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Module 9 - Thévenin and Norton Equivalent Circuits

In this module, we’ll learn about an important propertyof resistive circuits called Thévenin Equivalence.

M. Leon Thévenin (1857-1926), published his famous theorem in 1883.

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Thévenin’s Theorem applies to circuits containingresistors, voltage sources, and/or current sources

Thêvenin Equivalent Circuit

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Thévenin Equivalent Circuit

VTh

RTh

Thévenin’s Theorem: A resistive circuit can be represented by one voltage source and one resistor:

Resistive Circuit

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Definition of a “Port”

Resistive Circuit

Port: Set of any two terminals

PORT

PORT

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Simple resistive circuit

Vo

R1

R2vX

+

_

iX

Illustrate concept with a simple resistive circuit:

• Any two terminals can be designated as a port.

Define portvariables vX and iX

• Our objective: Find the equivalent circuit seen looking into the port

ix flows to some load

(not shown)

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Vo

R1

R2 vX

+

_

iX

Find an equation that relates vx to ix

i1

i2

KVL: i1R1 + i2R2 = Vo

(Each resistor voltage expressed using Ohm’s Law)

Also note: vX = i2R2

KCL: i1 = i2 + iX

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Solve these equations for vX versus iX :

i1R1 + i2R2 = Vo i1 = iX + i2 vX = i2R2

(iX + i2) R1 + i2R2 = Vo

(iX + vX/R2) R1 + vX = Vo

iX R1 + vX (R1 /R2 + 1) = Vo

Rearrange the variables…

or Vo – iX R1

vX = ––––––––– 1 + R1 /R2

R2 R1 R2

vX = Vo ––––––– – iX ––––––– R1 + R2 R1 + R2

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R1 R2 RTh = –––––––

R1 + R2

R2 R1 R2

vX = Vo ––––––– – iX ––––––– R1 + R2 R1 + R2

Vo

R1

R2 vX

+

_

iX

Examine this last equation:

It has the form vX = VTh – iX RTh

R2

VTh = Vo ––––––– R1/ + R2

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Constructing the Thévenin Equivalent Circuit

VTh

RTh

vX

+

_

iX

Write down KVL for this circuit:

vX = VTh – iXRTh

+ –iXRTh

“Output voltage = voltage source – voltage drop across RTh”

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vX = VTh – iXRTh

VTh

RTh

vX

+

_

iX

Vo

R1

R2 vX

+

_

iX

R2 R1 R2

vX = Vo ––––––– – iX ––––––– R1 + R2 R1 + R2

Choose model parameters VTh and RTh:

Actual Circuit: Model:

R2

VTh = Vo ––––––– R1 + R2

R1 R2 RTh = ––––––= R1 || R2 R1 + R2

and

• From the point of view of vX and iX, the Thévenin circuit models the actual circuit in every way.

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Vo

R1

R2

Actual Circuit:

R1 || R2

R2

Vo ––––––– R1 + R2

PORT

PORT

vX

vX

iX

iX

+

_

+

_

Thévenin Equivalent:

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Significance of RTh

Vo

R1

R2

• Set all independent sources in the actual circuit to zero.

• For a voltage source, that means substituting a short circuit.

Equivalent resistance

• Equivalent resistance RTh= R1||R2

RTh is the equivalent resistance seen looking into the port with all independent sources set to zero.

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Setting a Voltage Source to Zero

Voltage betweennodes fixed at Vo

Current determined bywhat’s connected…

Vo

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Setting a Voltage Source to Zero

Voltage betweennodes fixed at 0 Vby short circuit

LOAD

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Setting a Current Source to Zero

Voltage betweennodes determined bywhat’s connected

Current through branch set to Io

Ioopen circuit

x

x

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Significance of VTh

Vo

R1

R2

Open Circuit Voltage

• Connect nothing to the port

iX = 0

Open Circuit Voltage

• VTh represents the open circuit voltage of the actual circuit

iX = 0

• iX automatically set to zero.

• Port voltage is called the open circuit voltage.

+

_

VTh

RTh

+

_

+ 0 V –

KVL

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Example: Resistor Network

Balanced audio microphone system

R3 = 10 kR1=100 k

R2 = 30 k

R4 =10 k

Vmic

10 mV

What voltage is developed across a 50 k resistive load?

50 k = Input resistance of typical audio amplifier.

Microphone networkLoad

50 k

+vLOAD

–

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Solution Method: Find the Thévenin Equivalent of the Microphone Network

• Find Thevenin Equivalent remaining circuit.

• Reconnect the load.

• Find vLOAD from simplified circuit.

• Disconnect the load.

R1=100 k

R2 = 30 k

R3 = 10 k

R4 =10 k

Vmic

10 mV

Load Load

Find VTh

and RTh

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Step 1: Find the Equivalent Resistance

• Set the voltage source to zero. (Substitute a short circuit.)

R1=100 k

R2 = 30 k

R3 = 10 k

R4 =10 k

RThVmic

• Find the equivalent resistance RTh

• RTh = R3 + R1||R2 + R4 = 10 k + 23 k + 10 k = 43 k

43 k

RTh

Note: R1||R2 = (100 k)||(30 k) = 23 k

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Step 2: Find the Open Circuit Voltage

• Analyze the circuit under no-load conditions.

R1=100 k

R2 = 30 k

R3 = 10 k

R4 =10 k

Vmic

10 mV

• Voltage across port terminals will be VTh

• From KVL around the inner loop*: v2 = VmicR2/(R1 + R2) = 2.3 mV

• Note that no current flows through R3 and R4. Voltage across these resistors is zero.

VTh = 2.3 mV

+

–

*basically, voltage division

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Step 3: Reconnect the Load to the Thévenin Equivalent Model

RTh = 43 k

VTh

2.3 mV

Thévenin equivalent ofmicrophone network

From simple voltage division:

vLOAD = VTh (RLOAD/(RLOAD + RTh)

= 2.3 mV (50 k)/(93 k) = 0.54 mV Answer

50 k

+vLOAD

–RLOAD

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More Examples:The Norton Equivalent Circuit

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Short Circuit Current

Another important parameter of a circuit is its short circuit current

The short circuit current of a port is defined as the current that will flow if:

The load is disconnected

A short circuit is connected instead

RTh

VTh Isc = VTh /RTh

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Circuit Containing a Current Source

Consider the following simple circuit:

I1 R1

What is the Thévenin equivalent circuit seen looking intothe port?

Port

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Step 1: Find the open circuit voltage:

I1 R1

+VTh

–

• From Ohm’s Law:

VTh = I1R1

(That part is simple…)

+

–

Current is zero

• Open circuit conditions All of I1 flows through R1

• Voltage develops across R1 with polarity shown.

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Step 2: Find the equivalent resistance

I1R1

• Trivially, by inspection: RTh = R1

• Set the current source to zero.

• Find the resistance looking into the port.

RTh

• Set the current source to zero open circuit

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The Thévenin Equivalent Circuit:

R1

I1R1

Thévenin Equivalent

VTh = I1R1

RTh = R1

Actual Circuit:

Done!

I1 R1

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Norton Equivalent Circuit

• The Norton and Thévenin equivalents of a circuit are interchangeable.

• The equivalent resistance is the same: RN = RTh

• The open circuit voltage is the same: VTh = INRN

Norton Circuit.

RN

INRNIN RN

Thévenin Circuit.

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INRN Isc = IN

+VN = 0

–

What about the short-circuit current from a Norton Circuit?

• Apply a short circuit:

• The voltage across the Norton resistance becomes zero.

• No current flows through the Norton resistance (I = V/R).

• All the current flows through the short circuit.

• The short circuit current is the source current IN.

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Norton Equivalent Circuit

• The short circuit current is the same in each circuit: IN = VTh/RTh

Norton Circuit

VTh = INRN

IN =VTh/RTh

RTh = RN

Thévenin Circuit

RN = RTh

IN IN

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Example: Resistive Network

R1=100 k

R2 = 30 k

R3 = 10 k

R4 =10 k

Vmic

10 mV

Find the Norton Equivalent of the following circuit using theshort-circuit current method

Step 1: Find RTh (same as RN) by setting the source to zero.

RTh or RN

By inspection, RTh = R3 + R1||R2 + R4 = 10 k + 23 k + 10 k = 43 k

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R2 = 30 k

R3 = 10 k

R4 =10 k

R1=100 k

Vmic

10 mV

Step 2: Apply a short circuit to the port and compute the short-circuit current.

R1=100 k

Vmic

10 mV

RP = R2 || (R3 + R4) = 12 k

IP = Vmic/(R1 + RP) = 0.9 A

From current division:

R2

[R2 + (R3 + R4)]ISC = IP = 0.9 A

30 k50 k = 0.54 A = ISC

= 0.54 AISC

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Find the Norton Equivalent of the Circuit

ISC = 0.54 A

RN = 43 k

RN = 43 kIN = 0.54 A = 23 mV

+vOC

–

“Open Circuit Voltage”

vOC = IN RN = (0.54 A)(43 k) = 23 mV

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Construct the Thévenin Equivalent of the Circuit

ISC = 0.54 A

RTh = 43 k

VTh = ISC RTh = (0.54 A )(43 k) = 23 mV

VTh = 23 mV

RTh = 43 k

This result is the same one obtained in the previous example!

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A circuit that can be represented by a Thévenin Equivalentcan also be represented by its corresponding Norton circuit

VTh= INRN

RTh = RN

INRN

Norton Equivalent Thévenin Equivalent

RTh

VTh

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End of This Module

Do the Homework Exercises