Objective of LectureState Thévenin’s and Norton Theorems.

Chapter 4.5 and 4.6 Fundamentals of Electric Circuits

Demonstrate how Thévenin’s and Norton theorems ca be used to simplify a circuit to one that contains three components: a power source, equivalent resistor, and load.

Thévenin’s TheoremA linear two-terminal circuit can be replaced

with an equivalent circuit of an ideal voltage source, VTh, in series with a resistor, RTh.VTh is equal to the open-circuit voltage at the

terminals.RTh is the equivalent or input resistance when

the independent sources are turned off.

Circuit Schematic:Thévenin’s Theorem

Definitions for Thévenin’s Theorem

Linear circuit is a circuit where the voltage is directly proportional to the current (i.e., Ohm’s Law is followed).

Two terminals are the 2 nodes/2 wires that can make a connection between the circuit to the load.

Definitions for Thévenin’s Theorem

Open-circuit voltage Voc is the voltage, V, when the load is an open circuit (i.e., RL = ∞).

+Voc

_

ThOC VV

Definitions for Thévenin’s TheoremInput resistance is the resistance seen by

the load when VTh = 0V.

It is also the resistance of the linear circuit when the load is a short circuit (RL = 0).

SCThThin iVRR

Steps to Determine VTh and RTh1. Identify the load, which may be a resistor or a

part of the circuit.2. Replace the load with an open circuit .3. Calculate VOC. This is VTh.4. Turn off all independent voltage and currents

sources.5. Calculate the equivalent resistance of the

circuit. This is RTH. The current through and voltage across the load

in series with VTh and RTh is the load’s actual current and voltage in the originial circuit.

Norton’s TheoremA linear two-terminal circuit can be replaced

with an equivalent circuit of an ideal current source, IN, in series with a resistor, RN.IN is equal to the short-circuit current at the

terminals.RN is the equivalent or input resistance when

the independent sources are turned off.

Definitions for Norton’s Theorem

Open-circuit voltage Isc is the current, i, when the load is a short circuit (i.e., RL = 0).

NSC II

Definitions for Norton’s TheoremInput resistance is the resistance seen by

the load when IN = 0A.

It is also the resistance of the linear circuit when the load is an open circuit (RL = ∞).

NOCNin IVRR

Steps to Determine IN and RN1. Identify the load, which may be a resistor or a

part of the circuit.2. Replace the load with a short circuit .3. Calculate ISC. This is IN.4. Turn off all independent voltage and currents

sources.5. Calculate the equivalent resistance of the

circuit. This is RTH. The current through and voltage across the load

in parallel with IN and RN is the load’s actual current and voltage in the originial circuit.

Source ConversionA Thévenin equivalent circuit can easily be

transformed to a Norton equivalent circuit (or visa versa).If RTh = RN, then VTh = RNIN and IN = VTh/RTh

Value of TheoremsSimplification of complex circuits.

Used to predict the current through and voltage across any load attached to the two terminals.

Provides information to users of the circuit.

Example #1

Example #1 (con’t)Find IN and RN

Example #1 (con’t)Calculation for IN

Look at current divider equation:

If RTh = RN= 1k, then IN = 6mA

NN

N

NloadNload

NloadN

load

eqload

IRk

RmA

IRRR

RRI

R

RI

22

1

Why chose RTh = RN?Suppose VTh = 0V and IN = 0mA

Replace the voltage source with a short circuit.Replace the current source with an open

circuit.

Looking towards the source, both circuits have the identical resistance (1k).

Source TransformationEquations for Thévenin/Norton

Transformations

VTh = IN RTh

IN = VTh/RTh

RTh= RN

Alternative Approach: Example #1 IN is the current that flows when a short circuit

is used as the load with a voltage source

IN = VTh/RTh = 6mA

Alternative Approach VTh is the voltage across the load when an open

short circuit is used as the load with a current source

VTh = IN RTh = 6V

Example #2Simplification through Transformation

Example #2 (con’t)

Example #2 (con’t)Current Source to Voltage Source

Example #2 (con’t)

RTh = 3

VTh = 0.1A (3) = 0.3V

0.3V

Current Source to Voltage Source

Example #2 (con’t)

0.3V

Example #2 (con’t)

RTh = 2

IN = 3V/2 = 1.5A

Voltage Source to Current Source

0.3V

Example #2 - Solution 1Simplify to Minimum Number of Current

Sources

RTh = 6

IN = 0.3V/6 = 50.0mA

0.3V

Voltage Source to Current Source

Example #2 (con’t)

Current Sources in Parallel Add

Example #2 - Solution 2Simplify to Minimum Number of Voltage

Sources

0.3V

Example #2 (con’t)Transform solution for Norton circuit to Thévenin circuit to obtain single voltage source/single equivalent resistor in series with load.

PSpice

Example #2 - Solution 1

Example #2 – Solution 2

SummaryThévenin and Norton transformations are

performed to:Simplify a circuit for analysisAllow engineers to use a voltage source when a

current source is called out in the circuit schematic

Enable an engineer to determine the value of the load resistor for maximum power transfer/impedance matching.