# Lecture 8: 09/18/03 A.R. Neureuther Version Date 09/14/03 EECS 42 Introduction Digital Electronics Andrew R. Neureuther

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1 EECS ntroduction Digital Electronics ndrew. Neureuther Lecture #8 Node Equations Systematic Node Equations Example: oltage and Current Dividers Example 5 Element Circuit Schwarz and Oldham 5-58,.5, &.6 Quiz 9/5 0 min: asic Circuit nalysis and asic Transient Midterm 0/: Lectures # -9: Topics See slide Length/Credit eview T First Midterm Exam: Topics asic Circuit nalysis (KL, KCL) Equivalent Circuits and Graphical Solutions for Nonlinear Loads Transients in Single Capacitor Circuits Node nalysis Technique and Checking Solutions Exam is in class 9:0-0:5 M, Closed book, Closed notes, ring a calculator, Paper provided

2 Class nput on Midterm/eview Midterm Options Option : 65 min and % of grade in course (Final is 9%) Option : 80 min and 8% of grade in course (Final is 9%) eview Options Date Tu 9/0 or Wed 0/ Time 5-6:0 PM or Other FOML CCUT NLYSS Systematic approaches to writing down KCL and KL: Section. of Text - n particular use of KCL gives NODL NLYSS Mathematical foundation is rigorous: EE 0 Circuit Theory Nodal nalysis: Node voltages are the unknowns Mesh nalysis: ranch currents are the unknowns Use one or the other for circuit analysis We will do only nodal analysis (because voltages make more convenient variables than currents) Thus omit Text Section. ; it is redundant.

3 THE CHCTESTCS OF N ELEMENT LLOW ETHE NODL O MESH (LOOP) NLYSS EEW FO ESSTO: OHM S Law i f we use associated current and voltage (i.e., i is defined as into v terminal), then v = i (Ohm s law) i Z Y nother version of the same statement, and the one most important to us: i = ( Z Y )/ (Ohm s law) NOTE ODE OF NODES: Z Y! FOML CCUT NLYSS USNG KCL: NODL NLYSS (Memorize these steps and apply them rigorously!) Choose a eference Node Define unknown node voltages (those not fixed by voltage sources) Write KCL at each unknown node, expressing current in terms of the node voltages (using the constitutive relationships of branch elements*) Solve the set of equations (N equations for N unknown node voltages) * With inductors or floating voltages we will use a modified Step : The Supernode Method see Lecture #8

4 NODL NLYSS USNG KCL Example: The oltage Divider Choose reference node Define unknown node voltages Write KCL at unknown nodes i SS i Solve: SS 0 = = SS This is of course the voltage divider formula and is by itself very useful. ESSTOS N PLLEL X Select eference Node Define unknown node voltages SS Note: ss =, i.e., X X SS = X = SS = SS ESULT EQULENT ESSTNCE: = ESULT CUENT DDE: X = = SS X = = SS

5 GENELZED OLTGE DDE (solved without Nodal nalysis) Circuit with several resistors in series SS?? We know = SS /( ) Thus, = SS and = SS etc.. etc.. WHEN S OLTGE DDE FOMUL COECT? SS SS Z X i 5 = SS Correct if nothing else connected to nodes Z SS because 5 removes condition of resistors in series i.e. i What is Z? 5

6 DENTFYNG SEES ND PLLEL COMNTONS Use series/parallel equivalents to simplify a circuit before starting KL/KCL eq parallel 5 = = = 0 K = 5 = 5 K 6 6 = 0 K 0 K ( ) = eq X? Please note the order of math operators here! DENTFYNG SEES ND PLLEL COMNTONS (cont.) Some circuits must be analyzed (not amenable to simple inspection) and are not in 5 - and 5 are not in series Special cases: = 0 O = 5 Example: = 0 ; 5 in series; eq = 5 O F = ( 5 ) ( ) 6

7 NOTHE EXMPLE OF NODL NLYSS Choose a reference node and define the node voltages (except reference node and the one set by the voltage source); node voltage known a b - S reference node pply KCL: Note the systematic use of the present a a b a node minus each of the other nodes = 0 divided by the element resistance that connects them. b a b S = 0 You can solve for a, b. What if we used different ref node? EXMPLE : Thevenin/Norton Find the Thévenin and Norton equivalents of: Find = OC from voltage divider. : KΩ KΩ f open circuit, = OC = X / = f - is shorted, SC = - /K = - m (into ) 0 TH = = K equivalent to K and equivalent to m K Thévenin Norton 7

8 EXMPLE Thevenin/Norton Continued n what sense is this circuit KΩ equivalent to these? K KΩ - m K They have identical - characteristics and therefore have The same open circuit voltage The same short circuit current Find the Thévenin and Norton equivalents of: Find Norton equivalent of circuit leg on the left: KΩ K m EXMPLE Thevenin/Norton KΩ m equivalent to.k (m and K) Two Norton circuits in parallel So, TH = N N = TH = N = / K Thus combine current sources (=) = N and combine resistors in parallel: 8/6 K = N and equivalent to m. K Thévenin Norton 8

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