1. Thevenin + Norton Eq. circuits

2. Low distortion audio power amplifier
TO MATCH SPEAKERS AND
AMPLIFIER ONE SHOULD ANALYZE
THIS CIRCUIT
From PreAmp
(voltage )
To speakers
RTH
TO MATCH SPEAKERS AND AMPLIFIER
IT IS MUCH EASIER TO CONSIDER THIS
EQUIVALENT CIRCUIT!
VTH
+
-
REPLACE AMPLIFIER
BY SIMPLER
“EQUIVALENT”

3. Independent Sources (Thevenin)
RTh
Voc
Circuit with independent
sources
+
–
Thevenin equivalent circuit

4. 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.

5. Thevenin Theorem
Example Application

6. To create the Thevenin Equivalent Circuit we need:
1. Value of the Thevenin Voltage Source
2. Value of the Thevenin Resistance

7. Determination
of the Thevenin
Voltage
EThevenin = Open circuit voltage with load removed

8. Determination
of the Thevenin
Voltage
EThevenin = Open circuit voltage with load removed
EThevenin = 11.2

9. Determination
of the Thevenin
Resistance
RThevenin = Net resistance in network with sources set to zero
RThevenin = 0.8 ohms

10. Thevenin Theorem
Summary
EThevenin = 11.2 volts
RThevenin = 0.8 ohms

12. 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.

13. 5k
6V
VO
1k
(6V ) 1[V ]
1k 5k

14. THEVENIN & NORTON
THEVENIN’S THEOREM: Example
Find VX by first finding VTH and RTH to the left of A-B.
4
12
A
+
_
30 V +
6
VX
2
_
B
First remove everything to the right of A-B.

15. THEVENIN & NORTON
THEVENIN’S THEOREM: Example
4
12
_
30 V +
A
6
B
VAB
(30)(6)
6 12
10V
Notice that there is no current flowing in the 4 resistor
(A-B) is open. Thus there can be no voltage across the
resistor.

16. THEVENIN & NORTON
THEVENIN’S THEOREM: Example
We now deactivate the sources to the left of A-B and find
the resistance seen looking in these terminals.
4
12
A
RTH
6
B
We see,
RTH = 12||6 + 4 = 8

17. THEVENIN & NORTON
THEVENIN’S THEOREM: Example
After having found the Thevenin circuit, we connect this
to the load in order to find VX.
RTH
A
8
VTH
+
_
10 V
+
2
VX
_
B
VX
(10)( 2)
2 8
2V

18. EXAMPLE: SOLVE BY SOURCE TRANSFORMATION
In between the terminals we connect a voltage
source in series with the resistor
The equivalent current source will have the
value 12V/3k
The 3k and the 6k resistors now are in parallel
and can be combined
In between the terminals we connect a current
source and a resistance in parallel
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/4k
and 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 then
compute V_0 using Ohm’s law

19. LEARNING EXAMPLE
COMPUTE Vo USING THEVENIN
In the region shown, one could use source
transformation twice and reduce that part to
a single source with a resistor.
... Or we can apply Thevenin Equivalence
to that part (viewed as “Part A”)
RTH
The original circuit becomes...
VTH
6
3 6
4k
12[V ] 8[V ]
For the open loop voltage
the part outside the region
is eliminated
And one can apply Thevenin one more time!
For open loop voltage use KVL
1
VTH
1
TH
R
1
VTH
4k
...and we have a simple voltage divider!!
V0
8
8 8
16[V ] 8V
4k * 2mA 8V
16V

20. PROBLEM Compute V_0 using source transformation
EQUIVALENT CIRCUITS
I0
Or one more source transformation
R eq
Veq
+
-
3 current sources in parallel and
three resistors in parallel
R3
RTH
VTH
Veq Req I eq
R4
V0
Veq
Req I eq
V0
R4
R4
Veq
R3 Req