Circuits 1 EE T220
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Date Created: 09/12/15
Minimum Spanning Trees Yan Liu LDCSEE West Virginia University7 Morgantown7 WV yanliucseewvuedu 1 Statement of Problem Let G lt VE gt be a connected graph with realvalued edge weights w 39 E A R having n vertices and m edges A spanning tree in G is an acyclic subgraph ofG that includes every vertex of G and is connected every spanning tree has exactly n 7 l edges A minimum spanning tree lVlST is a spanning tree of minimum weight which is de ned to be the sum of the weights of all its edges Our problem is to nd the MST of G 2 Minimum Spanning Forests Some algorithms we present here will deal with subgraphs that are not necessarily connectedi Thus7 we introduce the de nition of minimum spanning foresti ll A forest F is an acyclic subgraph of G that consists of a collection of disjoint trees in Gi 2 A spanning forest F is a forest whose trees are spanning trees for the connected components of the graph Gt 3 A minimum spanning forest F V7 E is a subgraph of CV7 E such that the sum of all the weights we for all the edges e in E is minimal 3 Deterministic Algorithms 31 Kruskal7s Algorithm Algorithm MST Input GVE Output T7 the MST 1 T lt7 empty graph 2 for lto lV do 3 Let e be the minimum weight edge in G that does not form a cycle with Ti T lt7 T Ue 4 end for Algorithm 31 Kruskal s Algorithm The only difference between these two algorithms is that Kruskalls Algorithm doesnlt require that the edge e to be connected to the evolving tree T7 which makes T be a forest other than a tree The Kruskal7s Algorithm also runs in Omlogn time 32 Boruvka7s Algorithm There is another greedy strategy for MST called Borquals algorithm which also runs in Omlogn time Later we will show that using randomization in conjunction with the algorithm leads to a lineartime algorithm Borquals algorithm is based on the following lemma Lemma 31 Let v E V be any vertex in G The MST for G must contain the edge vw that is the minimum weight edge incident on v Proof Suppose that vw the minimum weight edge incident on v is not contained in the MST T of G then we must have another edge vu in T such that vertex v can be covered in T By adding vw into T we will form a path of cycle in T which passes v Since vw is the minimum weight edge incident on v we can remove vu and thus keep vw in T resulting in that the weights of T is less than that of T This reaches the contradiction with that T is the MST of G Therefore the MST for G must contain the edge vw D De nition 31 The basic idea in Boruvka s algorithm is to contract simultaneously the minimum weight edges incident on each of the vertices in G In a contraction only the minimum weight edge needs to be retained out of any set of multiple edges The process of contracting the minimum weight edge for each vertex in the graph is called the Boruvka phase A good implementation of a Borqua phase is the following 1 mark the edges to be contracted 2 determine the connected components formed by the marked edges 3 replace each connected component by a single vertex 4 nally eliminate the selfloops and multiple edges created by these contractions Lemma 32 The set of edges marked for contraction during a Boruvka phase induces a forest in G Proof From the steps listed above we can compute the running time of implementing such a Boruvka phase Step 1 takes Omn time in step 2 and step 3 we may use a DepthFirst Search DFS to nd the connected components and create a new vertex that correspond to each connected component and associate each vertex with that new vertex all these take Om n time nally step 4 takes 0m time Thus the whole process of a Boruvka phase needs Om n time D Lemma 33 The set of edges marked for contraction during a Boruvka phase induces a forest in G Proof By Lemma 31 each of the edges marked for contraction must be one of the edges in the MST of G During each Boruka phase the edges we marked is a subgraph of the MST of G Therefore these edges might not be connected but must be acyclic This makes them form a forest of G D We claim that the graph G obtained from the Borqua Phase has at most vertices This is because that each contracted edge can be the minimum incident edge on at most two vertices The number of marked edges is thus at least Since each vertex chooses exactly one edge to mark it is easy to verify that each marked edge must eliminate a distinct vertex The number of edges in G is no more than m since no new edges are created during this process By Lemma 31 we know that each of the contracted edges must belong to the MST of G In fact the forest induced by the edges marked for contracting is a subgraph of the MST Lemma 34 Let G be the graph obtained from C after a Boruvka phase The MST of G is the union of the edges marked for contraction during this phase with the edges in the MST of G Proof By Lemma 33 we know that during each Boruvka phase the edges we marked is a forest of G and these edges belong to the MST of G Each marked edge eliminate a distinct vertex of G thus the uncovered vertices are still remained in G The union of the MST of G with these marked edges will cover all the vertices with no cycle Furthermore since the MST of G will also be induced by Boruvka phases the MST of G is also a subgraph of the MST of G Therefore the union of the edges marked for contraction during this phase with the edges in the MST of G is the MST of G D Algorithm MST Input CV E Output T the MST T lt7 empty graph for each v in V do Let e be the minimum weight edge in G that is incident on v T HT e end for G lt7 G with all edges in T contracted T lt7 recursively compute the minimum spanning tree of G return T T 3999 57quot Algorithm 32 Boruvka s Algorithm Boruvkals algorithm reduces the MST problem in an n vertex graph with m edges to the MST problem in an gvertex graph with at most m edges The time required for the reduction is only Om The worst case running time is Omlognl 4 Heavy Edges and MST Veri cation Before describing how randomization can be used to decrease the running time of Boruvkals algorithm we develop a technical lemma on random sampling of edges from grap Let F be any forest in graph G and consider any pair of vertices uv E Vl Let wF u v denote the maximum weight of any edge along the unique path in F from u to v If there is no path exists then wFuv 00 Note that if an edge u v exists in G the normal weight of this edge is denoted as wuvl De nition 41 An edge u v E E is said to be Fheavy if wuv gt wFuvl The edge uv is said to be Flight if wu v S wFu vl In particular all the edges in F must be Flightl An edge u v is Fheavy if the forest F contains a path from u to v using only edges of weight smaller than that of uv itself Lemma 41 Let F be any forest in graph G If an edge uv is F heavy then it does not lie in the MST of G The converse is not true Proof By the de nition of Fheavy edges we know that F must contain a path from u to v using the edges of weight smaller than uv Suppose uv belongs to the MST of G then by replacing uv with the existed path in Ffrom u to v into the MST we can obtain another MST of G with the smaller weights Therefore there is contradiction thus uv cannot lie in the MST of G However F may contain isolated vertex say u There is no path existed in Ffrom any other vertex to u By the de nition wFuvZ 00 vi 6 Vu and u vi is F light It is easy to see that for a certain edge not in the MST of C it might be F light Thus the converse is not true D A veri cation algorithm for the MST can be viewed as taking as input a tree T in a graph G and checking that the only Tlight edges are the edges in T itself This is equivalent to verifying T is the MST There exists lineartime veri cation algorithms that can be adapted to identify all Fheavy and Flight edges with respect to any forest F The performance of these algorithms are summarized in the following theoreml Theorem 41 Given a graph G and a forest F all F heavy edges in G can be identi ed in time Onm 5 Random Sampling for MSTS In order to reduce the number of edges in the graph we will use the fact that a random subgraph of G has a minimum spanning tree To be precise consider a random graph Cp obtained by independently including each edge of G in Cp with probability p The graph Cp has n vertices and expected number of edges mpi Note that there is no guarantee that Cp is connected Let F be the minimum spanning tree for Cp We expect very few edges in G to be Flight such that F would be a good approximation to the MST of Gi This expectation is explained in details in the lemma presented belowi First we recall some probability de nition and theory De nition 51 Let X1 X2 in X7 be independent random variables whose common distribution is the geometric distribution with parameter p The random variable X X1 X2 X7 denotes the number of coin flips needed to obtain n HEADS The random variable X has the negative binomial distribution with parameters n and p The density function for this distribution is de ned only for z nn 1 n 2M PrX z I 1gtpnqrin nil The Characteristics are EX varX g and C2 0322 De nition 52 For independent variables X and Y we say that random variable X stochastically dominates variable Yif for all 2 E R PrX gt 2 2 PrY gt Proposition 51 Let X and Y be random variables with nite expectations If X stochastically dominates Y then EX 2 EY equality holds if and only if XY are identically distributed Lemma 51 Let X have the negative binomial distribution with parameters n1 and p and Y have the negative binomial distribution with parameters ng and p For n1 2 n2 X stochastically dominates Y Proof Without loss of generality we assume that 2 is a positive integer PrXgt2 17PrX 2 17PrX2PrXziluPrXn1 PrYgtz17PrY 217PrY2PrYziluPrYn2 For n1 2 n2 PrX 2 PrX 2 71 H PrX n1 PrY 2 PrY 2 7 l H PrY Therefore PrX gt 2 2 PrY gt D Lemma 52 Let F be the minimum spanning forest in the random graph Cp obtained by independently in cluding each edge of G with probability p Then the number of F light edges in G is stochastically dominated by a random variable X that has the negative binomial distribution with parameters n and p In particular the expected number of F light edges in G is at most Proof Let e1 e2 be the edges of G arranged in order of increasing weight Suppose that we construct Cp by traversing the list of edges in this order flipping a coin with probability of HEADS equal to p for each edge and including an edge ei in Cp if the ith coin flip turns up HEAD The minimum spanning forest Ffor Cp can be constructed online during this process as following 1 Initially F is empty 2 At step i after the coin is flipped for ei uv ei is chosen for Cp then ei is considered for inclusion in F 5 The edge is added to F and only u and v belong to di erent connected parts of F Given the order of the examination of the edges by the de nition of Flight edge an edge is F light when examined if and only if its endpoints lie in di erent connected components he crucial observations are 1 the F lightness of ei depends only on the outcome of the coin ips for the edges preceding it in the ordering 2 edges are never removed from F during this process 5 and the edge ei is Flight at the end and only it is Flight at the start of step i De ne phase k as starting after the forest F has h 7 l edges and continuing until it has k edges Every edge that is Flight during this phase has probability p of being included in Cp and hence of being added to F The phase ends exactly when an F light edge ia added to Cp for the rst time during the phase It follows that the number of Flight edges considered during this phase has the geometric distribution with parameter p Suppose the forest F grows in size from 0 to s Then the total number of Flight edges processed till the end of phase s is distributed as the sum of s independent geometrically distributed random variables each with parameter p To account for the Flight edges processed after that but not chosen for Cp we continue flipping coins until a total of n HEADS have appeared The total number of coin ips is a random variable which has the negative binomial distribution with parameters n and p Since s is at most n 7 1 it follows that the total number of Flight edges is stochastically dominated by the random variable which represents the total number of coin ips The expected number of Flight edges is bounded from above by the expectation of this random variable which is g D 6 The LinearTime MST Algorithm The randomized linea time MST algorithm interleaves Boruvka phases that reduce the number of vertices With random sampling phases that reduce the number of edges After a random sampling phase the minimum spanning forest F of the sampled edges is computed using recursion and the veri cation algorithm is used to eliminate all but the F light edgesl Then the MST With respect to the residual F light edges is computed using another recusive invocation of the algorithmi This is summarized in the following Algorithm MSTi Linear time algorithm Input Weighted undirected graph CV E With n vertices and m edgesl Output Minimum spanning forest F for Gi 1 Using three applications of Boruvka phases interleaved With simplication of the contracted graph compute a graph G1 With at most vertices and let C be the set of edges contracted during the three phases If G is empty then exit and return F Cl 2 Let G2 G1 p be a randomly sampled subgraph of G1 Where p 3 Recursively applying Algorithm MST compute the minimum spanning forest F2 of the graph Cgi 4 Using a lineartime veri cation algorithm identify the Fgheavy edges in G1 and deleted them to obtain a graph Cgi 5 Recursively applying Algorithm MST compute the minimum spanning forest F3 of the graph Cgi 6 Return forest F C Ung Algorithm 61 Algorithm MST Theorem 61 The expected running time of Algorithm MST is On Proof Let Tnm denote the expected running time of Algorithm MST for a graph CV E with n vertices and m edges Consider the cost of various steps in this algorithm for such input 1 At step 1 the three invocations of Boruvka algorithm which runs in On m time run in deterministic time On m After this step a graph G1 with at most vertices and m edges is produced 2 At step 2 the algorithm performs a random sampling to produce the graph G2 G1 with vertices and m an expected number of edges equal to 7 This step also takes Om n time 93 Finding the minimum spanning forest of G2 in step 3 has the expected running time of T A The lineartime algorithm veri cation in step 4 runs in time On m and produces a graph G3 with at most vertices and an expected number of edges at most by Lemma 5 5 Finding the minimum spanning forest of G3 in step 3 has the expected running time of T 6 Finally 0n time is needed for step 6 Adding up all these steps give us the recurrence n m n n T lt T 7 7 T 7 7 mm 7 lt8 2gt lt8 4gtcltnmgt for some constant c A solution satisfying this recurrence is 2cn m We have base case of n 1 and m 0 T10 01 Suppose that T S 24 and T S 24 g then by induction n m n n lt 7 7 7 7 Tnm S T8 2T8 4cnm Tnm 3 M2 E M2 E Cn m 8 2 8 4 Tnm 3 mg Z g Cn m Tnm S 2cn m This implies that the expected running time of the MST algorithm is On D References 1 Rajeev Motwani and Prabhakar Raghavani Randomized Algorithms Cambridge University Press Cambridge England June 1995 2 Ti Cormen Ci Leiserson and R Rivesti Introduction to Algorithms MCGraWHill 1999i E ect of light mass doping on ion velocity distribution functions in an electric double layer Magne c e d n he expansion egion o heHELIXLEIA sys em LEIA magnets g a E Mam E E I HELIX a magnets y quot gammaggg B a Zlt x a IN 29 quotNHH P2 scanning jxxg nx gag probe if h rf gas IE g Z E X g XI antenna inlet lt 45 m 15 m gt 39 39 I I i I 20 800 a 15 g A 700 g 10 9 E 600 E 399 500 5 2 LL 390 o 0 g 395 400 0 8 5 73 g 300 5 1039quot 2 200 L E 15 39I J 2 100 g lt 20 bng 0 25 2 600 560 460 360 260 160 0 Axial distance cm a H0r 0 ta r0 et0n fheH LIXL IAhe 0n ouc dfu 0n ham er ytmwh nepaddv wof he neton e0 ery b hemantcfed pofe nd hem gntc edgad nt nteaxs 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fetswreo s rvderepasm sfrwhcht e hermas nwasArAt 0u h 0Xe nbemwsobeved addt nofAr ed 0an nee se nth sp edo the ac gr und onp puat n rm13 ms npueXet022 ms oran 75Ar Ar Xe atoAsma ddton er nArXepamadont mpr vesgn can yth su er oncagon 0n xa ve cty rt eDLpoenta dope mpa ed 0 ure r onpasma F rthsp rt uarm t re Ar eamp puaton ompeeyds ppe rs psramanddwnsramtheDL tmor th n400Xemxngrt0 h tsrt hes vera eoc yra geo 5 Plasma Physics Laboratory CHAPTER 3 BIPOLAR JUNCTION TRANSISTORS RTT junctions However it cannot be made thh two independent backrtorbackdwdes BJTs PNP or as NPN wttt 7 base and collectorrepresented by E B and Crespecuvely The dAfference tn the cncmt normally As awayto remember which is which aformer student explanedto me that N39PN stands for Not Pomtmg N E c E C E B 5 an E PNP Fxgme 1 BJTsymbaIs and represenmtmns 1n dAgnal will concentrate on switching charactensacs ratherthan the unear propemes We will cncmts use N39PNtransxstors we shall concentrate pnmanly on them collect the base or orreverse bxased andthe baseremmerjuncuon can also be based exther way The four combination EVEJun tmn Reverse FEde Reverse arc Juncan A Funer Reva39se Acuve Fxgme z 7712 four npemang sandman Bipolar Junction Transistors 1 The reverse active region is seldom used but as we will see it occurs in TTL gates We begin our st dy nftran i tor lquot 39 L 39 ofo eration P N Figure 3 Common Bare Con guration Showing BimingArrangement for Cutof CUTOFF REGION Both junction reverse biased Let us start with the cutoffregion bothjunctions reverse biased as shown in Figure 3 quot 39 39 know that there are 39 39 junctions L L and I will be ignored Figure 4 Common bare con guration with biming for the forward active region FORWARDACTIVE REGION BE junction forward biased BC junction reverse biased The biasing condition for the forward active region of operation is shown in the Figure 4 The BEjunction is forward biased and the BC junction is reverse biased In this case the forwardbias of 39 39 quot 39 39 39 ofbothL 4 across thejunction quot L L fquot 4 r39 areadjustedto 39 39 39 quot L 39 ofinterest quot L 39j quot 39 wherethey 39 39 carrier even thoughthey greatly outnumberthe holes They quotpile up at the BE junction From there they diffuse across thebasere ion L 39 439 quot L electron recombination but the majority reach the BC junction At the BC junction the electrons encounter a potential gradient due to the depletion region and are swept across the junction into the collector region Bipolar Junction Transistors 2 Fxgure 5 Cammm Base Con guratth mfgrward Actxve tegmn Showmg Internal Currents As shown m Flgure 5 there are several components of current H es n ctedfrom BrE Thls ls ll 2 Electrons rmecteol fr ctro ecombrnataon current Thlsls 71 rec E 5 Collectorreverse saturataon current ICU whlch we wlll usually neglect The term a ls the forward current txansferrauo Thls term refers to the fractlon of the emltterjunctlon In most transrstors xFls close to unrty typrcally 0 970 99 The ll tr Thumm ttrnttr small and the collector and emltter currents are close to the same magmde must equal zero The sum of currents entenng the three termrnals 1 1 0 1 B Also from current components 3 and 5 above Ic quotXx391 1c0 forwardractlve reglon as shown m Flg 5 anol emrtter Slnce the BEJunctlon lsJust aPrNJunctlon thrs current ls essentaally dlode current Reduclng tlus blas v tage to zero wlll reoluce the emrtter current to zero In tlus case TB 1 Icg We wlll hang on to thrs leakage current for awhlle yet before we completely olrop rt Blpolar Junctron Transrstors SATURATION REGION Both junctions forward biased In the active region the collector current is proportional to the emitter current plus the leakage current ICO This implies that the voltage bias across the basecollector junction is unimportant But if we look again at Fig 5 we see a resistor in series with the collector lead If the current increases to a point that the voltage drop across the resistor plus the collector supply voltage begins to forward bias the collector junction then holes will be injected into the collector region from the base This hole current will counteract increases in electrons coming from the emitter effectively limiting the transistor current The basecollector voltage VBC at which this limiting effect begins is at about VBC 04 Volts and becomes fully limiting at about 06 Volts This region of operation is known as saturation Please note that our terminology can get a little confusing In this case saturation refers to the circuit not the transistor The transistor could carry more current but the external circuit the voltage source and resistor in the collector circuit limits the current REVERSE ACTIVE REGION BE junction reverse biased BC junction forward biased In this case the biasing arrangement is just the reverse of the forward active region The collector junction is forward biased while the emitter junction is reverse biased The operation is just the same as the forward active region except all voltage sources and hence collector and emitter currents are the reverse of the forward bias case Another difference is that IF is replaced by IR The equation corresponding to Eq 2 is 1E 0 RIC 1E0 This configuration is rarely used because most transistors are doped selectively to give forward current transfer ratios very near unity which automatically causes the reverse current transfer ratio to be very low NOTE Please be aware that the above discussion and resulting equations are only approximations to the total operation of a transistor A number of other phenomena are simultaneously occurring but for typical applications these phenomena cause only small errors in analysis Such phenomena are surface leakage currents current due to holes injected from base into the emitter holeelectron thermal generation body resistance of the transistor and avalanche multiplication It is not our intention to study all effects in the transistor in this course but to develop means to reduce the characteristics to simple but appropriate models which will allow us to easily analyze transistor switching circuits However in the process of developing and using a model an engineer must develop a feeling for when it is appropriate to use such a model and the limitations on the model Bipolar Junction Transistors 4 COMMON BASE AMPLIFIER that VEE ls rhereaseol by lrhv Thls change rh voltage woulol eause ah rherease u the eurrertt urr utu mamfestrtselfrrt a change 1 the voltage aeross the eolleetorresrstor RC Nowrfthe Common Base 5 Common Emitter Common Callecmr Fxgme g Busts Can gmatmns and Erasinga Lhe NPN Transsmrs COMMON EMIITER CONFIGURATION 39T L quota urtt quot n tll rh drgrtal eomparrsoh The name eorhes from the fact that the erurttertermrhal ls helol at ground or eo Iflo equauohs are wrrtteh orboth the rhput and the output erreurts the two eurrehts oflnterest wlll be the eolleetor eurreht1c and base eurreht IB Equauoh 2 above lrhto Equauoh 2 we can solve for the eolleetor eurrehtrh terms ofthe base eurreht l i1 3 C 11x 5 lb cu Itls convenlentto de ne anew varlable m u Thus Equataoh 3 beeorhes 1 5 1c e 813 Palm z The terms B or he relate a ehauge rh base eurrehtto a ehauge rh eolleetor eurreht or urrehtgath Ifwe lgnor ICC then the ole eurrehtgath hm ls the ratro ofeolleetorto base eurrehts Bls useolrhterehahgeably for both ae and ole eurrehtgath Many umes we use B and hFE rhterehahgeably Blpolar Juhetroh Trahsrstors 5 GRCUTT MODELS FOR A COMMON EMIITER TRANSISTOR BASEVEMJTTER CIRCUIT PNJunctlon Whlle there ls some elreet eaused by the collectorremmervoltage these eets are qurte small andfor the purposes ofthrs eourse wlll be hegleeted The eourse Thus de d th slmllarly When talldrrg about dlgltal logre gates we wlll assume that there wlll be no base currentlme baseremlttervoltage VB ls less than Vmo 50 Volts We wlll also VBEls equal to Vggm 0 80 Volts Sometrmes rtrs eorrvemerrt to use anlntermedlate value of 0 volts whm the trausrstorrs m the aeuve reglon We wlll also use voltages of 50 mv hrgrerfortheECl gates whreh are dseussedlater shown as ddscussed Next we wlll develop the eolleetormodels a Culuff b Acuve Sammuun VEEltEISEI 1EgtEI IEgtEI VEClt u 5n VCEgt n m cs315 FngIE 7 Common Emme ercmt Modelsat an NPN Tnmsxsmr COLLECTOR EMITTER GRCUTT MODELS three reglons of operatror The eolleetor eharaeterrsue ls aplot oflc vs ch for a few Flgure 9 Beeause the reverse aeuve reglon ls seldom used we wlll not develop that model here Blpolar Juuetror Trausrstors o TEETRDNIX 571 Curve Tracer Mm clrsors Pan 1 Matt Ibstep 1 u N of stgps 18 I 81084 25 Gum 5m IVdlv NE 117 Figure 8 F arward collector characteristics for 1 2N3903 Bipolar Junction Transistors TEkIZTFx39ONIX 571 Curve Tracer Hove cursors grim 1 Natt Ibstep 5 uR N of steps 18 cad 1 Ohm hFE 85 Figure 9 Reverse characteristics for a 2N3903 Bipolar Junction Transistors Cutoff Ifthe baseemitter voltage is less than cutin VBEy the base current is zero From Equation 3 we see that the only current is the leakage current which we usually consider negligible Thus we can consider the transistor in cutoff to be an open circuit between the three terminals as shown in Figure 7a The limit is of course that VBE lt 050 Volts Active Region The discussion of the active and saturation regions rests on Equation 5 and the collector characteristic graph which is shown in Figure 8 The cutoff region is shown on the collector characteristic as the single line where the base current is zero in this case the x axis or Ic 0 The active region is the large middle part of the family of curves where the collector current is essentially constant for a fixed base current The saturation region is the lefthand portion of the curves where the curves are almost vertical The curves in the active region correspond closely to Equation 5 The current is only a function of base current not a function of collectoremitter voltage VCE In this case a suitable model of the transistor is shown in Figure 7b This figure shows the baseemitter model as a voltage source and the collectoremitter model as a currentcontrolled current source BIB The defining limitations are 13 gt0 and VCE gt02 V Again you should be aware that this model is only an approximation to the real device For instance collector current does vary slightly with VCE and the spacing between the curves is not exactly constant B varies somewhat with collector current The course in linear electronics more closely models the first of these effects with an additional parameter r0 Saturation Region The saturation region model is perhaps the poorest approximation to the real device The saturation region is modeled simply as a fixed voltage of VCE 02 Volts In this region of operation the base circuit is modeled as VBE 080 Volts This model is shown in Figure 4c The limitations are that 13 gt0 and IC ltBIB From the curves in Figure 8 there is a substantial portion of the curve between the straightline portions in the linear region to the nearly vertical lines in the saturation region Probably the best way to view this transition is that B is decreasing A typical approach is to assume that when one is discussing saturation the appropriate 5 is approximately 85 that for the linear region In this course we will consider this problem in two ways First when we are working with problems in the early part of the course where we are considering operation in all three regions we will assume 5 to be the linear B for all regions Here we will assume that saturation is reached abruptly In the Bipolar Junction Transistors 9 second case when we are considering only cutoffand saturation regions we will use a saturation b only We will simply ignore the linear region The models we use in both es are the same and should cause little di iculty CONCLU SION The effect ofusing the models for the three regions can be seen in Figure 10 In gure the cutoffregion is represented by the straight line on the voltage axis indicating zero curren The linearre inn 39 39 39 lines The saturation region is represented by the vertical line at VCE 02 Volts A comparison between Figures 8 and 10 will show that ourmodels are only approximations th actual characteristic These approximations will however allowus to gre ly L L W u i i The39ywi117ru u to us when we attempt to understand circuit operation SATURATION r CUTOFF 1 3 VCE volts Fzgmz 10 A mmpuszte Characterzstzc mprzsemed by he szmtmudzls Bipolar Junction Transistors 10 THE TRANSISTOR AS AN INVERTERSWITCH indigim A current with just a small control current Thus we are primarilyinterestedin he cutoff 39 39 1n hissee ion 39quot 39 39 developed in he previous section To sun he discussion we will consider he circuit in Figure 11 This circuit is he classic switch The Opemim ofthe transistoris controlled by he cun39mt in he base eiieui lhusquotr 39 quot 39 39Wewill 39 quot39 iii cases Vin 4 and 10 Volts ID 500 5k V v m 620 Figme 11 Transistor 5mm simpli es Drawing Vin 0 u i i quot quot quot 39 39 Withno source the diode there will be no current ow With no base current the transistoris cutoff and there will be no collector current See the circuit in Figure 12 where the transistor has 39 39 quotL 39 39 39therewillbe r 39 quot u u H and emitter will be VCE Vo Vcc 10 Volts 1 Note he order ofthe subseiip s CE Apositivevoltage forVCE means that he collector is more positive than he mutter Figme 12 Cum With Vm u The transistor is replaced mm m ma madel Bipolar Junction Transistors 11 Fzgme 13 szmt Wm Vm 5 Volt Vin 5Volts Aquot L From Figure 13 v mVBE 57 El T u 86 mA 2 L 39 L L L quot current V 39 i L 39 39 L in Figure 13 We can solve this circuit for the collector cunent ci31B20 086172 mA 3 The nal information we would like to know is the output voltage the voltage at the collector We can ot get VCE directly we have to use the voltage drop across the collector resistor V0 VCE Vcc ICRC 10 172 rnAL 0500 K9 14 V 4 since VCE gtchsat 02V this result is consistent with the assumption that the transistor is operating in the active region Now let us look at the nal case Vin 10 Volts Le mt the previous case We can look at Figure 13 but with the input voltage at 10 volts v 7v EE m U7 15 RE 7 5K vIBEmA 5 r quot circuit we will mi in as di 1c IB 20186372mA o The output voltage is then Bipolar Junction Transistors V0 Vcc IR 10 372 it 500 86 Volts 77 7 T39 39 39 39 r 39 39 mm 39 uThereis no source for the negative voltage no negative power supply The transistor is modeled t quot A L 39 can only work withinthequot 39 r upplie i u quot L thaldklclilit u u ii is no longer valid I Bf c 7 500 in f 0 an 7 U 1 4 ID l liEgl l Fzgme 14 szmtzrl Saturanun wae go to the saturation region model the circuit is shown in Figure 14 Note that the only base circuit change is to change VBE to 080 volts This change makes the base current A lDDSIJ 8 137 0 SK 184mA We can also detennine the collector cuxrent 7V I s M g In 7 U 23 g c RC U W 19 0 mA 9 We already know that the output voltage is 02 Volts VCEsaI since the transistor is in aturation The only thing left is to verify that the model used is appropriate for the situation We alread know the active region model is inappropriate and that we are not in cutoff 1B gt0 To demonstrate that the saturation model is appropriate we need only show the Icsat lt0IB since 013 368 mA gt 196 the saturation model is appropriate SUMNIARY as r 39 39 39 r operation in each mode Cutoff VBE lt VBEV and VBC lt VBC Lfthis condition occurs the base cunent will be zero and the collector current will be zer ofcourse 39 39 L I I n t w i possibility ofusing the transistorbackwards with collector and emitter reversed Thus 39 39 quot 39 39 39 phage We will discuss the value ofVBEvlater Active Region 1E gt 0 VBE 070 L 315 VCE gt 02 V Bipolar Junction Transistors 13 Saturation Region IE gt 0 vBE 080 VCE 02 1C lt 313 The value of My 39 39 i in transition betwe ncutoffandthe activeregion For logic systems we will use VBEV 050 V0 5 39 39 39 tIansistoris cutoff Note that when the transistor is saturated VBE 080 Forlogic systems nnmever 39 39 39 39 u and saturation operation For we will 39 39 mat arie continuously between cutoff and saturation lNVERTJNG AMPLIFIER To get a better feel for how a tiansistor works with real signals we will discuss a tiansistor inverter with a sine wave input The tiansistor will operate in all three modes The circuit is shown in Figure 15 Temporarily we will assume VBEy 075 VBEact VBEsat Inother wnrd L 39 39 39 39 39 39 and nn u siuiiiiic adrcuit 39 39 niep 39 39 39 transistor will either be cutoff in active region or saturated From our previous example a i ran Ill ue cutoff above that value the transistor will be in the active region with positive base mnent 39 39 39 39 nn ansistor Our job here is to determine which region or regions are utilized First let us look to see ifthe transistor is cutoff Model VEEy VEEactm Vafsit 4375 VcEsat 02 340 Fing i 5 Transistor Invcrtirig Amplifier Cutoff Bipolar Junction Transistors 14 In order for the transistor to conduct at all the input voltage must be above VBEV Conversely if the input voltage is below VBEV the transistor will be cutoff Does that occur here es 39 p a Volts 1hquot 39 input voltage is below VBE 4175 Volts the transistor is cutoff The transistor is p 39 39Figurelo m L 39L 39 439 L39 time V0 VCC 10 Volts 10 Fig L L 17 shows the input and output waveforms during the cutoffperiod 1EI 1K lEIK D V V in Ssinbt l H F7ng 16 circuit mm Transistor Cuto F7ng 17 V0 Vth 7707151520715 cutoff Active Regi on Vm gt0 7s Flng 18 Transmm Invertzr Crmut DulrlgActlvz Region 39 wae 39 ureuar 39 lUl 39 39 39 Figure 18 we can then analyze the circuit The base current is I 11 m BE g Ssmmts 75 R an Thus the base current will have a sinusoidal component like the input voltage only shifted 39 39 39 39 39 ingtVBEv 39 nfrnllrp Bipolar Junction Transistors L quot 39 357 An39bthe output 39 39 39 A L re i tor Sslnwt r I 75 VD VccrlcRClUr4 gtltlKgtltT 12 Again the output voltage has a sinusoidal component This equation is valid only for Vin gt075 Volts and also as long as the circuit is not saturated V VCE gt0 2 v Thus the output voltage is sinusoidal only for aportion ofthe cycle It should also be noted that ifthe input voltage 075 Volts the output is equal to 10 Volts which is consistent with the cutoffcase no discontinuity We now need to determine ifthe circuit saturates Saturation te from our earlier example that the circuit will be in saturation ifthe collector emitter voltage drops to 02 Volts We could solve Equation 12 to nd what input i H0 r tc i as fr approach L tn imp 39 39 input voltage must be in order to reach saturation l a A c Fzgme 19 Transzstu Inverter CHEM m Samratzun The saturation model for the circuit is shown in Figure 19 We now work backwards from the collector circuit In Figure 19 the collector current is V 39V lu r u 20 r cc CEsat Z 53 RC 1K 9 0 mA 13 The minimum base current required to support this collector current is Csal u 3 155m B E I245ma l4 The minimum input voltage required to furnish this much base cunent is men VBEsaI 1353 RB 075 0245 mA 10 K9 32 Volts 15 Bipolar Junction Transistors Smce m mpul vulng nses alum mum m mm daesmdeedsmxau andwx be m smxananwhznzv v22 2 Ms In smxaum F a m mm c m wutyutvnltagus lvalts 11M 1gurel shawsthzvulpulvnlug2f yd F eZEI 5mm mm spmdsxelahvzlylmje ummm wave 29 Hawever mung hm m mpuz Mug has a snusmdal campmzntand39ha snusmdal pm amgm gimmmpmm m m vnlugz hmce 12nxxne uf mm Acmyhsmcunsm xemymu efm 1 u m wahauz suhst u ma cahms 1 vaung amph ens dlscmsedmmme dam mm hmueleckamcscwurse m t 2 x Mm aha ma am W V W 5 14 1n Amvtz may Va 1 I 2 Smuvmton t I I 7 F3502 20 Wmfomsfor n mptm cycleor m wemwgmplwev arm mug Tnns r Chanctzxisic mm w W the m xexmmmpmwemm mpuz vulng mam mpul vnlugz ca ed the wing mgr chmacunsuc a VTC Fax m m vow m 7 407 n 75 whnhxsane Hanaan 3mg mFngerl z 7 mu m Th2 vanguard1 charmnine cuve 15 Shawn Expnlaz unmnnmms n75 vm 32m 49 Fxgure 21 V51ng quotansst shnmstensnsm Lhe tmnsxsmr Inverter PNP Transislnrs of electrons and all the eunents and voltages are tn the opposite oneehon Fxgme 22 PNP vansszm thh mzetml currents Figure 22 is a drawing of aPN39P transistor shovnng the mtemal and extremal eunents The mtemal eunents eonespondto the same eunents as for the NPN transistor shown earhen The eunents ofxmponance are numberedZ 3 and 4 hole eunents injected from emitterto base2 the defuslon eunent that reaches the collector 3 and the ho1es1ost Currents V wd 1gnored IE eunent 2 1c e eunent 3 sale 15 e eunent 4 mun Bipolar Junction Transistors 18 However eurrerrts m terrrrs of tlre base eurerrts We strll get 7 1 a a but rrovv IE rs arregatave quantaty We also strll have Tc 15 1 0 g I 13 Tc h r makmg tlre voltage drops aeross Lhejuncuons m tlre opposrte drreetaorr Thus forthe A r values A typreal eorrrrrrorr errrrtter eorrfrgurataorr rs gverr m Flgure 23 From tlre erreurt rrroolel 17135K026mA The eolleetor eurrerrt rs B 10 0 26B 72 6 mA andtlre voltage attlre eolleetorrs Vo 5 le 2 orrrA 724volts Frgme 24 PNP appzrsanmr m a Iagx system mter zce Blpolar Jurretrorr Trarrsrstors 19 1r tlus ease vve wlll assume the logl gate output ls erther5 volts or 0 volts Later on we wlll use more aeeurate data forthe gate Kthe gate outputrs hlgh at 5 the voltage The erreurt wth rts cutoff model ls shown u Flgure 25 Note that we speclfy the emltterrbase voltage as VE 1r thrs ease VE 0 beeause both terrrurrals are eormeetedto 5 volts 5 5 Vm JE r c u 2 SK E v a Cliv 5 SK 1K 1K u i i I Fxgme 25 Cuta mldel Fxgme 25 Sammtmn made Ifthe gate outputrs low orzero volts the emltterrbasejuncnon ls tumed on We suspeet the trausrstorrs saturated so we use the saturatror model u Flgure 26 for the analysls 1t r a r a r eouuterparts Note agam we use reverse polanues for the the tvvo terrrurral voltages We an ealeulate the tvvo eurreuts 570 85K 0 84 mA 12 570 21K4 8 mA 7 the txanslstorls obvlously m saturatlorl and V0 4 8 volts Notlee that the voltage aeross the 1K loadreslstor ls 4 8 volts Blpolar Juuetror Trausrstors EXERCISES 1 For the fouowrng crrcurt assurne the fouowrng model fothe transrstor VBEV 0 5 Vt VBEact 0 7 Vt VBEsat 0 8 Vt B 3039 VcEsat 0 2 v I 12 V 49 t Vn Crrcurt Dmgam Voltage Transfer cnaractenstrc IfVm 5 0 volts a Draw arrows showrng the actual current dArecuon m each terrnrnal ofdne transrstor b Deterrnrne three terrnrnal currents c Deterrnrne Vo V d Fmd rnrnrrnurnvnsn and rnarnrnurn Vnenm we can use these pornts to draw the rtag uan fer two pornts transrstor rs Vuct Vacant Vases 0 7 volts Veg 0 2 volts 5 25 Brpolar Junctron Transrstors 21 Val1 VEEaM Val2m 70 70 VCBA 70 20 B 20 vam r 0 volts a Draw arrows snowrng the actual eurrent dArecnon m eaen terrnrnal ofthe transrstor b Deterrnrne tnree terrnrna1 eurrents IE rr rr e Deterrnrne Vo V d Fmd mm and vmm Vmm the 39 e Draw the voltage transfer enaraetensae Brpolar Junetron Transrstors DESIGN EXAMPLE TRANSISTOR SWITCH We have a slgnal from a 74HC00 CMOS logl gate thatrs supposeolto turn on an LED when the output of the gate ls ahlgh The speclflcatlons oflnterest are 74HC00 Vo mn 3 84 volts when 1OH 74 00 rnA LED VD180 volts when TD 30 mA TM 60 mA supplythe neeessarv eurrent ABJT swrteh ls agoool candldate forthlsjob Ln ths ease n t b t A ls low the transrstor wlll be turneol off The hlgh output ease ls shown 5 Il to n 9 V z I 2 VEEsat 8 49 3 ID 74ch0 1 74 mA 5339 thute 27 Cum m be desgned Flr t Lnthrs ease zero and the transrstor wlll be cutoff There wlll also be no dlode e urrent saturated Kthe dlode eurrentls 30 rnA and the dlode voltage 15 180 volts the voltage 1 t 3 0 volts Thus be 100 1 Assumlng the transrstor But 10the base eurrent rnust be greaterthan 3 mA Ln thrs ops as low as 3 84 volts Ifthe baseremlnervoltage ls 0 8 volts the voltage aeross the resrstorrnustbe 3 04 volts At abase eurrent of4 mA the resrstorvalue ls 760 1 Note that the base eurrentln saturauon needs only to be 3 rnA or greater Thus the base 110 to 1013 1 are 820910 and 10009 Any ofthese values 15 sausfa t rylf e Eonslder only nomlnal ealeulatron e o w s wae rnust guarantee operauon wth 5 resrstors 800 lt 12mm 964 Wrth thrs deslgn the LED wlll be off when the gate outputrs low and the LED wlll be on when the gate output ls hlgh Blpolar Junetron Transrstors Prnhlems 1 Tu th l h r t N From these curveS curves g m dels Tum m the o lglnal eolleetor eharaeteustae eurvesthe erreult moolels auolthe curves represented by these mo els 2 trausrstor to be cutoffandsaturated Use the followlng moolel eomportertt values 550 VBEV 0 so VBEM 0 80 Vc m 0 20 7 Problem 2 Problem 3 and 4 For Problems 3 and 4 use the followlng model for the trausrstor B 3039 VBE1VBEact VBEsat 0 7039 VcEsat 0 20 aw the voltage transfer eharaeteustae vo vs Vm for the erreult glven Let Vm go hlgh enough forthe erreultto satura e 4 eterm e the tame lttakes the output waveform go from one ratl to the otherrfthe frequencyls 60 Hz 5 Use a eomputer srmulataor to venfy the results from Problem 4 Use the llbrary moolel for a 21mm vvrth Beta 30 Note that the results may uotbe qulte the same make the trausrtaor from cutoffto saturatror o Fmolthe mput voltage requlreolto eause the pnp trausrstor erreult to saturate Blpolar Juuetror Trausrstors PN39PTranslstorModel V sn08 Veer 2 Betaeat20 Problem 6 7 the mput and output voltages at eaeh breakpomt The trahsrstor has the followlng model parameters VElem V ml VERA 0 7 V0115 VCEm 0 2 volts Beta 25 a n 12 Vm l K In K V U 7 Vm 8 A PN39P translstorls usedm the followlng elreultto dnve a load as shown Determme andln the other state the pnp trahsrstorrs to be saturated The pnp trahsrstor has the followlng model parameters mem 0 5 volts VESA 0 8 volts Vacs 0 2 volts Beta 20 Blpolar Juhetroh Trahsrstors 25 12 Rh KUU Vin Probfem 8 V n u 39 7 What s M for the output txansxsto V i R Bipolar Juncuon Transistors Elements of Information Theory Introduction Course number EE 568 Instructor Natalia Schmid Januan 13 2009 WVU EE 568 Concepts Information Theory IT is a subject suggesting a common mathematical approach to the study of data collection and manipulation of information It provides theoretical basis for activities such as Observation measurement Data compression data storage Communication Estimation Decision making Pattern recognition and learning Januan 13 2009 WVU EE 568 Concepts Increasing System Complexity Performance requirements Mathematical Models IT is the study of how the laws of probability and mathematics in general describe limits on the design of information processing and transmission systems ex lT tells how one may design strategies in a communication system by using the laws of large numbers to overcome the noise imposed by nature and the following errors that may occur in the system Januan 13 2009 WVU EE 568 Communication Channel The primary scope of the IT are communication systems Two fundamental questions What is the data compression limit What is the transmission limit in a noisy media Januan 13 2009 WVU EE 568 Communication System Ex noise disturbances Timevarying frequency response Crosstalk from other channels Thermal noise MOdUlatiOh Distortions rotation blur illumination etc Data reduction I Inserted redundancies l NO39Se I Source gtl Encoder lgtl Channel lgtlDecoder lgtl Destinationl Ex Voice waveform EXJ Telephone line SUf Cim Processing Bits from magnetic tape Wireless line to replicate the Input Optical bits Storage medium Sensor output erjagirtg system Target in radar system 83903909390al organism Input to biological organism AchiSition enVironmem Januan 13 2009 WVU EE 568 5 The need More complex communication systems ex Persontoperson communication wireless links Broadcast communications lntracomputer and intercomputer communications Multiple access Designer is pressed to improve performance The goal is to maximize transmission subject to some reliability requirements ex Detect a signal that is close in its level to the background thermal noise Approach Involve some Optimality Theory to design waveforms or messages to ensure reliable communication of signals at the level close to noise Januan 13 2009 WVU EE 568 6 Basic Questions What is information How do we measure it What are the fundamental limits on the transmission of information What are the limits on extraction information from the environment What are the limits on compression and refinement of information January 13 2009 WVU EE 568 7 Examples ModelBased Approach If probabilistic model is well fitted to describe experiment fundamental limits 3D world channel transformation measurement acquisition device etc January 13 2009 WVU EE 568 8 Example CT Imaging Ex Brachytherapy FBP Modeer ased Approach will provide better estimation results WVU EE SEE 9 January ta 2009 Example Radar Dimensions 1 Range Send pulse and listen to echo Resolution pulse width 2 Azimth Require large antenna hundreds ol meters Synthetic Aperture Fine resolution is synthesized while airplane is llying http WWW Sandie govRADARWhatls html January ta 2009 WVU EE 503 Example ATR SAR images displayed as variance images Given orientation complex Gaussian model provides a good t to SAR data and high recognition performance January 13 2009 WVU EE 568 Information Theory Claude Shannon 1948 Mathematical Theory about limits of Communication Theory Characteristics Emphasis on probability theory Optimization Theory Concerns with encoder and decoder Proves existence of encoders and decoders Limits but no indication how to achieve them involves computationally complex procedures to prove results January 13 2009 WVU EE 568 Information Theory Deals with theory mathematical models Appropriate way 7 Physical 39 Mathematical gt Our approach Study a family of simple sources and channels More complicated Theory is Useful Framework to model real sources and channels Points to types of tradeoffs in encodersdecoders Januan 13 2009 WVU EE 568 13 Source and Channel Coding Study source and channel models separately Binarydata Source Source Channel encoder encoder 1 D t I Source Channel d es ma Ion decoder decoder Binary data Source encoder binary representation numberof bits Channel encoder reliable communication and reproduction Januan 13 2009 WVU EE 568 14 Sources and Source Coding 39 Sources are random processes ex Discrete memoryless source finite alphabet al az aK iid drawing probability assignment pa1 pa2paK Random process model is meaningful ex Repeated measurements alphabet Store sequence in binary form 011 a 2 a 3 a 4 Two methods Effectiveness depends on frequency of different events Januan 13 2009 WVU EE 568 15 Entropy Selfinformation uncertainty Iak710g ak If ak 1 then Iak0 Entropy discrete sources mp 7i pmk og pm Ch 5 CT 91 H Ratedistortion continuous sources A A mm A IXX 13y pXXdXXltD Januan 13 2009 WVU EE 568 16 Channels Setofinputs alazaK Set of outputs Probability measure on the set of outputs pbj iak Discrete memoryless channels ex binary symmetric o 1quot o 1D Discrete channels with memory Additive white Gaussian channel Januan 13 2009 WVU EE 568 17 Channel Coding Separate the channel encoderdecoder in two parts ex FSK PSK Alphabet a1a2aK sit52tSat I One letter er139sec Ema data Digital data Waveforms modulator Discrete channel encoder Discrete channel d deco er Discrete Digital data J demodulator channel Alphabet b1b2b Januan 13 2009 WVU EE 568 18 Capacrty Discrete channel capacity Ch 8 CT 91 Continuous channel capacity Ch 10 CT 91 Capacity maximum average number of bits that can be transmitted with probability close to zero Capacity of continuous channel can be approached proper modulator and demodulator ex turbo codingdecoding Channel Coding Theorem If data are communicated at RltC then the input sequence can be reconstructed with asymptotically small Perror Joint SourceChannel Coding If HltC then the source sequence can be reconstructed on the output of decoder with arbitrary small Perror Januan 13 2009 WVU EE 568 19 Modern Information Theory IT contributes and gets contribution from Shannon Huffman Gallager Cover Kullbuck Sanov Sanov Csiszar Chernott Probability and Statistics Decision Theory Information Theory Physics Ami Economies I Computer Science L Shannon McMillan 39 y Breiman Cover Breiman Algoet Barron Kolmogorov Chaitin Januan 13 2009 WVU EE 568 AEP Law of Large Numbers EE51 3 applied to the sequence 1 n iizlog 170k quot k1 Typical Set set of long sequences with probability close to 27m and cardinality close to ZWH Applications efficient elegant proves of source and channel coding theorems Januan 13 2009 WVU EE 568 21 Entropy Rates AEP results are established for iid sequences lf variables X1 X2 are not iid define entropy rate lim lHX1X2Xn namn Januan 13 2009 WVU EE 568 22 IT and Statistics Sanov s Theorem Large Deviations Given sequences 1 quot Xkltgtltlld k1 1 n P Bythe Law of Large Numbers ZZXk AEX1 k1 What is the probability p 1 n Zxk EX1 nkl for large n It converges exponentially fast to zero Sanov s Theorem finds the exponent Januan 13 2009 WVU EE 568 23 IT and Statistics Hypothesis Testing Problem R is an observed random variable H1Rp1 H22Rp2 Optimal test statistics EE591K likelihood ratio NeymanPearson minimum probability of error Conditional probabilities of error False Alarm and Missed Detection Sanov s theorem provides the rates of convergence for the FA and MD Januan 13 2009 WVU EE 568 24 Network Information Theory Multiple Access Channel TDMA FDMA CDMA Broadcast Channel January 13 2009 WVU EE 568 25 Maximum Entropy and Spectral Estimation From physics Given temperature find distribution of molecular velocity MaxwellBoltzmann Maximum Entropy distribution We will derive maximum entropy distribution under some constraints Application maximum entropy spectral density estimation January 13 2009 WVU EE 568 26 IT Questions What is information That is how do we measure it What are the fundamental limits on transmission of information What are the fundamental limits on the extraction of information from the environment What are the fundamental limits on the compression and refinement of information Januan 13 2009 WVU EE 568 27 IT Discipline Theoretical Basis for activities as Observation Measurement Data compression Data storage Communication Estimation Decision making Pattern recognition Januan 13 2009 WVU EE 568 28 Definition of IT Modern IT is the science of quantifying information content information loss in transmission and quantizations fundamental limits in recovery corrupted data and signals and design of systems to approach these limits Januan 13 2009 WVU EE 568 29 Related Classes EE561 Communication Theory EE513 Stochastic Systems Theory Probability and Processes prerequisite EE591 H Coding Theory EE591 K Estimation and Detection EE562 Wireless Communication Systems Januan 13 2009 WVU EE 568 30
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