Download S04_EXAM_Final_Sol - Engineering Class s

U niversity of S outhern
C alifornia
School Of Engineering
Department Of Electrical
Engineering
EE 348:
Final Examination
(Solutions)
May 6, 2004
11:00 –to–1:00
Problem #1: (35%)
(a).
Since capacitor C is an open circuit for quiescent operating conditions, I3Q = I4Q . Note
that the gate-source voltages of transistors M3 and M2 are identical. Since the gate
aspect ratio of transistor M2 is K-times larger than the gate aspect ratio of transistor M3,
and since drain currents are directly proportional to gate aspect ratio, the drain current
conducted by M2 is KI3Q = KI4Q. It follows that I1Q = KI4Q .
 ANS. #1a
(b).
In view of the stipulated simplifying approximations, the small signal current flowing
into any one of the MOSFETs is simply the pertinent transistor transconductance (gm)
times its gate -to- source signal voltage (Vgss). By inspection, I4s = –gm4Vs , and I3s =
gm3Vgs3s = I4s = –gm4Vs . Now I1s = gm2Vgs2s = gm2Vgs3s. But transconductance is, in
general, proportional to the square root of the product of gate aspect ratio and quiescent
drain current. We already know from Part (a), that the quiescent current conducted by
the drain terminal of M2 is K-times larger than that of M3, and we also know that the gate
aspect ratio of M2 is likewise K-times the gate aspect ratio of transistor M3. It follows
that I1s = gm2Vgs3s = Kgm3Vgs3s = –Kgm4Vs .
 ANS. #1b
(c).
The signal voltage, Vos, is developed across the source-drain terminals of transistor M5,
which functions as a resistance of value 1/gm5. Thus, Vos = –I1s/gm5 = Kgm4Vs/gm5. Since
gm4 = gm5, the resultant gain is Vos/Vs = K.
(d).
This one is a proverbial slam dunk. In particular, Rout = 1/gm5.
 ANS. #1d
Problem #2: (50%)
(a).
The amplifier is rendered balanced by open circuiting the gate of transistor M5 . This
means that at low signal frequencies, the signal voltage response, Vy2, is effectively a
Thévenin equivalent output response of the resultantly balanced pair. Note that the gate
of transistor M6 need not be open circuited because by disconnecting the gate of
transistor M5, no signal current can be conducted by transistor M6.
 ANS. #2a
(b).
For the resultantly balanced circuit, the differential input signal voltage, Vdi, is Vs1 – Vs2 =
–Vs, while the common mode input signal, Vci, is (Vs1 + Vs2)/2 = Vs/2. Under differential
signal conditions,
V
V
I1s  g m1 di   g m1 s
2
2
Vdi
Vs
I 2s   g m1
 g m1
.
2
2
(c).
 ANS. #2b
Transistor M2 functions as a resistance, 1/gm2, as does the identical transistor, M2a. The
gate aspect ratio of M1 is 100-times larger than that of M2, whence gm1/gm2 = 10 owing to
the square root dependence of transconductance on gate aspect ratio. Note further that
M3 and M3a, which are terminated at their source terminals in current sinks formed of
transistors M4 and M4a, respectively, operate as source followers having, within the
context of the invoked approximations, unity voltage gain. Thus,
Vo1s 
 g V
Vdo
I
  1s   m1  s  5Vs
2
g m2
 g m2  2
 g V
V
I
Vo2s   do   2s   m1  s   5Vs
2
g m2
 g m2  2
V y1s  Vo1s  5Vs
 ANS. #2c
V y2s  Vo2s   5Vs .
(d).
Since transistor M8 emulates an ideal constant current sink that accordingly carries zero
signal current, the common mode components of all four voltages found in the preceding
part of this problem solution are zero .
 ANS. #2d
(e).
The gate aspect ratio of transistor M5 is 4-times larger than that of transistor M7, which
means that gm5/gm7 =2. Since M5 operates as a common source amplifier having a source
degeneration resistance of 1/gm7, the signal component, I3s, to the indicated current, I3 is
I 3s 
(f).
g m5V y2s
1  g m5 g m7
 
3
.
 ANS. #2e
The gate aspect ratio of transistor M5 is 9-times larger than that of transistor M6, which
means that gm5/gm6 =3. Since M5 operates as a common source amplifier having an
effective load resistance of 1/gm6, the small signal output voltage response is
Vos   I 3s g m6 
g m5  5Vs 
Vos
 5.
Vs
(g).
g m5  5Vs 
3g m6
 5Vs ;
 ANS. #2f
Capacitance C faces an effective resistance of 1/gm3. Accordingly, the 3-dB bandwidth, B,
is (in units of radians -per- second)
B  g m3 C .
 ANS. #2g
Problem #3: (15%)
(a).
Since the forward transconductance is proportional to the square root of gate aspect ratio
and since the bandwidth is directly proportional to the transconductance, large
(b).
(c).
(d).
(e).
bandwidth demands large gate aspect ratio .
 ANS. #3a
The response speed of a bipolar junction transistors is inversely dependent, to first order,
on the square of base region width .
 ANS. #3b
The base-emitter junction is forward biased in the linear regime, thereby giving rise to
both a depletion component and a diffusion component of net junction capacitance. On
the other hand, the base-collector junction is reverse biased, thereby rendering it
divorced of a diffusion capacitance component .
 ANS. #3c
The indicated transistors establish a voltage divider with respect to the two supply lines
to bias the gates of transistors M4, M4a, and M8 .
 ANS. #3c
Transistor M3 is part of a feedback loop established by M3, M1, and M2 . In particular,
the gate-source voltage of transistor M3 is influenced by the gate-source voltage of M1,
whose source terminal drives the gate of M3.
 ANS. #3e
Bonus Problem: (30%)
The small signal model pertinent to determining the effective resistance seen by
capacitance C is given below. From this model, and using the fact that M1 and M2 are
identical transistors conducting identical quiescent currents,
g V
V1  Vx  V3  Vx  m1 1  Vx  V1 ;
g m2
Vx
.
2
Observing that
V1 
g V 
V
I x  gm3V3  gm3  m1 1   gm3 x ,
2
 gm2 
the 3-dB bandwidth follows as B = gm3/2C .
 ANS. BONUS
Vx

V1
gm1V1


Ix
gm3V3
1/gm2
V3

1/gm5