EE602 1

Nodal Analysis

L EE602 2

Steps of Nodal Analysis

1. Choose a reference (ground) node.

2. Assign node voltages to the other nodes.

3. Apply KCL to each node other than the reference node; express currents in terms of node voltages.

4. Solve the resulting system of linear equations for the nodal voltages.

EE602 3

Example: A Summing Circuit

• The output voltage V of this circuit is proportional to the sum of the two input currents I1 and I2

• This circuit could be useful in audio applications or in instrumentation

• The output of this circuit would probably be connected to an amplifier

EE602 4

1. Reference Node

The reference node is called the ground node where V = 0

+

–

V 500Ω

500Ω

1kΩ

500Ω

500ΩI1 I2

EE 602 5

Steps of Nodal Analysis

1. Choose a reference (ground) node.

2. Assign node voltages to the other nodes.

3. Apply KCL to each node other than the reference node; express currents in terms of node voltages.

4. Solve the resulting system of linear equations for the nodal voltages.

EE602 6

2. Node Voltages

V1, V2, and V3 are unknowns for which we solve using KCL

500Ω

500Ω

1kΩ

500Ω

500ΩI1 I2

1 2 3

V1 V2 V3

EE602 7

Steps of Nodal Analysis

1. Choose a reference (ground) node.

2. Assign node voltages to the other nodes.

3. Apply KCL to each node other than the reference node; express currents in terms of node voltages.

4. Solve the resulting system of linear equations for the nodal voltages.

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Currents and Node Voltages

500Ω

V1500ΩV1 V2

Ω−

50021 VV

Ω5001V

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3. KCL at Node 1

500Ω

500ΩI1

V1 V2

Ω+

Ω−=

500500121

1

VVVI

EE602 10

3. KCL at Node 2

500Ω

1kΩ

500Ω V2 V3V1

0500k1500

32212 =Ω

−+Ω

+Ω

− VVVVV

Lect4 EEE 202 11

3. KCL at Node 3

2323

500500I

VVV =Ω

+Ω

−500Ω

500Ω

I2

V2 V3

EE602 12

Steps of Nodal Analysis

1. Choose a reference (ground) node.

2. Assign node voltages to the other nodes.

3. Apply KCL to each node other than the reference node; express currents in terms of node voltages.

4. Solve the resulting system of linear equations for the nodal voltages.

Lect4 EEE 202 13

+

–

V 500Ω

500Ω

1kΩ

500Ω

500ΩI1 I2

4. Summing Circuit Solution

Solution: V = 167I1 + 167I2

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A Linear Large Signal Equivalent to a Transistor

5V100Ib

+

–

Vo

50Ω

Ib

2kΩ1kΩ+–

+ –

0.7V

EE602 15

Steps of Nodal Analysis

1. Choose a reference (ground) node.

2. Assign node voltages to the other nodes.

3. Apply KCL to each node other than the reference node; express currents in terms of node voltages.

4. Solve the resulting system of linear equations for the nodal voltages.

Lect4 EEE 202 16

Linear Large Signal Equivalent

5V 100Ib

+

–

Vo

50Ω

Ib

2kΩ

1kΩ

0.7V

12 3 4

V1V2 V3 V4

+–

+ –

EE602 17

Steps of Nodal Analysis

1. Choose a reference (ground) node.

2. Assign node voltages to the other nodes.

3. Apply KCL to each node other than the reference node; express currents in terms of node voltages.

4. Solve the resulting system of linear equations for the nodal voltages.

Lect4 EEE 202 18

KCL @ Node 4

Ω=+

Ω−

k2100

50443 V

IVV

b

100Ib

+

–

Vo

50Ω

Ib

2kΩ

1kΩ+–

0.7V

12 3 4

V1 V2 V3 V4

5V

+ –

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The Dependent Source

• We must express Ib in terms of the node voltages:

• Equation from Node 4 becomes

Ω−=k1

21 VVIb

0k2k1

10050

42143 =Ω

−Ω

−+Ω

− VVVVV

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How to Proceed?

• The 0.7-V voltage supply makes it impossible to apply KCL to nodes 2 and 3, since we don’t know what current is passing through the supply

• We do know that

V2 – V3 = 0.7 V

• The above is a needed constraint equation

EE602 21

100Ib

+

–

Vo

50ΩIb

2kΩ

1kΩ

0.7V

14

V1 V2 V3 V4

+–

+ –

050k1

4312 =Ω

−+

Ω− VVVV

KCL at Supernode