ECE 3050A – Spring 2003 Page 1
QUIZ NO. 9 - SOLUTION
(Average score = 7.1/10 of the number of students taking this quiz.)
A NMOS amplifier is shown. Assume that the
small-signal parameters of the MOSFET are gm =
1mS, rds = ∞, Cgs = 9pF, and Cgd = 1pF.
a.) Find the midband voltage gain of this
amplifier, Vout/Vin.
b.) Find the value of the upper -3dB frequency,
fH, in Hz, first using the Miller approximation
and secondly using the open-circuit time constant
approach.
c.) Which of the two answers for fL in part b.) is
the most accurate and why?
Solution
a.) The small-signal model for all
three parts of this problem is shown.
The MBG is easily found by
inspection as,
Vout(0)
Vin(0) = -gm(RD||RL) = -10 V/V
b.) The Miller approximation gives the following capacitance between gate and source.
Ceq = Cgs + (1-MBG) Cgd = 9pF + (1+10)1pF = 20pF.
∴
ω
H = 1
RSCeq = 1
1K·20pF = 50 Mrads/sec. →fH = 50x106
2
π
= 7.96MHz
The OCTC approach requires finding RcgsO and
RcgdO. These are found as,
RcgsO = Ri = 1kΩ
RcgdO = ?
Vt = Vgs + (It + gmVgs)10kΩ
= ItRi + (It + gmItRi)10kΩ
∴RcgdO =
Vt
It = Ri + (1+ gmRi)10kΩ = 1kΩ + (1+1)10kΩ = 21kΩ
∴
ω
H = 1
RcgsOCgs + RcgdOCgd = 1
1K·9pF + 21K·1pF = 1000
30 x106 = 33.3x106 rads/sec.
Thus, fH = 33.3x106
2
π
= 5.3MHz
c.) The answer given by the OCTC method is more correct because the impedance of
Cgd at
ω
H for the Miller approach turns out to be (1/50 x106·10-12) = 20kΩ which is not
that much greater than RL||RD = 10kΩ.
V
in
V
out
V
DD
R
i
=1kΩ
R
D
=
20kΩ
R
L
=
20kΩ
R
S
=
2kΩ
C =
∞
C=
∞
+
-
V
SS S03Q09P1
Vin Vout
Ri=1kΩ
RD =
20kΩ
RL =
20kΩ
+
-
Cgd =1pF
Cgd =
9pF gmVgs
Vgs
+
-
S03Q09S1
V
t
R
i
=
1kΩR
D
=
20kΩ
R
L
=
20kΩ
+-
g
m
V
gs
V
gs
+
-
S03Q09S2
I
t