ECE 3050A – Fall 2004 Page 1
QUIZ NO. 4 - SOLUTION
(Average score = 7.5/10 of those taking the quiz)
(a.) Replace the transistor in the
circuit shown with a npn BJT that has
a
β
o = 100, VT = 25mV, and VA = ∞.
Assume that ICQ = 0.5mA and find
the numerical values of voltage gain,
vout/vin, Rin. and Rout.
(b.) Replace the transistor in the
circuit shown with a NMOS FET that
has a Kn = 1mA/V2 and
λ
= 0.
Assume that IDQ = 0.5mA and find
the numerical values of voltage gain,
vout/vin, Rin. and Rout.. (Hint: let r
π
of
part (a.) be ∞.)
c.) In your own words tell why the small-signal voltage gain of the BJT CE amplifier is
greater (roughly x10) than the small-signal voltage gain of the NMOS CS amplifier when
the currents are the same and the external circuit is the same.
Solution
a.) The small-signal model for the case of the BJT is shown below (RB = R1||R2).
gm =
ICQ
VT = 0.5mA
25mV = 1
50
r
π
=
β
VT
ICQ = 100·50 = 5kΩ
(ro = ∞)
Rin = Rs + RB||r
π
= 1kΩ + 100kΩ||5kΩ = 1kΩ+4.762kΩ = 5.762kΩ Rout = R3 = 20kΩ
vout
vin = (-gm·R3||RL)
r
π
||RB
Rs + RB||r
π
=
-10K
50
4.762K
50 = -200·0.826 = -165.3 V/V
b.) If we let r
π
= ∞, then the above results are applicable to the MOSFET.
gm = 2KNIDQ = 2·1·0.5 = 1mS
Rin = Rs + RB = 1kΩ + 100kΩ = 101kΩRout = R3 = 20kΩ
and
vout
vin = (-gm·R3||RL)
RB
Rs + RB = (-1·10)
100
101 = -9.9 V/V
c.) The difference is due to the small-signal transconductances. The transconductance is
the slope of the collector/drain current as a function of base-emitter/gate source voltage.
This function for the BJT is an exponential and for the MOSFET is a parabola. At any
equivalent value of collector/drain current the slope of the exponential is at least 10 times
that of a parabola. One could also say that it is due to the difference between diffusion
current (BJT) and drift current (FET).
Transistor
B/G
C/D
R
3
=
20kΩ
R
2
=
200kΩ
V
PP
V
PP
E/S
R
4
=
1kΩ
R
s
=1kΩ
C
1
= ∞
C
3
= ∞
C
2
= ∞
R
L
=
20kΩ
R
1
=
200kΩ
v
in
v
ou
t
+
-
R
out
R
in
F04Q04P1
R
3
=
20kΩ
R
s
=1kΩ
R
L
=
20kΩ
v
in
v
out
+
-
R
out
R
in
F04Q04S1
v
be
+
-
i
b
βi
b
or
g
m
v
be
R
B
=
100kΩ
r
π
Transistor