,n
+ g
m,n
r
o,n
2
)]
≈
D
OV
A
D
A
OV
D
o
m
I
V
V
I
V
V
I
r
g
2
2
2
2
2
2
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
(since g
m
r
o
>> 1)
A
d
=G
M
R
o
=
2
2
2
OV
A
V
V
.
v
in
+
v
in

V
C
V
bias
V
DD
v
out
+
v
out

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At V
OV
=0.25V and
V
A
=20V,
A
d
= 12,800.
This is a large value even BJT hard to match, but you do
need a large
V
DD
in order to make the fivetransistor stack saturated.
For sure whenever you work
with a highgain system, everything else has to be precise to avoid gain saturation.
2.
(Frequency response of amplifiers)
For an NMOSFET in an amplifier setup with
C
gs
=0.5pF,
C
gd
=0.1pF,
C
db
=0.1pF,
C
L
=1pF (including
C
db
),
g
m
=5mA/V,
r
o
= 20k
Ω
, and
R
sig
=
R
L
= 20k
Ω
,
(a)
In the CS amplifier of Fig. 6.20, use the Miller theorem to find the quasistatic gain
A
M
and the
3dB corner frequency
f
H
. (8 pts)
The quasistatic gain for this simple firstorder (single pole) system is just
A
M
= g
m
(r
o
 R
L
)
or
A
M
= 50.
The Miller Theorem allows us to substitute the feedback capacitor,
C
gd
, with two separate capacitors
from the gate and the drain, both going to ground. The Miller Theorem estimates the equivalent
capacitance on the gate as
C
in,miller
= C
gd
[1 + g
m
(r
o
 R
L
)] = 5.1 pF
The equivalent capacitance at the output at the output is given by
C
out,miller
= C
gd
[1 + (g
m
(r
o
 R
L
))
1
]
≈
C
gd
Therefore, the total input capacitance is
C
in
= C
in,miller
+ C
gs
= 5.6 pF
And we obtain the 3dB corner frequency:
f
H
=
in
sig
C
R
π
2
1
= 1.42 MHz
(b)
Repeat (a) with the opencircuit time constant method.
Is the answer different? Give the
percentage contribution to
τ
H
by each of the three capacitances (
C
gs
,
C
gd
and
C
L
).
Give also the
gainbandwidth product. (8 pts)
The opencircuit time constant method of finding the 3dB frequency can be done by:
1)
Find the equivalent resistance seen by each capacitor using Thevenin/Norton equivalent circuits
(note that all other capacitors become open circuits, which is the origin of the name; don’t forget
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 Fall '07
 SPENCER
 Amplifier, Microelectronics, Transistor, Vin, Vin =2V Vin, Vin =1V

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