Homework 2 - Department of Electrical and Electronic Engineering

ENGG1203: Introduction to Electrical and Electronic Engineering
First Semester, 2014–15
Homework 2
Due:
(r1.2)
Oct 20, 2014, 11:55pm
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Instruction: Submit your answers electronically through Moodle. In Moodle, you must submit
a total of ONE file:
1. Under PDF Submission, submit ONE (1) PDF file containing answers to all 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).
Short Questions
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Question 1
Part(a)
Use KVL to compute the (i) voltage across each ressistor and (ii) current flowing through each
voltage source.
+
I1
−
10 Ω
SO
−
+
10 V
V R1
10 Ω
+
+
V R2
−
10 Ω
V R3
−
− 10 V
+
I2
Using KVL we have:
10 − VR1 − VR3 = 0
(1)
VR3 − VR2 + 10 = 0
(2)
(3)
Also, using KCL, we have:
VR2
VR3
VR1
=
+
10
10
10
(4)
Now,
VR2 + VR3 = 10 − VR3
(4) & (1)
(VR3 + 10) + VR3 = 10 − VR3
from (2)
Therefore, VR3 = 0, VR2 = 10 − 2VR3 = 10 and VR1 = 10.
VR2
VR1
= 1, I2 = 10
= 1.
Also, I1 = 10
ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Part(b)
Use KCL to compute the (i) voltage across each ressistor and (ii) current flowing through each
voltage source.
+
+
V R1
VR2
10 Ω
−
0.5 A
5Ω
−
I R2
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IR1
−
+
5V
−
+
10 V
Using KCL at the + node of R1 , we have:
0.5 = IR1 + IR2
(5)
Also, the voltage of each branch must be equal as they are parallel, therefore,
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10 + IR1 R1 = 5 + IR2 R2
(6)
Combining (5) and (6), we have:
1
6
and
I2 =
5
3
and
VR2 =
I1 = −
and
VR1 = −
SO
Part(c)
2
3
10
3
Current to Voltage Converter
In the following circuit, determine Vo in terms of Iin .
R
Iin
R
R/2
Vo
+
According to the ideal op-amp model, there is no current flowing into the op-amp and
v+ = v− . Therefore,
Vo = −Iin R
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ENGG1203: Introduction to Electrical and Electronic Engineering
Part(d)
Homework 2
Voltage to Current Converter
In the following circuit, determine Io in terms of Vin .
R
-
Io
RL
+
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Vin
R
R1
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SO
According to the ideal op-amp model, there is no current flowing into the op-amp and
v+ = v− . Therefore,
Vin
Io =
R1
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ENGG1203: Introduction to Electrical and Electronic Engineering
Question 2
Homework 2
Superposition Theorem
When analyzing complex circuits, it is sometimes useful to make use of the superposition theorem.
Using this theorem, it is possible to decompose a circuit with multiple input sources into multiple
circuits with only 1 source. By doing so, it simplifies the analysis and may help to understand
the behavior of a circuit.
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In particular, the superposition theorem states that the response of a circuit with multiple input
sources is the sum of the responses from each independent source when acting alone. In practice,
to obtain the response due to one particular source S, one would turn off all the sources (voltage
sources or current sources) in the circuit except S. Turning off a source can be achieved by:
• For voltage sources, replace the source with a short circuit.
• For current sources, replace the source with an open circuit.
The overall response of the circuit can then be obtained by superpositioning each independent
response on top of each other. This question illustrates how superposition theorem works and
you will use it to analyze some complex op-amp circuits.
Part(a)
Circuit Analysis without using Superposition
I1
−
+
10 V
I2
1Ω
I3
1Ω
5V
−
+
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Compute the current I1 , I2 and I3 in the following circuit.
SO
Figure 1: Circuit with 2 voltage sources
Based on KVL and KCL, we have:


I1 + I2 + I3 = 0
10 + I3 = 0


I2 = 5 + I3
Solving the equations yeild
I1 = 15,
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I2 = −5,
I3 = −10
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ENGG1203: Introduction to Electrical and Electronic Engineering
Part(b)
Homework 2
Circuit Analysis with Superposition
It is possible to analyze the same circuit in Figure 1 by decomposing it into the following 2
circuits:
1Ω
I2a
I1b
1Ω
I3b
I3a
1Ω
1Ω
5V
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−
+
10 V
I2b
−
+
I1a
(a) Decomposition A – Turn off right voltage source
(b) Decomposition B – Turn off left voltage source
Figure 2: Two decompositions of circuit in Figure 1 with each of the voltage source replaced by
a short circuit.
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In Figure 2(a), the 5 V voltage source is replaced by a short circuit while keeping the 10 V source
intact. On the other hand in Figure 2(b), the 10 V is replaced by a short circuit, leaving the
original 5 V source intact. Calculate the values I1a , I2a and I3a from Figure 2(a), and I1b , I2b
and I3b from Figure 2(b).
I1a = 20 A
I2a = −10 A
I3a = −10 A
SO
I1b = −5 A
I2b = 5 A
I3b = 0 A
Part(c)
According to the superposition theorem, the overall response of the circuit can be obtained by
summing the contribution from each independent sources. Compute the 3 values I1 = I1a + I1b ,
I2 = I2a + I2b , I3 = I3a + I3b . Are they the same as the results you have obtained from Part (a)?
I1 = I1a + I1b = 20 + (−5) = 15
I2 = I2a + I2b = −10 + 5 = −5
I3 = I3a + I3b = −10 + 0 = −10 The resulst are the same as Part (a).
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Part(d)
In the following circuit, calculate the values of R1 and R4 in terms of R2 and R3 such that
Vo = V1 − 10V2 . To take advantage of the superposition theorem, set V1 = 0 and V2 = 0 in turn.
R1
V1
R2
R3
+
VO = V1 - 10V2
R4
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Vo
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V2
Replace the voltage source V1 with a short circuit, i.e., set V1 = 0, then:
V2 −0
0−Vo
R2 = R1
Vo = −10V2
SO
Solving the above gives:
R1 = 10R2
(7)
Now, replace voltage source V2 with a short circuit, i.e., set V2 = 0, then:
(
R4
R4
0− R +R
V1
R3 +R4 V −Vo
3
4
=
R2
R1
Vo = V1
Solving the above yields:
R1 R4 = R2 R3
Now, combinging (7) and (8) results in
(
R1 = 10R2
R3 = 10R4
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ENGG1203: Introduction to Electrical and Electronic Engineering
Question 3
Homework 2
Delta Star Transformation
When simplifying complex resistance network, it is sometimes useful to recognize and transform
between a delta topology by a star topology, which are shown below.
A
A
R
3
R1
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Ra
Rc
C
B
R2
(a) Delta
C
R
b
B
(b) Star
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Figure 3: Delta and star resistor networks are equivalent if the resistor values are chosen correctly.
While the 2 topologies may look quite different, by choosing the correct resistor values, it can be
shown that the two networks can be equivalent. This exercise helps you to deduce the correct
resistor values.
Part(a)
Delta
SO
Consider the delta topology in Figure 3(a). Define Rab , Rbc , and Rca as the equivalent resistances between terminal A and B, B and C, and C and A respectively. Express Rab , Rbc , and
Rca in terms of the 3 resistors R1 , R2 and R3 .
Using the notation Rx k Ry to means a parallel combination of Rx and Ry , then
Rab = (R3 + R2 ) k R1
=
(R3 + R2 )R1
R1 + R2 + R3
Rbc = (R1 + R3 ) k R2
=
(R1 + R3 )R2
R1 + R2 + R3
Rca = (R1 + R2 ) k R3
=
r1.2
(R1 + R2 )R3
R1 + R2 + R3
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ENGG1203: Introduction to Electrical and Electronic Engineering
Part(b)
Homework 2
Star
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Now, consider the star topology in Figure 3(b). Similarly define Rab , Rbc , and Rca as the
equivalent resistances between terminal A and B, B and C, and C and A respectively. Express
Rab , Rbc , and Rca in terms of the 3 resistors Ra , Rb and Rc .
The resistances can be simply be obatined by the series combination of the resistors, i.e.:
Rab = Ra + Rb
Rbc = Rb + Rc
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Rca = Ra + Rc
Part(c)
Transformation
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For the delta and star topology to be equivalent, the values for Rab , Rbc , and Rca must be the
same in both cases. Using your results from above, express Ra , Rb , Rc in terms of R1 , R2 , R3 .
Combining the results from Part (a) and (b), equating on the values of Rab , Rbc and Rca
gives:
(R3 + R2 )R1
= Ra + Rb
R1 + R2 + R3
(R1 + R3 )R2
=
= Rb + Rc
R1 + R2 + R3
(R1 + R2 )R3
=
= Ra + Rc
R1 + R2 + R3
Rab =
Rbc
Rca
Subtracting (10) from (11) and adding the result to (9):
(R1 + R2 )R3 − (R1 + R3 )R2 + (R3 + R2 )R1
R1 + R2 + R3
R1 R3
Ra =
R1 + R2 + R3
2Ra =
Similarly, we can obtain:
Rb =
R1 R2
R1 + R2 + R3
Rc =
R2 R3
R1 + R2 + R3
and
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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◦
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light source
θ
RL
RR
tracker head
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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=
(12)
0
otherwise,
where km is a motor dependent constant.
SO
When Vm is positive, the motor turns clockwise. When it is negative, the motor turns counterclockwise.
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 4: 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 4 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
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
light-sensitive resistor is 100 Ω 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)
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When the light is on the right, RR ≈ 100, RL ≈ 10000. Therefore, Vmp ≈ 2.6 V and
Vm ≈ 0.1 V.
When the light is on the left, RR ≈ 10000, RL ≈ 100. Therefore, Vmp ≈ 2.4 V and
Vm ≈ −0.1 V.
As a result, the motor WILL NOT WORK according to what is expected because |Vm | <
2.
Second Attempt
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You search the laboratory again find some op-amps. You then modify the circuit as shown below:
Vdd
Vdd
RR
Vp
+
Vmp
+ −
Vmn
−
SO
−
+
RL
Does this circuit function correctly? That is, does it correctly track the direction of the light
source? Explain your answer.
This circuit functions correctly. The introduction of a buffer using the op-amp allows Vp
to change as expected – When light is on right, RR RL , therefore, Vp ≈ Vdd . When
light is on left, RL RR , therefore Vp ≈ 0. Using the op-amp as a voltage follower,
Vmp = Vp . Therefore, light is on right, Vm > 0 and the motor turns clockwise toward the
light. When light is on the left, Vm < 0 and the motor turns counter-clockwise toward
the light.
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:
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ENGG1203: Introduction to Electrical and Electronic Engineering
Homework 2
Vdd
Vdd
Vdd
RR
R
+
Vp
Vmp
+ −
Vmn
−
RL
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R
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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?
Because of the lack of buffer on the negative end of the motor, the voltage at Vmn tends
to follow that of Vmp . As a result, there’s still not enough voltage (power) delivered into
the motor.
SO
Part(d)
Show how you may further revise the design above so it functions as desired.
Vdd
Vdd
Vdd
RL
Vdd
R
+
Vmp
+ −
Vmn
−
+
Vp
Vn
−
RR
Your Circuit
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R

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