1、Chapter 3 Methods of Analysis,要求深刻理解与熟练掌握的重点内容有:节点电压法和网孔电流法。 要求一般理解与掌握的内容有:支路电流法与回路电流法。 难点：独立方程数、回路电流法。,3-1 Introduction,3-2 Nodal Analysis,3-3 Nodal Analysis with Voltage Sources,3-4 Mesh Analysis,3-5 Mesh Analysis with current Sources,3-6 Nodal and Mesh Analyses by Inspection,3-7 Nodal Versus Mesh
2、 Analysis,3-10 Summary,3.1 Introduction,simultaneous equations 联立方程,determinant 行列式,quotient 商，比值，系数,simulation 仿真,transistor 晶体管,datum node 基准结点,planar 平面的,nonplanar work 非平面网络,software package 软件包,a chassis ground 框架接地线,substitution method 替代法,elimination method 消去法,NETWORK TOPOLOGY,A given circui
3、t and its linear graph (or simply a graph) Node：A point at which two or more elements have a common connection. Branch：A single path, containing one simple element, which connects one node to any other node.,连通图 有向拓扑图,Tree（树）：A set of branches which does not contain any loops and connects every node
4、 to every other nodes.Cotree（余树）：After a tree has been specified, those branches that not part of the tree form the cotree.Link（连支）：Any branch belonging to the cotree.,Check Your Understanding： 一个有n个节点、b条支路的连通图，连支数l为？ l=b(n1),3-2 结点电压法 Node-Voltage Analysis / Nodal Analysis,一、Steps： 1. Select a node
5、 as the reference node. Assign voltages V1,V2,Vn-1 to the remaining n-1 nodes . 2. Apply KCL to each of the n-1 nonreference nodes. Use Ohms law to express the branch currents in terms of node voltages. 3. Solve the resulting simultaneous equations to obtain the unknown node voltages.,Node 1： -I1-IS
6、2-I3+IS4+I5+I6=0 I1=G1(U-U+US1)， I3=G3(U-U)， I5=G5(U-U)， I6=G6(U-U+US6),(G1+G3+G5+G6)U-G1U- G3U-( G5+G6)U= G1US1- G6US6+IS2-IS4,G11U+G12U+G13U+G15U=,G11U+G12U+G13U+G15U=,Gij（i=j）：自导，Sum of the conductances connected to node i . Gij（ij）：互导，Negative of the sum of the conductances directly connecting n
7、odes i and j . （Turn OFF all sources.）,：Sum of all sources directly connected to node 1, with currents entering the node treated as positive.,(G1+G3+G4+G5+G6)U-G1U-G3U-G4U-(G5+G6)U = G1US1- G6US6+IS2-IS4+G4US4 (G1+G3+G5+G6)U-G1U- G3U-( G5+G6)U= G1US1- G6US6+IS2-IS4,(G1+G3+ G4+G5+G6)U-G1U- G3U- G4U-(
8、 G5+G6)U= G1US1- G6US6+IS2-IS4+G4US4,Turn OFF all sources ! 电源置零,Example：R1=R1=0.5，R3=R4=R5=R6=1，US1=1V，US3=3V，IS2=2A，IS6=6A. Calculate the currents I1 through I4.,Solution：Select the node as the reference node.,At node ,At node , U=1.2V， U=3.8V。 I1=(US1-U)/(R1+R1)= -0.2A，I3=(U- U +US3)/R3=0.4A I4=(
9、U- U)/R4= -2.6A，I5= U /R5=3.8A,改进节点法 MODIFIED NODAL ANALYSIS,US2：无伴电压源 Assign an unknown current to the branch which contains the voltage source.,a：Ua=US1 b：-G3Ua+(G3+G5)Ub= -IS2 c：-G4Ua+(G4+G6)UC= IS2Ub -Uc= US2,3-3 Nodal Analysis with Voltage Sources,a：Ua=US1 (1) b&c：I3+I4= I5+ I6,Ub -Uc= US2 (3),
10、(2),The other method is to treat node b, node c and the voltage source together as a sort of supernode and apply KCL to both nodes at the same time.,A supernode may be regarded as a closed surface enclosing the voltage source and its two nodes.,Example：R1=R2=R3=R4=R6=2，R7=6，US1=4V，US5=10V，IS7=1A，IS8
11、=4A，2，2. Find the node voltages.,UC=10V a：(G1+G3+G4+G6)Ua-G1Ub-(G3+G4)Uc=G1US1-U-I+IS7 b：-G1Ua+(G1+G2)Ub-G2Uc=U - G1US1+IS8,Solution：Select the node d as the reference node.,U=Ub-Ua+4； I=0.5(Ub-Uc)=0.5(Ub-10),Ub=10V Ua=14V,fundamental loop（单连支回路）：The loop contains only one link-branch and several tr
12、ee-branches, it is conventional to assume that each loop current flows the link direction.,L1：Ub1+Ub2Ub5=0 L2：Ub1Ub3+Ub4=0 L3：Ub1Ub3+Ub5+Ub6=0,3-4 Loop Analysis and Mesh Analysis 3-5 Mesh Analysis with Current Sources,Mesh(网孔)：A loop which does not contain any other loops within it.,m1：Ub1Ub2+Ub5=0
13、m2：Ub1+Ub3Ub4=0 m3：Ub2+Ub6Ub3=0,支路电流法 BRANCH ANALYSIS,以支路电流为未知量，直接应用KCL和KVL建立电路方程，从方程中解出各支路电流。,aI1+I2+I4=0 bI2I3I5=0 cI4+I5I6=0 d I1+I3+I6=0,独立节点数：全部节点数1 An independent node is a node whose voltage cannot be derived from the voltage of another node.,m1：US1-U S3=R2I2-R3I3+R1I1； m2： US4=R5I5-R2I2+R4I4
14、； m3： US3-US6= -R5I5-R6I6+R3I3,Example： R1=1，R3=2，US1=2V，IS2=0.5A，find I3，PU，PI。,Solution： At node a：I1+I3=IS2 (1) Uab=R1I1-US1，Uab=R3I3 R1I1- US1=R3I3 US1= R1I1- R3I3 (2) I1=1A，I3=0.5A Uab=R3I31V PU=US1I1=2W（ Power supply ）， PI= UabIS2= -0.5W（ Power absorb ）,Example： R2=2，R3=3，US3=3V，gm=1S，find the
15、 currents I2，I3。,Solution： a：gmU3=I2+I3， U3=R3I3， I2 -2I3=0 Mesh-voltage equation： US3= - R2I2+ R3I3， -2I2+3I3=3I2= -6A，I3= -3A,以基本回路电流为独立变量，在基本回路中建立KVL方程，解方程组得回路电流，随后确定支路电流。基本回路电流：在基本回路（单连支回路）中，由连支电流形成的环流，简称回路电流。,Steps： constructing a suitable tree； assigning a current reference to each link； KVL e
16、quations must be written around each of loops.,Loop Analysis,I1：-US1+US6=R1I1+R6I6+R5I5+ R4I4I4= I1+I2，I5=I4-I3=I1+I2-I3，I6=I1-I3 I1：(R1+R4+R5+R6)I1+(R4+R5) I2 -(R5+R6)I3= -US1+US6I2：(R4+R5)I1+(R4+R5)I2-R5I3= -US2I3：-(R5+R6) I1-R5I2+(R3+R5+R6) I3= -US6,Thus：R11I1+R12I2+R13I3=,Rij (i=j ) ：自电阻, Sum of
17、 the resistances in loop i Rij (ij )：互电阻, Sum of the resistances in common with loops i and j.,：Sum of all independent voltage sources in loop 1，with voltage rise treated as positive.,Example： R1=1，R4=4，R5=5，R6=6，IS2=2A，IS3=3A，US4=4V，find the branch currents。,Solution： Select the branch 4,5,6 as the
18、 tree. I2=IS2=2A，I3=IS3=3A Loop I1：(R1+R4+R6)I1+R4I2-R6I3=US4 I1=14/11=1.27A， I4=I1+I2=36/11A， I5=I2+I3=5A， I6=I1-I3= -19/11A。,Example： R1=1，R2=2，R3=3，R4=4，IS5=5A. What value of gm is required to give the current I3 of 0A.,Solution：Select the tree in the network of Figure. Loop I4：(R1+R2+R3+R4)I4+(R
19、1+R2) gmU4 -(R2+R3)IS5=0 U4=R4I4,网孔电流法 Mesh Analysis,A loop is any closed path, and a mesh is a loop which does not contain any other loops within it. A mesh current is a current that flows only around the perimeter of a mesh, may often be identified as a branch current. Mesh analysis is not quite a
20、s general as nodal analysis and loop analysis because it is only applicable to a circuit that is planar.,Im1：(R1+R2+R3)Im1-R2Im2-R3Im3=US1-US3 Im2 ：-R2Im1+(R2+R4+R5)Im2-R5Im3=US4 Im3 ：-R3Im1-R5Im2+(R3+R5+R6)Im3=US3-US6 I1=Im1 ，I2=Im1 -Im2，I3= -Im1+Im3， I4=Im2，I5=Im2 -Im3，I6=-Im3,Example： E1=1V，E3=6V
21、，IS=6A，R1=3，R2=2，R3=1，R4=4. Find the mesh current.,Solution： Im3=IS=6A Mesh Im1： (R1+R2)Im1-R2Im2=E1 Mesh Im2： -R2Im1+(R2+R3+R4)Im2-R3Im3=E3 Thus：5Im1-2Im2=1，-2Im1+7Im216=6 Im1=1A ，Im2=2A 。,Example： RS=R1=1，R2=2，R3=3，=3，What value of US1 is required to give the current I3 of 3A.,Solution：Mesh Im1： (
22、R1+R2+RS)Im1-R2Im2=US1-UMesh Im2： -R2Im1+(R2+R3) Im2=U U=US1-RSIm1，,Substituting the value into Eqs： (1-)RS+R1+R2Im1-R2Im2=(1-)US1； (RSR2)Im1+(R2+R3)Im2=US1 7Im2=5US1 I3= Im2=3A， US1=4.2V,Check Your Understanding：Calculate the mesh currents.,Im1：Im1=IS5 Im2：-R2Im1+(R1+R2) Im2=0,R1I1+R2I2+UgU4=0 UgU4
23、R3I3+R4I4 A supermesh results when two meshes have a dependent or independent current source in common.,U4=R4I4 I1=Im2，I4=Im3,R1I1+R2I2+ R3I3+R4I4=0，I2=I1-IS5，I3=I4-IS5,I1=gmU4+I4,Which method is the best or most efficient? The choice is dictated by two factors. The first factor is the nature of the
24、 particular network. The key is to select the method that results in the smallest number of equations. The second factor is the information required. By the way, nodal analysis is more amenable to solution by computer, as it is easy to program.,3-7 Nodal Versus Mesh Analysis,3-10 Summary,Exercises：,
25、We set the mesh currents Im1, Im2, Im3, Im4,gU=gR1Im1 I=(Im3Im1),1. Write the mesh-current equations for the circuit.,Solution： write its linear graph,We set the loop currents il1， il2， il3， il4,gmu1=gmR1il3,u1=R1il3,2. Write the loop-current equations for the circuit.,Solution： Write its linear gra
26、ph in Fig. (b),iu,We assume the current iu passed by the voltage source uS7.,3. Write the node-voltage equations for the circuit.,Solution： The circuit has four nonreference nodes. The voltages are un1，un2，un3，un4 individually, and the node 0 is reference node.,四、设法只用一个方程求出电流I。,解：绘有向拓扑图,回路I,(22)I212
27、223I =25 I=0.3A,五、求I。,解：对封闭面列广义KCL方程,Example： Determine the voltage labeled v in the circuit of Fig. 4.32 using nodal analysis techniques.,Solution：,We need concern ourselves with the bottom part of this circuit only. Writing a single nodal equation,-4 + 2 = v/ 50,We find that v = -100 V.,Example： C
28、onsider the circuit of Fig. 4.37. Determine the current labeled i1.,i1 = 1.636 A.,Solution：,Example： Make use of the supernode concept to assist in the determination of the voltage labeled v20 in Fig. 4.38. Crossed wires not marked by a solid dot are not in physical contact.,1,2,3,4,0,v20 = -90.9 V,
