Transcript ENGG 1015 Tutorial - University of Hong Kong

ENGG 1203 Tutorial
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Electrical Circuit (II) and Project
1 Mar
Learning Objectives
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News
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Analyze circuits with resistors
Illustrate stages and components used in the project
Mid Term (TBD)
Revision tutorial (TBD)
Project Brief plan (8 Mar)
Ack.: HKU ELEC1008 and MIT OCW 6.01
1
Quick Checking
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Assuming the voltage at node N0 = 0, compute
the voltage at node N1 in each of these circuits.
IR
V/2
2
Quick Checking
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Assuming the voltage at node N0 = 0, compute
the voltage at node N1 in each of these circuits.
3 IR
3
Analyzing Circuits
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Assign node voltage variables to every node except ground
(whose voltage is arbitrarily taken as zero)
Assign component current variables to every component in
the circuit
Write one constructive relation for each component in terms
of the component current variable and the component voltage
Express KCL at each node except ground in terms of the
component currents
Solve the resulting equations
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Power = IV = I2R = V2/R
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4
Question: Finding Resistances via
Circuit Analysis
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Determine the indicated parameters for
each of the following circuits.
Because the resistors are in series, the
resistance between successive nodes will
be proportional to the voltage between the
nodes.
R1 ∝ V1 = 1R;
R2 ∝ V2 − V1 = 1R;
R3 ∝ V3 −V2 = 2R;
R4 ∝ V4 −V3 = 4R;
R5 ∝ V5 −V4 = 2R.
5
Question:
Another Example
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Determine the indicated
parameters for each of
the following circuits.
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KCL at the left-center node determines that a 1A
current flows rightward through R2.
To make V1 −V2 = 2V, it follows that R2 = 2Ω.
KCL at the right-center node then determines that a 3A
current flows downward through R1.
To make 10V-V2 = 6V, it follows that R1 = 2Ω.
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Question: Another Example
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If VAB = 4V, determine R1, R2, R3 and R4.
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Solution
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VAB = 4V
If VB = -1.5V
 VA = 2.5V
By potential divider,
R1:R2 = 1:1, R3:R4 = 1:1
VA  5 
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R2
R1  R 2
 2.5V
VB  3 
R4
R3  R4
  1.5V
You can pick any value for resistances.
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Solution
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If VB = -1V  VA = 3V
By potential divider,
R1:R2 = 2:3,
R3:R4 = 2:1
VA  5 
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R2
R1  R 2
 3V
VB  3 
R4
R3  R4
  1V
If VB = -2V  VA = 2V
By potential divider,
R1:R2 = 3:2, R3:R4 = 1:2
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Question: Resistance Calculation using
Parallel/Series Combinations
Find Req and io in the circuit of the figure.
10
Solution
(i)
60Ω
i0
12Ω
1 2  // 6   4 
5Ω
6Ω
40V
15Ω
20Ω
80Ω
2 0  // 8 0   1 6 
Req
(ii)
60Ω
i0
5Ω
40V
4Ω
15Ω
16Ω
4  16  20
Req
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Solution
i0
(iii)
5Ω
40V
15Ω
20Ω
60Ω
Req
R eq   15 // 20 // 60    7.5 
V  IR  40  i0  7.5  5   i0  3.2 A
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Question: Circuit Analysis with
multiple sources
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Find vo in the circuit of the figure.
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Solution
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Step 1: Define the node voltage (v1,v2,v3)
Step 2: Define the current direction
5A
v2
v1
2Ω
1Ω
4Ω
v3
8Ω
+
v0
--
40V
20V
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Solution
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Apply: 1) V = IR 2) KCL
Step 3: Consider node 1
v1  v 2
5
40  v1
2
1
1 
 3 v1  v 2  70
5A
5A
v2
v1
2Ω
v1
(v1-v2)/2
(40-v1)/1
1Ω
4Ω
v3
8Ω
+
v0
--
40V
20V
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Solution
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Step 3: Consider node 2
v1  v 2
2
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5
v2
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4
v 2  v3
8
 4 v1  7 v 2   20
2
5A
Step 4, 5: From (1) and (2),
v1 = 30V, v2 = 20V, v0 = v2 = 20V
v2
v1
2Ω
5A
v2
(v22-v
-v03)/8
(v
(v1-v2)/2
1Ω
4Ω
v3
8Ω
+
v0
--
40V
20V
(v1-0)/4
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Rube Goldberg Machine
Actuator to
pop the
balloon
Input
Output
Input
Output
Input
Output
Output
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(Tentative) At least five distinct stages with
different triggering mechanisms
The machine is started with pushing a
button/switch, and is ended by popping a
balloon.
Input
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START button
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Probing Questions for the Project
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Big questions
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How to design an (complicated) electrical system?
How do you (as a team) build a multi-stage Rube Goldberg
Machine that is functional and creative?
Small questions
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How do you describe the stages that are involved in the
machine?
How do you describe the electrical components in the machine?
How do you demonstrate your skills of technical design and
implementation?
How do you demonstrate your ability to work effectively with
diverse teams?
How do you demonstrate your originality and inventiveness?
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Brief Plan for the Project
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A brief plan for the project construction
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No technical requirement
A way to start thinking about the project
A block diagram with illustrations
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Showing how the machine works after pressing the
START button
No fire/chemicals/heat/explosion/knife
No water  Use green beans
Lego bricks are fine…
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Design a reliable and robust machine
because of the “Professor’s Negation Field”
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Systems that You Have Built in Lab
Sessions
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Ball counting
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Lab 1 – Lab 4
The tunnel increments its internal counter every
time a ball rolls through the tunnel.
When three balls have
rolled through the
tunnel, it raises a
digital DONE signal.
Not necessarily to use
straight tube
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Systems that You Will Build in Lab
Sessions
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Light tracking
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Lab 5 – Lab 7
The head can follow the direction of a light source
Decide your own light source triggering mechanism
Light
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Sensors and Actuators in the Project
Generation of
sound
Circuit isolation
Generation of
air flow
Circuit isolation
Generation of
rotation
Generation of
push force
Contact switch
Contact switch
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Sensors and Actuators in the Project
Sensing of
rotation
Non-contact
switch
Non-contact
switch
Time counter
START button
LED
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Mechanical Parts in the Project
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Metal ball
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Stages triggering
Metal pulley, sloted wheel
eye hook, Wheeled castor
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Pulley, conveyor belt, sling …
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Mechanical Parts in the Project
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Aluminum cage (New Requirement)
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Plastic board
Bracket
Hinge
Cable tie and cable mount
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Weak fix joint between
components
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A Stage in the Rube Goldberg
Machine
Mechanical
part
Electrical signal
Actuator
Mechanical signal
from previous stage
Buffer
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In a stage, an electrical sensor is triggered by an
external mechanical input, the sensor then
switches on the actuator(s) through buffers.
Electrical actuator then moves mechanical parts,
which finally trigger the electrical sensor in the
next stage.
Sensor
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Mechanical
signal for
next stage
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A Partial Stage in the Rube Goldberg
Machine
Sensor
(Push)
Actuator
(Push)
Sensor
Buffer Actuator
(Rotation)
(Push)
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A Partial Stage in the Rube Goldberg
Machine
Sensor
Buffer
(Non contact)
Actuator
(Light)
Actuator
(Fan)
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A Partial Stage in the Rube Goldberg
Machine
Mechanical
part
Actuator
Buffer
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Ball counting circuit:
A sensor only
Light tracking circuit:
Sensor + Buffer +
Actuator (Not yet a
complete stage)
Sensor
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Stages in the Rube Goldberg Machine
Actuator
(Rotation)
Buffer
Sensor
(Button)
Mechanical Sensor
Buffer Actuator
parts
(Rotation)
(Fan)
(Mechanical parts in the second stage are missed.)
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(Appendix) Discussions about the
thermistor in Lecture 5 pp. 52 – 55
that page
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(Appendix) Question: Circuit Analysis
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R1 = 80Ω, R2 = 10Ω, R3 = 20Ω,
R4 = 90Ω, R5 = 100Ω
Battery: V1 = 12V, V2 = 24V, V3 = 36V
Resistor: I1, I2, …, I5 = ?
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Solution a
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VN = 0
I1: M  R5  V1  R1  B
I2: M  V3  R3  R2  B
I4: M  V2  R4  B
Step 1, Step 2
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Solution b
Let’s try
another
reference
ground
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VM = 0
I1: B  R1  V1  R5  M
I2: B  R2  R3  V3  M
I4: B  R4  V2  M
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Solution b
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I1: B  R1  V1  R5  M
I2: B  R2  R3  V3  M
I4: B  R4  V2  M
Different direction, different result?
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Solution b
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KCL of Node B: I1 + I2 + I4 = 0
VB – VM = R1I1 – V1 + R5I1
I1 = (VB – VM + V1)/(R1 + R5) = (VB + 12)/180
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Solution b
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VB – VM = R2I2 + R3I2 – V3
I2 = (VB – VM + V3)/(R2 + R3) = (VB + 36)/30
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Solution b
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VB – VM = R4I4 – V2
I4 = (VB – VM + V2)/R4 = (VB + 24)/90
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Solution b
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KCL of Node B: I1 + I2 + I4 = 0
(VB + 12)/180 + (VB + 36)/30 + (VB + 24)/90 = 0
 VB = – 92/3 V
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Solution b
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I1 = (VB + 12)/180 = –14/135 A = – 0.104A
I2 = (VB + 36)/30 = 8/45 A = 0.178A
I4 = (VB + 24)/90 = –2/27 A = – 0.074A
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(Appendix) Notes about Multimeters
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Our multimeters allow you to measure current, voltage, and
resistance. You connect the multimeter to a circuit using two
leads. You can use The black lead should be plugged into the
ground (common) jack. The red lead should be plugged into a
jack labeled “V-Ω-mA,”.
Because the meter probes are large, they can bridge, and
thereby make unwanted electrical connections (“short circuits”)
between adjacent pins of small components. Such short circuits
can damage your circuit. To avoid this, you can measure the
resistance or voltage across points in your breadboard by using
short wires that are connected to the meter probes.
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(Appendix) Notes about Breadboard
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The breadboards have holes into which wires and components can
be inserted. Holes in the long top row (labeled +) are connected
internally (as are those in the second row, bottom row and next-tobottom row), as indicated by the horizontal (green) boxes (above).
These rows are convenient for distributing power (+10 V) and
ground. Each column of 5 holes in the center areas is connected
internally, as indicated by two representative vertical (blue) boxes
(above). Note that the columns of 5 are NOT connected across the
central divide.
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(Appendix) Notes about Wire and
Resistors
Wire
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We have a lot of wire kits that contained wires of different lengths
that are pre-cut and pre-stripped. Use these if you can. Try to
select wires that are just the right length, so they can lie flat on
the board. Messes of loopy wires are harder to debug and more
likely to fall apart. If you need a longer wire, cut what you need
from a spool. Use one of the pre-stripped wires for guidance on
how much to strip: too little and it won’t go deep enough into the
breadboard; too much, and you’ll have a lot of bare wire showing,
risking shorts against other wires and components.
Resistors
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We use quarter-watt resistors, which means that they can
dissipate as much as 250mWunder normal circumstances.
Dissipating more than 250mW will cause the resistor to overheat
and destroy itself.
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(Appendix) Notes about
Potentiometer (or Pot)
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It is a three terminal device whose electrical properties depend on
the angle of its mechanical shaft. The following figure shows a
picture of the pot that we will use in lab (left), the electrical symbol
used for a pot (center), and an equivalent circuit (right).
The resistance between the bottom and middle terminals increases
in proportion to the angle of the input shaft (θ) and the resistance
between the middle and top terminal decreases, so that the sum of
the top and bottom resistors is constant. We define a proportionality
constant α, which varies between 0 and 1 as the angle of the
potentiometer shaft turns from 0 to its maximum angle Θ, which is
approximately 270◦ for the potentiometers that we use in lab.
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By connecting a pot as a variable resistor (using top and middle
terminals), the resistance across those terminals is proportional to
the angle of the shaft. By connecting a pot as a voltage divider (top
terminal to a voltage source and bottom terminal to ground), the
voltage at the middle terminal is made proportional to the angle of
the shaft.
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(Appendix) Notes about Photoresistor
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A photoresistor is a two terminal device whose electrical
resistance depends on the intensity of light incident on
its surface. A photoresistor is made from a high
resistance material. Incident photons excite the electrons
– liberating them from the atoms to which they are
normally held tightly – so that the electrons can move
freely through the material and thereby conduct current.
The net effect can be characterized by plotting electrical
resistance as a function of incident light intensity, as in
the following plot (notice that the axes are logarithmically
scaled).
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Normal room lighting is between 10 and 100 lux.
Illuminance near a 60 watt light bulb (as we will use in
lab) can be greater than 10,000 lux.
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