ELG 2136
Electronics I / lectronique I
Winter / Hiver 2014
MIDTERM EXAMINATION (90 minutes)
EXAMEN DE MISESSION (90 minutes)
Professeur/Professor E. Gad
Nom/Name : _
Date
Feb. 14th 2014 St./tudiant #: _
This is a closed book exam
Answer all questions
Ex
ELG3120C Winter 2002
Initial _
ELG 3120C
SIGNAL AND SYSTEM ANALYSIS
Final Exam Winter 2002
Wednesday, 24 April 2002
Time: 14:00 17:00
Vanier Hall
Room: 231
Prof. Jianping Yao
Time allowed: 3 hours
Plain calculator permitted
Textbook and notes not allowed
ELG 2136 B Winter 2017
Feb 4th 2017
Due date: Feb 10th 2017
Assignment 4
General Instructions:
You have options:
A: To have your equations and solutions hand written, scanned into a
PDF format, and uploaded to your account on the Blackboard
B: Typed solu
Current Limiting Circuits
March 17, 2017
University of Ottawa
School of Electrical Engineering and
Computer Science
Table of Contents
1.0 Introduction
1
2.0 Transforming 120 VAC to 10 VDC 1
3.0 First Current Limiting Circuit Design
3
4.0 Second Current Li
Assignment 12
1) From Assignment 8
Rc = 5.666k and I = 0.5mA
Using DC analysis we open circuit capacitors
IE = 0.5mA = (BIB + IB) IB = 4.95uA IC = 0.495 mA
Using these values we solve rpi = 5.05k
AC analysis, short circuit capacitors and DC voltage source
Assignment 11
1) Common Source
2) By doing DC analysis we remove the AC voltage source and open circuit all of the capacitors
Vov = Vgs  Vt = 0.3V
(0=Vs)  (1) = 0.3
Vs = 0.7
iD = iS = 1mA
(2.5  0.7)/Rs = 1mA
Rs = (2.5  0.7)/1mA = 1.8 k
3)
Gv =
Assignment # 6
Q1 is PNP and Q2 is NPN
Assume Q1 and Q2 are both in active mode
KVL at BE Loop on Q1:
3V + (9.1)(iE1) + 0.7V + (100)(iB1) = 0
iB1 = (2.3)/(1019.1) = 0.002257 mA
iE1 = (101)(0.002257mA) = 0.227 mA
iC1 = (100)(0.002257mA) = 0.225 mA
V1 = (
Cross Layer Design for Wireless Multimedia Networks
Assignment 1
1. Which of the following uses the widest spectrum?
(a) 802.16 (b) 802.11 (c) CDMA2000 (d) UWB
2. Which of the following is true?
(a) Smallscale multipath fading is the primary driver behi
ME 115(a): Notes on Rotations
1
Spherical Kinematics
Motions of a 3dimensional rigid body where one point of the body remains xed are termed
spherical motions. A spherical displacement is a rigid body displacement where there is
a xed point in the initia
Robotics: Control, Sensing, and Intelligence
University of Ottawa
Assignment #1
Due (see course web site)
Dr. W. Gueaieb
Total (120 points)
Instructions
Individual work only. Show your work in detail.
No computation software is to be used. All calculation
Technical Concepts
Orientation, Rotation, Velocity and Acceleration,
and the SRM
Version 2.0
20 June 2008
Author:
Paul Berner, PhD
Contributors:
Ralph Toms, PhD, Kevin Trott, Farid Mamaghani, David Shen,
Craig Rollins, Edward Powell, PhD
Copyright 2008 SE
ELG2136 Winter 2017
Mavungu Kambu
7230579
Assignment 1
Solution:
The Exponential model (Shockley diode model) is represented as the following:
= exp 1
Since n is assumed to be 1 and the subtracted 1 is considered to be negligible, it can be rewritten
as
ELG2136 Winter 2017
Mavungu Kambu
7230579
Assignment 1
Q: A bridge rectifier circuit, shown in figure 1 below, is fed by a sinusoidal source of
frequency 60 Hz and RMS value of 120 V, and is loaded with a resistor RL of 1 k.
Figure 2 provides the waveform
ELG2136 Winter 2017
Mavungu Kambu
7230579
Assignment 1
A) In this case, it will be assumed that the diode is on (this will be replaced with a short
circuit).
After replacing the diode in its respective position, we can start calculating the current and
vo
ELG2136
Assignment 4
1.A: Find the value of Vzo in the Zener model.
Vz= 10 volts
Iz = 10 mA
Rz = 50 ohms
Vz = Vzo + Iz x Rz
10V = Vzo + (0.01) (50 ohms)
Vzo = 10 0.5
Vzo = 9.5 Votls
1.B: Calculate the voltage when the diode current is halved.
Vzo= 9.5 vol
Assignment 5
ELG 2136
IE = 1 mA
IB = 50 A, 10 A, 20 A
Case 1: IB =10 A
Case 2: IB =20 A
Case 3: IB =50 A
IC= IE IB
IC= 1x103 10x106
IC= 9.9x104 A
IC= IE IB
IC= 1x103 20x106
IC= 9.8x104 A
IC= IE IB
IC= 1x103 50x106
IC= 9.5x104 A
= IC/ IB
= 9.9x1
1
(2)1/6
BJT Transistors
SmallSignal Analysis of the
CommonCollector (Emitter
Follower) Configuration
An EXAMPLE
1
2
(2)1/6
BJT configurations
Figure 7.33 The basic configurations of transistor amplifiers. (d)(f) for the BJT.
Microelectronic Circuits, S
1
(2)1/6
Lecture outlines
Real Diodes
Analysis of diode circuits
Introduction to diode models
VI curve
Example
2
(2)1/6
Real
Diode
3
(2)2/6
Analysis of Diode Circuits
Problems requiring plotting
Y vs. X
4
(2)1/6
Ideal vs. real diode
Reverse region
Real d
1
(2)4/6
Applications of real diode
(PN junction)
The power supply
3. The Regulator component
and the Zener Diode
2
120V (rms)
60 Hz
(2)4/6
The Application
Power Supply
15 V
dc
Ripples minimized
Transformer
Stepped down, e.g.,
15 V
Polarity is changing
Z
1
(2)3/6
Applications of real diode
(PN junction)
The power supply
2. The Filter component
2
(2)3/6
The filter component
The combination of the Rectifier and the
filter component is called the
PEAK
RECTIFIER
3
(2)3/6
The rectifier with a filter
The peak
1
(2)1/6
BJT Transistors
A Complete SmallSignal
Analysis of BJT amplifier circuit
An EXAMPLE
2
(2)1/6
Example
For the circuit shown, calculate:
assuming that
+

P6.100 From Sixth Edition Sedra & Smith
2/27/2017
Prepared by Dr. Emad Gad
2
3
(2)1/6
Split
Assignment 10
Friday, April 1, 2016
6:42 PM
Donald Dang
7747085
a) Voltage divider on
= 7.5V
Assignments Page 1
= 7.5V
Take 2.1 since it's satisfies the saturation theorem
b) See bottom of the page.
c) From the equivalent circuit shown at the bottom we se
Assignment 9
Wednesday, March 23, 2016
7:20 PM
Donald Dang
7747085
a)
If saturation is assumed, then:
0.1
Assignments Page 1
b)
If saturation, then:
Therefore, they are both saturated.
Assignments Page 2
Assignment 5
Saturday, February 20, 2016
8:35 PM
Donald Dang
7747085
The current is experiencing diffusion/natural. There is no electric field.
Since the answer is positive, it is moving positively along the x axis.
Assignments Page 1