Analyzing a Complex NetworkIdentifying the Reference NodeNode Equations for Resistive BranchesKCL Example 1 Example 1 KCL EquationsKCL Example 1 (cont)Dealing with ``Floating" Voltage SourcesKCL at a ``Super Node"KCL Example 2 (Super Node)Elimination Trivial NodesDependent SourcesDependent Source (cont)Dependent Voltage SourceKCL Example 3 (Dependent I Source)KCL Example 4: (Dependent V Source)EE 42/100Lecture 5: Analyzing a Complex Circuit: Nodal AnalysisELECTRONICSRev B 1/30/2012 (8:56PM)Prof. Ali M. NiknejadUniversity of California, BerkeleyCopyrightc 2012 by Ali M. NiknejadA. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 5 p. 1/?? – p. 1/Analyzing a Complex Network++•Simple circuit can be solved using repeated applications of series-parallel formulasand current/voltage dividers (left schematic). Next lecture we’ll learn a few moretricks to add to our toolbox.•For now we’d like to develop a systematic methodology to solve for the currentsand voltages (and hence power) in any arbitrary circuit (right schematic).•Nodal Analysis: Repeated application of KCL to each node of the circuit. It can beused to solve for all of the voltages (and subsequently currents) in the circuit. Thiswill be our prime analysis technique in this class.•Mesh and Loop Analysis: Repeated application of KVL can be used to find thecurrents (and subsequently voltages) in a circuit. In this class we will not usemesh/loop analysis.A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 5 p. 2/?? – p. 2/Identifying the Reference Node+VjVkVgnd= 0•Recall that voltage is defined as a quantity that measures the potential differencebetween two nodes in a circuit, VAB.•We can arbitrarily pick one node of the circuit and define all node voltages inreference to this node. Call this node ground, or node ‘0’. In other words, define Vkas the node voltage at node k which is the energy gained per unit charge as itmoves from node gnd to node k, or in more cumbersome notation, Vk,gnd.•If we subtract the two node voltages, we getVj,gnd− Vk,gnd= Vj− Vgnd− (Vk− Vgnd) = Vj− Vk= Vj,k•In other words Vj,k= Vj,gnd− Vk,gnd, which makes sense since they are definedwith respect to the same reference. Note that the reference potential is bydefinition at zero potential, Vgnd= 0.A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 5 p. 3/?? – p. 3/Node Equations for Resistive BranchesVjVkRxIxI1I2I3•Since the potential at the ground node is known, only n − 1 unknown node voltagesremain. Writing KCL at each other node results in an equation in the following formI1+ I2− I3+ Ix= 0•In a resistive branch, the current can be written in terms of the node voltages. Ifcurrent Ikflows from node a to node b, it’s given byIx=Vk− VjRx= Gx(Vk− Vj)•If an independent current source is connected to the node, then the current isknown.A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 5 p. 4/?? – p. 4/KCL Example 1+R1R2R3R4R5vsis•We are now in a position to solve many circuit problems. For example, in the abovecircuit we proceed as follows.•First we choose the reference ground potential at the negative terminal of thevoltage source. Why? Because then we eliminate one unknown node voltage fromthe circuit. The node connected to the positive terminal of the source is at avoltage of vs.•We have eliminated two nodes from the list of unknowns. Now there are only twonodes left in the circuit.A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 5 p. 5/?? – p. 5/Example 1 KCL Equations+R1R2R3R4R5vsisA. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 5 p. 6/?? – p. 6/KCL Example 1 (cont)•For each node we write KCL equations. We setup two equations with twounknowns.•It’s good to write these equations in standard form. Put the unknowns on the lefthand side and the known currents (due to the independent sources) on the righthand side. Then we have a matrix equationAv = bwhich we solve by matrix inversion (Gaussian elimination, Cramer’s rule, matlab,etc.)v = A−1bA. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 5 p. 7/?? – p. 7/Dealing with “Floating" Voltage Sources+VsR1R2R3R4R6AB+VsR5•In the above example, two voltage sources appear so that we cannot label bothnegative terminals as ground. So we pick one.•In this circuit, the number of unknowns is smaller because if we know the voltageat one node connected to the voltage source, the voltage at the other has a fixedrelation, since VAB= Vs.•But in writing down KCL at one of the nodes, we encounter a problem. The currentthrough the voltage source can take on any value, which means that other circuitelements determine the current through it.A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 5 p. 8/?? – p. 8/KCL at a “Super Node"+VsR1R2R3R4R6+VsR5II1I2IsI4•But note that if we write the KCL equations for both terminals of the voltagesource, this unknown current appears twice.At the positive terminal:I1+ I2+ I3+ Is= 0At the negative terminal: I4− Is= 0•If we add these two equations, Iscancels outI1+ I2+ I3+ I4= 0•This is an independent equation we can use. It’s actually KCL for a “super node",which is what we call nodes a and b together. We have just shown that currentcontinuity applies to a node and also to a super node.A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 5 p. 9/?? – p. 9/KCL Example 2 (Super Node)+VsR1R2R3R4R6+VsR5•Using the steps that we have learned, we first identify a ground node at aconvenient location (negative terminal of a source). Then we identify two supernodes in the circuit. For each super node we write KCL, put the equation intostandard form, and solve.A. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 5 p. 10/?? – pElimination Trivial Nodes+VsR1R2R3R4R6+VsR5•Any node with less that three elements is in some sense trivial. That’s because wecan find the node voltage for such a node from the branch current. Thus, it’s smartto avoid writing equations for these nodes and to deal with them later.•In the above example the number of unknowns has been reduced from 2 to 1 byusing this techniqueA. M. Niknejad University of California, Berkeley EE 100 / 42 Lecture 5 p. 11/?? – pDependent SourcesR1R2R3R4R6+VsR5Ix= KVR2+VR2−VjVk•Dependent sources require a bit more work
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