TABLE A3 Precision (1% ) Resistors 36.5 0'0L 10.2 8'69 71.5 EEI 255 3.40K it'9 12,K IE6 23,7K 45.3K 84.5K 15KK 294K RL'I 9%1 37.4 хиТ 'E 3.57K 24,3K 46.4K 162K 62K 10.5 EYE 73.2 75.0 13.0K 24,9K 47.5K 576K 205 39.2 M16'1 3,74K XE'EI 26.1K 25.5K 487K X691 316K 324K X99'E хоS 896. L'34 O'Ll 13.7K 9.1K 174K 9ES хо1 MST'L XI'IS XE'ZS SIZ ЖОТ хот E8'E 221 42.2 43.2 44.2 11.5 NE'SS 거619 Ж02 11.8 82.5 N0'RZ 54.9K LOE 89'L 232 X9'ES 12,1 х001 84,5 068 1.15K EI'I 2,21K 4.1K хя 15.4K 291 28.7K 357K хSL 665K яо LEZ 45.3 46.4 12,4 4.22K 29.4K 56.2K 998 196K BIK 12.7 91E 243 619 ÞE9 1.21K 699 1.24K 4.32K 4.42K 825K 30.1K 57.6K 200K 13.0 249 47.5 606 l'E6 332 2,32K &45K 162K L'It 669 N6DE 60.4K 13.3 XERE У 8'3 16SK 113K $99 189 L'EI 32,4K 61.9K 115K 21SK 40RK DE'I 2,4K XG0'6 ЛЕб XES'6 18,2K 18 7K I'IS 33.2K 63.4K 118K 412K 9'26 LSE EE'I MLE'I 2,61K 1.40K 4,87K я89 EZE 34.0K 64.9K 001 121K 422K 161 869 14.7 961 314 66.5K 124K 232K 9'ES 201 206K 6'4 56.2 15.0 ZEL OSL 2.67K 5.11K 35.7K XIY9 127K 2.74K хOт 19. IK LOE EZ'S 8'SE 16.2 MEST 2.8CK 60E 91E 6'91 324 BE'S t'S хо жо 71.SK XYLE XE'BE XZ'6E 75.0K NEEI XZ'EL 137K 140K 16.5 t'09 115 6'19 118 ЯL8 908 825 S'I RS'I 3.0 IK X9'S 487K 432 я60 21.0K 40.2K 17.4 X8'9L ZEE 64.9 121 жот S'S 6.04K ME'IL 21.5K 22,1K NS'I1 хО1 42,2K XOSI 154K NS9'I 998 S99 I'89 19 6.19K NZ'E 82.5K 18.7 6.34K 2SE DEL 606 XI'ZI

Introductory Circuit Analysis (13th Edition)

13th Edition

ISBN:9780133923605

Author:Robert L. Boylestad

Publisher:Robert L. Boylestad

Chapter1: Introduction

Section: Chapter Questions

Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...

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Related questions

Concept explainers

Question

(a) Design a low-pass amplifier (i.e., choose R₁,
R₂, and C) to have a low-frequency input resistance
of 10 k ohm, a midband gain of 20 dB, and a
bandwidth of 20 kHz. (b) Choose element values
from the table.

TABLE A3
Precision (1% ) Resistors
36.5
0'0L
10.2
8'69
71.5
EEI
255
3.40K
it'9
12,K
IE6
23,7K
45.3K
84.5K
15KK
294K
RL'I
9%1
37.4
хиТ
'E
3.57K
24,3K
46.4K
162K
62K
10.5
EYE
73.2
75.0
13.0K
24,9K
47.5K
576K
205
39.2
M16'1
3,74K
XE'EI
26.1K
25.5K
487K
X691
316K
324K
X99'E
хоS
896.
L'34
O'Ll
13.7K
9.1K
174K
9ES
хо1
MST'L
XI'IS
XE'ZS
SIZ
ЖОТ
хот
E8'E
221
42.2
43.2
44.2
11.5
NE'SS
거619
Ж02
11.8
82.5
N0'RZ
54.9K
LOE
89'L
232
X9'ES
12,1
х001
84,5
068
1.15K
EI'I
2,21K
4.1K
хя
15.4K
291
28.7K
357K
хSL
665K
яо
LEZ
45.3
46.4
12,4
4.22K
29.4K 56.2K
998
196K
BIK
12.7
91E
243
619
ÞE9
1.21K
699
1.24K
4.32K
4.42K
825K
30.1K 57.6K
200K
13.0
249
47.5
606
l'E6
332
2,32K
&45K
162K
L'It
669
N6DE
60.4K
13.3
XERE
У
8'3
16SK
113K
$99
189
L'EI
32,4K
61.9K
115K
21SK
40RK
DE'I
2,4K
XG0'6
ЛЕб
XES'6
18,2K
18 7K
I'IS
33.2K
63.4K
118K
412K
9'26
LSE
EE'I
MLE'I
2,61K
1.40K
4,87K
я89
EZE
34.0K
64.9K
001
121K
422K
161
869
14.7
961
314
66.5K
124K
232K
9'ES
201
206K
6'4
56.2
15.0
ZEL
OSL
2.67K
5.11K
35.7K
XIY9
127K
2.74K
хOт
19. IK
LOE
EZ'S
8'SE
16.2
MEST
2.8CK
60E
91E
6'91
324
BE'S
t'S
хо
жо
71.SK
XYLE
XE'BE
XZ'6E
75.0K
NEEI
XZ'EL
137K
140K
16.5
t'09
115
6'19
118
ЯL8
908
825
S'I
RS'I
3.0 IK
X9'S
487K
432
я60
21.0K
40.2K
17.4
X8'9L
ZEE
64.9
121
жот
S'S
6.04K
ME'IL
21.5K
22,1K
NS'I1
хО1
42,2K
XOSI
154K
NS9'I 998
S99
I'89
19
6.19K
NZ'E
82.5K
18.7
6.34K
2SE
DEL
606
XI'ZI

Expert Solution

Step 1

(a) Given

Low-frequency input resistance is $10 kΩ$

Midband gain is 20 dB.

The mid-band voltage gain value in linear is given by

$\begin{array}{rcl} 20 \log (A_{m i d}) & = & 20 dB \\ A_{m i d} & = & 10^{\frac{20}{20}} \\ = & 10 \end{array}$

The expression for voltage gain is given by

$|A_{L}| = \frac{R_{2}}{R_{1}}$

Put $A_{L} = 10 and R_{1} = 10 kΩ$ in the above equation

$\begin{array}{rcl} \frac{R_{2}}{10 \times 10^{3}} & = & 10 \\ R_{2} & = & 100 kΩ \end{array}$

Step 2

The expression for upper cut-off frequency is given by

$\begin{array}{rcl} f_{H} & = & \frac{1}{2 π R_{2} C} \\ C & = & \frac{1}{2 π R_{2} f_{H}} \end{array}$

Put $f_{H} = 20 kHz and R_{2} = 100 kΩ$

$\begin{array}{rcl} C & = & \frac{1}{2 π \times 100 \times 10^{3} \times 20 \times 10^{3}} \\ = & 0.795 \times 10^{- 9} \\ = & 795 pF \end{array}$

Hence, the design values of the low pass filter are $R_{1} = 10 kΩ, R_{2} = 100 kΩ and C = 795 pF$

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