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Mechatronic Sys Design

by: Princess Rolfson

Mechatronic Sys Design ME 456

Princess Rolfson
GPA 3.57

Clark Radcliffe

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Clark Radcliffe
Class Notes
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This 18 page Class Notes was uploaded by Princess Rolfson on Saturday September 19, 2015. The Class Notes belongs to ME 456 at Michigan State University taught by Clark Radcliffe in Fall. Since its upload, it has received 65 views. For similar materials see /class/207528/me-456-michigan-state-university in Mechanical Engineering at Michigan State University.


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Date Created: 09/19/15
Input vs Output Last Chapter you experienced output 7 Turned an LED on and off 7 Saw active high vs active low control 7 Saw SSR for high DC amp AC power 7 Discussed low impedence output pins 0 In this chapter we experience input 7 high impedance77 is safer A First Pushbutton Circuit WM 0 Pushbutton pullup circuit 6065666 Hymn 3 thumw Test cum Bibiooiolt Your First Sensor 7 the Pushbutton Many highlow SVOV sensors 7 Proximity sensors 7 Switches 7 Tachometer 7 The pushbutton emulates them all 7 You can prototypetest with it The Pushbutton 4 pins but only 2 connections Flvum 11 V Nuvma 0901 h I 4 hmmu m Siluqu synod mm m par and w mm 0 4 pins provide Mechanical Stability whom B82 Pullup Circuit 0 Two States open and closed Flnurn 16 Puihhmmn Cum Cunnoce we no Pm ra my mummy Hm uummm cram o m 0 Open 10k pulls P3 to ground 0v Bsz Pullup Circuit 0 Closed Switch pulls P3 up to 5v Flaum 36 thbmun claw nocied o m Pm a nun connecth m uusnmm mm m vs 0 l 5vZZOQ 23 mA max to P3 usually 70 0 l 5v10kQ 05 mA to ground 0v CC 77 39 39 6 B82 Pulldown Clrcult 7O J 70 Two States open and closed 70 70 7 o 7Q 7 C e 17 Fblum m 0 mama Pushlmllnn Cllcull 7 r 0 Closed switch pulls P3 down to 0v o L L 7175v10k070 5 mA to ground 0v rlt3 74C Ca c ControlU 74 74 4c ibbbbbbb Now you can input and output cu 4mm rm 0 Use pushbutton to tell BS2 to ash LED 7 What s missing is the software 35600601 BSZ Pulldown Circuit 0 Two States open and closed 500M MMWM mm 39 mama thLulkm Clvzuu l v noun 0 Open 10k pulls P3 up to 5v Project Development Project development is stepbystep Develop to Debug 7 DO Go one change ata time 7 DON T Put it all together at once Look at this chapter 7 LED pushbutton test circuits 2 7 Pushbutton gt B82 test circuits amp software 7 Finally LED Pushbutton circuit amp software Flash LED Program Start with a flow chart Write separate code segments for each box 7 Andtest them toiiooibun l 006556 l w G 0 713 i 1 bbbbbbhtbl The Transistor N PN 0 A current amplifier 7 Makes m currentsBIGGER The 2N3904 Transistor 8 30400 0 u C Collector n B B T Base E mew E 39 e Emitter 66566 2N3904 Datasheet Posted on the ME456 Website 65565 66656 MMBT3904 PZTSQM v 35600003 NFN Gnnoral Puma Ampll or quotWild 1 NGUEWN I NGENI 740 quotC f2 Those L1ttle Black Chips iii Integrated Circuim 7 Expandpossible tasks for amicrocontroller 7 Substitute hardware for so ware 0 7 This chapter includes 7 The Transistor NPN 3 The Digital Potentiometer 5w quotHutu bat Important 2N3 904 Specifications it hhhh ote high maximum ratings NPN General Purpose Amplifier bbo blo 4y Humour BUT 200mA maximum current Maximum Thermal Dissipation 39 Imam V 5 Your package max dissipation P 065 W OO I Electrical Characteristics en the transistor is off y lbu Pawn 39IEI Cami Ilium Min Max Unlts 39 ERI I Ll SHCS rrliouivmm nuiwv 5 WM WW 4quot n I ITransistor will tolerate supply voltage on Collector v gt 4060 volts IExpect current leakage through collector 13912 ng lt 50 nA Galn39 lcallectar base ext says B 416 but that is a maximum Vypiul Puln oermm mm vs Coll r curnm mun Hugome lam cmcmkumznr a n B 100 is more reasonable lt3 4C Still A powerful driver 1 00 mA load driven by 1 mA i3 AND load powered by unregulated supply 9 Here 10 mA is driven by 01 mA Max mmmx rm 10 0 lmA r 60665 mmamm mmi nobioobout NOTE This IS NOT dependent on collector supply voltage fuse your 9v battery instead ofBSZ supply MIX Unlu Min Current Gain 30 lt lt100 Max CollectorEmitter Saturation Voltage 0 3 volt Note Power dissipated PVIlt 0 3volt0 2A 0 06 W O 2 AND load poweredby unregulated supply 75 g 9 Here 10 mA is driven by 01 mA zmr Amp lmA oten ometer produces voltages lt Vlt 5V01t humMr I With Transistor drive you can t 39 use the 9v battery instead of B 82 supply 7 Saves your 11mm 50 mA 1352 Supply Move Vdd connection to Vin j 7 anyuumdrivingal m More Important IO Move LED connection from Vdd gt to Vin 7 Save IO pin power 0 Circuit will work the same Now you are sugarchmgzd BE CAREFUL This can t be done with all circuits Care il analysis required it ll btcbbobo reboot rwyw hunt The Digital Potentiometer The Digital Potentiometer Rememe acts the same as a standard New potentiometer but controlled digitally The chip has 3 potentiometer connections 2 power connections amp 3 control connections The Digital Potentiometer he Digital Potentiometer cm the same as a standard potentiometer djusts the wiper to change the resistance to adjust the voltage at the wiper The tap is digitally controlled opens and closes l of 128 possible switches really transistors AD5220 Pin Summary A low signal to this pin enables the chip What do the bars over the symbols mean How it works Each element 78 125 ohms 128 78 125 10K ohm total With any single tap closed 10K ohm resistance is split amp varying Wiper voltage w mm m mm The Digital Potentiometer Circuit Which Vdd could be Vin Why i oobib babe l r w lo bbbbtthi bit Basic BOEbot Maneuvers Forward travel requires 7 Right CW lt750 andLeft CCW gt750 mi 1 Backward Forwnm Movement Timing TimeLoop FOR counter 1 TO 122 17mSec PULSOUT13850 13 1360 39 PULSOUT 12 650 20 0mSec lt pAUSE 20 1 7 mSec NEXT 24 7mSec bbthbb bb 05000 obiooiiui 3000 instSec gt 0 33 mSecirist 5 lines 0 33 mSecline 30 sec00247 SecLoop122 Loops AaNHySummaw 1 Program the BoeBot for basic maneuvers forward backward rotate leit rotate right amp pivoting turns 2 Tune the maneuvers for more precision 3 Calculate pulsesdistance for BoeBot travel 4 Make gradual BoeBot acceleration 5 Write basic maneuver subroutines lt3 do lt Go Forward for 3 seconds 70 2 2 EBUG quotProgram Running Signal program j ounter VAR Word startreset quotc3 REQOUT 4 2000 3000 1 3 Lg OR counter 1 TO 122 PULSOUT 13 850 Run BOEbot Forward What About This 39 PULSOUT12 650 Le Side13 CCW PAUSE 20 Right Side12 CW o NEXT 3 END coal What if you want another time l39i 3 Seconds gt 122 00247 Secloop ibbbblcbi 10 Seconds 10 Sec00247 Secloop 405 Loops Parallax uses 407 loops My estimate is off a little quotmum 003be What is Wrong with This Code ri 150 Hard to tell What it is doing Us es Constants rather than Symbols I SC r bi Let s make it Clearer FOR counter 1 TO 122 PULSOUT13850 PULSOUT 12 650 NJ b 306quotan or Better Forward 0 TIME CON 3 40510 Seconds 405 LoopsSec ON 650 Leit Servo Forward lt750 j o RFwd CON 1500650 Right Servo Forward gt750 Pulse rate is 50 pulsessec EEPROM Navigation Characters stored in EEPROM with do DATA quotFLFFRBLBBQ 39Navigation instructions 5 F In byte 0 O L in byte 1 F In byte 2 7 Etc 0 Retrieved with READ address instruction 39 Data at address in instruction If address 5 then instruction B nobuooiout A Few Constants Better Code T ME CON 3 41 41 LoopsSec LServo CON 12 Le Servo Port A RServo CON 13 Right Servo Port LFwd CO 650 Le Servo Forward lt750 s RFwd CON 1500650 Right Servo Forward gt750 FOR counter 1 TO TIME SOUT RServo RFwd PULSOUT LServo LFwd PAUSE 20 Pulse rate is 50 pulsessec NEXT END Going Straight f Your BOE bot drifts to the Left or Right Two Choices i L j g 7 Slow Down Fast Servo 7 Speedup Slow Servo Lo At High Speed 7 Better to Slow Down Faster Servo b s o Change Faster Servo s PULSEOUT closer to 750 s u 7 7 Or do both 0 PULSOUT LServo LFvvd1 Slows down Le Servo z PULSOUT RServo RFvvd1 Speeds up Right Servo o s EEPROM Navigation Varlahlei 17 p quot0rd 025 11 s u an s on Byte 1 Scores mason address nscrucclun VAR Byte 1 Scores mason nscrucclun 777 mason Dar 77 7 1 Address amassan Thsss mo tormented Jlnes Show 1 1111mm EEPROM address of ssoh datum on oquot usn mun nscruccluns i 7774 nsm nourms 1777777777777777 Address u no new address nstructlun i Data at address 111 nstructlon ss address 1 i Add 1 to address for rer read quottenth Digital Displays boooblb The 7Segrnent Display 7 Has 7 LED segmenm plus DP decimal point m R ii i l bonhacbocbbobo w G 0 J z pl ioinoiuio Cbbbbbbtbi no quotG Display a digit Programmed Display lt2 lt2 3 The digit 3 is formed by segments Wire Display to 10 pins 5 D c B A G i2 g r 1 0 0 yo 7 o J r Q gt 39 392 391 39 1 i A Z 3 Output Current Limit 56mm SWPMWD Max Current Sourced to Pms MW 0 Compute Source Current IvR5vl kQ5mA Sink 25mA Source 20mA bbbb hbhb x 0 So that the maximum source current 7 ach of two 8 bit ports OUTH OUTL can Sink a total of 50 mA Source a total of 40 mA i40ma 3560060 niini39nt IO pins are the OUTH byte OUTHXXXXXXXX Pins 151413121110 9 8 LED BAFGCDE Pin 15 Most Signi cant Bit MSB Pin 8 Least Significant Bit LSB The Lookup Function 0 LOOKUP Index Value0 Value ValueN Variable Find the value at location Index and store it in Variable If 7 Index exceeds the highest index value of the items in the list Variable is led unaffected iiib ttoOtttttiob Index is a variableconstantexpression byte Values are variablesconstantsexpressions Word Variable will be set to the Value at the Index location ompare Target value to a list of values and store the index number of the first value that matches into Variable unaffecte Z Onthe B82 BSZe BSst andBSZp the optional ComparisonOp is used as criteria for the match 0 the default criteria is quotequal toquot o quot uent Character set by 10 Pattern OUTH byte sets character OUTH 00000000 39OUTH initialized to lo DIRH 11111111 39Set P871315 to all outp 39 Digit 39 OUTH 11100111 PAUSE 1000 UTH 10000100 PAUSE 1000 OUTH 11010011 0 H N Lookup Example index VAR Nib result VAR Byte Init index 3 result 255 Main LOOKUP index 26 177 13 1 0 17 99 result DEBUG quotItem DEC index is DEC result END DEBUG prints quotItem Note that the rst Inthe 1ist above item 0 is 26 item 1 is 177 etc Lookdown Example value VAR Byte result VAR Nib Init value 17 result 15 Main LOOKDOWN value 26 177 13 1 0 17 99 result DEBUG quotValue matches item DEC result quotin listquot END DEBUG prints Value matehes item 5 in list because the value 17 matches item 5 of26l7713 10 1799 itbtbblbbo bbbbbthV w G 0 713 i 1 bit Audio Response 0 Small devices are not loud SPL varies with frequency 0 They are efficient small power required b bb 65 66 60655 i i More than a Square Wave PWM used to form analog signal level The speaker Acoustics of enclosure can be predicted Helmholtz Resonator tliabbbota l l Piezo Element til Programming Tones 393 1 2 REQOUT Pin Duration Freq1 i2 Frqu 4 f8 enerates a high speed PWM sinewave on 10 PM for period Duration at Frequency Freq1 Hz Simultaneous second tone at Freq2 Him OOOIVOUIOI DATA quotCquot quotc quotA39 y quotGquot etc at location Notes 0 9 m m E m gt H mm 9 3 o a m W O G Frequencles DATA Word 21293 Ward 21393 Ward 3135 Ward 3136 Ward 3520 Ward 3520 Ward 3135 quotbiotin Stores the wordlength frequencies ofnotes at location Frequencies 006066 Reading Stored Data 50 Use the READ statement and defined location symbols to retrieve data 1 66 t t r N FOR index I to 6 READ Notes index noteLetter READ Frequencies index 2 Word noteE req DEBUG noteLetter quot quot DEC noteE req CR NEXT NJ b Prints on the Debug screen ioinocuunb 0 L is 00 1n 7 p g 9 The For Next Statement A Hohh iobbbbbb 7 Loops with anindex FOR 1 o to 6 DEBUG 7 i The DO UNTIL Statement 7 Loops until acondition is satis ed no UNTIL letter v quot AD noteindex letter I index index1 66666 DCD computes 2 As you go down one octave 55565 7 Note frequency is divided by 2 m mt 60666 As you go down 11 octaves 7 Note frequency is divided by 2quot nobcooiout Complete Program a chxacancxallex e haamemm demaJasZ even notes mm mime mime Little Star mam 3135 Hum 3135 a 3136 mus quot mum s naceDuxaman naceFxE NEXT m4 naceFxEur n END 600 DCD 12 DEBUG bin 3 WI Displays w1960001000000000000 Note that 25 sem the Most Signi cant Bit MSB Where 15m 1111 is the largest 4 bit value niinibn DEBUGIN DEBUG Enter A DEBUGIN DEC A DEBUG The value of A is DEC A Indicator Lights l oobtb Ahtbb First external device 1 r An Output to human operator 7 Indicates binary condition I OnOff TrueFalse OKFault etc Your Indicator Light 7 the LED 7 Light Emitting Diode Makes other outputs possible w 739 G 0 3 i bbbbhth39 bit quot2 72 LED Current Limiting Resistor 9 9 Electrical Properties E Assume 5 volt supply Constant diode drop Of 0714 volts 7 Resistance Ohm Potentialcurrent Depends on color semiconductor used 72 R VI 5 volt10milliamp 5101073 7 Current Max about 15 ma Min about 5 ma 72gt 39 O I Get light anywhere inthat mnge 7 R VI 500 Ohm 7 Q TOO muCh Of ember39 j z I But there is no 500 Ohm Resistor 39 SMOKE 7 Use 470 Ohm Closestto 500 r c quotC dcquot Wrong Current Limit The Board of Education 2 With 5 volt supply and 1 volt diode drop Has connections directly to BS2 pins R 2201 I V43 51220 182 IRA No current protection for IO pins R 4703 I VR 5391470 85 mA f8 7 Too much current over 2025 mA 2 7R 690 I VR 5l690 58 mA 2 burns outIO pin I Either 220 or 470 are commonly used Be Careful 7 The B82 you save may be yours nobiooiout quotInfant The Homework Board 666666 66 Builtin 220 Ohm current limiters 7 Why 66 r 7 I To limit max current short on any pin 1 1 7 WR 7 5 volt220 Ohm23 mA 7 D 7 This protects the B82 components from an inadvertent short I 2 I When can this happen 7 7 Let s talk IO pins 7 O v Basic Stamp IO Pins Two Operating Conditions 7 Input or High Impedance I Used to sense level 7 Near 5 volts above 3 5 volts 7 True 1 7 Near 0 volts below 1 5 volts 7 False 1 I Input impedance is about 10M Ohm very high 7 I 7 5 Volts10X106 ohm 7 5 10x10397 Amp very small 7 Output or Low Impedance I Usedto set level lquot 5 volts 0 0 volts I No change in potential with current 7 UNTIL you overload pin 66666 Turning LEDs OnOff Two Methods Active High what the book does 66666 l 66666 LED goes on When P14 is High 6066606066 6663666666666 66 Hinton The Homework Board 220 Ohm resistors protect IO pins Serial Connedor 9 volt Bate Connector Reset valtch IO pin current limited to 5 volt220 Ohm 23 mA 66 666666 6666666 ill Huntor BSZ IO pin Overload Set IO pin to output then 7 Pin 0 connected to 5 volts gt OVERLOAD 7 Pin 1 connected to 0 volts gt OVERLOAD In either case IO pin is competing with power supply and one will lose 7 Usually the IO pin I Power supply 2 A pin 25 mA max 66666 66666666 quot66006quot Turning LED s OnOff Active Low Often Recommended Vdd 5v LED goes on Current 470 Q when P14 is LoW P14 0v lemmam Boom Many microcontroller pins WDI k best as sinks 36666666666 6 w 0 7 713 i 4 b bhth 39 bit The Piezoelectric Speaker Low current audio device 7 Best at high frequency gt 2k Hz World s simplest wiring diagram The Piezoelectric Speaker 0 How does it work 7 Let s look at the system Electrical Power Room Ac oustics quotdonut Beeps in Everyday life 0 All the Beeepps around you come from microcontrollers 7 Your phone 7 The microwave 7 Your PDA 0 Your kit has an inexpensive piezoelectric crystal speaker 7C P1ezoelectr1c Speaker 397 2 0 Low of sizes and shapes 0 Ceramic crystals de ect with charge 2 8 P q 0 5 mechanical strain q electrical charge As 4y bunHr 3900 Refa39mce Piezuelectm Suund Cumpunmts 7Apphmuun Manual Mum Manu ctunng CampanyFlSE pdf 63 a m 6666 The Piezoelectric Speaker Electrical Power is PotentialCuIrent What s the Speaker input What s the Speaker output 6 666666666 quot66006quot Power The Piezoelectric Speaker Electrical Power is n PotentialCurrent What s the Speaker input v What s the Speaker output I 7 O 7 4C 76 C C gt 50 tt r ttbbt Pvi V i Electrical Power totnttutt The Room Acoustics Acoustic Power is PressureFlowrate What s the Speaker input What s the Speaker output P pQ O H P Acousti c Power 7 4c P1ezoelectr1c Speaker Geometry 2 Structural Bimorph g 7 piezoelectric cemmic layer 7 7 metallic plate f2 Electric Charge gt Plate Bending J totttotttt The Room Acoustics Acoustic Power is PressureFlowrate What s the Speaker input What s the Speaker output PpQ Illlll ttbot iotnouto Acoustic wer The Room Acoustics Acoustic Power is wer 4 PressureFlowrate 7 Why is the small hole there s V g 7 An acoustic resonator ampli es Q response As i V P pQ 39 7 o P 0 Acoustic o Piezoelectric defection Piezoelectric strain deforms metallic plate De ection pumps air gt owrate Q into room 7 Room acoustics respond with Pressure p output Sound 054m t A my at1 c Chap 1 7 Act1v1t1es 7O I 75 743 Mostly a rev1ew Review of what you saw in What s a Microcoutroller 3 Look through it quickly 8 al wwy n 62an we Seamus yo lwlty 12 mum m a samw V mny u Scum up In Puldwaxu and mum uu sum 7 154 hulanu PA 15 Wm numtzqustllcma J m 7 391 39 I39 I 3 3 auw mm T 39 Might be new Look more carefully Then answer the Questions and Exercises at the r o end of the Chapter No Projects are required ac Connectmg Servos Book lo Basic Stamp and Servos g 7 6 volt power from 4 AA 7 Servos USE POWER 7 Watch your batteries MN iobbootott Today s lecture Robotics Chapter 1 7 Very short Robotics Chapter 2 7Anewservo toott tcbbtbb vb H Quinton Chapter 2 New Stuff ew material on special servos y g 41 mock up ma Emanuel munan Semo a m at Hum u Tune mid Roma Acton 42 Act 57 may Tm and Rapaamq Achequot um I Cram 45 mm 3 5 mm um Serw Maw 55 AchW M cameth the 5mm 6 mm 5 Heme Slwu anuui um own 7x mm m mm m 5mg 7 Swimmy us a z This servo output is r rotational velocity NOT o a rotational displacement Continuous Rotation Servo External Parts Proportional Control Control uses measured error to drive Motor Angle changes until error is zero The bigger the error the bigger the motor drive Some Important Facts No Feedback of Rotation 7 Not for Displacement Not for Velocity Desired Angle Controls Motor Drive 7 Motor Speedisn t controlled Because it is a si effect ofMotor Drive 39 O Q MotorD L Where D is drive and L is load 0 Speed is a Function ofDrive ONLY WhenLoad is a constant 0 Changing the Load Changes the Speed 0 Flat Floors and the Gearbox Minimize Load changes 600000 Displacement Servo Load L Control Drive 0 High Control Gain gt Small Proportional error K GH G Output P Desired7 Load 1KPGH 1KPGH u nn nu Displacement vs Velocity Velocity Servo Control Drive Out ut K I Angle Proportional Control edback Measured angle never changes 0 Control uses Desired Angle to drive Motor The bigger the DesiredAngle the bigger the motor drive 0 Velocity Servo Performance Gain rm 6606M i g 3 3 Mituu OUOIVOIOOI Linear Region about 725 775 PUL SOUT counts Velocity Servo LoadL d ControlD O L t BSWE Error rive u g1eR ie Kg 0 Angie 1 MotorGears Trim Sensor Hs 0 No Feedback gt Load Drives Output Directly Output KPG1Desired G1Load 7 KPGHlTrim quotintuit do 6 123 BOEbot The Screwdrlver Other Tools Requlred O 5 c Needlenose hers Optional Ugeful g 0 18 1 Phillips head required 40 77 a 14 2 won twork o 14 combination wrench Optional 6 r r r7 7 gt 18 dia r z I 39 SmalampSharp End B11intEnd Big BluntEnd r Buy th se at gout fgvor e tool store is1 1s2 ism Meuers ome epo 39


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