EE 204
Lecture 28
Maximum Power Transfer in Sinusoidal
Circuits
Important Note:
It is customary in power engineering to refer to “average power”, using the single
word “power”.
In this course, whenever we deal with sinusoidal circuits, the term “power” will be
understood to mean “average power”.
Maximum Power Transfer in Sinusoidal Circuits:
Given the Thevenin’s Equivalent Circuit (TEC) [ Vth in series with Z th ], what value of
load impedance Z L results in maximum power absorption?
Z th
ZL
V th
Figure 1
It can be shown (no proof is given) that maximum power is absorbed by Z L if the
following condition is satisfied:
Z L = Z th∗
( Z L = the complex conjugate of Z th )
Z th
Z L = Z th*
V th
Figure 2
Which means that:
RL + jX L = ( Rth + jX th )∗ = Rth − jX th
The above equation leads to the following conditions:
RL = Rth
[Load resistance = Thevenin’s resistance]
X L = − X th
[Load reactance = Minus the Thevenin’s reactance]
Z th = Rth + jX th
Z L = Rth − jX th
V th
Figure 3
The maximum power transferred to Z L can be obtained as follows:
I =
Vth
Vth
Vth
=
=
∗
Z th + Z L Z th + Z th ( Rth + jX th ) + ( Rth + jX th )∗
I =
Vth
Vth
V
=
= th
( Rth + jX th ) + ( Rth − jX th ) ( Rth + Rth ) + j ( X th − X th ) 2 Rth
Since only the resistive part of Z L absorbs power, then:
Pmax = 0.5 I 2 Rth = 0.5(
Pmax = I 2 Rth = (
Vth 2
V2
) Rth = th [When peak values are assumed]
2 Rth
8 Rth
Vth 2
V2
) Rth = th
2 Rth
4 Rth
[When RMS units are used]
Z th = Rth + jX th
I
Z L = Rth − jX th
V th
Figure 4
Summary:
For maximum power transfer Z L = Z th∗
And the maximum power is given by:
Pmax =
Vth2
V2
[peak values used] or Pmax = th [RMS units used]
8 Rth
4 Rth
Example 1:
The load Z L is connected to the circuit.
a) Calculate Z L for maximum power transfer.
b) Calculate the maximum power transferred to Z L .
c) If ω = 100 [rad / s ] , calculate the values of the resistance and inductance (or
capacitance) required to realize the load impedance.
4
-j7
IS = 5 0
[ A rms ]
o
j5
ZL
ω = 100[rad s ]
Figure 5
Solution:
a) The TEC can easily be obtained using source transformation:
ZL
V S = 25 90o
ω = 100[rad s ]
[V rms ]
Figure 6
Z th = 4 − j 2 Ω
V S = 25 90o
[V rms ]
ZL
Figure 7
For maximum power transfer:
Z L = Z th∗
⇒ Z L = (4 − j 2)∗ = 4 + j 2 Ω
b) The maximum power transferred to Z L :
Pmax =
Vth2
[because we have RMS units]
4 Rth
Pmax =
252
= 39.063 W
4× 4
Z th = 4 − j 2 Ω
ZL =4+ j 2 Ω
V S = 25 90
o
[V rms ]
Figure 8
c) Since Z L = 4 + j 2 Ω
⇒
RL = 4Ω
Z L has a positive reactance ( X L = 2 > 0 ) ⇒ The impedance is inductive
ω LL = 2
⇒
100 LL = 2
⇒
LL = 20 mH
The load consists of a 4Ω resistor in series with a 20 mH inductor
R L = 4Ω
ZL =4+ j 2 Ω
L L = 20 mH
Figure 9
Maximum Power Transfer for Purely Resistive Loads:
For maximum power transfer the condition Z L = Z th∗ must be satisfied.
Z th
V th
Z L = Z th*
When the load is restricted to be purely resistive so that Z L = RL , then in general, the
load cannot absorb the maximum possible power no matter what RL is.
Z th
RL
V th
Figure 10
However, it can be shown that the following condition:
RL = Z th
( RL = Z th = Rth2 + X th2 )
Insures that maximum power is absorbed by the resistive load, which is less than the
maximum possible power (why?)
It can also be shown that the maximum power absorbed in this case is given by:
Pmax =
Vth2
[When peak values are assumed]
4( Z th + Rth )
Pmax =
Vth2
[When RMS units are used]
2( Z th + Rth )
This type of problem belongs to the class of conditional or constrained maxima
Summary:
When the load impedance is restricted to be purely resistive, maximum power is
transferred to the load when RL = Z th
And the maximum power in this case is given by:
2) Pmax
Vth2
=
[peak values used]
4( Z th + Rth )
or Pmax
Vth2
=
[RMS units used]
2( Z th + Rth )
Z th
R L = Z th = Z th
V th
Figure 11
Example 2:
A resistive load RL is connected to circuit as shown.
a) Calculate RL for maximum power absorption.
b) Calculate the maximum power absorbed by RL .
c) Show that the maximum power calculated in b) is less than the maximum possible
power, which can only be obtained if the load impedance is allowed to be complex.
RL
I S = 5 0o
ω = 100[rad s ]
[ A rms ]
Figure 12
Solution:
a) The TEC seen by the load has been calculated in the previous example.
Z th = 4 − j 2 Ω
RL
V S = 25 90o
[V rms ]
Figure 13
The purely resistive load required for maximum power transfer:
RL = Z th
⇒
RL = 4 − j 2 = 42 + 22 = 4.472 Ω
b) The maximum power transferred to RL :
Pmax =
Vth2
⇒
2( Z th + Rth )
Pmax =
252
= 36.886 W
2 × (4.472 + 4)
c) In the previous example, the maximum possible power transfer is calculated to be
39.063 W, thus:
( PZ
L=
4.472 Ω
= 36.886) < ( PZ
L=
4+ j 2 Ω
= 39.063)
Z th = 4 − j 2 Ω
R L = 4.472 Ω
V S = 25 90o
[V rms ]
Figure 14