Thevenin + Norton Eq. circuits

Low distortion audio power amplifier

From PreAmp(voltage ) To speakers

TO MATCH SPEAKERS AND AMPLIFIERIT IS MUCH EASIER TO CONSIDER THISEQUIVALENT CIRCUIT!

TO MATCH SPEAKERS ANDAMPLIFIER ONE SHOULD ANALYZETHIS CIRCUIT

+-

R T H

V T H

REPLACE AMPLIFIERBY SIMPLER “EQUIVALENT”

Independent Sources (Thevenin)

Circuit with independent sources

RTh

Voc

Thevenin equivalent circuit

+–

Thevenin's Theorem states that it is possible to simplify any linear circuit, no matter how complex, to an equivalent circuit with just a single voltage source and series resistance connected to a load.

Thevenin Theorem Example Application

To create the Thevenin Equivalent Circuit we need:

1. Value of the Thevenin Voltage Source

2. Value of the Thevenin Resistance

Determination of the Thevenin

Voltage

EThevenin = Open circuit voltage with load removed

Determination of the Thevenin

Voltage

EThevenin = Open circuit voltage with load removed

EThevenin = 11.2

Determination of the Thevenin

Resistance

RThevenin = Net resistance in network with sources set to zero

RThevenin = 0.8 ohms

Thevenin Theorem Summary

EThevenin = 11.2 volts

RThevenin = 0.8 ohms

The Maximum Power Transfer Theorem simply states, the maximum amount of power will be dissipated by a load resistance when that load resistance is equal to the Thevenin/Norton resistance of the network supplying the power.

V6

k5

][1)6(51

1VV

kk

kVO

THEVENIN & NORTONTHEVENIN’S THEOREM: Example

Find VX by first finding VTH and RTH to the left of A-B.

1 2 4

6 2 V X3 0 V +_

+

_

A

B

First remove everything to the right of A-B.

THEVENIN & NORTONTHEVENIN’S THEOREM: Example

1 2 4

6 3 0 V +_

A

B

(30)(6)10

6 12ABV V

Notice that there is no current flowing in the 4 resistor(A-B) is open. Thus there can be no voltage across theresistor.

THEVENIN & NORTONTHEVENIN’S THEOREM: Example

We now deactivate the sources to the left of A-B and findthe resistance seen looking in these terminals.

1 2 4

6

A

B

RTH

We see, RTH = 12||6 + 4 = 8

THEVENIN & NORTONTHEVENIN’S THEOREM: Example

After having found the Thevenin circuit, we connect thisto the load in order to find VX.

8

1 0 VV TH

R TH

2 V X

+

_

+_

A

B

10 22

2 8

( )( )

XV V

EXAMPLE: SOLVE BY SOURCE TRANSFORMATION

The equivalent current source will have the value 12V/3k

The 3k and the 6k resistors now are in paralleland can be combined

In between the terminals we connect a currentsource and a resistance in parallel

In between the terminals we connect a voltagesource in series with the resistor

The equivalent source has value 4mA*2k

The 2k and the 2k resistor become connected in series and can be combined

After the transformation the sources can be combined

The equivalent current source has value 8V/4kand the combined current source has value 4mA

Options at this point

1. Do another source transformation and get a single loop circuit

2. Use current divider to compute I_0 and thencompute V_0 using Ohm’s law

LEARNING EXAMPLE

In the region shown, one could use source transformation twice and reduce that part toa single source with a resistor.

... Or we can apply Thevenin Equivalenceto that part (viewed as “Part A”)

kRTH 4 For the open loop voltagethe part outside the regionis eliminated][8][12

63

6VVVTH

The original circuit becomes...

And one can apply Thevenin one more time!

kR TH 41

1THV

For open loop voltage use KVL

VVmAkVTH 1682*41

...and we have a simple voltage divider!!

VVV 8][1688

80

COMPUTE Vo USING THEVENIN

Or one more source transformation

eqeqeq IRV +-V e q

R e q R 3

R 4

0V

PROBLEM Compute V_0 using source transformation

3 current sources in parallel and three resistors in parallel

0I

eqeqeq IRV eqeq

VRRR

RV

34

40

EQUIVALENT CIRCUITS

TH

TH

V

R