Thevenin + Norton Eq. circuits
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