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Microelectronic
Circuits
Adel S. Sedra
University of Waterloo
Kenneth C. Smith
University of Toronto
Tonny Chan Carusone
University of Toronto
Vincent Gaudet
University of Waterloo
New York Oxford
OXFORD UNIVERSITY PRESS
Sedra_FM_BM.indd 3 9/30/2014 9:36:34 PM
, Exercise 1–1
Chapter 1 Rs
Solutions to Exercises within the Chapter
vs (t)
vo RL
Ex: 1.1 When output terminals are
open-circuited, as in Fig. 1.1a:
For circuit a. v oc = v s (t)
Given v s (t) = 10 mV and Rs = 1 k.
For circuit b. v oc = is (t) × Rs
If RL = 100 k
When output terminals are short-circuited, as in
100
Fig. 1.1b: v o = 10 mV × = 9.9 mV
100 + 1
v s (t)
For circuit a. isc = If RL = 10 k
Rs
10
For circuit b. isc = is (t) v o = 10 mV × 9.1 mV
10 + 1
For equivalency If RL = 1 k
Rs is (t) = v s (t) 1
v o = 10 mV × = 5 mV
1+1
Rs a If RL = 100
100
v o = 10 mV × 0.91 mV
vs (t) 100 + 1 K
For v o = 0.8v s ,
RL
= 0.8
b RL + Rs
Figure 1.1a Since Rs = 1 k,
RL = 4 k
a
Ex: 1.4 Using current divider:
is (t) Rs io
is 10 A Rs RL
b
Figure 1.1b
Rs
io = i s ×
Rs + RL
Ex: 1.2
Given is = 10 µA, Rs = 100 k.
For
Rs isc 100
RL = 1 k, io = 10 µA × = 9.9 µA
voc vs 100 + 1
For
100
RL = 10 k, io = 10 µA × 9.1 µA
100 + 10
For
v oc = 10 mV 100
RL = 100 k, io = 10 µA × = 5 µA
100 + 100
isc = 10 µA
100 K
v oc 10 mV For RL = 1 M, io = 10 µA ×
Rs = = = 1 k 100 K + 1 M
isc 10 µA
0.9 µA
Ex: 1.3 Using voltage divider: 100
For io = 0.8is , = 0.8
100 + RL
RL
v o (t) = v s (t) × ⇒ RL = 25 k
Rs + RL
1 1 Ex: 1.9 (a) D can represent 15 equally-spaced
Ex: 1.5 f = = −3 = 1000 Hz
T 10 values between 0 and 3.75 V. Thus, the values are
ω = 2π f = 2π × 103 rad/s spaced 0.25 V apart.
1 1 v A = 0 V ⇒ D = 0000
Ex: 1.6 (a) T = = s = 16.7 ms
f 60 v A = 0.25 V ⇒ D = 0000
1 1 v A = 1 V ⇒ D = 0000
(b) T = = −3 = 1000 s
f 10
v A = 3.75 V ⇒ D = 0000
1 1
(c) T = = 6 s = 1 µs (b) (i) 1 level spacing: 20 × +0.25 = +0.25 V
f 10
(ii) 2 level spacings: 21 × +0.25 = +0.5 V
Ex: 1.7 If 6 MHz is allocated for each channel,
then 470 MHz to 608 MHz will accommodate (iii) 4 level spacings: 22 × +0.25 = +1.0 V
806 − 470 (iv) 8 level spacings: 23 × +0.25 = +2.0 V
= 23 channels
6
(c) The closest discrete value represented by D is
Since the broadcast band starts with channel 14, it
+1.25 V; thus D = 0101. The error is -0.05 V, or
will go from channel 14 to channel 36.
−0.05/1.3 × 100 = −4%.
T
1 v2 Ex: 1.10 Voltage gain = 20 log 100 = 40 dB
Ex: 1.8 P = dt
T R Current gain = 20 log 1000 = 60 dB
0
Power gain = 10 log Ap = 10 log (Av Ai )
1 V2 V2
= × ×T =
T R R = 10 log 105 = 50 dB
Alternatively,
P = P1 + P3 + P5 + · · · Ex: 1.11 Pdc = 15 × 8 = 120 mW
√
2 (6/ 2)2
4V 2 1 4V 1 PL = = 18 mW
= √ + √ 1
2π R 3 2π R
2 Pdissipated = 120 − 18 = 102 mW
4V 1
+ √ + ··· PL 18
5 2π R η= × 100 = × 100 = 15%
Pdc 120
V2 8 1 1 1
= × 2 × 1+ + + + ··· Ex: 1.12 v o = 1 × 6
10
10−5 V = 10 µV
R π 9 25 49
10 + 10
It can be shown by direct calculation that the (10 × 10−6 )2
infinite series in the parentheses has a sum that PL = v 2o /RL = = 10−11 W
10
approaches π 2 /8; thus P becomes V 2/R as found
from direct calculation. With the buffer amplifier:
Ri RL
Fraction of energy in fundamental vo = 1 × × Av o ×
Ri + Rs RL + Ro
= 8/π 2 = 0.81 1 10
=1× ×1× = 0.25 V
Fraction of energy in first five harmonics 1+1 10 + 10
v 2o 0.252
8 1 1 PL = = = 6.25 mW
= 2 1+ + = 0.93 RL 10
π 9 25
Fraction of energy in first seven harmonics vo 0.25 V
Voltage gain = = = 0.25 V/V
vs 1V
8 1 1 1
= 2 1+ + + = 0.95 = −12 dB
π 9 25 49
PL
Fraction of energy in first nine harmonics Power gain (Ap ) ≡
Pi
8 1 1 1 1
= 2 1+ + + + = 0.96 where PL = 6.25 mW and Pi = v i i1 ,
π 9 25 49 81
v i = 0.5 V and
Note that 90% of the energy of the square wave is
in the first three harmonics, that is, in the 1V
ii = = 0.5 µA
fundamental and the third harmonic. 1 M + 1 M
Thus, Ex: 1.16 Refer the solution to Example 1.3 in the
text.
Pi = 0.5 × 0.5 = 0.25 µW
v i1
and = 0.909 V/V
vs
6.25 × 10−3
Ap = = 25 × 103 v i1 = 0.909 v s = 0.909 × 1 = 0.909 mV
0.25 × 10−6
10 log Ap = 44 dB v i2 v i2 v i1
= × = 9.9 × 0.909 = 9 V/V
vs v i1 vs
Ex: 1.13 Open-circuit (no load) output voltage = v i2 = 9 × v S = 9 × 1 = 9 mV
Av o v i
v i3 v i3 v i2 v i1
Output voltage with load connected = × × = 90.9 × 9.9 × 0.909
vs v i2 v i1 vs
RL
= Av o v i = 818 V/V
RL + Ro
1 v i3 = 818 v s = 818 × 1 = 818 mV
0.8 = ⇒ Ro = 0.25 k = 250
Ro + 1
vL vL v i3 v i2 v i1
= × × ×
vs v i3 v i2 v i1 vs
Ex: 1.14 Av o = 40 dB = 100 V/V
2 = 0.909 × 90.9 × 9.9 × 0.909 744 V/V
v2 RL
P L = o = Av o v i RL
RL RL + Ro v L = 744 × 1 mV = 744 mV
2
1
= v i × 100 ×
2
1000 = 2.5 v 2i Ex: 1.17 Using voltage amplifier model, the
1+1
three-stage amplifier can be represented as
v 2i v 2i
Pi = =
Ri 10,000
Ro
PL 2.5v 2
Ap ≡ = −4 i 2 = 2.5 × 104 W/W
Pi 10 v i
10 log Ap = 44 dB vi Ri Avovi
Ex: 1.15 Without stage 3 (see figure)
vL
= Ri = 1 M
v s
1 M 100 k Ro = 10
(10)
100 k + 1 M 100 k + 1 k
Av o = Av 1 ×Av 2 ×Av 3 = 9.9×90.9×1 = 900 V/V
100
×(100) The overall voltage gain
100 + 1 k
vo Ri RL
vL = × Av o ×
= (0.909)(10)(0.9901)(100)(0.0909) vs Ri + Rs RL + Ro
vs
= 81.8 V/V
This figure belongs to Exercise 1.15.
Stage 1 Stage 2
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