NODE ANALYSIS. One of the systematic ways to determine every voltage and current in a circuit

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1 NODE ANALYSIS One of the systematic ways to determine eery oltage and current in a circuit The ariables used to describe the circuit will be Node oltages -- The oltages of each node with respect to a pre-selected reference node

2 IT IS INSTRUCTIE TO START THE PRESENTATION WITH A RECAP OF A PROBLEM SOLED BEFORE USING SERIES/ PARALLEL RESISTOR COMBINATIONS COMPUTE ALL THE OLTAGES AND CURRENTS IN THIS CIRCUIT

3 4k k k SECOND: BACKTRACK USING KL, KCL OHM S I 3 a OHM' S : I KCL : I I I3 OHM' S : b 3k * I3 OTHER OPTIONS... I4 I3 4 + FIRST REDUCE TO A SINGLE LOOP CIRCUIT b 4k * I4 KCL : I5 + I4 I3 OHM'S : C 3k * I5 I k a 3 () 3 + 9

4 THE NODE ANALYSIS PERSPECTIE + + S S S + KL a a a a KL a 3 + b b b + 5 KL c 5 + b c REFERENCE WHAT IS THE PATTERN??? THERE ARE FIE NODES. IF ONE NODE IS SELECTED AS REFERENCE THEN THERE ARE FOUR OLTAGES WITH RESPECT TO THE REFERENCE NODE 5 b c ONCE THE OLTAGES ARE KNOWN THE CURRENTS CAN BE COMPUTED USING OHM S LAW THEOREM: IF ALL NODE OLTAGES WITH RESPECT TO A COMMON REFERENCE NODE ARE KNOWN THEN ONE CAN DETERMINE ANY OTHER ELECTRICAL ARIABLE FOR THE CIRCUIT R m N + R DRILL QUESTION ca A GENERAL IEW

5 THE REFERENCE DIRECTION FOR CURRENTS IS IRRELEANT + R ' R + i' USING THE LEFT-RIGHT REFERENCE DIRECTION THE OLTAGE DROP ACROSS THE RESISTOR MUST HAE THE POLARITY SHOWN ' i i OHM'S LAW i m R PASSIE SIGN CONENTION RULES! N IF THE CURRENT REFERENCE DIRECTION IS REERSED... THE PASSIE SIGN CONENTION WILL ASSIGN THE REERSE REFERENCE POLARITY TO THE OLTAGE ACROSS THE RESISTOR OHM'S LAW i ' R N m

6 DEFINING THE REFERENCE NODE IS ITAL THESTATEMENT 4 IS MEANINGLES UNTIL THE REFERENCE POINT IS DEFINED BY CONENTION THE GROUND SYMBOL SPECIFIES THE REFERENCE POINT. ALL NODE OLTAGES ARE MEASURED WITH RESPECT TO THAT REFERENCE POINT?

7 THE STRATEGY FOR NODE ANALYSIS S a b c. IDENTIFY ALL NODES AND SELECT A REFERENCE NODE. IDENTIFY KNOWN NODE a : I + I + I 3 a s a a b + + 9k b : I + I + I 4 5 b a b b c + + 3k 4k c : I + I 6 c b c + 9k 3k 3 5 REFERENCE 3. AT EACH NODE WITH UNKNOWN OLTAGE WRITE A KCL EQUATION (e.g.,sum OF CURRENT LEAING ) 4. REPLACE CURRENTS IN TERMS OF NODE OLTAGES AND GET ALGEBRAIC EQUATIONS IN THE NODE OLTAGES... SHORTCUT: SKIP WRITING THESE EQUATIONS... AND PRACTICE WRITING THESE DIRECTLY

8 WHEN WRITING A NODE EQUATION... AT EACH NODE ONE CAN CHOSE ARBITRARY DIRECTIONS FOR THE CURRENTS b a R b R3 c a c I R I 3 I a b a R b 3 c ' I R d d R ' I 3 ' I c d d AND SELECT ANY FORM OF KCL. WHEN THE CURRENTS ARE REPLACED IN TERMS OF THE NODE OLTAGES THE NODE EQUATIONS THAT RESULT ARE THE SAME OR EQUIALENT CURRENTS INTO NODE a b b d b c I I I3 R R R 3 3 CURRENTS LEAING ' ' ' b a b d c b I + I I3 + R R R CURRENTS INTO NODE ' ' ' b a b d c b CURRENTS LEAING I I + I3 + R R R3 a b b d b c I + I + I3 + + R R R WHEN WRITING THE NODE EQUATIONS WRITE THE EQUATION DIRECTLY IN TERMS OF THE NODE OLTAGES. BY DEFAULT USE KCL IN THE FORM SUM-OF-CURRENTS-LEAING THE REFERENCE DIRECTION FOR THE CURRENTS DOES NOT AFFECT THE NODE EQUATION 3

9 CIRCUITS WITH ONLY INDEPENDENT SOURCES HINT: THE FORMAL MANIPULATION OF EQUATIONS MAY BE SIMPLER IF ONE USES CONDUCTANCES INSTEAD OF NODE USING RESISTANCES i A + R + R WITH CONDUCTANCES REORDERING TERMS i A + G + G( NODE REORDERING TERMS THE MODEL FOR THE CIRCUIT IS A SYSTEM OF ALGEBRAIC EQUATIONS

10 EXAMPLE WRITE THE KCL NODE WE ISUALIZE THE CURRENTS LEAING AND WRITE THE KCL EQUATION REPEAT THE PROCESS AT NODE i + + R4 R3 OR ISUALIZE CURRENTS GOING INTO NODE

11 ANOTHER EXAMPLE OF WRITING NODE EQUATIONS B B 5mA A MARK THE NODES (TO INSURE THAT NONE IS MISSING) A 8k k 8k k C SELECT AS REFERENCE k B B B + 5mA 8k k WRITE KCL AT EACH NODE IN TERMS OF NODE A A A + + 5mA

12 LEARNING : 4mA + + : ma + + k BY INSPECTION + k k + + k k USING KCL 4mA ma

13 LEARNING EXTENSION 6mA I I 3 I Node : ma 6mA 6 k + : 6mA + + 3k IN MOST CASES THERE ARE SEERAL DIFFERENT WAYS OF SOLING A PROBLEM I 8mA 3k I (6 ma) ma 3k + I (6 ma) 4mA 3 3k + NODE EQS. BY INSPECTION k ( ) ( 6)mA + + 3k ( ) ma Once node oltages are known I I I3 k 3k CURRENTS COULD BE COMPUTED DIRECTLY USING KCL AND CURRENT DIIDER!!

14 CIRCUITS WITH INDEPENDENT OLTAGE SOURCES 3 nodes plus the reference. In principle one needs 3 equations... but two nodes are connected to the reference through oltage sources. Hence those node oltages are known!!! Hint: Each oltage source connected to the reference node saes one node equation Only one KCL is necessary + k [ ] + k THESE ARE THE REMAINING TWO NODE EQUATIONS 6[ ] SOLING THE EQUATIONS + ( 3 ) 6[ ] + (.5[ ] )

15 THE SUPERNODE TECHNIQUE We will use this example to introduce the concept of a SUPERNODE SUPERNODE I S Conentional node analysis requires all currents at a 6mA ma + I S S k eqs, 3 unknowns...panic!! The current through the source is not related to the oltage of the source Math solution: add one equation 6[ ] Efficient solution: enclose the source, and all elements in parallel, inside a surface. Apply KCL to the surface!!! + + 4mA k 6 ma + The source current is interior to the surface and is not required We STILL need one more equation 6[ ] Only eqs in two unknowns!!!

16 ALGEBRAIC DETAILS The Equations () () Solution + 6mA k 6[ ] + 4mA. Eliminate denominators in Eq(). Multiply by [ ] 6[ ]. Add equations to eliminate 3[ ] [ ] 3. Use Eq() to compute 6[ ] 4[ ]

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18 SUPERNODE 4 6 SOURCES CONNECTED TO THE REFERENCE 4 CONSTRAINT EQUATION SUPERNODE k IS NOT NEEDED FOR I O ( 4) k k k * / k * /3 and add OHM'S LAW 3 I O 3. 8mA k