Fall 2001ENGR201 Nodal Analysis1 Read pages 65 - 80 Nodal Analysis: Nodal analysis is a systematic...

Fall 2001 ENGR201 Nodal Analysis 1

Read pages 65 - 80

Nodal Analysis: Nodal analysis is a systematic application of KCL that generates a system of equations which can be solved to find voltage at each node in a circuit. (We sum currents at each node to find the node voltages.)

Homework:•online HW, Nodal #1 and Nodal #2•3FE-1 and 3FE-3•Due 9/24/01

Chapter 3 – Nodal Analysis


1. Label all nodes in the circuit, 2. Select one node as the reference node (also called common).

The voltage at every other other node in the circuit is measured with respect to the reference node.

3. Write a KCL equation (i = 0) at each node.4. Solve the resulting set of equations for the node voltages.

Nodal Analysis Steps:

Branches connected to a node will have one of three types of elements:

current sources (independent or dependent)resistorsvoltage sources (independent or dependent)


Since we are applying KCL, current sources (either independent or dependent) connected to a node provide terms for our KCL equation that we can write down by inspection.

The next step is to write each resistive current in terms of the node voltages.If a current source is dependent, we must also write the dependent current in terms of the node voltages.

Nodal Analysis – Branches With Curent Sources

IS = IR1 + IR2 + IR3


Consider a single resistor connected between two arbitrary nodes:

A BR

+ VAB -

By KVL, the voltage drop from node-A to node-B is the difference between the voltage at node-A (VA0 = VA) and the voltage at node-B (VB0 = VB) .

+V A

-

+V

B-

The current leaving node-A going toward node-B, IAB, is:

IAB

AB A BAB

V V VI

R R

The current leaving node-B going toward node-A is:

BA B ABA

V V VI

R R

Nodal Analysis – Resistive Branches

0 V


If we apply the previous techniques to the resistors connected to node-X in the following circuit and apply KCL at node-X, we get the following equation. Note that the equation should have five terms since there are five branches connected to node-X and each branch will have a corresponding current

A

B

C D0

EX

I1

I2R1

R2

R32 1

1 2

01 2 3

01 2 3

X CX A X E

X CX A X E

V VV V V VI I

R R RV VV V V V

I IR R R

Example 1

currents leaving node-Xresistive branches

currents entering node-Xcurrent Sources


12 k

2 mA4 mAIx

6 k 6 k

Use nodal analysis to find Ix.

Ix

Step 1, Label nodes:

Example 2

4 mA 6 k

12 k

6 k 2 mA

V1 V2


Use nodal analysis to find Ix.Example 2 - continued

12 k

2 mA4 mAIx

6 k 6 k

V1 V2

Step 2: Write KCL equations at each node (except reference node):

1 2 1

2 1 2

04

12 60

212 6

V V VmA

k kV V V

mAk k

1 2

1 2

1 1 14V

12 6 12

1 1 12V

12 12 6

V V

V V



12 k

2 mA4 mAIx

6 k 6 k

V1 V2

In matrix form:1

2

1 1 1412 6 12

V21 1 1

12 12 6

V

V

Solving these equations (shown on the following slide) yields:V1 = -15 V and V2 = 3 V.



12 k

2 mA4 mAIx

6 k 6 k

V1 = -15 V V2 = 3 V

In terms of the node voltages: Ix = (V1 - V2)/12k = (-15 – 3)/12 k = -18v/ 12k Ix = -1.5mA


1 2

1 2

1 1 14V

12 6 12

1 1 12V

12 12 6

V V

V V

1

2

3

4

TI-86 Solution


1 2

1 2

1 1 14V

12 6 12

1 1 12V

12 12 6

V V

V V

TI-86 Solution

6

5

7


• Circuits containing dependent sources generally introduce another unknown - the parameter (voltage or current) that controls the dependent source.

• This requires that the additional unknown be eliminated by writing an equation that expresses the controlling parameter in terms of the node voltages.

• The resulting equations, with the additional unknown eliminated, are solved in a conventional manner.

• The following example illustrates.

Dependent Sources


IoIo

R1

R2

R3 IS

V1 V2 The nodal equations are:

Node #1: 1 1 2

1 2

0o

V V VI

R R

Node #2: 2 2 1

3 2S

V V VI

R R

• There are three unknowns in the equations, V1, V2 and Io. • Another equation is needed that relates Io to V1 and/or V2.

The additional equation can be formed by noting the Io is the current through R3, and by Ohm’s law Io = V2/R3. This relation can be used to form a system of three- equations or to eliminate Io from the first equation, leaving a two-by-two system to solve.

Dependent Source Example


Use nodal analysis to find node voltages V1 and V2.

2Io

Io

V1 V2 The node equations are:

1 1 2

2 1

410 10

22 0

10 10 o

V V Vma

k kV V V

Ik k

The “extra” unknown, Io, can be expressed as:

k10

VI 1oThe equations become:

0k10

V2

k10

VV

k10

V

ma4k10

VV

k10

V

1122

211

v8Vandv16V0V2V

volts40VV221

21

21

Example 3

Fall 2001ENGR201 Nodal Analysis1 Read pages 65 - 80 Nodal Analysis: Nodal analysis is a systematic...

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