EEE 4309C Electronics II Homework 4
Due 11/4 (Monday)
Please work on the problems in the following sequence:
1.
P15.9
2.
P15.15 & 15.16
3.
P15.29
4.
P15.36
5.
P15.40 & 15.41
6.
P15.46 & 15.47
7.
P15.51
8.
P15.55(a)
9.
P15.60
10. P15.74
Grading:
5 x 0.5 (a
EXPERIMENT #2
More Linear Operational Amplifier Applications
Goals:
To introduce the concepts of an operational amplifier (op-amp) used as a difference
amplifier, an instrumentation amplifier, and a voltage-to-current converter. Data
collected during this
Table of Contents
Safety:
Safety Rules and Operating Procedures
Introduction:
When in Doubt, Read This
Experiment #1:
Linear Operational Amplifier Applications - 2.5 Wks
Experiment #2:
More Linear Operational Amplifier Applications - 2.5Wks
Experiment #3:
Introduction
When in Doubt, Read This
Laboratory Requirements:
This laboratory requires that each student obtain a copy of this manual, a bound quadruled engineering notebook and have access to a Spice simulation program such as
Multisim, Workbench or LTS
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
_
11.4
a.
I1 =
10 2 ( 0.7 )
IC 2 =
I1 = 1.01 mA
8.5
I1
2
1+
(1 + )
=
1.01
I C 2 1.01 mA
2
1+
(100 )(101)
100 1.01
IC 4 =
I C 4 0.50 mA
101 2
V
EEE 4309C Electronics II
Department of Electrical Engineering and Computer Science
College of Engineering and Computer Science, University of Central Florida
COURSE SYLLABUS
Instructor:
Office:
Phone:
E-Mail:
Office Hours:
Chung Yong Chan
HEC 407
407-882-
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
_
9.4
Ri
vid =
vI
Ri + 25
Ri
0.790 =
( 0.80 )
Ri + 25
0.9875 ( Ri + 25 ) = Ri
24.6875 = 0.0125 Ri
Ri = 1975 K
_
9.5
(a) A =
R 2 200
=
= 10
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
_
15.8
12 dB T = 0.2512
(1.333)
1
0.2512 =
2N
2
2N
1
=
1 = 14.85
0.2512
4
1+
3
th
N=5, 5 order filter
_
15.9
1
T =
f
1+
f 3dB
2N
T = 0.9
At
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
_
12.16
A f =
A
1 + A
=
Rif = Ri (1 + A
5 10 3
Af = 121.95
1 + (0.0080 ) 5 10 3
(
)
) = (10)[1 + (0.0080)(5 10 )] R
3
if
= 410 k
1 10
R of = 24.4
1
Microelectronics: Circuit Analysis and Design, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
_
vi
vo1 = 4.504495 103
452.5495
Or
Avf 1 =
vo1
= 9.9536
vi
For the second stage, RL =
5 103
1
1
100
vo 2 =
v1 = 4.950485 103 v1
1 1
+
1 10
EEE 4309C Electronics II Homework 3
Due 10/21 (Monday)
Please work on the problems in the following sequence:
1.
P12.18
2.
P12.26 & 12.27
3.
P12.35
4.
P12.37 ( For (b), finding the gain value is not required, just write down
the required equations. )
5.
P
EXPERIMENT #3
Linear Voltage Regulators
Goals:
To introduce the concepts of a linear voltage regulator and their use to maintain a
regulated voltage output. Data collected during this laboratory experiment will be
compared to the datasheet for the parts u