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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 1 CIRCUIT ELEMENTS AND CIRCUIT ANALYSIS Last time: Terminology: Nodes and branches Introduce the implicit reference (common) node – defines node voltages Introduce fundamental circuit laws: Kirchoff’s Current and Voltage Laws Today: Formal nodal analysis Resistors in series and parallel Voltage and current dividers +

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 2 NOTATION: NODE VOLTAGES The voltage drop from node X to a reference node (ground) is called the node voltage V x. Example: inout ground R C Vin Vout + - + -

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 3 FORMAL CIRCUIT ANALYSIS USING KCL: NODAL ANALYSIS 2 Define unknown node voltages (those not fixed by voltage sources ) 3Write KCL at each unknown node, expressing current in terms of the node voltages (using the constitutive relationships of branch elements*) * with inductors or floating voltages we will use a modified Step 3 1 Choose a Reference Node (ground, node 0) (look for the one with the most connections!) 4 Solve the set of equations (N equations for N unknown node voltages) (Memorize these steps and apply them rigorously!)

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 4 EXAMPLE OF NODE ANALYSIS Choose a reference node and define the node voltages (except reference node and the one set by the voltage source); node voltage known VaVa V b reference node Apply KCL: R 4 V 1 R 2 + - ISIS R 3 R1R1 You can solve for V a, V b. What if we used different ref node?

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 5 PARALLEL ELEMENTS KVL tells us that any set of elements which are connected at both ends carry the same voltage. We say these elements are in parallel. KVL clockwise, start at top: Vb – Va = 0 Va = Vb

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 6 RESISTORS IN PARALLEL (and the Current Divider) R 2 R 1 I SS I 2 I 1 2 Define unknown node voltages V X 1 Select Reference Node Note: I ss = I 1 + I 2, i.e., RESULT 1EQUIVALENT RESISTANCE: RESULT 2CURRENT DIVIDER:

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 7 SERIES ELEMENTS KCL tells us that all of the elements in a single branch carry the same current. We say these elements are in series. Current entering node = Current leaving node i 1 = i 2

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 8 RESISTORS IN SERIES Circuit with several resistors in series – Can we find an equivalent resistance? KCL tells us same current flows through every resistor KVL tells us Clearly, Thus, equivalent resistance of resistors in series is the simple sum R 2 R 1 V SS I ? R 3 R 4 + (Here its more convenient to use KVL than node analysis)

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 9 NODAL ANALYSIS USING KCL – The Voltage Divider – 1 Choose reference node 2 Define unknown node voltages V2V2 3 Write KCL at unknown nodes 4 Solve: +V2+V2 R1R1 V SS + i2i2 i1i1 R2R2

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 10 GENERALIZED VOLTAGE DIVIDER Circuit with several resistors in series R 2 R 1 V SS I R 3 R 4 + + + V1V1 V3V3 We know Thus, and etc..

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 11 WHEN IS VOLTAGE DIVIDER FORMULA CORRECT? R 2 R 1 V SS I R 3 R 4 + + 4321 2 2 V RRRR R V Correct if nothing else connected to nodes 3 V SS i R 2 R 1 R 3 R 4 + + 2 V R 5 i X 4321 2 2 V RRRR R V because R 5 removes condition of resistors in series Ii 3 What is V 2 in the second case? Answer: SS 43521 2 V )RR(RRR R V2V2

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 12 NODAL ANALYSIS WITH “FLOATING” VOLTAGE SOURCES A “floating” voltage source is a voltage source for which neither side is connected to the reference node. V LL in the circuit below is an example. What is the problem? We cannot write KCL at node a or b because there is no way to express the current through the voltage source in terms of. Solution: Define a “supernode” – that chunk of the circuit containing nodes a and b. Express KCL at this supernode. R 4 R 2 I 2 V a V b + - V LL I 1

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 13 FLOATING VOLTAGE SOURCES (cont.) Use a Gaussian surface to enclose the floating voltage source; write KCL for that surface R 4 R 2 I 2 V a V b + - V LL I 1 supernode We have two unknowns: V a and V b. We obtain one equation from KCL at supernode: 2 Equations & 2 Unknowns We obtain a second “auxiliary” equation from the property of the voltage source: (often called the “constraint”)

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 14 ANOTHER EXAMPLE 1Choose reference node (can it be chosen to avoid floating voltage source?) + + a b 2Label all unknown node voltages 4Auxiliary equation: R V R V ) R 1 R 1 (V 1 1 4 b 21 a Solve: 3Equation at supernode:

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 15 NODAL ANALYSIS WITH INDUCTORS Problem:The i-v relationship for inductors expresses voltage in terms of current, not current in terms of voltage. Its use would lead to integral differential equations, a messy business. We would like to see only derivatives. Solution:a)Enclose inductor in a supernode b)Add one variable, the inductor current c)Write equations as follows c.1KCL at supernode c.2 c.3KCL to express i L in terms of other currents TWO auxiliary equations because one extra variable is introduced

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S. Ross and W. G. OldhamEECS 40 Fall 2002 Lecture 8 Copyright, Regents University of California 16 EXAMPLE WITH INDUCTOR + + v 1 (t) R2R2 R3R3 V SS R4R4 L1L1 R1R1 Define Reference Define unknowns (including inductor current) : v a, v b, i L At supernode: inductor: KCL for i L : (1) (2) (3) Results in differential equation (substitute 3 into 2 and (1) into (2) to produce ODE for v a.)

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