Nodal Analysis - Lecture notes 2 PDF

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EET 300 || Chapter 2(C) Lesson Notes

Nodal Analysis

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9/6/2011

Circuit Analysis using Nodal Analysis While Mesh analysis works pretty well, it has the problem of only working within a flat plane circuit with obvious “window panes.” We can develop another method that works in more instances that uses KCL to formally analyze a circuit. This method, known as Nodal Analysis, starts out with the following general steps: 1.

Determine the number of nodes within a network.

2.

Pick a reference node, and label each remaining node with a subscripted value of voltage: V1, V2, etc.

3.

Apply KCL at each node except the reference node. Assume all currents are LEAVING each node (unless the current is the result of a current source). Each node is treated as a separate entity, independent of the application of KCL at the other nodes.

4.

Solve the resulting equations for the node voltages.

EET 300 || Chapter 2(C) Lesson Notes

Nodal Analysis

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Example: Determine the 2 node voltages and the current thru R2. 2A 

The 1st step is to identify a “common” or

R1

“Ground” reference node (the node the other nodes will be ‘with reference to’) and then label

 R2

64v

+ -

R3



the other nodes with subscripted voltages.

2A V1 8  64v

V2 R1

+ -

 R2

R3 10 

(The circuit was redrawn just a bit to make the node choice more obvious.)

The next step would be to perform KCL at each node, again assuming that all currents are leaving the node unless they are current sources.

@ V1 :

0  I8  I4  2A V  64v V  V2   8    2A  1  1  8 4    16V  V1  64v  2  V1  V2  48V  V1  2V1  2V2 48V  3V1  2V2

(Continued on the next page)

EET 300 || Chapter 2(C) Lesson Notes

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(Example continues)

@ V2 :

0  I10  I 4  2A  V2  0v V  V1   2 2A    20 10 4   40v  2  V2  0v   5  V2  V1  40v  2V2  5V2  5V1 40   5V1  7V2

Place the two equations into MATRIX FORM:

 48  3 2  V1         40  5 7   V2 

Now solve the two equations for the two unknowns.  3 2 D    3  7     2   5   21  10  11  5 7  48  2 D1   48  7   40   2   336  80  416 40 7 D2 

3 48  40  3  48   5  120  240  360 5 40 D1 416   37.818v  D 11 D 360 V2  2   32. 727v  11 D V1 

Check :  Assumes the current direction is changed on I1 so the negative 0  I1  I2  2A    sign is removed from the calculated I1.  0  I2  2A  I3 0    3.273A  1.273A  2 0    1.273A  2  3.273A 0    3.273A  3.273A 0    1.273A   1.27 3A 0  0 0  0

Example continues on the next page)

EET 300 || Chapter 2(C) Lesson Notes

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Example continues) Now all that is left is to determine the current thru R2.

IR2 

V1  V2 37.818v  32.727v  1.273A  R2 4

2A V1

Let’s take a second look at the circuit and the

 R1

equations we derived:

@V1:

1  1  1 6A   V1  V   4 2 4   8

V2  R2

64v

+ ‐

R3 

Note that V1 is multiplied by the sum of the Conductance’s of all attached resistors to that node. Then note that V2 is multiplied by the negative of the Conductance which

separates the two nodes.

@V2:

 1  1 1 2A     V1     V2  10 4   4

Again, note that in the equation for V2, the sum of the Conductance’s of the resistors

attached to V2 is multiplied by V2, while the negative Conductance of the resistor separating the two nodes is multiplied by V1. This enables us to define some formalized rules for creating a Nodal analysis equation set.

EET 300 || Chapter 2(C) Lesson Notes

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Formal Nodal Analysis I.

Choose a reference node and label the remaining (N-1) nodes with a subscripted voltage. Convert all voltage sources to current sources if possible.

II.

The number of equations required is equal to the # of nodes - 1 (N-1). Column

1 of each equation is formed by summing the Conductance’s tied to the node of interest and multiplying the result by the subscripted nodal voltage of interest. III.

We must now consider the mutual terms that are subtracted from the 1st column. It’s possible to have more than one mutual term since there may be more than one branch off of a node. Each mutual term is the product of the mutual conductance and the other node voltage tied to the node in question.

IV.

The other side of the equal sign holds the algebraic sum of the current sources attached to the node of interest. A.

A current source is assigned a positive sign if it supplies current to a node (goes into the node) and a

B.

V.

Negative sign if it draws current from a node (leaves a node).

Reorder the equations and solve. Again, nothing was said of what to do if there

is a dependent source in the circuit.

EET 300 || Chapter 2(C) Lesson Notes

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Let’s use this formalized method to solve the same circuit that we solved with a previous MESH example.



i1 

Transform the voltage sources, identify the ground node and label the nodes. 20v  2.5A 8

i2 

10v  625mA 10  6

15v  500mA 30  This is the circuit after the source transformations. i3 



Now, we can use the formal method to determine the node equations.

@ V1

 1 1 1    V1  0  .5A  2.5A   30 8 8   8 30 30  30   2A   V2   V1  240 240 240 240  

1 1 V  V 8 2 8 3 30 V 240 3

 480  68V1  30V2  30V3

@ V2

1 1 1  1 1  0    V3  625mA  2.5A  V2  V1  8 12  8 12 16   6 4 3  6 4   3.125A   V1  V3  V2  48 48 48 48 48   150   6V1  13V2  4V3

@ V3

1 1 1  1 1 0    V  V3  V1  8 12 2  8 5 12   15 15 10 24 10    0   V1  V2    V3 120 120  120 120 120  0  15V1  10V2  49V3

Example continues on the next page)

EET 300 || Chapter 2(C) Lesson Notes

Nodal Analysis

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Example continues) 

Next, place the three equations into Matrix form: 480  68V1  30 V2  30 V3 150  6V1  13V2  4V3 0  15V1  10V2  49V3



 480   68 30    13   150     6   15  10  0    

30   V1    4   V2  49  V3 

Now, solve for the three node voltages:

First, solve for V1:

68  30  30 13 4 13 6 4 6 D  6 13    30     30   4  68  10 49  15 49  15 10  15  10 49  68   13  49    4   10   30   6   49    4   15   30   6   10   13  15         68  637  40  30  294  60  30  60  195  68  597  30  354  30 2 55  40.596k  10.6 2k  7.65k   480 D1  150 0

D  22.32 6k

 30  30 13 4 150  4 150 13 13   30     30  4  480  10 49  10 49 0 0  10 49

  48 0 13   49    4   10    30  150  49   0   30 150   10   0    480 637  40   30 7.35k   30  1.5k    480 597   30 7.3 5k   30 1.5k   286.56k  220.5k  45k  V1 

D1 D



21.06k   943.29m V 22.326k

D1  21.06k

EET 300 || Chapter 2(C) Lesson Notes

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Now with D already known, we can solve for V2 and V3: 68  480 150 D2  6 0 15 D2  68

150 0

30 4 49

4 6 4  6 150    480    30  49  15 49  15 0

 68  150   49   0   480     6   49     4    15   30   0  150   15      68  7.35k    480  294  60    30  2.25k 

 499.8k   480 354   67.5k  432.3k  169.92 k  26 2.38k V2 

D2 D



262.38k  11 .752v 22.326k

68 30 480 D3   6 13 150 0  15  10   68

13

150

 10

0

   30 

13 6 150 6   480   15 0  15  10

  68 0  150    10    30  0  150    15      480  6   10    13   15       68 1.5k   30  2.25k   480 60  195    102k  67.5k  4 80 255  D3  169.5k  122.4k  47.1k V3 

D3 47.1k   2.109v D 22.326 k

Example continues on the next page)

EET 300 || Chapter 2(C) Lesson Notes

Nodal Analysis

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Example continues) Finally, let’s check the results and see how they compare with our Mesh results. Remember from the MESH example, the currents were:

IR1  913.061mA  IR2  109.514mA  IR3  803.547mA  IR4  109.514mA  IR5  381.618mA  IR6  421.929mA  IR7  531.443mA  Applying a series of KVL equations:

0   V3  R6 IR6 V3  R6IR6

V3  5  421.929mA  V3  2.11v checks 0=-V2 +R2I2++10v+R 4I 4 but I4 =I2 V2 =R2I2 ++10v+R4I2

V2 =  R2  R4  I2 +10v

V2 =16  109.514mA   10v V2  11.752v

checks

0   V1  20v  8I1  V2 V1  20v  8I1  11.752v

V1   8.248v  8 9 13.06 1mA  V1   943.512mv checks...