ENGG1203: Introduction to Electrical and Electronic Engineering
First Semester, 2015–16
Homework 2
Due:
Oct 19, 2015, 11:55pm
Instruction: Submit your answers electronically through Moodle. In Moodle, you must submit under
the PDF Submission link with ONE PDF file containing answers to all remaining questions.
Your homework will be graded electronically so you must submit your work as a PDF file. To generate
PDF file from your computer, you may use one of the many free PDF creators available, e.g. PDFCreator
(http://sourceforge.net/projects/pdfcreator), CutePDF Writer (http://www.cutepdf.com/Products/
CutePDF/Writer.asp).
Question 1
Short Questions
Part(a)
In the following circuit, a huge number (e.g., more than 100) of resistors are connected in parallel, where
each successive resistor has double the resistance compared to the current one. Express the current i in
terms of v and R.
i
−
+
v
R
2R
Part(b)
Find the currents i1 and i2 for the following circuit:
4R
8R
ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
3 kΩ
20 kΩ
5 kΩ
−
+
12 V
20 kΩ
5 kΩ
3 kΩ
10 kΩ
i1
10 kΩ
i2
Part(c)
How does closing the switch (which connects the lamp to the rest of the circuit) affect vo and io ?
Explain whether each quantity will increase, decrease, or stay the same. The lamp resistance has not
been measured.
io
2 kΩ
+
−
+
5V
vo
1 kΩ
−
r1.0
Page 2 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Part(d)
Find the current i in the following circuit. Also find the power generated/dissipated at each component.
What happens to the voltage source?
i
−
+
2V
3 mA
1 kΩ
2 kΩ
Part(e)
In the following circuit, either compute the values of R1 and R3 in terms of R2 and R4 , such that vo is
always equal to v1 + 5v2 , or explain why this is not possible.
R4
R3
v2
−
vo
R1
v1
+
R2
r1.0
Page 3 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Part(f )
In the following circuit, find Rf in terms of R such that vo = −15vi .
Rf
R
R
+
vi
−
vo
−
+
Rf
R
r1.0
Page 4 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Part(g)
For the circuit below, (a) is a typical buffer, where the output Vo is equal to the input, Vi . What will
happen if you connect two additional resistors with resistance R1 and R2 , as shown in (b)?
R1
−
Vi
Vo
+
R2
Vi
(a)
r1.0
−
Vo
+
(b)
Page 5 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Question 2
Part(a)
Homework 2
Circuit Theorems
Thévenin’s Theorem
A useful tool to simplify the analysis of eletrical circuits is the Thévenin’s theorem. It converts any
complex circuits with voltage sources, current sources, and resistors into one — from the point of view
of a load — with only a single voltage source in series with a resistor. This is illustrated below:
A
A
network of sources
and resistors
vL
−
+
+
iL
RL
+
Rth
vth
vL
−
iL
RL
−
B
B
RL is the load resistor connected to the terminals A and B. We can maintain the same voltage vL and
current iL in the transformation by the following procedure:
• The Thévenin equivalent voltage, vth , is equal to the voltage across AB if there is no load, i.e., an
open circuit;
• Calculate the current I when AB is short-ciruited. Then, the Thévenin equivalent resistance, Rth ,
is equal to vth /I.
Now consider the circuit below. Find the Thévenin equivalent voltage and resistance. What is the
resultant voltage across AB if they are connected by a RL = 1 kΩ resistor?
1 kΩ
1 kΩ
5V
A
2 kΩ
1 mA
−
+
B
r1.0
Page 6 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Part(b)
Homework 2
Norton’s Theorem
There is a “dual” to the Thévenin’s theorem above, called the Norton’s theorem. Instead of having a
voltage source and a resistor in series, we can convert any complex circuits with voltage sources, current
sources, and resistors into one — from the point of view of a load — with only a single current source in
parallel with a resistor. This is illustrated below:
A
A
+
network of sources
and resistors
vL
iL
+
in
RL
Rn
vL
−
iL
RL
−
B
B
RL is the load resistor connected to the terminals A and B. We can maintain the same voltage vL and
current iL in the transformation by the following procedure:
• The Norton equivalent current, iN , is equal to the current across AB if they are short-circuited. In
other words, that’s I from the Thévenin equivalent above!
• The Norton equivalent resistance is the same as the Thévenin equivalent resistance, i.e., RN = Rth .
For the same circuit as in Part(a), find the Norton equivalent current and resistance. What is the
resultant voltage across AB if they are connected by a RL = 1 kΩ resistor?
r1.0
Page 7 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Part(c)
Homework 2
Power Dissipation
Find the power dissipation in a resistor RL connected across AB in Part(a) if:
1. RL = 1 kΩ;
2. RL = 2 kΩ;
3. RL = 3 kΩ.
(Bonus) If RL can be any resistance, what should be the value that would cause maximum power
dissipation in that resistor?
r1.0
Page 8 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
r1.0
Page 9 of 21
Homework 2
ENGG1203: Introduction to Electrical and Electronic Engineering
Part(d)
Homework 2
Applying Thévenin Theorem
Using your knowledge developed from previous parts, determine in the voltage across AB in the following
circuit:
1 kΩ
1 kΩ
A
1 kΩ
2 kΩ
1 kΩ
2 kΩ
2 kΩ
−
+
5V
1 mA
−
+
B
r1.0
Page 10 of 21
5V
1 mA
ENGG1203: Introduction to Electrical and Electronic Engineering
Question 3
Homework 2
Op-Amp Arithmetics
As its name suggests, an operational amplifier (Op-Amp) is a versatile circuit that can be used in
many mathematical operations, such as addition and subtraction. In this question, you will explore how
different operations can be achieved by changing the way an op-amp is connected.
Part(a)
The following op-amp circuit performs a simple operation on the two input v1 and v2 .
R
R
v1
R
v−
−
vo
v2
v+
+
Determine the output voltage vo in terms of v1 and v2 . You can then derive the relationship by letting
the input voltage at the negative input of the op-amp as v− and set up KCL at that node.
r1.0
Page 11 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Part(b)
In the following answer boxes, design circuit connections similar to that in Part(a) to perform the
following functions:
1. vo = −v1 + v2
2. vo = v1 − v2
3. vo = v1 + v2
4. vo = −v1 + v2 − v3
You may leave a resistor unconnected if it is not needed. If the required function is not possible to
construct, indicate so and explain why it cannot be achieved.
1. vo = −v1 + v2
R
R
v1
R
−
v2
vo
+
R
R
R
R
v3
2. vo = v1 − v2
v1
R
−
v2
vo
+
R
R
v3
r1.0
Page 12 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
3. vo = v1 + v2
R
R
v1
R
−
v2
vo
+
R
R
R
R
v3
4. vo = −v1 + v2 − v3
v1
R
−
v2
vo
+
R
R
v3
r1.0
Page 13 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Question 4
Homework 2
Rotating Motor
One key component of your project is the light tracker. In this question, you will explore some of its
circuit function.
Your light tracker is mount on top of a motor and is able to rotate left or right such that it is always
facing directly toward the light source. To detect the direction of light, it uses 2 light-sensitive resistors
that are placed at ±45◦ with respect to centerline of the device as shown below.
0◦
light source
θ
RL
RR
tracker head
The resistance of the light-sensitive resistor decreases when light is shined on it and increases when there
is no light. Therefore, when the light is on the left of the centerline, RL decreases and RR increases.
Similarly, when the light is on the right of the centerline, RR decreases and RL increases.
In addition, the motion of the tracker head is controlled by the motor it is attached to. The rotational
speed of the motor, ω, depends linearly on the voltage across its two input terminals Vm = Vmp − Vmn ,
provided that it exceeds a certain threshold. That is,
(
km Vm if Vm > 2
ω=
(1)
0
otherwise,
where km is a motor dependent constant.
When Vm is positive, the motor turns clockwise. When it is negative, the motor turns counter-clockwise.
Part(a)
First Attempt
Armed with the above information about the light tracker head, your project partner has designed the
following circuit to control the light tracker:
Vdd
RR
Vmp
motor
+−
Vmn
−
+
RL
Vdd /2
Figure 1: First Attempt Motor Control Circuit
Your partner’s idea is that when the light is on the right of the centerline, RR decreases, making Vm
increases, turning the motor clockwise and toward the light. Similarly, when the light is on the left of
the centerline, RL decreases, making Vm decreases, turning the motor counter-clockwise so it will point
back to the light source.
While the design of the circuit in Figure 1 may seem fine, it does not work as expected when you test
it in the lab. Explain why it does not work according to the design by considering the voltage Vm when
light is on left and on right of the centerline. Assume the resistance of the light-sensitive resistor is 100 Ω
r1.0
Page 14 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
when there is light and 10 kΩ when there is no light. Also assume the voltage sources has 0 resistance,
and the internal resistance of the motor is 4 Ω. Vdd is 5 V.
Part(b)
Second Attempt
You search the laboratory again find some op-amps. You then modify the circuit as shown below:
Vdd
Vdd
RR
Vp
+
Vmp
+ −
Vmn
−
RL
−
+
Does this circuit function correctly? That is, does it correctly track the direction of the light source?
Explain your answer.
Part(c)
Armed with your experience in constructing potential divider using resistors, you try to produce the
voltage Vdd /2 using two identical resistors R as shown below:
r1.0
Page 15 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Vdd
Vdd
Vdd
RR
R
Vp
+
Vmp
+ −
Vmn
−
RL
R
Unfortunately, the circuit is no longer behaving the same as before. Explain why the motor does not
rotate the same way as before. Hint: What is the volgate Vmn in the circuit?
r1.0
Page 16 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Part(d)
Show how you may further revise the design above so it functions as desired.
Vdd
Vdd
RR
Vp
+
Vmp
+ −
Vmn
−
RL
Your Circuit
r1.0
Page 17 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Question 5
Homework 2
Semiconductor Lighting
You work at a lighting company that develops next-generation lighting systems that are friendly to the
environment. With years of research, a highly energy efficient semiconductor lamp, called SiliLamp, has
been invented. You are charged with the task to design circuits that control this next-generation lamp.
Because of the special materials used to produce SiliLamps, they exhibit a unique current-voltage characteristic (I-V characteristic) that is different from a traditional incandescent lamp, which behaves as a
simple resistor. The relationship between the current through a device and the voltage across its terminals is called the I-V characteristic of that device. For example, a resistor has a simple linear relationship
between the current flowing through (IR ) and the voltage (VR ) across it, namely, Ohm’s law:
IR =
VR
R
In the case of SiliLamp, let IL be the current flowing through and VL be the voltage across it while it
operates, then
IL = kVL3
where k = 0.5 for this particular model. To illustrate, the Figure 2 shows the I-V relationship for (a) a
simple resistor, and (b) a SiliLamp.
Semiconductor Lamp
Current (A)
Current (A)
Incandescent lamp
4
3.5
3
2.5
2
1.5
1
0.5
0
0
0.5
1
1.5
2
4
3.5
3
2.5
2
1.5
1
0.5
0
0
Voltage (V)
(a) Resistor: IR =
VR
R
0.5
1
1.5
Voltage (V)
2
(b) SiliLamp: IL = kVL3
Figure 2: I-V characteristics
Part(a)
Simple Driver
You start with driving a SiliLamp with the simple circuit below:
R
−
+
10 V
SiliLamp
Find the resistance R in the above circuit such that the voltage across SiliLamp (VL ) is 1.5 V. (Hint:
KCL and KVL remains true regardless of the I-V characteristics of the circuit components involved.)
r1.0
Page 18 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Part(b)
Homework 2
Efficiency
An important aspect of designing a lamp driving circuit is its efficiency. The efficiency of your driver
circuit is defined as:
PL
ηS =
PL + PR
where PR is the power dissipated at the resistor, and PL is the power dissipated at the lamp. The power
dissipated at SiliLamp, in turn, determines how bright the lamp is.
Based on your result in , assuming VL = 1.5 V, determine PR , PL and ηS .
r1.0
Page 19 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Part(c)
Homework 2
Series Combination
To increase brightness the resulting system, you try to connect 2 SiliLamps in series as shown below:
R
SiliLamp
−
+
10 V
SiliLamp
While keeping the source voltage as 10 V, and keeping the operating voltage of each SiliLamp at 1.5 V,
determine (i) total power of the 2 SiliLamps (PL,tot ) in combination, and (ii) efficiency of the resulting
system.
Part(d)
Parallel Combination
Alternatively, you try connecting two SiliLamps in parallel as shown below:
R
−
+
10 V
R
SiliLamp
SiliLamp
Again, while keeping the source voltage as 10 V, and keeping the operating voltage of each SiliLamp
at 1.5 V, determine (i) total power of the 2 SiliLamps (PL,tot ) in combination, and (ii) efficiency of the
resulting system.
r1.0
Page 20 of 21
ENGG1203: Introduction to Electrical and Electronic Engineering
Part(e)
Homework 2
Design Problem
Armed with your observations from the above series and parallel circuits, you are asked to develop a
product with the following specifications:
•
•
•
•
Total lamp power: 25 W ± 1 W
Minimum system efficiency: 85 %
Supply voltage: 10 V
The brightness of each lamp should be the same
This is an open-ended question. You are free to make any design decisions in combining multiple
SiliLamps as long as your solution meet the above specification. Submit your final circuit and determine
your resulting total lamp power and system efficiency.
r1.0
Page 21 of 21