# 1 ENGG 1203 Tutorial Electrical Circuit (II) and Project 1 Mar Learning Objectives  Analyze circuits with resistors  Illustrate stages and components.

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1 ENGG 1203 Tutorial Electrical Circuit (II) and Project 1 Mar Learning Objectives  Analyze circuits with resistors  Illustrate stages and components used in the project News  Mid Term (TBD)  Revision tutorial (TBD)  Project Brief plan (8 Mar) Ack.: HKU ELEC1008 and MIT OCW 6.01

Quick Checking Assuming the voltage at node N 0 = 0, compute the voltage at node N 1 in each of these circuits. 2 IR V/2

Quick Checking Assuming the voltage at node N 0 = 0, compute the voltage at node N 1 in each of these circuits. 3 3 IR

4 Analyzing Circuits 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 Power = IV = I 2 R = V 2 /R

Question: Finding Resistances via Circuit Analysis 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. R 1 ∝ V 1 = 1R; R 2 ∝ V 2 − V 1 = 1R; R 3 ∝ V 3 −V 2 = 2R; R 4 ∝ V 4 −V 3 = 4R; R 5 ∝ V 5 −V 4 = 2R. 5

Question: Another Example Determine the indicated parameters for each of the following circuits. KCL at the left-center node determines that a 1A current flows rightward through R 2. To make V 1 −V 2 = 2V, it follows that R 2 = 2Ω. KCL at the right-center node then determines that a 3A current flows downward through R 1. To make 10V-V 2 = 6 V, it follows that R 1 = 2Ω. 6

7 Question: Another Example If V AB = 4V, determine R1, R2, R3 and R4.

8 Solution V AB = 4V If V B = -1.5V  V A = 2.5V By potential divider, R1:R2 = 1:1, R3:R4 = 1:1 You can pick any value for resistances.

9 Solution If V B = -1V  V A = 3V By potential divider, R1:R2 = 2:3, R3:R4 = 2:1 If V B = -2V  V A = 2V By potential divider, R1:R2 = 3:2, R3:R4 = 1:2

10 Question: Resistance Calculation using Parallel/Series Combinations Find R eq and i o in the circuit of the figure.

11 Solution (i) (ii)

12 Solution (iii)

13 Question: Circuit Analysis with multiple sources Find v o in the circuit of the figure.

14 Solution Step 1: Define the node voltage (v 1,v 2,v 3 ) Step 2: Define the current direction

15 Solution Apply: 1) V = IR 2) KCL Step 3: Consider node 1

16 Solution Step 3: Consider node 2 Step 4, 5: From (1) and (2), v 1 = 30V, v 2 = 20V, v 0 = v 2 = 20V (v 2 -v 3 )/8

Rube Goldberg Machine (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. 17 START button Output Input Output Input Output Input Actuator to pop the balloon Input

Probing Questions for the Project Big questions  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  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? 18

Brief Plan for the Project A brief plan for the project construction  No technical requirement  A way to start thinking about the project  A block diagram with illustrations 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… 19

Design a reliable and robust machine because of the “Professor’s Negation Field” 20

Systems that You Have Built in Lab Sessions 21 Ball counting  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

Systems that You Will Build in Lab Sessions Light tracking  Lab 5 – Lab 7  The head can follow the direction of a light source  Decide your own light source triggering mechanism 22 Light

Sensors and Actuators in the Project 23 Circuit isolation Generation of air flow Generation of rotation Circuit isolationContact switch Generation of sound Generation of push force

Sensors and Actuators in the Project 24 Sensing of rotation Non-contact switch Time counter START button Non-contact switch LED

25 Mechanical Parts in the Project Metal ball  Stages triggering Metal pulley, sloted wheel eye hook, Wheeled castor  Pulley, conveyor belt, sling …

26 Mechanical Parts in the Project Aluminum cage (New Requirement) Plastic board Bracket Hinge Cable tie and cable mount  Weak fix joint between components

A Stage in the Rube Goldberg Machine 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. 27 Actuator Sensor Buffer Electrical signal Mechanical signal for next stage Mechanical signal from previous stage Mechanical part

A Partial Stage in the Rube Goldberg Machine 28 Sensor (Push) Actuator (Push) Sensor (Rotation) Actuator (Push) Buffer

A Partial Stage in the Rube Goldberg Machine 29 Sensor (Non contact) Actuator (Fan) Buffer Actuator (Light)

A Partial Stage in the Rube Goldberg Machine Ball counting circuit: A sensor only Light tracking circuit: Sensor + Buffer + Actuator (Not yet a complete stage) 30 Actuator Sensor Buffer Mechanical part

Stages in the Rube Goldberg Machine 31 Sensor (Button) Actuator (Fan) Buffer Actuator (Rotation) Mechanical parts Sensor (Rotation) Buffer (Mechanical parts in the second stage are missed.)

(Appendix) Discussions about the thermistor in Lecture 5 pp. 52 – 55 32 that page

33 R 1 = 80Ω, R 2 = 10Ω, R 3 = 20Ω, R 4 = 90Ω, R 5 = 100Ω Battery: V 1 = 12V, V 2 = 24V, V 3 = 36V Resistor: I 1, I 2, …, I 5 = ? (Appendix) Question: Circuit Analysis

34 Solution a V N = 0 I 1 : M  R 5  V1  R 1  B I 2 : M  V 3  R 3  R 2  B I 4 : M  V 2  R 4  B Step 1, Step 2

35 Solution b V M = 0 I 1 : B  R 1  V 1  R 5  M I 2 : B  R 2  R 3  V 3  M I 4 : B  R 4  V 2  M Let’s try another reference ground

36 Solution b I 1 : B  R 1  V 1  R 5  M I 2 : B  R 2  R 3  V 3  M I 4 : B  R 4  V 2  M Different direction, different result?

37 Solution b KCL of Node B: I 1 + I 2 + I 4 = 0 V B – V M = R 1 I 1 – V 1 + R 5 I 1 I 1 = (V B – V M + V 1 )/(R 1 + R 5 ) = (V B + 12)/180

38 Solution b V B – V M = R 2 I 2 + R 3 I 2 – V 3 I 2 = (V B – V M + V 3 )/(R 2 + R 3 ) = (V B + 36)/30

39 Solution b V B – V M = R 4 I 4 – V 2 I 4 = (V B – V M + V 2 )/R 4 = (V B + 24)/90

40 Solution b KCL of Node B: I 1 + I 2 + I 4 = 0 (V B + 12)/180 + (V B + 36)/30 + (V B + 24)/90 = 0  V B = – 92/3 V

41 Solution b I 1 = (V B + 12)/180 = –14/135 A = – 0.104A I 2 = (V B + 36)/30 = 8/45 A = 0.178A I 4 = (V B + 24)/90 = –2/27 A = – 0.074A

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(Appendix) Notes about Multimeters  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. 46

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-to- bottom 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. 47 (Appendix) Notes about Breadboard

(Appendix) Notes about Wire and Resistors Wire  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  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. 48

(Appendix) Notes about Potentiometer (or Pot) 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. 49

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. 50

(Appendix) Notes about Photoresistor 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). 51

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. 52

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