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by: Daron Kemmer IV


Daron Kemmer IV
GPA 3.63


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Class Notes
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This 38 page Class Notes was uploaded by Daron Kemmer IV on Thursday October 22, 2015. The Class Notes belongs to SPAN 6 at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 11 views. For similar materials see /class/226848/span-6-university-of-california-santa-barbara in Spanish at University of California Santa Barbara.




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Date Created: 10/22/15
ll ammo 105311 Eauouatiiluu lm i small am 3 The ATPG Problem Problem definiTion Given a logical faulT model and a circuiT deTermine a small seT of TesT vecTors ThaT deTecT all faulTs in The circuiT Problem complexiTy Under The sTuckaT faulT model The problem is NPcompleTe even for combinaTional circuiTs However commercial TesT generaTors ThaT efficienle generaTe TesTs for gt10MgaTe ckTs are in use Today uKYYmElug usmu inn Basics Path Sensitization Method ATomic operaTions 1 AcTivaTion fault exciTaTion Specifyinputs so as to generate the appropriate value at fault site for fault excitation Ie set 5 to 1 for SstuckatO fault 2 Error propagation specify additional signal values to propagate the fault effect from the fault site to the outputsobservation points 3 LinejusTificaTion Specify nputvalues so as to produce the signal values specified n 1 or 2 4 Value implication un que determination of values at other signals due to value assignments made in 1 2 or 3 uKYYmElug usmu inn Outline Automatic Test Pattern Generation TesT generaTion sysTems TesT generaTion for combinaTional ckTs DAlgorithm PODEM Boolean Satisfiability approach Test compaction TesT generaTion for sequenTial ckTs T niefranie expansion a Extended DAlgorithm on Issues of sequential ATPG uKYYmElug lumps um Test Generation Systems Ack Bushnell and Agrawal Essenllcll uf EleclrunlcTesllng 2000 BKY quotlining osmos inn 1 Fault activation A B c 1 2 Have a cho ce of error propagat ng through 65 or 66 a Propagat ng through G requ res 52 1 A b Propagating through 66 requires 54 126 gtTest ABCDE111x0 conTr39ud diequot 0 BKY quotlining osmos inn Line Justification a E sa12 E 0 Q C D 1 to propagate through 61 To propagate 64 need 62 63 1 Attempt to line justify 62 G3 1 631possibeifAF1orBH1 39 If A F 1 2 inconsistency since C 1so62 0 2Therefore 6312BH1 621needAOorF0 2 Tests arzlecoEl l BCDEH nKYYmElug mm m m Single Path Sensitization Is NOT Complete d saO 2 AB1 Propagate along 63 66 requires C 1 62 64 65 derfor641eitherEOor6102 inconsistency B 12 65 0 2 inconsistency Propagate along 64 66 2 E 1amp 62 63 65 1 62 12 C 1 ABC 1112 G3 0 inconsistency 2 No test WWW nsm u ll n T o p ths G3 62 and 64 66 are sensitized ie error is propagated along boil paths nKYYmElug mm m Completeness of ATPG Algorithms it will find a test for a fault if exists or prove that there exists no test given sufficient time Major complete algorithms for comb ckts Dulgorithm koth 1966 PODEM Goel 1981 FAN Fujiwara 1983 Socrates Schulz 1988 BooleanSATbmed ATPG Lurr39ubee 1992 m quotusing ammag m A test generation algorithm is deemed comp2f iff Complete algorithms can identify untestable faults But Boolean Difference Method Finds Test ABCD1111 slamming aim g m The D Algebra Need to be able to deal with multiple quoterrorsquot at the inputs to a gate D represents a signal which has value 1 in normal faultfree good ckt and value 0 in faulty ckt I e D 210 Similarly 5 201 0 0 0 Ul ZI UUl Cl G Cl C Behaves like a Boolean variable m quotusing ammag m PrimiTive Dcubes Specifies The minimal inpuT condiTions which musT be applies To a logic elemenT Ein order To produce an error signal aT The oquuT of E Propagation Dcubes The propagaTion Dcubes of a logic elemenT E specify minimal inpuT condiTions which are required To propagaTe an error signal on an inpuT or inpuTs To The oquuT ofThaT elemenT uKYYmElug assmu inn An Example GlsaO Primitive cubes 01 35 exnenx ekr quotholeg Hammig inn e e1u11121a 1 2 a 5 a 7 WlmltlveDrcube 1 1 D primniyeauueseiea u gtlt gtlt 1 gtlt u gtlt 1 gtlt gtlt u 1 1 1 1 u enmnweauueseies 1 gtlt 1 gtlt 1 1 u u u Primitive cubes 039 e Primitive cubes 039 es The DAlgoriThm 1 Select a primitive Dcube of the fault 2 Implication and checking for inconsistency If inconsistency occurs go to D in its inputs to its outp t D 39onfierconsisTs of set of all elements whose output values are unspecified but inputs have some signas with D or D Ddrive is donegy intersecting the test cube with a propagation Dcube of the selected element Backtrack ie select another propagation Dcube if ntersection is null 4 Imgicafion ofDdrilz perform mplication for the new test cube 5 kepeuT 3 54 until faulty signal propagated to an out ut 6 Line gustmotion Consistency check on input conditions required uKYYmElug assmu inn 1 3 Ddrilz selects an element in Dfrontier 51 attempts to propagate D or 39 39 39 u DAlgoriThm Example ekrrnciew nsrwmh a inn 12345678910111213 lrlltlaltestcllbetl 1 1 D propagation Dcube ofGa D 0 D testcubeafterDdrlve 1 1 D 0 D through 33 tc performlrnpllcatlorl 1 1 1 1 D 0 D Check implicaTion Dcube of 61 does noT imply any oTher signa DfronTier 6 GeT propagaTion Dcube for G3 uKYYmElug assmu inn Now DfronTier is 65 At 66 SelecT 65 and a propagaTion Dcube of 65 1234 910111213 perform implication 1 1 1 propagation Dcllbe of 35 Test cube afterDdrlve 1 1 1 Llrlelllstlflcatlorl1 1 1 1 1 1 1 D 1 D 1 D 1 1 D Flowchart for D algorithm Initialize test Cube tc Select a primitive Drcube at that as c Deintersect c With previous test cube tc and erform imlication onSiStEnt is there D 3 HQ on anyP Backtrack to Select a gate from gation a ate and a pro 3 W 5 5 of the selec e uKYYmclug ism iiii immune Line Justification C to us if the line Intersect C with previous test cube c consistent Backtmck to the last Oint a choice exists Line ustification impossible ism g iiii i 2 3 a 5 annNeDmbEU lmPllc marl 2 A D Algorithm SeleciDlmnillszc l 1 1 g D 11 11 implement i i D i i i D D D implicaiiaMS X n l i i i D i D seieeiDnaniieicxtD 1 D 1 1 E i i i D D i Di i D lmpllcailan i D i tineiusineeiianm i i i D D D i D i i D 2 l Pllcaila D D i LlllelMSWlEBllWlUl i i D D D D i i i lmpllcailari l 1 i D DDe i D i i D i D i meansisienie backivackia51 i i D D i D seeeiDtmniei D D 66 i i i i D D i D D iwiieeiiantiu 1 1 i i i i D Tesilsimmd iiii immune ism in ii ii immune Potential Problems with D Algorithm Since the assignment of values is allowed to internal lines more than one choice is available at such internal linegate and backtracking could occur at each gate Could result in inefficiency for large ckts and some special classes o Example An ECAT errorcorrectiondttranslation ckt ism g iiii The PODEM Algorithm Goel 1981 Only allows assignment of values to primary inputs The values assigned to primary inputs are then propagated toward internal lines by the impi a on Example First a binary value is assigned to an unassigned PI to provide a fault effect at fault site A 1 z Determine the implications of assigne P15 0 y AB1implyHD uKYYmclug ism iiii Test has be Successfully generated immune PODEM Decision Tree for the Example Start ism g iiii PODEM EssenTially a process of finding a PI At a binary value for iniTial assignmenT error is being propagaTed To oquuTsa1quotTer each PI assignmenT perform forward implicaTion If aT any sTage eiTher The faulT cannoT be exciTed or The error cannoT be propagaTed furTher backfrackTo The mosT recenT PI assignmenT and change iT uKYYmelug osmium viii ConTinue assigning PI values checking To see if The Biathlong STeps in P0 DEM 1 DeTerrnine an initial obiecfile If The faulT effecT has noT appeared aT faulT siTe The iniTial objecTive is direcTed Toward providing The faulT effecT on The faulTy line 2 Given The iniTial objecTive a PI At a logic value are chosen ThaT have a good likelihood of meeTing The objecT Done using The battrace procedure ascm on vi n Biathlong A POD EM Example Initial objective 0 62 Backtrace to PI x2 1 Initial objective 0 62 Backtrace x 1 Implication 52 6 Dfrontier is 65 66 Attempt to propagate through 65 kequire X 1 Implication e 0 64 Attempt to propagate D through GE ascm on vi n slamming FlowcharT of PODEM ssign a inary w Lie to an unassined PI Determine im lication of all PIs sther an Lkntriled Com ination o Va ue on assigned P157 Set untried combination of values on assigned PI nsrwmh g viii ob ective slamming Flowchart of Backfrace egm value OR NAND 1 V0 1C A DNOR 1V1 and is the easiest to Control Next Ob line is the in ut of G W K is at X anoi is the hardest to control ext Ob wine is the same as the Current objective value nsrwmh g viii A PODEM Example Con r39d Initial objective 1 66 BuckTr uce To seT X Impl cation63 1e Impl cation 53 o 66 slamming 4 ea 1 failed in propagating error Backtrack to most recent PI assignment reassign xi D ea D Test is generated 0 X l A test is uund ggg uggc nsrwmh g viii Cost of ATPG A How long Time complexity B How much RAM Space complexity C How many vectors generated Test application time Theoretical result Ibarra amp Sahni 1975 Generating atest for comb ckt is NPcomplete Worstcase time cc constant5 6 of gates Emperical result averagetime behavior Total ATPG time cc constant 62 Test length cc 6 immune osmium um Socrates Static Lea rnlng Preprocessing the ckt 1 Assign a logic value to a certain signal of tlie ckt 2 Perform all implications from that assignment 3 Learn from the results of implications Using law of contraposition A s B 12IB IA a X b t 1 e Learned a1f1lt2f0 1nformation s X b Preprocess rig During ATPG immune osmium um Step 1 Extracting the formula Each node of the ckt is tagged with the logic formula in 3element conjunctive normal form or 3CNF The formula is true iff the values assigned are consistent with the truth for the logic element C c EA EE CAE 39 0K CE EAE cA cX ammonia nsrcm no it n Accelerating Comb ATPG Basic goals 39 Reduce number of backtracks 39 Reduce processing between backtracking Basic tools 39 Topological analysis 39 Multiple backtrace 39 Learning eta quotlining manna um Boolean Satisfiability Approach Given a fault it consists of two steps Step 1 Construct a formula expressing the Boolean Difference of a circuit with respect to the fault Step2 Apply a Boolean Satisfiability SAT solver to the resulting formula eta quotlining manna um Step 1 Extracting the formula Cont39d a Construct the formula of the good circuit output x 63 faulty c rcult output X tor fault D bConstruct ormula of o no need to repeat tne part identical to a stuckeate Blamingg nsrwmh g um Step 1 Extracting the formula Cont39d c Construct the formula of the Boolean Difference XOR of a At b and the output of XOR should be 1 X l Bb7gtltgtlt7vVZ1 39 From a x a XEgtltEampDE5A5BDEjclza 39 From b XEagtcEYb39eb39 39 From c 35 ii if rm BDVV2BD I I noteBD 7gtltgtltvv2 uKtnnelug n53me up Conjunctive Normal Form CNF f W n lallh tc r 00mm n l uKttmelug osmium inn Basic Backtracking Search 00mm n l Step 2 Satisfying the Formula Boolean Satisfiability Given a suitable representation for a Boolean function x Find an assignment Xsucll that 4 1 Or prove that such an assignment does not exist Le x o for all possible assignments In the classical39SAT problem rm is represented in productof sums P05 or conjunctive normal form CNF J a u u i Y I r Y or indirectly in terms of Boolean Satisfiability uKYYhElug asean up Literal amp Clause Classification Sauioe Ju ulhlquesr lvzznd KalemA Sakollon quotBoolean Sallsllzbllllylrl EDA cszooolulaiiol BKY mains osmium inn Unit Clause Rule Implications An unresolved clause is ml if it has exactly one unassigned literal 4 11 cb A unit clause has exactly one option for being satisfied abaac ie cmust be set too r 000ml n l uKt mains osmium inn Testability Analysis Test To glvE an Early Warnlng about thE Estlng problEms that llE ahEad lt pro 125 guldanCE tn mpto ng thE Estab ty mquot a I ISJJI 5 To b2 usEful Establllty analysts s ould b than actual Est gEnEratlon ot fault 2 slmplEr lrnulatlon a Topologlcal analysts only thE struc tum of ckt ts analyzEd NotEstvECtorsarE g at nEr 2d a LmEarComplEXlty T hE analysts should b21ln2ar or almost llnE at tn ottoutt sle Ismmmm 2cm SCOAP Sandia Cnnlrnlabilily Observabilily Analy 39 ccn tax CC N thE numbEr of oombtnattonal nodEs that 515 Program musth asslgnEd valuEs to Justlfy a0 ot a 1 on no 2 N so tgzsom thE numbst of sEquEntlal nodEs that GoldstEln IEEETrans oncAs1979 mustsm Ass th o Justlfy a o ot a 1 onnodE N Establllty of a ottoutt ts ElatEd to thE dttttoulty of oonttolltng thE loglcal valuEs on nodEs from Z D valuEs atPOs EnCmEasurEs of thE dlfflculty SCOAP ComputEs slgtlt numbEr tot CC N ampCC N cornbtnattonalo 391 Eac nodE tax 1 oonttollabtltty Y Z mm5Cquotgtlt1t5Cquotgtltz no 2 N sc m sc gtlt sc gtltz scum ampSC N sEquEntlal 0 tax 1 oonttollabtltty of nodE N CON ComblnatlonalobsErvablllty soN sEquEntlal obsErvablllty woman 2cm Ismmmm 2cm ccnm com com ccnm 1 Si ComZmmmowcowGem 1 Controllablllfy Examples X Xt Xt sew sc J 50st sc J ma Cm a sc m mmlsc lxt sc lxtltsc lxtll z momsmtcoatst mom I 00111 cm i o GOV Dbl ml Th2 obsErvabllly of x ts afunctlon of ths output am 10 obsEvablllty and of thE cost of holdlng all othEr tnputs at o DLCW 139 m 7 9 quot Wquot CCzzmmycclyscC11n 1 colxt com 1009 ccnw 1 n a 1 cculzmmln zacc yb Cm an act DIM ti 50th 5000 5009 SC X1 7 col 1zzmmlccy la con volt can i CE bl t Boundary condtttons n 1 1 5 1 0002ka 121 551 15 1 cc cc 1 t2 50 SC 0 for all P15 7 5012mmlccaaltvcalblv t 2cosoo forallPOs Tam Anus 2255 s madam 2255 t More Comrollabilify Examples 5 1 can 1 7 mm 1551 a 51 my 1 n cctlzlomtajoccamjn a 2 one 121 a mm cm M ca 1 55013 cm mu c 1 p 521117 mm 155a lap cm 12 cm 151 mun1 1 1 66012 55115 cm 121 05171 1 11 Tammany 2cm Observabilify Examples To obszrvz agavz m u Obszrvz output and cm a on a 7 co 1 cm a c 1 a cc 51 compcmzpccuapx 9mm 011 15 so a 7 ca 1 4 can u c 1 m quot on m a on 1 can a 1 1 0015 so 1 mm cm W 00110 1 a I cum ca 1 5515 1 p 1 mm mm mpm vamcs newcomrong CO 5 2 wbmmncm 31 DED CO 5 00 I o 001 b v 1 al 1 b an I z More Observabi lify Exam ples 0 Observe a fanou 1 Stem Observe 11 through branch Wm best obscrvabmy c015 7002 com 1 5 2 mm mm 1 DDo 0015 ca 11 11w can M 511 a v 1 5 Error Due To Sfems amp Reconverging FanouTs measures wrong y assume that comraHMg or are 1nd2p2nd2m 2v2ms X 0 zjcorw atz 7 m x any zcorr2a2 7 0 x mm 0 z corrz We 1 comcmgcmm1a 2 W1 cum CDfalCOzvY s mmmmucom cm mm y mm m Correlafion Error Exam ple Exact compuranon of mcasurcs 1s NP7comp212 and 1mprac11ca Imhmzzd green measures Show correctva uzs 7 scow measures are m red or b0 d CCU C51 CO W51 1 t x 234 c 2m 1N1 cum 115 2cm Applicafions of Tesfnbilify Measures Can b2 usEd to gde mg 5212mm of Est pomts for mpmvmg Estabxlxty Can b2 usEd to gde mg sEarch m Est gEnEranon 7 mp121u52dto d c NotE that Estabxhty mEasurEs am only appmmm Con ErgEnt fanouts CausE maccu racy nmmm 2cm Atmospheric and Oceanic Circulations con nued Chapter 6 Lecture 15 9 February 2005 The oceanic conveyor belt L 39 quot1quot the time the gulf stream has lelivered warmth to Europe it is ready to sink Why The G8 water is colder it released a lot of heat amp saltier due to evaporation as it travels northward It is therefore more dense than surrounding water amp sinks The oceanic conveyor belt Labrador 53953 Heat transfer Gum wa tar In alr dmvnwelling Cu rr en t Deep cmd circulation Absurbs heal lmrn air Warm shalluw current Winter sea ice cover u If Stream North Atlantic deep water V A Figure Credit Earth s Iimate by W Ruddiman Depth km Figure Credit Earth s Iimatequot by W Ruddiman Latitude Ocean Weather is Important Too T 55 5mm Hm nm 2 25 Mesoscale 25 eddies 24 23 Dominate KE ofoceans 22 21 El 19 Hiquot 152 451139 5 155 454 AL HA Gutrent 22 Aug 01 tn 03 Nov 04 weekly imagest must HHDT IIIIIIIIIIIIIIIiiIIIIIIIIIII Hill Ell 22 24 Chapter 7 Water amp Atmospheric Moisture Water on Earth the big picture Water is most abundant compound on Earth s surface Covers 72 of the area or 35 x 1014 m2 Total volume of water on Earth s surface and in the atmosphere is 14 billion km3 or 14x1024 9 where did all the water come from Credit I39mpllpegasusphas1umassedula101Amagesearthalmsourcesjpg Water on Earth the big picture ll water cycles through the atmosphere very quickly the residence time is on the order of a week or two how to calculate this Well how long is a student is at UCSB 20000 students 5000 new studentsyear 4 years time stock flux for atmosphere 13x104 km3 water 49x105 km3year rainfall 003 year 11 days Where is it AIII wamar 1111033 More than 97 of all water is in the ocean Water in the world s oceans Earth39s cean Area V39uluirne Mean Damn uf Ewan Areaut la mmmil km3mi3 Main Basin m 111 Pacific nilEl 1 E39Q a EILBED 428C aa mi 1 tamm 14040 Atlantic 28 13543 355250 393i 4111130 55200 1290 Indian 20 M 292310 3950 23930 if 1 BIG 129w Arctic 4 391 4094 11DD 1335 5440 4100 3950 Data in thousands 300 includes all marginal aaaa Pacific Ocean covers more than 13 of Earth and nearly 12 the total water Hydrological Cyclel Clouds drift inland From the ocean V 1 1 e l A Evaporation from nvers lakes Prec39P39mt39Dquot as Snow and glaciers and transpir on animals Evaporation am the ocean l In ltration below V ground into soil and rock features Groundwater ow Where is it All water Freshwater Surface water 100 273 of all percentage if v H Ice and glaciers 1 x 9935 Freshwater lakes quot Ocean x quot K I 033 g 9722 x quot quotaquot 7 39quotquot Fresh Saline Lakes 7 39 39 39 r 28cy h Io 39 39 Atmosphere 39I 003 l I Soil moisture Groundwater Boar 39HWETS 3392 Streams 313 1 1 2 groundwater 1093 6 J 1102 va l Percentage of 39 J surface water Table 71 ml less than 001 of Earth s water is found in rivers streams and freshwater lakes there is about 10x as much water in the atmosphere as in all the rivers and streams Freshwater on Earth Fernanng Percentage umeshwnter ofTuul Water Location Amullm 1900mm Surface Wm k4 shuns an TAIWAN 39 U Fruslnuu Armmphum RN er a and 0mm Tum mr we Lam39 70sx400 ubsurfnc arel Groumlwmerisu ram 0 712 m 0 m 0 21 70000 1000000 0 300 n 00 100770 In 1 m 100 m H000 Ir 0010 0 I 70000 0000000 00 s Imnnmmmmx 07000 10000 000 Torn suhmr guntgr 7000 r0000 om Tn F shwmrrmun0e0 37000000 ma0000 10000 173 swam Am kmlh39nizI szor Fmshwnm Lakm Vniume rkm hni n 17000 n Physical Chemistry of Water water weighs 1 g cubic centimeter 1 gmL about 8 13 lbsgallon 780x more dense than air Water is the universal solvent dissolves many compounds rare to find in its pure state Water s unique properties are due to hydrogen bonding liquid press ure Mint Iltllt39l39llilill rmding murmlul tniliuu ruti m UHIEL39HJ trim phase diagram for water 113m no I I H Liquid water Water vapor 0 Pressure atm 001 0001 00001 Mercury daylight sideLL O 300 200 1OO O 100 200 300 400 500 A Temperature C Figure Credit Earth s Iimatequot by W Ruddiman boiling point water boils when the saturation vapor pressure equals the air pressure P of Surroundrhgs P of evapratrhg hqurd much 1 r Ehrrr g mm 92 c In M Eumrum mum r m I 4 I 9 1 9 2 b A r A X v0 rm m mmr u H I I I Hwhur r r qr w r m uh uh I run http 4 w 1 mmrmmrmw w 1 Credit jumwgeogucsbedu39oeIg110w03chaptO4humidiw g05st1jpg Water is a polar or charged molecule charge on the hydrogen side amp a charge on oxygen hydrogen bonding is the attraction between oxygen and hydrogen atoms on adjacent molecules ji Cin V Hydrogen Bonding 5 H amp pdarcovabnt bonds v W 5 H 5 Credlt http bl0 Wlnona rnSuS eduberglLLUSTHybond5 glf Phenomena due to hydrogen bonding surface tension cohesive force that pulls a surface closer together water has highest surface tension 7 cagilla action ascent of water up a tube thanks to the pull of water molecules on one another W a H Water mutecute amp polarity a Gas V vapor Freezmg Q Mening I JV Latent heat changes due to ph changes are very important for atmospheric circulatipn Latent Heat of Phase Change Called that as heat energy is latent within the water latent heat of vaporizationevaporation the energy required to convert liquid water to water vapor latent heat of condensation the energy released when water vapor condenses to liquid water latent heat of freezingmelting energy required released when ice melts to liquid or vice versa latent heat of sublimation the energy required to convert solid water to water vapor 720 calories required gt Lamean heat 01 Latent heai DI malming uapo zatiun 7 30 caknries abswbedl HUD calories absnmedi 540 calories absurbEd 1 U salaries released 54 calmies released Latent heat at condensation 80 calun las released Latent heat of freezing lt 720 calories released 39 Latent Heat Exchanges from Liquid to Vapor my evaporatiun Inf 1 gram 1 warm 140 COOLING CLIMATE Calories released Calories warervapcr released Calories Va 120 condenses to water r abmbed Immhear vaporcouls 100 vaporforms Calm Calories absurbed turning water into water vapor so Amie 2 latent heatofvaporization o 17 w r g 39 and Calories absorbed e V as wemr warms 2 PM 40 from 0 010039C w 39 20 Ice forms 5 WARMING CLIMATE u gt Calories absorbed meldng 39 ice latent heatofmelring Ice 3920 Calories absorbed warmlngsubfreezlng Ice 40 O 100 200 300 600 700 800 900 400 500 Calories per gram gure Credit Earth s Climatequot by W uddiman Measures of Humidity humidity refers to water vapor in air relative humidity ratio of the water vapor content of air to the maximum possible water vapor content at that temperature dew point temperature temperature at which air is saturated with water vapor specific humiditv mass of water vapor 9 divided by the mass of air kg independent of changes in pressure volume and temperature r airv greater capacny lo hol water Cooler air lesser capacny to hold waier vapor 20 50 100 relative relative relative humidity humidity humidity 5 PM 11 AM 5 AM How much water can air hold think of this like the amount of sugar you can dissolve in coffee or tea before it precipitates out Saturation Vapor Pressure vs T 80 saturation vapor pressure is the maximum amount of water air can hold 70 60 50 rule of thumb slope is 15 mbar C in typical ambient rang 20 15 10 5 0 5 10 15 20 25 30 35 40 Temperature C 40 30 20 10


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