Answers to Problems for Microelectronic Circuits, 8th Edition - Adel Sedra & Kenneth Carless Smith

Master microelectronics with detailed, step-by-step solutions to exercises from Microelectronic Circuits, 8th Edition by Sedra & Smith. This comprehensive resource helps students understand semiconductor devices, analog and digital circuits, and amplifier designs. Each solution is meticulously explained, making it ideal for self-study and exam preparation. Whether you're tackling MOSFETs, BJTs, operational amplifiers, or frequency response, this guide will strengthen your understanding and problem-solving skills. Get expert guidance and ensure academic success with this essential study aid!

1. Table of Contents
Problems Solutions 2
Exercise Solutions [email protected]@gmail.com
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2. Chapter 1–1
Solutions to End-of-Chapter Problems
1.1(a)I=
V
R
=
0.5V
1k
=0.5mA
(b)R=
V
I
=
2V
1mA
=2k
(c)V=IR=0.1mA×20 k=2V
(d)I=
V
R
=
5V
100
=0.05 A=50 mA
Note:Volts, milliamps, and kilohms constitute a
consistent set of units.
1.2(a)P=I
2
R=(20×10
−3
)
2
×1×10
3
=0.4W
Thus,Rshould have a
1
2
-W rating.
(b)P=I
2
R=(40×10
−3
)
2
×1×10
3
=1.6W
Thus, the resistor should have a 2-W rating.
(c)P=I
2
R=(1×10
−3
)
2
×100×10
3
=0.1W
Thus, the resistor should have a
1
8
-W rating.
(d)P=I
2
R=(4×10
−3
)
2
×10×10
3
=0.16 W
Thus, the resistor should have a
14
-W rating.
(e)P=V
2
/R=20
2
/(1×10
3
)=0.4W
Thus, the resistor should have a
1 2
-W rating.
(f)P=V
2
/R=11
2
/(1×10
3
)=0.121 W
Thus, a rating of
1 8
W should theoretically
suffice, though
1
4
W would be prudent to allow
for inevitable tolerances and measurement errors.
1.3(a)V=IR=2mA×1k=2V
P=I
2
R=(2mA)
2
×1k=4mW
(b)R=V/I=1V/20 mA=50 k
P=VI=1V×20 mA=20 mW
(c)I=P/V=100 mW/1V=100 mA
R=V/I=1V/100 mA=10
(d)V=P/I=2mW/0.1mA
=20 V
R=V/I=20 V/0.1mA=200 k
(e)P=I
2
R⇒I=
φ
P/R
I=
φ
100 mW/1k=10 mA
V=IR=10 mA×1k=10 V
Note:V, mA,k,and mW constitute a
consistent set of units.
1.4See figure on next page, which shows how to
realize the required resistance values.
1.5Shunting the10 k by a resistor of value of
Rresult in the combination having a resistance
R
eq,
R
eq=
10R
R+10
Thus, for a 1% reduction,
R
R+10
=0.99⇒R=990 k
For a 5% reduction,
R
R+10
=0.95⇒R=190 k
For a 10% reduction,
R
R+10
=0.90⇒R=90 k
For a 50% reduction,
R
R+10
=0.50⇒R=10 k
Shunting the10 kby
(a)1Mresults in
R
eq=
10×1000
1000+10
=
10
1.01
=9.9k
a 1% reduction;
(b)100 k results in
R
eq=
10×100
100+10
=
10
1.1
=9.09 k
a 9.1% reduction;
(c)10 k results in
R
eq=
10
10+10
=5k
a 50% reduction.
1.6V
O=VDD
R2
R1+R2
To findR O,we short-circuitV DDand look back
into node X,
R
O=R2βR1=
R
1R2
R1+R2
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3. Chapter 1–2
This figure belongs to Problem1.4.
All resistors are 5 kΩ
12.5 kΩ
23.75 kΩ
1.67 kΩ
20 kΩ
1.7Use voltage divider to findV O
VO=1
2
2+1
=2V
Equivalent output resistanceR
Ois
R
O=(2kΩβ1kΩ)=0.667 kΩ
The extreme values ofV
Ofor±5% tolerance
resistor are
V
Omin=3
2(1−0.05)
2(1−0.05)+1(1+0.05)
=1.93 V
V
Omax=3
2(1+0.05)
2(1+0.05)+1(1−0.05)
=2.07 V
Ω3 V
V
O
R
O
1 kβ
2 kβ
The extreme values ofR Ofor±5% tolerance
resistors are667×1.05=700 kΩand
667×0.95=633 kΩ.
1.8
Ω3 V
Ω6 V
10 kβ
(a)
10 kβ
10 kβ
Ω9 V
R π 10 kβ // 20 kβ
π 6.67 kβ
R π 20 // 10 kβ
π 6.67 kβ
Ω4.5 V
(b)
10 kβ
10 kβ
R π 10 // 10
π 5 kβ
Ω9 V

4. Chapter 1–3
3 V
9 V
10 kβ
(c)
10 kβ 10 kβ
R π 10 // 10 // 10
π 3.33 kβ
10 kβ
10 kβ 10 kβ
9 V
6 V
(d)
R π 10 // 10 // 10
π 3.33 kβ
Voltage generated:
+3V [two ways: (a) and (c) with (c) having lower
output resistance]
+4.5 V (b)
+6V [two ways: (a) and (d) with (d) having a
lower output resistance]
1.9
1 kβ
470β
3 V
V
O
VO=3
0.47
1+0.47
=0.96 V
To increaseV
Oto 1.00 V, we shunt the 1-k
resistor by a resistorRwhose value is such that
1βR=2×0.47.
Thus
11
+
1
R
=
1
0.94
15 V
5.00 V
10 kβ
4.7 kβ
R
⇒R=15.67 k
Now,
R
O=1kβRβ0.47 k
=0.94β4.7=
0.94
3
=313
To makeR
O=333, we add a series resistance
of approximately20,as shown below,
3 V
1 kβ15.7 kβ
20 β
ω

R
O
V
O0.47 kβ
1.10
I
2I
1
I V

ω
R
2R
1
V=I(R 1βR2)
=I
R
1R2
R1+R
2
I1=
V
R1
=I
R
2
R1+R
2
I2=
V
R2
=I
R
1
R1+R
2
1.11The parallel combination of the resistors is
R
βwhere

5. Chapter 1–4
1
Rβ
=
N
3
i=1
1/Ri
The voltage across them is
V=I×R
β=
I
4
N
i=1
1/Ri
Thus, the current in resistorR kis
I
k=V/R k=
I/R
k
4
N
i=1
1/Ri
1.12Connect a resistorRin parallel withR L.
To makeI
L=I/4(and thus the current through
R,3I/4),Rshould be such that
6I/4=3IR/4
⇒R=2k
I
I RRL π 6 kβ
3
4
I
L π
I
4
1.13
I R
1
R
in
R
0.2I 0.8I
To make the current throughRequal to0.2I,we
shuntRby a resistanceR
1having a value such
that the current through it will be0.8I; thus
0.2IR=0.8IR
1⇒R 1=
R
4
The input resistance of the divider,R
in,is
R
in=RβR 1=Rβ
R
4
=
1
5
R
Now ifR
1is 10% too high, that is, if
R
1=1.1
R
4
the problem can be solved in two ways:
(a) Connect a resistorR
2acrossR 1of value such
thatR
2βR1=R/4, thus
R
2(1.1R/4)
R2+(1.1R/4)
=
R
4
1.1R
2=R2+
1.1R
4
⇒R
2=
11R
4
=2.75 R
R
in=Rβ
1.1R
4
β
11R
4
=Rβ
R
4
=
R
5
I R
R
in
1.1R
4
R
4
11R
4
}
0.2I
(b) Connect a resistor in series with the load
resistorRso as to raise the resistance of the load
branch by 10%, thereby restoring the current
division ratio to its desired value. The added
series resistance must be 10% ofR(i.e., 0.1R).
0.1R
1.1R
4
R
in
R
0.8I
0.2I
I
Rin=1.1Rβ
1.1R
4
=
1.1R
5
that is, 10% higher than in case (a).
1.14The source current is
i
S=0.5sinωtmA
having peak amplitude
I
s=0.5mA
To ensure a peak voltage across the source of
V
s,max=1V, a load must be connected with
maximum value
R
L,max=Vs,max/Is=0.5/1=2k
IfR
L=10k, a second resistorRmust be
connected in parallel so that
RβR
L=2k
⇒
1
R
+
1
10
=
1
2
⇒R=
1
1/2−1/10
=2.5k

6. Chapter 1–5
In this case,
I
l=Is×
R
RL+R
=0.5×
2.5
10+2.5
=0.1mA
Thus,i
L=0.1sinωtmA.
1.15(a) Between terminals 1 and 2:
3 V V
Th π 2..25
R
Th
π 75
R
Th
2
1
2
1
300 β
100 β
100β
2.25 V
75 β
ω

300 β
(b) Same procedure is used for (b) to obtain
0.75 V
2
3
75 β
(c) Between terminals 1 and 3, the open-circuit
voltage is 3 V. When we short-circuit the voltage
source, we see that the Thévenin resistance will
be zero. The equivalent circuit is then
3 V
1
3
1.16
3 kβ
12.31 kβ
0.77 V I
Now, when a resistance of3kis connected
between node 4 and ground,
I=
0.77
12.31+3
=0.05 mA
1.17(a) Node equation at the common mode
yields
I
3=I1+I2
Using the fact that the sum of the voltage drops
acrossR
1andR 3equals 10 V, we write
10=I
1R1+I3R3
=10I 1+(I1+I2)×2
=12I
1+2I 2

7. Chapter 1–6
5 V
10 V
5 kβ
10 kβ
2 kβ
R
2
R
1
R
3
I
2 I
1
I
3
V
That is,
12I
1+2I 2=10 (1)
Similarly, the voltage drops acrossR
2andR 3add
up to 5 V, thus
5=I
2R2+I3R3
=5I 2+(I1+I2)×2
which yields
2I
1+7I 2=5 (2)
Equations (1) and (2) can be solved together by
multiplying Eq. (2) by 6:
12I
1+42I 2=30 (3)
Now, subtracting Eq. (1) from Eq. (3) yields
40I
2=20
⇒I
2=0.5mA
Substituting in Eq. (2) gives
2I
1=5−7×0.5mA
⇒I
1=0.75 mA
I
3=I1+I2
=0.75+0.5
=1.25 mA
V=I
3R3
=1.25×2=2.5V
To summarize:
I
1=0.75 mAI 2=0.5mA
I
3=1.25 mAV=2.5V
(b) A node equation at the common node can be
written in terms ofVas
10−V
R1
+
5−V
R2
=
V
R3
Thus,
10−V
10
+
5−V
5
=
V
2
⇒0.8V=2
⇒V=2.5V
Now,I
1,I2,andI 3can be easily found as
I
1=
10−V
10
=
10−2.5
10
=0.75 mA
I
2=
5−V
5
=
5−2.5
5
=0.5mA
I
3=
V
R3
=
2.5
2
=1.25 mA
Method (b) is much preferred, being faster, more
insightful, and less prone to errors. In general,
one attempts to identify the lowest possible
number of variables and write the corresponding
minimum number of equations.
1.18Find the Thévenin equivalent of the circuit
to the left of node 1.
Between node 1 and ground,
R
Th=(1kβ1.2k)=0.545 k
V
Th=10×
1.2
1+1.2
=5.45 V
Find the Thévenin equivalent of the circuit to the
right of node 2.
Between node 2 and ground,

8. Chapter 1–7
R
Th=9.1kβ11 k=4.98 k
V
Th=10×
11
11+9.1
=5.47 V
The resulting simplified circuit is
R
5 π 2 kβ
V
5
12
5.45 V 5.47 V
I
5
0.545 kβ 4.98 kβ
ω
I5=
5.47−5.45
4.98+2+0.545
=2.66µA
V
5=2.66µA×2k
=5.32 mV
1.19We first find the Thévenin equivalent of the
source to the right ofv
O.
V=4×1=4V
Then, we may redraw the circuit in Fig. P1.19 as
shown below
5 V 1 V
3 kβ 1 kβ v
o
Then, the voltage atv Ois found from a simple
voltage division.
v
O=1+(5−1)×
1
3+1
=2V
1.20Refer to Fig. P1.20. Using the voltage
divider rule at the input side, we obtain
v
π
vs
=
r
π
rπ+Rs
(1)
At the output side, we findv
oby multiplying the
currentg
mvπby the parallel equivalent ofr o
andR L,
v
o=−g mvπ(roβRL)( 2)
Finally,v
o/vscan be obtained by combining Eqs.
(1) and (2) as
v
o
vs
=−
r
π
rπ+Rs
gm(roβRL)
1.21(a)T=10
−4
µs=10
−11
s
f=
1
T
=10
11
Hz
ω=2πf=6.28×10
11
rad/s
(b)f=3 GHz=3×10
9
Hz
T=
1
f
=3.33×10
−10
s
ω=2πf=1.88×10
10
rad/s
(c)ω=6.28×10
4
rad/s
f=
ω
2π
=10
4
Hz
T=
1
f
=10
−4
s
(d)T=10
−7
s
f=
1
T
=10
7
Hz
ω=2πf=6.28×10
7
rad/s
(e)f=60 Hz
T=
1f
=1.67×10
−2
s
ω=2πf=3.77×10
2
rad/s
(f)ω=100 krad/s=10
5
rad/s
f=
ω
2π
=1.59×10
4
Hz
T=
1
f
=6.28×10
−5
s
(g)f=270 MHz=2.7×10
8
Hz
T=
1
f
=3.70×10
−9
s
ω=2πf=1.696×10
9
rad/s
1.22(a)Z=1kat all frequencies
(b)Z=1/jωC=−j
1
2πf×10×10
−9
Atf=60 Hz,Z=−j265 k
Atf=100 kHz,Z=−j159
Atf=1 GHz,Z=−j0.016
(c)Z=1/jωC=−j
1
2πf×10×10
−12
Atf=60 Hz,Z=−j0.265 G
Atf=100 kHz,Z=−j0.16 M
Atf=1 GHz,Z=−j15.9
(d)Z=jωL=j2πfL=j2πf×10×10
−3
Atf=60 Hz,Z=j3.77
Atf=100 kHz,Z=j6.28 k
Atf=1 GHz,Z=j62.8M

9. Chapter 1–8
(e)Z=jωL=j2πfL=j2πf(1×10
−6
)
f=60 Hz,Z=j0.377 m∴
f=100 kHz,Z=j0.628∴
f=1 GHz,Z=j6.28 k∴
1.23(a)Z=R+
1
jωC
=10
3
+
1
j2π×10×10
3
×10
−9
=(1−j15.9)k∴
(b)Y=
1
R
+jωC
=
1
10
3
+j2π×10×10
3
×0.01×10
−6
=10
−3
(1+j0.628)∴
Z=
1
Y
=
10
3
1+j0.628
=
10
3
(1−j0.628)1+0.628
2
=(717−j450)∴
(c)Y=
1
R
+jωC
=
1
10×10
3
+j2π×10×10
3
×100×10
−12
=10
−4
(1+j0.0628)
Z=
10
41+j0.0628
=(9.96−j0.626)k∴
(d)Z=R+jωL
=100×10
3
+j2π×10×10
3
×10×10
−3
=10
5
+j6.28×100
=(10
5
+j628)∴
1.24Y=
1
jωL
+jωC
=
1−ω
2
LC
jωL
⇒Z=
1
Y
=
jωL
1−ω
2
LC
The frequency at which|Z|=∞is found letting
the denominator equal zero:
1−ω
2
LC=0
⇒ω=
1
√
LC
At frequencies just below this,∠Z=+90
◦
.
At frequencies just above this,∠Z=−90
◦
.
Since the impedance is infinite at this frequency,
the current drawn from an ideal voltage source is
zero.
1.25
i
s
R
s
R
s
∴
ω
v
s
Thévenin
equivalent
Norton
equivalent
voc=vs
isc=is
vs=isRs
Thus,
R
s=
v
ocisc
(a)v s=voc=3V
i
s=isc=3mA
R
s=
v
oc
isc
=
1V
3mA
=1k∴
(b)v
s=voc=0.5V
i
s=isc=50µA
R
s=
v
oc
isc
=
0.5V
50µA
=0.1M∴=10 k∴
1.26
R
s
R
L
∴
ω
v
s i
o
iO
vS
=
1
RL+RS
⇒v S=iO(RL+RS)
Thus,
v
S=10
−4
×
γ
10
5
+RS
υ
=10+10
−4
×RS(1)
and,
v
S=5×10
−4
γ
10
4
+RS
υ
=5+5×10
−4
×RS(2)
Subtracting equation (2) from equation (1) gives
0=5−4×10
−4
×RS
⇒R S=
5
4×10
−4
=12.5k∴
Substituting into equation (1),
v
S=10+10
−4
×12.5×10
3
=11.25V
Finally, the Norton equivalent source is found as
follows,
i
S=vS/RS=900µA

10. Chapter 1–9
1.27
P
L=v
2
O
×
1
RL
=v
2
S
R
2
L
(RL+RS)
2
×
1
RL
=v
2
S
RL
(RL+RS)
2
Since we are told that the power delivered to a
16speaker load is 75% of the power delivered
to a 32speaker load,
P
L(RL=16)=0.75×P L(RL=32)
16
(RS+32)
2
=0.75×
32
(RS+32)
2
√
16
RS+32
=
√
24
RS+32
⇒(
√
24=
√
16)R S=
√
16×32−
√
24×16
0.9R
S=49.6
R
S=55.2
v
s v
O
R
s
R
L


ω
ω
1.28The observed output voltage is 0.5 mV/
◦
C,
which is one quarter the voltage specified by the
sensor, presumably under open-circuit conditions:
that is, without a load connected. It follows that
that sensor internal resistance must be equal to 3
timesR
L,thatis,30 k.
1.29

ω

ω
v
s v
o R
s
R
s
i
o

ω
i
s v
o
i
o
vo=vs−ioRs
Open-circuit
(
i
o π 0)
voltage
0
R
s
v
s
v
s
v
o
π i
s
i
o
Slope π ωR
s
Short-circuit (v
o π 0) current
1.30The nominal values ofV LandI Lare
given by
V
L=
R
L
RS+RL
VS
IL=
V
S
RS+RL
After a 10% increase inR L, the new values
will be
V
L=
1.1R
L
RS+1.1R L
VS
IL=
V
S
RS+1.1R L
(a) The nominal values are
V
L=
200
5+200
×1=0.976V
I
L=
1
5+200
=4.88µA
After a 10% increase inR
L, the new values
will be
V
L=
1.1×200
5+1.1×200
=0.978V
I
L=
1
5+1.1×200
=4.44µA
These values represent a 2% and 9% change,
respectively. Since the load voltage remains
relatively more constant than the load current, a
Thévenin source is more appropriate here.
(b) The nominal values are
V
L=
50
5+50
×1=0.909V
I
L=
1
5+50
=18.18mA
After a 10% increase inR
L, the new values
will be
V
L=
1.1×50
5+1.1×50
=0.917V
I
L=
1
5+1.1×50
=16.67mA
These values represent a 1% and 8% change,
respectively. Since the load voltage remains
relatively more constant than the load current, a
Thévenin source is more appropriate here.
(c) The nominal values are
V
L=
0.1
2+0.1
×1=47.6mV
I
L=
1
2+0.1
=0.476mA
After a 10% increase inR
L, the new values
will be
V
L=
1.1×0.1
2+1.1×0.1
=52.1mV
I
L=
1
2+1.1×0.1
=0.474mA
These values represent a 9% and 0.4% change,
respectively. Since the load current remains

11. Chapter 1–10
relatively more constant than the load voltage, a
Norton source is more appropriate here. The
Norton equivalent current source is
I
S=
V
S
RS
=
1
2
=0.5mA
(d) The nominal values are
V
L=
16
150+16
×1=96.4mV
I
L=
1
150+16
=6.02mA
After a 10% increase inR
L, the new values
will be
V
L=
1.1×16
150+1.1×16
=105mV
I
L=
1
150+1.1×16
=5.97mA
These values represent a 9% and 1% change,
respectively. Since the load current remains
relatively more constant than the load voltage, a
Norton source is more appropriate here. The
Norton equivalent current source is
I
S=
V
S
RS
=
1
150
=6.67mA
1.31
R
s
R
L
R
L
i
s

⇒

⇒
v
s v
o
v
o
R
s
i
oi
o

⇒
RLrepresents the input resistance of the processor
Forv
o=0.95v s
0.95=
R
L
RL+Rs
⇒R L=19R s
Fori o=0.95i s
0.95=
R
s
Rs+RL
⇒R L=RS/19
1.32
Caseω(rad/s) f(Hz) T(s)
a3.14×10
10
5×10
9
0.2×10
−9
b 2×10
9
3.18×10
8
3.14×10
−9
c6.28×10
10
1×10
10
1×10
−10
d3.77×10
2
60 1.67×10
−2
e6.28×10
4
1×10
4
1×10
−4
f6.28×10
5
1×10
5
1×10
−5
1.33(a)V peak=117×
√
2=165 V
(b)V
rms=33.9/
√
2=24 V
(c)V
peak=220×
√
2=311 V
(d)V
peak=220×
√
2=311 kV
1.34(a)v=3sin(2π×20×10
3
t)V
(b)v=120
√
2sin(2π×60)V
(c)v=0.1sin(10
8
t)V
(d)v=0.1sin(2π×10
9
t)V
1.35The waveform is symmetrical, and therefore
has an average value of 0 V. Its lowest and
highest values are -1 V and +1 V respectively. Its
frequency is1/0.5ms=2kHz. It may be
expressed as a sum of sinusoids using Eq. (1.2),
1
π
⇒
sin 2π×2000t+
1
3
sin 6π×2000t
+
1
5
sin 10π×2000t+...
√
V
0.5 ms
1 V
⇒1 V
t
v
1.36The two harmonics have the ratio
126/98=9/7. Thus, these are the 7th and 9th
harmonics. From Eq. (1.2), we note that the
amplitudes of these two harmonics will have the
ratio 7 to 9, which is confirmed by the
measurement reported. Thus the fundamental will
have a frequency of 98/7, or 14 kHz, and peak
amplitude of 63×7=441mV. The rms value of
the fundamental will be441/
√
2=312mV. To
find the peak-to-peak amplitude of the square
wave, we note that4V/π=441 mV.Thus,
Peak-to-peak amplitude
=2V=441×
π
2
=693 mV
PeriodT=
1
f
=
1
14×10
3
=71.4µs
1.37The rms value of a symmetrical square
wave with peak amplitudeˆVis simplyˆV. Taking
the root-mean-square of the first 5 sinusoidal
terms in Eq. (1.2) gives an rms value of,
4ˆV
π
√
2
≥
1
2
+
⇒
1
3
√
2
+
⇒
1
5
√
2
+
⇒
1
7
√
2
+
⇒
1
9
√
2
=0.980ˆV
which is 2% lower than the rms value of the
square wave.

12. Chapter 1–11
1.38If the amplitude of the square wave isV
sq,
then the power delivered by the square wave to a
resistanceRwill beV
2
sq
/R.If this power is to be
equal to that delivered by a sine wave of peak
amplitudeˆV,then
V
sq
V
sq
0
T
V
2
sq
R
=
(ˆV/
√
2)
2
R
Thus,V
sq=ˆV/
√
2.This result is independent of
frequency.
1.39
DecimalBinary
0 0
5 101
13 1101
32 10000
63 111111
1.40
b3b2b1b0Value Represented
0 0 0 0 +0
0 0 0 1 +1
0 0 1 0 +2
0 0 1 1 +3
0 1 0 0 +4
0 1 0 1 +5
0 1 1 0 +6
0 1 1 1 +7
1 0 0 0 –0
1 0 0 1 –1
1 0 1 0 –2
1 0 1 1 –3
1 1 0 0 –4
1 1 0 1 –5
1 1 1 0 –6
1 1 1 1 –7
Note that there are two possible representations
of zero: 0000 and 1000. For a 0.5-V step size,
analog signals in the range±3.5Vcan be
represented.
InputStepsCode
+2.5 V+50101
−3.0 V−61110
+2.7 +50101
−2.8 −61110
1.41(a) AnN-bit DAC produces2
N
discrete
equally spaced levels over a range of
V
FS−0=V FS. Thus, the levels are spaced
V
LSB=
V
FS
(2
N
−1)
(1)
(b) Any voltage within the range 0 toV
FScan be
approximated by the nearest DAC output level,
which will be up toV
LSB/2above or below the
desired voltage. Thus, the maximum quantization
error of the converter is
V
LSB/2=V FS/2(2
N
−1).
(c) To obtain a resolution of 2 mV or less, using
Eq. (1)
0.002≤
5
(2
N
−1)
⇒N≥log
2
ω
5
0.002
+1
π
=11.3
Taking the minimum integer satisfying this
condition, we require a 12-bit DAC which
provides a resolution of
V
LSB=
5
2
12
−1
=1.22mV
1.42(a) Whenb
i=1, theith switch is in
position 1 and a current (V
ref/2
i
R) flows to the
output. Thusi
Owill be the sum of all the currents
corresponding to “1” bits, that is,
i
O=
V
ref
R
ω
b
1
2
1
+
b
2
2
2
+···+
b
N
2
N
π
(b)b
Nis the LSB
b
1is the MSB
(c)i
Omax=
10 V
10 k
ω
1
2
1
+
1
2
2
+
1
2
3
+
1
2
4
+
1
2
5
+
1
2
6
+
1
2
7
+
1
2
8
π
=0.99609375 mA
Corresponding to the LSB changing from 0 to 1
the output changes by(10/10)×1/2
8
=
3.91µA.
1.43There will be 44,100 samples per second
with each sample represented by 16 bits. Thus the

13. Chapter 1–12
throughput or speed will be44,100×16=
7.056×10
5
bits per second.
1.44Each pixel requires8+8+8=24bits to
represent it. We will approximate a megapixel as
10
6
pixels, and a Gbit as10
9
bits. Thus, each
image requires24×10×10
6
=2.4×10
8
bits.
The number of such images that fit in 16 Gbits of
memory is
Γ
2.4×10
8
16×10
9
=66.7=66
1.45(a)A
v=
v
O
vI
=
10 V
100 mV
=100 V/V
or 20 log 100=40 dB
A
i=
i
O
iI
=
v
O/RL
iI
=
10 V/100
100µA
=
0.1A
100µA
=1000 A/A
or 20 log 1000=60 dB
A
p=
v
OiO
vIiI
=
v
O
vI
×
i
O
iI
=100×1000
=10
5
W/W
or 10 log10
5
=50dB
(b)A
v=
v
O
vI
=
1V
10µV
=1×10
5
V/V
or 20 log1×10
5
=100 dB
A
i=
i
O
iI
=
v
O/RL
iI
=
1V/10 k
100 nA
=
0.1mA
100 nA
=
0.1×10
−3
100×10
−9
=1000 A/A
or 20 logA
i=60 dB
A
p=
v
OiO
vIiI
=
v
O
vI
×
i
O
iI
=1×10
5
×1000
=1×10
8
W/W
or 10 logA
P=80 dB
(c)A
v=
v
O
vi
=
5V
1V
=5V/V
or 20 log 5=14 dB
A
i=
i
O
iI
=
v
O/RL
iI
=
5V/10
1mA
=
0.5A
1mA
=500 A/A
or 20 log 500=54 dB
A
p=
v
OiO
vIiI
=
v
O
vI
×
i
O
iI
=5×500=2500 W/W
or 10 logA
p=34 dB
1.46
0.2 V
2.2 V
+3 V
3 V
ω

v
o
t
1 mA
100 π
i
i
R
L
v
i
The voltage gain is
V
o
Vi
=
2.2
0.2
=11V/V=20.8dB
The current gain is
I
o
Ii
=
2.2/0.1
1.0
=22A/A=26.8dB
The power gain is
V
oIo/2
ViIi/2
=11×22=242W/W=23.8dB
The supply power is
V
2
o
2RL
×
1
η
=
2.2
2
2×0.1×0.1
=242mW
Since the power is drawn from±3V supplies, the
supply current must be
242
3−(−3)
=40.3mA
The power dissipated in the amplifier is the total
power drawn from the supply, less the power
dissipated in the load.
242−
2.2
2
2×0.1
=217.8mW
1.47For±5V supplies:
The largest undistorted sine-wave output is of
4-V peak amplitude or4/
√
2=2.8V rms.Input
needed is14 mV
rms.
For±10-V supplies, the largest undistorted
sine-wave output is of 9-V peak amplitude or
6.4 V
rms. Input needed is 32 mVrms.

14. Chapter 1–13
v
O
v
I
200 V/V
V
DD 1.0
V
DD ω 1.0
For±15-V supplies, the largest undistorted
sine-wave output is of 14-V peak amplitude or
9.9 V
rms. The input needed is 9.9 V/200=
49.5 mV
rms.
1.48v
o=Avovi
RL
RL+Ro
=Avo
ω
v
s
RiRi+Rs
π
R
L
RL+Ro
ω

ω

v
i
ω

v
i
v
s R
i
R
s
ω

R
L
A
vov
i
R
o
Thus,
v
o
vs
=Avo
Ri
Ri+Rs
RL
RL+Ro
(a)A vo=100,R i=10R s,RL=10R o
vo
vs
=100×
10R
s
10Rs+Rs
×
10R
o
10Ro+Ro
=82.6V/V or 20 log82.6=38.3dB
(b)A
vo=100,R i=Rs,RL=Ro
vo
vs
=100×
1
2
×
1
2
=25V/Vor20log25=28dB
(c)A vo=100V/V,R i=Rs/10,R L=Ro/10
v
o
vs
=100
R
s/10
(Rs/10)+R s
Ro/10
(Ro/10)+R o
=0.826V/V or 20 log0.826=−1.7dB
1.49
ω

ω

ω

v
i
v
o
100 π
500 π1 Mπ 100v
i
20 logA vo=40 dB⇒A vo=100 V/V
A
v=
v
ovi
=100×
500
500+100
=83.3 V/V
or 20 log 83.3=38.4 dB
A
p=
v
2
o
/500
v
2
i
/1M
=A
2
v
×10
4
=1.39×10
7
W/W
or 10 log(1.39×10
7
)=71.4dB.
For a peak output sine-wave current of 20 mA,
the peak output voltage will be 20 mA×500
=10V. Correspondinglyv
iwill be a sine wave
with a peak value of10 V/A
v=10/83.3,oran
rms value of10/(83.3×
√
2)=0.085 V.
Corresponding output power=(10/
√
2)
2
/500
=0.1 W
1.50(a)
v
o
vs
=
v
i
vs
×
v
o
vi
=
1
5+1
×100×
100
200+100
=5.56V/V
Much of the amplifier’s 100 V/V gain is lost in
the source resistance and amplifier’s output
resistance. If the source were connected directly
to the load, the gain would be
v
o
vs
=
0.1
5+0.1
=0.0196V/V
This is a factor of284×smaller than the gain
with the amplifier in place!
(b)
The equivalent current amplifier has a dependent
current source with a value of
100V/V
200
×i
i=
100V/V
200
×1000×v
i
=500×i i
Thus,
i
o
is
=
i
i
is
×
i
o
ii
=
5
5+1
×500×
200
200+100
=277.8A/A
Using the voltage amplifier model, the current
gain can be found as follows,
i
o
is
=
i
i
is
×
v
i
ii
×
i
o
vi
=
5
5+1
×1000×
100V/V
200+100
=277.8A/A

15. Chapter 1–14
This figure belongs to Problem1.50.
ω

ω

v
i
v
s
i
o5 kπ
1 kπ
200 kπ
ω

v
o 100 π 100 η vi
ω

Figure 1
5 kπ 1 kπ
i
i
i
s 200π500 η i
i 100π
i
o
Figure 2
1.51
ω

ω

v
i
200 kπ
1 Mπ1 V
ω

ω

v
o
20 π
100 π
1ηv
i
vo=1V×
1M
1M+200 k
×1×
100
100+20
=
1
1.2
×
100
120
=0.69 V
Voltage gain=
v
ovs
=0.69 V/Vor−3.2 dB
Current gain=
v
o/100
vs/1.2M
=0.69×1.2×10
4
=8280 A/A or 78.4 dB
Power gain=
v
2
o
/100
v
2
s
/1.2M
=5713 W/W
or 10 log 5713=37.6 dB
(This takes into account the power dissipated in
the internal resistance of the source.)
This figure belongs to Problem1.52.
1.52In Example 1.3, when the first and the
second stages are interchanged, the circuit looks
like the figure above, and
v
i1
vs
=
100 k
100 k+100 k
=0.5V/V
A
v1=
v
i2
vi1
=100×
1M
1M+1k
=99.9 V/V
A
v2=
v
i3
vi2
=10×
10 k
10 k+1k
=9.09 V/V
A
v3=
v
L
vi3
=1×
100
100+10
=0.909 V/V
Total gain=A
v=
v
L
vi1
=Av1×Av2×Av3
=99.9×9.09×0.909=825.5 V/V
The voltage gain from source to load is
v
L
vs
=
v
L
vi1
×
v
i1
vS
=Av·
v
i1
vS
=825.5×0.5
=412.7 V/V
The overall voltage has reduced appreciably. This
is because the input resistance of the first stage,

16. Chapter 1–15
R
in, is comparable to the source resistanceR s.In
Example 1.3 the input resistance of the first stage
is much larger than the source resistance.
1.53(a) Case S-A-B-L (see figure below):
v
o
vs
=
v
o
vib
×
v
ib
via
×
v
ia
vs
=
ω
10×
100
100+1000
π
×
ω
100×
10
10+10
π
×
ω
100
100+100
π
v
ovs
=22.7V/Vand gain in dB 20 log 22.7=
27.1 dB
(b) Case S-B-A-L (see figure below):
v
o
vs
=
v
o
via
·
v
ia
vib
·
v
ib
vs
=
ω
100×
100
100+10 K
π
×
ω
10×
100 K
100 K+1K
π
×
ω
10 K
10 K+100 K
π
v
ovs
=0.89 V/Vand gain in dB is 20 log 0.89=
−1dB. Obviously, case a is preferred because it
provides higher voltage gain.
1.54Each of stages#1,2, ..., (n−1)can be
represented by the equivalent circuit:
v
o
vs
=
v
i1
vs
×
v
i2
vi1
×
v
i3
vi2
×···×
v
in
vi(n−1)
×
v
o
vin
where
v
i1vs
=
100k
100k+50k
=0.667V/V
This figure belongs to1.53, part (a).
This figure belongs to1.53,part(b).
vo
vin
=15×
200
1k+200
=2.5V/V
vi2
vi1
=
vi3
vi2
= ··· =
vin
vi(n−1)
=15×
100k
100k+1k
=14.85V/V
Thus,
v
o
vs
=0.667×(14.85)
n−1
×2.5=
1.667×(14.85)
n−1
Forv s=5mV andv o=3V, the gain
v
ovs
must
be≥600, thus
1.667×(14.85)
n−1
≥600
⇒n=4
Thus four amplifier stages are needed, resulting in
v
o
vs
=1.667×(14.85)
3
=5458V/V
and correspondingly
v
o=5458×5mV=16.37V
1.55Deliver 0.5 W to a 100-load.
Source is 30 mV rms with 0.5-Msource
resistance. Choose from these three amplifier
types:
A B C
R
i β 1 Mπ
A
v β 10 V/ V
R
o β 10 kπ
R
i β 10 kπ
A
v β 100 V/ V
R
o β 1 kπ
R
i β 10 kπ
A
v β 1 V/ V
R
o β 20 π

17. Chapter 1–16
This figure belongs to1.54.
ω
∼
R
s β 50 kπ
200 π
R
i1 β 10 kπ R
on β 1 kπ
5 mV
v
s
#1 #2 # n
v
i1
ω
∼
v
i2
ω
∼
v
in
ω
∼
v
o
ω
∼
R
L
100 kπ
1 kπ
100 kπ
#m
R
i(m ω1)
v
i(m ω1)
ω
∼
v
im
ω
∼
v
i(m ω1)
10v
im
ω
∼
v
im
ω
∼
ω
∼
This figure belongs to1.55.
ω
∼
v
i1
ω
∼
v
i2
ω
∼
v
i3
ω
∼
v
o
100 π
20 π1 kπ10 kπ0.5 Mπ
10 kπ
100v
i2 1v
i310v
i1
1 Mπ
30 mV
rms
10 kπ
ω
∼
ω
∼
ω
∼
∼
Choose order to eliminate loading on input and
output:
A, first, to minimize loading on 0.5-M∼source
B, second, to boost gain
C, third, to minimize loading at 100-∼output.
We first attempt a cascade of the three stages in
the orderA,B,C(see figure above), and obtain
v
i1
vs
=
1M∼
1M∼+0.5M∼
=
1
1.5
⇒v
i1=30×
1
1.5
=20mV
v
i2
vi1
=10×
10k∼
10k∼+10k∼
=5
⇒v
i2=20×5=100mV
v
i3
vi2
=100×
10k∼
10k∼+1k∼
=90.9
⇒v
i3=100mV×90.9=9.09V
v
o
vi3
=1×
100∼
100∼+20∼
=0.833
⇒v
o=9.09×0.833=7.6V
P
o=
v
2
orms
RL
=
7.6
2
100
=0.57W
which exceeds the required 0.5 W. Also, the
signal throughout the amplifier chain never drops
below 20 mV (which is greater than the required
minimum of 10 mV).
1.56(a) Required voltage gain≡
v
o
vs
=
2V
0.005 V
=400 V/V
(b) The smallestR
iallowed is obtained from
0.1µA=
5mV
Rs+Ri
⇒R s+Ri=50 k∼
ThusR
i=40 k∼.
ForR
i=40 k∼.i i=0.1µApeak, and
Overall current gain=
v
o/RL
ii
=
2V/1k∼
0.1µA
=
2mA
0.1µA
=2×10
4
A/A
Overall power gain≡
v
2
orms
/RL
vs(rms)×ii(rms)

18. Chapter 1–17
=
ω
2
√
2
π
2η
1000
ω
5×10
−3
√
2
π
×
ω
0.1×10
−6
√
2
π
=8×10
6
W/W
(This takes into account the power dissipated in
the internal resistance of the source.)
(c) If (A
vovi) has its peak value limited to 3 V,
the largest value ofR
Ois found from
R
s β 10 kπ
ω

ω

v
s v
i
v
o
i
i
R
i
5-mV
peak
R
o
R
L
ω

A
vov
i 1 kπ
=3×
R
L
RL+Ro
=2⇒R o=
1
2
R
L=500
(IfR
owere greater than this value, the output
voltage acrossR
Lwould be less than 2 V.)
(d) ForR
i=40 kandR o=500,the
required valueA
vocan be found from
400 V/V=
40
40+10
×A
vo×
1
1+0.5
⇒A
vo=750 V/V
(e)R
i=100 k(1×10
5
)
R
o=100(1×10
2
)
400=
100
100+10
×A
vo×
1000
1000+100
⇒A
vo=484 V/V
1.57
(a)
v
o=10 mV×
40
40+10
×300×
100
100+100
=1.2 V
(b)
v
o
vs
=
1200 mV
10 mV
=120 V/V
(c)
v
o
vi
=300×
100
100+100
=150 V/V
(d)
R
s
R
p R
i
...
Connect a resistanceR Pin parallel with the input
and select its value from
(R
pβRi)
(RpβRi)+R s
=
1
2
R
i
Ri+Rs
⇒1+
R
s
RpβRi
=2.5⇒R pβRi=
R
S
1.5
=
10
1.5
⇒
1
Rp
+
1
Ri
=
1.5
10
R
p=
1
0.15−0.025
=8k
1.58The equivalent circuit at the output side of a
current amplifier loaded with a resistanceR
Lis
shown. Since
R
LR
oA
isi
i
i
o
io=(A isii)
R
o
Ro+RL
we can write
1=(A
isii)
R
oRo+1
(1)
and
0.5=(A
isii)
R
o
Ro+12
(2)
Dividing Eq. (1) by Eq. (2), we have
2=
R
o+12
Ro+1
⇒R
o=10k
A
isii=1×
10+1
10
=1.1mA
1.59
The current gain is
i
o
ii
=
R
m
Ro+RL
=
5000
10+1000
=4.95A/A=13.9dB

19. Chapter 1–18
This figure belongs to Problem1.59.
ω

ω

v
i
v
s
i
oR
s
R
i
R
o β 10π
R
LR
mi
i
1 kπ
100π
i
i
ω

v
o
1 kπ
ω

The voltage gain is
v
o
vs
=
i
i
vs
×
i
o
ii
×
v
o
io
=
1
Rs+Ri
×
i
o
ii
×RL
=
1
1000+100
×4.95×1000
=4.90V/V=13.8dB
The power gain is
v
oio
vsii
=4.95×4.90
=24.3W/W=27.7dB
1.60
R
s β 500 π
ω

ω

v
s v
i
R
i
G
mv
i

v
o
R
LR
o
ω

2 kπ
Gm=20 mA/V
R
o=5k
R
L=1k
v
i=vs
Ri
Rs+Ri
=vs
2
0.5+2
=0.8v
s
vo=G mvi(RLβRo)
=20
5×1
5+1
v
i
=20
5
6
×0.8v
s
Overall voltage gain≡
v
o
vs
=13.3V/V
1.61To obtain the weighted sum ofv
1andv 2
vo=10v 1+20v 2
we use two transconductance amplifiers and sum
their output currents. Each transconductance
amplifier has the following equivalent circuit:
R
i
10 kπ
R
o
G
mv
i
G
m β 20 mA/V
10 kπ
v
i
ω

Consider first the path for the signal requiring
higher gain, namelyv
2. See figure at top of next
page.
The parallel connection of the two amplifiers at
the output and the connection ofR
Lmeans that
the total resistance at the output is
10kβ10kβ10k=
10
3
k. Thus the
component ofv
odue tov 2will be
v
o2=v2
10
10+10
×G
m2×
10
3
=v
2×0.5×20×
10
3
=33.3v
2
To reduce the gain seen byv 2from 33.3 to 20, we
connect a resistanceR
pin parallel withR L,
ω
10
3
βR
p
π
=2k
⇒R
p=5k
We next consider the path forv
1.Sincev 1must
see a gain factor of only 10, which is half that
seen byv
2,wehavetoreducethefractionofv 1
that appears at the input of its transconductance
amplifier to half that that appears at the input of
thev
2transconductance amplifier. We just saw
that 0.5v
2appears at the input of thev 2
transconductance amplifier. Thus, for thev 1
transconductance amplifier, we want 0.25v 1to
appear at the input. This can be achieved by
shunting the input of thev
1transconductance

20. Chapter 1–19
This figure belongs to Problem1.61.
amplifier by a resistanceR p1as in the following
figure.
v
1
R
s1 β 10 kπ
v
i1
ω

R
p1 R
i1
10 kπ
ω

The value ofR p1can be found from
(R
p1βRi1)
(Rp1βRi1)+R s1
=0.25
Thus,
1+
R
s1
(Rp1βRi1)
=4
⇒R
p1βRi1=
R
s1
3
=
10
3
R
p1β10=
10
3
⇒R
p1=5k
The final circuit will be as follows:

21. Chapter 1–20
1.62(a)
Ω
∼
v
i
g
mv
i
R
i
i
1
i
2
i
x
v
x
Ω
∼
ix=i1+i2
i1=vi/Ri
i2=gmvi
vi=vx
⎫
⎪
⎪
⎪
⎪
⎪
⎪
⎬
⎪
⎪
⎪
⎪
⎪
⎪
⎭
i
x=vx/Ri+gmvx
ix=vx
Ω
1
Ri
+gm
Δ
v
x
ix
=
1
1/Ri+gm
=
R
i
1+g mRi
=Rin
(b)
Ω
∼
Ω
∼
v
i ≡
Ω
∼
v
o
v
s
v
s
R
s ≡ R
in
g
mv
iR
in
R
i
2
When driven by a source with source resistance
R
inas shown in the figure above,
v
i=
R
in
Rs+Rin
×vs=
R
in
Rin+Rin
×vs=0.5×v s
Thus,
v
ovs
=0.5
v
o
vi
1.63Voltage amplifier:
v
i v
oR
i
R
s
R
o1 to 10 kΔ
R
L
1 kΔ to
10 kΔ
Avo v
i
Ω
∼
Ω
Ω
∼
Ω
∼
∼
i
o
v
s
ForR svarying in the range 1 k∼to10 k∼and
v
olimited to 10%, selectR ito be sufficiently
large:
R
i≥10R smax
Ri=10×10 k∼=100 k∼=1×10
5
∼
ForR
Lvarying in the range 1 k∼to10 k∼,the
load voltage variation limited to 10%, selectR
o
sufficiently low:
R
o≤
R
Lmin
10
R
o=
1k∼
10
=100∼=1×10
2
∼
Now findA
vo:
v
omin=10 mV×
R
i
Ri+Rsmax
×Avo
RLmin
Ro+RLmin
1=10×10
−3
×
100 k∼
100 k∼+10 k∼
×A
vo×
1k∼
100∼+1k∼
⇒A
vo=121 V/V
v
o
R
o
A
vov
i
Ω
∼
Ω
∼
Ω
∼
v
iR
i
Values for the voltage amplifier equivalent circuit
are
R
i=1×10
5
∼,A vo=121 V/V,and
R
o=1×10
2
∼
1.64Transresistance amplifier:
To limitv
oto 10% corresponding toR svarying
in the range 1 k∼to10 k∼, we selectR
i
sufficiently low;
R
i≤
R
smin
10
Thus,R
i=100∼=1×10
2
∼
To limitv
oto 10% whileR Lvaries over the
range 1 k∼to10 k∼, we selectR
osufficiently
low;
R
o≤
R
Lmin
10
Thus,R
o=100∼=1×10
2
∼
Now, fori
s=10µA,
10 μA
1 to
10
kΔ
1 to
10
kΔ
Ω
Ω
∼
∼
Ri
i
i
R
R
R
R
v

22. Chapter 1–21
v
omin=10
−5
Rsmin
Rsmin+Ri
Rm
RLmin
RLmin+Ro
1=10
−5
1000
1000+100
R
m
1000
1000+100
⇒R
m=1.21×10
5

=121 k
1.65
The node equation atEyields the current through
R
Eas(βi b+ib)=(β+1)i b. The voltagev ccan
be found in terms ofi
bas
v
c=−βi bRL (1)
The voltagev
bcan be related toi bby writing for
the input loop:
v
b=ibrπ+(β+1)i bRE
Thus,
v
b=rπ+(β+1)R Eib (2)
Dividing Eq. (1) by Eq. (2) yields
v
c
vb
=−
βR
L
rπ+(β+1)R E
Q.E.D
The voltagev
eis related toi bby
v
e=(β+1)i bRE
That is,
v
e=(β+1)R Eib (3)
Dividing Eq. (3) by Eq. (2) yields
v
e
vb
=
(β+1)R
E
(β+1)R E+rπ
Dividing the numerator and denominator by
(β+1)gives
v
e
vb
=
R
E
RE+rπ/(β+1)
Q.E.D
1.66
v
A
R
L
WhenR L=0,
i
o=isc=5mA=
v
A
Ro
⇒v A−5R o=0 (1)
WhenR
L=2k,
v
o=5V=v A
2
Ro+2
⇒2v
A−5R o=10 (2)
Subtracting Eqs. (2) - (1),
v
A=10V
Substituting into Eq. (1)
R
o=
10
5
=2k
WhenR
L=1k,
Av=
v
o
vi
=
10×
1
2+1
0.001×200β5
=683.3V/V=56.7dB
The overall current gain is,
A
i=
i
o
is
=
10
2+1
0.001
=3333.3A/A=70.5dB
The power gain of the amplifier is,
A
p=
v
2
o
/RL
v
2
i
/Ri
=
3.33
2
/1
0.00487
2
/5
=2.34×10
6
W/W=127.4dB

23. Chapter 1–22
1.67
R
ω
∴
v
1
g
mv
1
g
m β 100 mA/V
R
β 5 kπ
g
mv
2
ω
∴
v
2
i
o
v
o
io=gmv1−gmv2
vo=ioRL=gmR(v1−v2)
v
1=v2=1V∴v o=0V
v
1=1.01 V
v
2=0.99 V
5
∴v
o=100×5×0.02=10 V
1.68
I
1
V
1
g
12I
2
1/g
11
ω
∴
g
21V
1
g
22
ω
∴
V
2
I
2
∴
ω
I
1
V
2
I
2
A
voV
1
V
1 R
i
R
o
∴
ω
ω
∴
ω
∴
The correspondences between the current and
voltage variables are indicated by comparing the
two equivalent-circuit models above. At the
outset we observe that at the input side of the
g-parameter model, we have the controlled
current sourceg
12I2. This has no correspondence
in the equivalent-circuit model of Fig. 1.16(a). It
represents internal feedback, internal to the
amplifier circuit. In developing the model of
Fig. 1.16(a), we assumed that the amplifier is
unilateral (i.e., has no internal feedback, or that
the input side does not know what happens at the
output side). If we neglect this internal feedback,
that is, assumeg
12=0, we can compare the two
models and thus obtain:
R
i=1/g 11
Avo=g21
Ro=g22
1.69Circuits of Fig. 1.22:
ω
∴
V
i V
o
R
C
(a) (b)
ω
∴
ω
∴
V
i V
o
C
Rω
∴
For (a)V o=Vi
ω
1/sC
1/sC+R
π
V
o
Vi
=
1
1+sCR
which is of the form shown for the low-pass
function in Table 1.2 withK=1and
ω
0=1/RC.
For (b)V
o=Vi
⎛
⎜
⎝
R
R+
1
sC
⎞
⎟
⎠
V
o
Vi
=
sRC
1+sCR
V
o
Vi
=
s
s+
1
RC
which is of the form shown in Table 1.2 for the
high-pass function, withK=1andω
0=1/RC.
1.70
ω
∴
V
i
ω ∴
R
s
R
i
V
s C
i

24. Chapter 1–23
V
i
Vs
=
R
i
1
sCi
Ri+
1
sCi
Rs+
⎛
⎜
⎜
⎝
R
i
1
sCi
Ri+
1
sCi
⎞
⎟
⎟
⎠
=
R
i
1+sC iRi
Rs+
ω
R
i
1+sC iRi
π
=
R
i
Rs+sCiRiRs+Ri
Vi
Vs
=
R
i
(Rs+Ri)+sC iRiRs
=
R
i
(Rs+Ri)
1+s
ω
C
iR
iRs
Rs+Ri
π
which is a low-pass STC function with
K=
R
i
Rs+Ri
andω 0=1/C i(RiβRs).
ForR
s=10 k∴,R i=40 k∴,andC i=5pF,
ω
0=
1
5×10
−12
×(40β10)×10
3
=25 Mrad/s
f
0=
25
2π
=4MHz
The dc gain is
K=
40
10+40
=0.8V/V
1.71Using the voltage-divider rule.
ω
∴
V
oV
i
R
1
R
2
C
ω
∴
T(s)=
V
o
Vi
=
R
2
R2+R1+
1
sC
T(s)=
ω
R
2
R1+R2
π
⎛
⎜
⎜
⎝
s
s+
1
C(R1+R2)
⎞
⎟
⎟
⎠
which from Table 1.2 is of the high-pass type
with
K=
R
2
R1+R2
ω0=
1
C(R1+R2)
As a further verification that this is a high-pass
network andT(s) is a high-pass transfer function,
see that ass⇒0,T(s)⇒0;andass→∞,
T(s)=R
2/(R1+R2). Also, from the circuit,
observe ass→∞,(1/sC)→0and
V
o/Vi=R2/(R1+R2). Now, forR 1=20 k∴,
R
2=100 k∴andC=0.1µF,
f
0=
ω
0
2π
=
1
2π×0.1×10
−6
(20+100)×10
3
=13.26 Hz
|T(jω
0)|=
K
√
2
=
100
20+100
1
√
2
=0.59 V/V
1.72Using the voltage divider rule,
R
L
R
s
C
V
l
V
s
ω
∴
ω
∴
Vl
Vs
=
R
L
RL+Rs+
1
sC
=
R
L RL+Rs
s
s+
1
C(R L+Rs)
which is of the high-pass STC type (see Table
1.2) with
K=
R
L
RL+Rs
ω0=
1
C(R L+Rs)
Forf
0≤200 Hz
1 2πC(R L+Rs)
≤200
⇒C≥
1
2π×200(10+4)×10
3
Thus, the smallest value ofCthat will do the job
isC=0.057µFor 57 nF.
1.73The given measured data indicate that this
amplifier has a low-pass STC frequency response
with a low-frequency gain of 60 dB, and a 3-dB
frequency of 1000 Hz. From our knowledge of
the Bode plots for low-pass STC networks [Fig.
1.23(a)], we can complete the table entries and
sketch the amplifier frequency response.
f(Hz)|T|(dB)∠T(
◦
)
0 60 0
100 60 0
1000 57 −45
◦
10
4
40 −90
◦
10
5
20 −90
◦
10
6
0 −90
◦

25. Chapter 1–24
f (Hz)
20 dB/decade
3 dB
10
0
10
20
30
40
50
60
T (dB)
10
2
10
3
10
4
10
5
10
6
1.74From our knowledge of the Bode plots of
STC low-pass and high-pass networks, we see
that this amplifier has a midband gain of 80 dB, a
low-frequency response of the high-pass STC
type withf
3dB=10
2
Hz, and a high-frequency
response of the low-pass STC type with
f
3dB=10
5
Hz. We thus can sketch the amplifier
frequency response and complete the table entries
as follows.
f (Hz)
3 dB
10
2
0
20
40
60
80
10
0
10
2
10
4
10
6
10
8
10
10
20 dB/decade
ω20
dB/decade
T (dB)
f(Hz)10
−2
110
2
10
3
10
4
10
5
10
6
10
7
10
9
|T|(dB)0407780807760400
1.75Since the overall transfer function is that of
three identical STC LP circuits in cascade (but
with no loading effects, since the buffer
amplifiers have infinite input and zero output
resistances) the overall gain will drop by 3 dB
below the value at dc at the frequency for which
the gain of each STC circuit is 1 dB down. This
frequency is found as follows: The transfer
function of each STC circuit is
T(s)=
1
1+
s
ω0
where
ω
0=1/CR
Thus,
|T(jω)|=
1
≥
1+
ω
ω
ω0
π
2
20 log
1
≥
1+
ω
ω
1dB
ω0
π
2
=−1
⇒1+
ω
ω
1dB
ω0
π
2
=10
0,1
ω1dB=0.51ω 0
ω1dB=0.51/CR
1.76R
s=100 k, since the 3-dB frequency is
reduced by a very high factor (from 5 MHz to
100 kHz)C
2must be much larger thanC 1. Thus,
neglectingC
1we findC 2from
100 kHz
1
2πC 2Rs
C
2
Shunt
capacitor
Initial
capacitor
C
1
R
s
=
1
2πC 2×10
5
⇒C 2=15.9pF
If the original 3-dB frequency (5 MHz) is
attributable toC
1,then
5MHz=
1
2πC 1Rs
⇒C 1=
1
2π×5×10
6
×10
5
=0.32 pF
1.77Since whenCis connected to node A the
3-dB frequency is reduced by a large factor, the
value ofCmust be much larger than whatever
parasitic capacitance originally existed at node A
(i.e., between A and ground). Furthermore,
it must be thatCis now the dominant determinant
of the amplifier 3-dB frequency (i.e., it is
dominating over whatever may be happening at

26. Chapter 1–25
node B or anywhere else in the amplifier). Thus,
we can write
200 kHz=
1
2πC(R o1βRi2)
⇒(R
o1βRi2)=
1
2π×200×10
3
×1×10
−9
=0.8k
Cv
o1 R
i2
R
o1
ω

A
1 nF
# 1 # 2 # 3
B
C
NowR i2=100 k.
ThusR
o10.8k
Similarly, for node B,
40 kHz=
1
2πC(R o2βRi3)
⇒R
o2βRi3=
1
2π×40×10
3
×1×10
−9
=3.98 k
R
o2=4.14 k
The designer should connect a capacitor of value
C
pto node B whereC pcan be found from
5kHz=
1
2πC p(Ro2βRi3)
⇒C
p=
1
2π×5×10
3
×3.98×10
3
=8nF
Note that if she chooses to use node A, she would
need to connect a capacitor 5 times larger!
1.78For the input circuit, the corner frequency
f
01is found from
f
01=
1
2πC 1(Rs+Ri)
ω

ω

ω

V V
R
100 kπ
100V
R
10 kπ
C
V
R
1 kπ
R
1 kπ
C
ω

Forf 01≤100 Hz,
1
2πC 1(10+100)×10
3
≤100
⇒C
1≥
1
2π×110×10
3
×10
2
=1.4×10
−8
F
Thus we selectC
1=1×10
−7
F=0.1µF.The
actual corner frequency resulting fromC
1will be
f
01=
1
2π×10
−7
×110×10
3
=14.5Hz
For the output circuit,
f
02=
1
2πC 2(Ro+RL)
Forf
02≤100 Hz,
1
2πC 2(1+1)×10
3
≤100
⇒C
2≥
1
2π×2×10
3
×10
2
=0.8×10
−6
SelectC 2=1×10
−6
=1µF.
This will place the corner frequency at
f
02=
1
2π×10
−6
×2×10
3
=80 Hz
T(s)=100
s
ω
1+
s
2πf01
πω
1+
s
2πf02
π
1.79The LP factor1/(1+jf/10
5
)results in a
Bode plot like that in Fig. 1.23(a) with the
3-dB frequencyf
0=10
5
Hz.The high-pass
factor1/(1+10
2
/jf)results in a Bode plot like
that in Fig. 1.24(a) with the 3-dB frequency
f
0=10
2
Hz.
The Bode plot for the overall transfer function
can be obtained by summing the dB values of the
two individual plots and then shifting the

27. Chapter 1–26
resulting plot vertically by 60 dB (corresponding
to the factor 1000 in the numerator). The result is
as follows:
60
50
40
30
20
10
0
110 10
2
10
3
10
4
10
5
10
6
10
7
10
8f (Hz)
(Hz)
f = 10 10
2
10
3
10
4
10
5
10
6
10
7
10
8
(dB)–
~
40 60
57 57
60 60 60 40 20 0
∴20 dB/decade
ω20 dB/decadeω20 dB/decade
≥A
v
≥ (dB)
≥A
v
≥
Bandwidth=10
5
−10
2
=99,900 Hz
1.80
T
i(s)=
V
i(s)
Vs(s)
=
1/sC
1
1/sC
1+R1
=
1
sC1R1+1
LP with a 3-dB frequency
f
0i=
1
2πC 1R1
=
1
2π10
−11
10
5
=159 kHz
ForT
o(s), the following equivalent circuit can be
used:
ω
∴
∴G
mR
2V
i
V
oR
3
R
2
C
2
φ
To(s)=
V
o
Vi
=−G mR2
R3
R2+R3+1/sC
2
=−G m(R2βR3)
s
s+
1
C2(R2+R3)
which is an HP, with
3-dB frequency=
1
2πC 2(R2+R3)
=
1
2π100×10
−9
×110×10
3
=14.5Hz
∴T(s)=T
i(s)To(s)
=
1
1+
s
2π×159×10
3
×−909.1×
s
s+(2π×14.5)
∴20 dB/decadeω20 dB/decade
14.5 Hz 159 kHz
59.2 dB
Bandwidth=159 kHz – 14.5 Hz∴159 kHz
1.81V
i=Vs
Ri
Rs+Ri
(1)
(a) To satisfy constraint (1), namely,
V
i≥
α
1−
x
100
∂
V
s
we substitute in Eq. (1) to obtain
R
i
Rs+Ri
≥1−
x
100
(2)
Thus
R
s+Ri
Ri
≤
1
1−
x
100
R
s
Ri
≤
1
1−
x
100
−1=
x
100
1−
x
100
which can be expressed as
R
i
Rs
≥
1−
x
100
x
100
resulting in
V
i
V
o
R
i R
oR
L
C
L
V
s
R
s
G
m V
i
ω
ω
∴
ω
∴
∴
Ri≥Rs
ω
100
x
−1
π
(3)
(b) The 3-dB frequency is determined by the
parallel RC circuit at the output
f
0=
1
2π
ω
0=
1
2π
1
CL(RLβRo)
Thus,
f
0=
1
2πC L
ω
1
RL
+
1
R0
π

28. Chapter 1–27
To obtain a value forf
0greater than a specified
valuef
3dBwe selectR oso that
1
2πC L
ω
1
RL
+
1
Ro
π
≥f
3dB
1RL
+
1
Ro
≥2πC Lf3dB
1
Ro
≥2πC Lf3dB−
1
RL
Ro≤
1
2πf3dBCL−
1
RL
(4)
(c) To satisfy constraint (c), we first determine
the dc gain as
dc gain=
R
i
Rs+Ri
Gm(RoβRL)
For the dc gain to be greater than a specified
valueA
0,
R
i
Rs+Ri
Gm(RoβRL)≥A 0
The first factor on the left-hand side is (from
constraint (2)) greater or equal to(1−x/100).
Thus
G
m≥
A
0
α
1−
x
100
∂
(R
oβRL)
(5)
SubstitutingR
s=10 kandx= 10% in (3)
results in
R
i≥10
ω
100
100
−1
π
=90 k
Substitutingf
3dB=2 MHz,C L=20pF, and
R
L=10kin Eq. (4) results in
R
o≤
1
2π×2×10
6
×20×10
−12
−
1
10
4
=6.61 k
SubstitutingA
0= 100,x= 10%,R L=10k,and
R
o= 6.61 k, Eq. (5) results in
G
m≥
100
ω
1−
10
100
π
(10β6.61)×10
3
=27.9mA/V
1.82Using the voltage divider rule, we obtain
V
o
Vi
=
Z
2
Z1+Z2
where
Z
1=R1β
1
sC1
andZ 2=R2β
1
sC2
It is obviously more convenient to work in terms
of admittances. Therefore we expressV
o/Viin the
alternate form
V
o
Vi
=
Y
1
Y1+Y2
and substituteY 1=(1/R 1)+sC 1and
Y
2=(1/R 2)+sC 2to obtain
V
o
Vi
=
1
R1
+sC1
1
R1
+
1
R2
+s(C 1+C2)
=
C
1
C1+C2
s+
1
C1R1
s+
1
(C1+C2)
ω
1
R1
+
1
R2
π
This transfer function will be independent of
frequency (s) if the second factor reduces to unity.
This in turn will happen if
1
C1R1
=
1
C1+C2
ω
1
R1
+
1
R2
π
which can be simplified as follows:
C
1+C2
C1
=R1
ω
1
R1
+
1
R2
π
(1)
1+
C
2
C1
=1+
R
1
R2
or
C
1R1=C2R2⇒C 1=C2(R2/R1)
When this condition applies, the attenuator is said
to be compensated, and its transfer function is
given by
V
o
Vi
=
C
1
C1+C2
which, using Eq. (1), can be expressed in the
alternate form
V
o
Vi
=
1
1+
R
1
R2
=
R
2
R1+R2
Thus when the attenuator is compensated
(C
1R1=C2R2), its transmission can be
determined either by its two resistorsR
1,R2or by
its two capacitors.C
1,C2, and the transmission is
nota function of frequency.

29. Chapter 1–28
1.83The HP STC circuit whose response
determines the frequency response of the
amplifier in the low-frequency range has a phase
angle of5.7
◦
atf=100 Hz. Using the equation
for∠T(jω)from Table 1.2, we obtain
tan
−1
f0
100
=5.7
◦
⇒f 0=10 Hz⇒τ=
1
2π10
=
15.9ms
The LP STC circuit whose response determines
the amplifier response at the high-frequency end
has a phase angle of−5.7
◦
atf=1kHz.Using
the relationship for∠T(jω)given in Table 1.2,
we obtain for the LP STC circuit.
−tan
−1
10
3
f0
=−5.7
◦
⇒f 0∴10 kHz⇒τ=
1
2π10
4
=15.9µs
Atf=100 Hz, the drop in gain is due to the HP
STC network, and thus its value is
20 log
1
≥
1+
ω
10
100
π
2
=−0.04 dB
Similarly, at the drop in gainf=1 kHz is caused
by the LP STC network. The drop in gain is
20 log
1
≥
1+
ω
1000
10,000
π
2
=−0.04 dB
The gain drops by 3 dB at the corner frequencies
of the two STC networks, that is, atf=10 Hz and
f=10 kHz.

30. Exercise 1–1
Chapter 1
Solutions to Exercises within the Chapter
Ex: 1.1When output terminals are
open-circuited, as in Fig. 1.1a:
For circuit a.v
oc=vs(t)
For circuit b.v
oc=is(t)×R s
When output terminals are short-circuited, as in
Fig. 1.1b:
For circuit a.i
sc=
v
s(t)
Rs
For circuit b.i sc=is(t)
For equivalency
R
sis(t)=v s(t)
R
s a
b
∼
ω
v
s (t)
Figure 1.1a
i
s (t)
a
b
R
s
Figure 1.1b
Ex: 1.2
v
oc π v
s
R
s
∼
ω
i
sc
voc=10 mV
i
sc=10µA
R
s=
v
oc
isc
=
10 mV
10µA
=1k∼
Ex: 1.3Using voltage divider:
v
o(t)=v s(t)×
R
L
Rs+RL
v
s (t)
v
o
R
s
∼
ω
∼
ω
R
L
Givenv s(t)=10 mVandR s=1k∼.
IfR
L=100 k∼
v
o=10 mV×
100
100+1
=9.9mV
IfR
L=10 k∼
v
o=10 mV×
10
10+1
∼9.1mV
IfR
L=1k∼
v
o=10 mV×
1
1+1
=5mV
IfR
L=100∼
v
o=10 mV×
100
100+1K
∼0.91 mV
Forv
o=0.8v s,
R
L
RL+Rs
=0.8
SinceR
s=1k∼,
R
L=4k∼
Ex: 1.4Using current divider:
R
s
i
s π 10 ηA R L
i
o
io=is×
R
s
Rs+RL
Giveni s=10µA,R s=100 k∼.
For
R
L=1k∼,i o=10µA×
100
100+1
=9.9µA
For
R
L=10 k∼,i o=10µA×
100
100+10
∼9.1µA
For
R
L=100 k∼,i o=10µA×
100
100+100
=5µA
ForR
L=1M∼,i o=10µA×
100 K
100 K+1M
∼0.9µA
Fori
o=0.8i s,
100
100+R L
=0.8
⇒R
L=25 k∼

31. Exercise 1–2
Ex: 1.5f=
1
T
=
1
10
−3
=1000 Hz
ω=2πf=2π×10
3
rad/s
Ex: 1.6(a)T=
1
f
=
1
60
s=16.7ms
(b)T=
1
f
=
1
10
−3
=1000 s
(c)T=
1
f
=
1
10
6
s=1µs
Ex: 1.7If 6 MHz is allocated for each channel,
then 470 MHz to 806 MHz will accommodate
806−470
6
=56 channels
Since the broadcast band starts with channel 14, it
will go from channel 14 to channel 69.
Ex: 1.8P=
1
T
T
0
v
2
R
dt
=
1
T
×
V
2
R
×T=
V
2
R
Alternatively,
P=P
1+P3+P5+···
=
ω
4V
√
2π
π
2
1
R
+
ω
4V
3
√
2π
π
2
1
R
+
ω
4V
5
√
2π
π
2
1
R
+···
=
V
2 R
×
8
π
2
×
ω
1+
1
9
+
1
25
+
1
49
+···
π
It can be shown by direct calculation that the
infinite series in the parentheses has a sum that
approachesπ
2
/8;thusPbecomesV
2
/Ras found
from direct calculation.
Fraction of energy in fundamental
=8/π
2
=0.81
Fraction of energy in first five harmonics
=
8
π
2
ω
1+
1
9
+
1
25
π
=0.93
Fraction of energy in first seven harmonics
=
8
π
2
ω
1+
1
9
+
1
25
+
1
49
π
=0.95
Fraction of energy in first nine harmonics
=
8
π
2
ω
1+
1
9
+
1
25
+
1
49
+
1
81
π
=0.96
Note that 90% of the energy of the square wave is
in the first three harmonics, that is, in the
fundamental and the third harmonic.
Ex: 1.9(a)Dcan represent 15 equally-spaced
values between 0 and 3.75 V. Thus, the values are
spaced 0.25 V apart.
v
A=0V⇒D=0000
v
A=0.25V⇒D=0000
v
A=1V⇒D=0000
v
A=3.75V⇒D=0000
(b) (i) 1 level spacing:2
0
×+0.25=+0.25V
(ii) 2 level spacings:2
1
×+0.25=+0.5V
(iii) 4 level spacings:2
2
×+0.25=+1.0V
(iv) 8 level spacings:2
3
×+0.25=+2.0V
(c) The closest discrete value represented byDis
+1.25 V; thusD=0101. The error is -0.05 V, or
−0.05/1.3×100=−4%.
Ex: 1.10
Voltage gain=20log100=40 dB
Current gain=20log1000=60 dB
Power gain=10logA
p=10log(A vAi)
=10log10
5
=50 dB
Ex: 1.11P dc=15×8=120 mW
P
L=
(6/
√
2)
2
1
=18 mW
P
dissipated=120−18=102 mW
η=
P
L
Pdc
×100=
18
120
×100=15%
Ex: 1.12v
o=1×
10
10
6
+10
10
−5
V=10µV
P
L=v
2
o
/RL=
(10×10
−6
)
2
10
=10
−11
W
With the buffer amplifier:
v
o=1×
R
i
Ri+Rs
×Avo×
R
L
RL+Ro
=1×
1
1+1
×1×
10
10+10
=0.25 V
P
L=
v
2
o
RL
=
0.25
2
10
=6.25 mW
Voltage gain =
v
o
vs
=
0.25 V
1V
=0.25 V/V
=−12 dB
Power gain(A
p)≡
P
L
Pi
whereP L= 6.25 mW andP i=vii1,
v
i=0.5Vand
i
i=
1V
1M+1M
=0.5µA

32. Exercise 1–3
This figure belongs to Exercise 1.15.

ω

ω

ω

ω

ω
v
s v
i1
10v
i1
100 kβ
1 Mβ
v
i2
1 kβ
100 kβ
100
v
i2
v
L
1 kβ
100 β
Stage 1 Stage 2

ω
Thus,
P
i=0.5×0.5=0.25µW
and
A
p=
6.25×10
−3
0.25×10
−6
=25×10
3
10 logA p=44 dB
Ex: 1.13Open-circuit (no load) output voltage =
A
vovi
Output voltage with load connected
=A
vovi
RL
RL+Ro
0.8=
1
Ro+1
⇒R
o=0.25 k=250
Ex: 1.14A
vo=40 dB=100 V/V
P
L=
v
2
o
RL
=
ω
A vovi
RL
RL+Ro
π
2η
R
L
=v
2
i
×
ω
100×
1
1+1
π
2η
1000=2.5v
2
i
Pi=
v
2
i
Ri
=
v
2
i
10,000
A
p≡
P
L
Pi
=
2.5v
2
i
10
−4
v
2
i
=2.5×10
4
W/W
10 logA
p=44 dB
Ex: 1.15Without stage 3 (see figure above)
v
L
vs
=
ω
1M
100 k+1M
π
(10)
ω
100 k
100 k+1k
π
×(100)
ω
100
100+1k
π
v
L vs
=(0.909)(10)(0.9901)(100)(0.0909)
=81.8V/V
Ex: 1.16Refer the solution to Example 1.3 in the
text.
v
i1
vs
=0.909 V/V
v
i1=0.909v s=0.909×1=0.909 mV
v
i2
vs
=
v
i2
vi1
×
v
i1
vs
=9.9×0.909=9V/V
v
i2=9×v S=9×1=9mV
v
i3
vs
=
v
i3
vi2
×
v
i2
vi1
×
v
i1
vs
=90.9×9.9×0.909
=818 V/V
v
i3=818v s=818×1=818 mV
v
L
vs
=
v
L
vi3
×
v
i3
vi2
×
v
i2
vi1
×
v
i1
vs
=0.909×90.9×9.9×0.909744 V/V
v
L=744×1mV=744 mV
Ex: 1.17Using voltage amplifier model, the
three-stage amplifier can be represented as
v
i
R
i
R
o
A
vov
i

ω

ω
Ri=1M
R
o=10
A
vo=Av1×Av2×Av3=9.9×90.9×1=
900 V/V
The overall voltage gain
v
o
vs
=
R
i
Ri+Rs
×Avo×
R
L
RL+Ro

33. Exercise 1–4
ForR
L=10∴
Overall voltage gain
=
1M
1M+100 K
×900×
10
10+10
=409 V/V
ForR
L=1000∴
Overall voltage gain
=
1M 1M+100 K
×900×
1000
1000+10
=810 V/V
∴Range of voltage gain is from 409V/Vto
810V/V.
Ex: 1.18
i
i i
o
A
is i
i
R
LR
oR
s R
ii
s
ii=is
Rs
Rs+Ri
io=Aisii
Ro
Ro+RL
=Aisis
Rs
Rs+Ri
Ro
Ro+RL
Thus,
i
o
is
=Ais
Rs
Rs+Ri
Ro
Ro+RL
Ex: 1.19
R
i
R
o
G
mv
i
R
L
R
i
∴
ω
v
i
∴
ω
∴
ω
v
o
v
s
vi=vs
Ri
Ri+Rs
vo=G mvi(RoβRL)
=G
mvs
Ri
Ri+Rs
(RoβRL)
Thus,
v
o
vs
=G m
Ri
Ri+Rs
(RoβRL)
Ex: 1.20Using the transresistance circuit model,
the circuit will be
R
iR
si
s
i
i R
o
R
L
v
oR
mi
i
∴
∴
ω
ω
ii
is
=
R
s
Ri+Rs
vo=Rmii×
R
L
RL+Ro
vo
ii
=Rm
RL
RL+Ro
Now
v
o
is
=
v
o
ii
×
i
i
is
=Rm
RL
RL+Ro
×
R
s
Ri+Rs
=Rm
Rs
Rs+Ri
×
R
L
RL+Ro
Ex: 1.21
vb=ibrπ+(β+1)i bRe
=ibrπ+(β+1)R e
Butv b=vxandi b=ix,thus
R
in≡
v
xix
=
v
b
ib
=rπ+(β+1)R e
Ex: 1.22
f Gain
10 Hz 60 dB
10 kHz 40 dB
100 kHz 20 dB
1MHz 0dB

34. Exercise 1–5
Gain (dB)
ω20 dB/decade
3 dB
frequency
0
20
40
60
110 101010101010f (Hz)
Ex: 1.23
R
LR
o
R
iV
i
V
i G
m V
o
C
L

ω

ω
Vo=G mViRoβRLβCL
=
G
mVi
1
Ro
+
1
RL
+sCL
Thus,
V
o
Vi
=
G
m
1
Ro
+
1
RL
×
1
1+
sC
L
1
Ro
+
1
RL
Vo
Vi
=
G
m(RLβRo)
1+sC L(RLβRo)
which is of the STC LP type.
ω
0=
1
CL(RLβRo)
=
1
4.5×10
−9
(10
3
βRo)
Forω
0to be at leastwπ×40×10
3
, the highest
value allowed forR
ois
R
o=
10
3
2π×40×10
3
×10
3
×4.5×10
−9
−1
=
10
31.131−1
=7.64k
Thedcgainis
G
m(RLβRo)
To ensure a dc gain of at least 40 dB (i.e., 100),
the minimum value ofG
mis
⇒R
L≥100/(10
3
β7.64×10
3
)=113.1mA/V
Ex: 1.24Refer to Fig. E1.24
V
2
Vs
=
R
i
Rs+
1
sC
+R
i
=
R
i
Rs+Ri
s
s+
1
C(Rs+Ri)
which is an HP STC function.
f
3dB=
1
2πC(R s+Ri)
≤100 Hz
C≥
1
2π(1+9)10
3
×100
=0.16µF