基本电路理论课程论文
2006-2007 第一学期
A discussion about maximum power transfer
夏海颖，包雪娜，李光北，张引玉，杨阳
Outline:
In this record, we will discuss the problem of maximum power transfer in resistor
circuit and sinusoidal steady-state circuit. And for the cases of three-phase circuit, we will
compare it with other circuits to show that it is the most efficient circuit for maximum
power transfer. Finally, we will give some applications in maximum power transformation to
verify that this issue is quite useful in real life.
Key words:
maximum power transfer, resistor circuit, sinusoidal steady-state circuit, three-phase
circuit, Applications.
Chapter One:
Linear Circuit
The Thevenin equivalent is useful to find the maximum power a linear circuit can deliver to
a load. As shown in the figure (1), the power delivered to the load is
⎛ V Th
p = i RL = ⎜⎜
⎝ RTh + R L
2
Figure
2
⎞
⎟⎟ R L
⎠
…………………… ①
Figure (2)
（1）
For a given circuit , VTh and RTh are fixed .By varying the load resistance RL ,the power
delivered to the load varies as sketched in figure(2).We can see that when RL is equal to
RTh ,the maximum power occurs. We take differential about p in Eq. ① with respect to RL
and set the result to zero.
Then we can get RTh = R L , and the maximum power transferred is
p max =
VTh2
.
4 RTh
But if the domain of RL is limited and RL can’t reach RTh (i.e. RL > RTh or
RL < RTh ), the result will change.
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基本电路理论课程论文
2006-2007 第一学期
Case 1:
RL ∈ [R1 , R2 ] Where R1 > RTh .
2
When RL = R1 , the power reaches to the maximum and p max
⎛ V Th ⎞
⎟⎟ R1 .
= ⎜⎜
⎝ RTh + R1 ⎠
Case 2:
RL ∈ [R1 , R2 ] where R1 < RTh .
When RL = R2 ,the power reaches the maximum p max
⎛ V Th
= ⎜⎜
⎝ RTh + R2
2
⎞
⎟⎟ R2 .
⎠
The two cases above can also be solved from the figure (2).In a word, the smaller the
difference between RL & RTh ,the the power is larger.
There is another situation when the circuit is Norton equivalent, then we can use the
similar process to solve the problems. As shown in figure (3).
When
G L = G N ,the power of
GL
reaches the maximum.
Example:
Figure (3)
Find the maximum power transferred to resistor R in the circuit of figure (4).
Figure (4)
We need the Thevenin equivalent across the resistor R.
below.
22 kΩ
10 kΩ
+
vo
To find RTh, consider the circuit
v1
40 kΩ
30 kΩ
3vo
2
1mA
基本电路理论课程论文
2006-2007 第一学期
Assume that all resistances are in k ohms and all currents are in mA.
10||40 = 8, and 8 + 22 = 30
1 + 3vo = (v1/30) + (v1/30) = (v1/15)
15 + 45vo = v1
But vo = (8/30)v1, hence,
15 + 45x(8v1/30) v1, which leads to v1 = 1.3636
RTh = v1/1 = -1.3636 k ohms
To find VTh, consider the circuit below.
10 kΩ vo 22 kΩ
+
100V
+
vo
−
v1
40 kΩ
30 kΩ
3vo
(100 – vo)/10 = (vo/40) + (vo – v1)/22
[(vo – v1)/22] + 3vo = (v1/30)
+
VTh
(1)
(2)
Solving (1) and (2),
v1
p
=
VTh2/(4RTh)
= VTh = -243.6 volts
= (243.6)2/[4(-1363.6)] = -10.882 watts
Chapter Two: Sinusoidal steady state
Draw the Thevenin equivalent circuit like Figure 1, then
1
2
RL I
2
VT
VT
I =
=
ZT + ZL
( R T + R L ) + j ( XT + XL )
P=
Zt
VT
Zl
DC
=
Figure 1
So
VT
( RT + RL ) + ( XT + XL ) 2
2
2
RL VT
1
P=
2 ( RT + RL )2 + ( XT + XL )2
1. The common situation ( no restriction )
1 RL VT
Let XL = − XT
then
P=
2
2 ( RT + RL) 2
From what has been discussed in chapter one ,we can simply let RL = RT To make
sure that P reach its maximum
So
RL = RT
X L = − XT
ZL = ZT * .
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基本电路理论课程论文
2006-2007 第一学期
And the maximum average power is
2
PMAX
VT
=
8RT
VT
where
is the peak amplitude of the Thevenin
equivalent circuit voltage source.
2. When the load is a resistor .
At this situation,
X L =0 , then
2
1
RL VT
P=
2 ( RT + RL) 2 + XT 2
2
1 2 ( RT + RL ) + X T − 2 RL( RT + RL)
P = VT
=0
2
2
2
2
⎡⎣( RT + RL) + XT ⎤⎦
2
'
(R
→ RL =
3. When
Adjust
2
T
RL
and
XL
as near to
P' =
+ XT 2 )
X L are restricted to a limited range of values
− XT
and
dP
dRL
2
2
1
2 ( RT + R L ) + ( XT + XL ) − 2 RL ( RT + RL )
= VT
=0
2
2 2
2
⎡⎣( RT + RL ) + ( XT + XL ) ⎤⎦
→ ( R T + R L ) 2 + ( X T + X L ) 2 = 2 R L ( R T + RL )
→ RL = RT 2 + ( XT + XL ) 2
So in order to have the max power ,we should adjust
RL
as near to
4. When Z L = Z L ∠φL
R T 2 + ( X T + XL ) 2
ZL
XL
as near to
− X T , and
.
can be changed but its phase angle cannot, which means
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基本电路理论课程论文
2006-2007 第一学期
φL is a constant, then
I=
VS
( RT + ZL cos φ ) + j ( XT + ZL sin φ )
P = I ∗ Re ( ZL )
2
VS ZL cos φ L
2
=
(R
+ ZL cos φ L ) + ( XT + ZL sin φ L )
2
T
2
So
dP
2
= 0 → ZL = RT2 + X T2
d ZL
→
That is
¾
ZL = RT2 + X T2 = ZT
Z L = ZT
Example 1:
As in the figure below, Vrms=212∠0°,Under the following 4 conditions , ask what
value of ZL can make the maximum power of ZL, and P max= ?
a). no restriction;
b). ZL is a resistor;
c). RL∈(1, 3) , XL∈(-1.5, -1);
d). the phase angle of ZL is fixed to 45° but the magnitude can be changed.
Sol:
2k
212<0°
mA
J4k Z
2k
Figure 2
a).
Voc =
2
× 212∠0D × j 4
2 + 2 + j4
= 300∠45DV
ZT =
(2 + 2) × j 4
= 2 + j 2k
2 + 2 + j4
= 2 2∠45D
∴ Z L = 2 − j 2k Ω
2
Pmax =
5
2
Voc
300
=
= 11.25W
4 RT
4× 2
基本电路理论课程论文
b). XL=0,
c).
2006-2007 第一学期
− RT + RT 2 + 2 −2 + 4 + 2
=
= −1 + 6 = 1.45Ω
RL =
2
2
ZT = 2 + j 2Ω
∴ X L = −1.5Ω
d).
RL = RT 2 + ( X L + X T ) 2 = 4 + 0.25 = 2.06Ω
RL 2 + X L 2 = 4 + 4 = 8
XL
= tan φL = tan 45D = 1
RL
→ Z L = RL + X L = 2 − j 2Ω
Chapter Three:
Three-Phase Circuits:
Then we discuss the power in the special balanced three-Phase Circuits.
Since all the balanced circuits can be converted to wye-connections, here we only
discuss the Y-connected load.
For a Y-connected load:
The phase voltages are:
VAN = 2Vp cos(wt )
VBN = 2Vp cos(wt −
2π
)
3
VCN = 2Vp cos(wt +
2π
)
3
ia = 2 cos(wt − θ )
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基本电路理论课程论文
ib = 2 Ip cos(wt − θ −
2π
)
3
ic = 2 Ip cos(wt − θ +
2π
)
3
2006-2007 第一学期
The instantaneous power:
P=Pa+Pb+Pc=VANia =VBNib =VCNic = 3VpIp cos θ
Then we can see that the instantaneous power remains constant.By this, we can also get
other powers(total):
The total everage power:
P= 3VpIp cos θ = 3VLIL cos θ
The total reactive power:
Q=3 Qp =3VpIp sin θ =
The total complex power:
S=P+jQ= 3VLIL ∠θ
3VLIL cos θ
After so much has been done, we can discuss the advantage of the three-phase system by
comparing it with the single-phase system.
First, for the two-wire single-phase system, suppose the resister of each line is R, then we
IL = PL /VL I
have:
So the power loss:
Ploss = 2 IL2 R = 2 RPL2 /VL2
Then for the three-wire three-phase system, we also suppose the resister of each line is
R ' ： IL ' = Ia = Ib = Ic = PL / 3VL
So the power loss:
P ' loss = 3( IL ')2 R ' = 3R ' PL2 / 3VL2 = R ' PL2 /VL2
For R= pl / π r ,for the same material and length of line, we have:
2
Ploss / Ploss ' = 2 R / R ' = 2 r '2 / r 2 ,
If the two types of system have the same power loss:
2
Material for single-phase/Material for three-phase= 2 r / 3 r
'
2
=1.33.
From the above we can see that if the same power loss is tolerated in both system, then
using three-phase system can reduce the material that we actually need. And if we use the
material with the same R, the three-phase system can reduce half the power loss ! All in
all, the three-phase circuit can save not only the power but also the material, that’s why
three-phase circuit is so popular in many fields today.
Applications:
1. Microphone.
The impedance of each microphone will be listed as a specification. According to the
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基本电路理论课程论文
2006-2007 第一学期
Maximum Power Transfer Theorem, when the load impedance matches the
microphone impedance, the power transferred will reach the maximum. However, in
most cases, the two impedances don’t match each other. Despite of this , the
microphone still can be used. If the micro’s impedance is lower than 600 Ω , it is
considered as low impedance; when its value is in a range from 600 to 10,000 Ω , it is
considered as middle impedance; if the value is greater than 10,000, it will be regarded
as high impedance.
2. A model of pocket headphone.
e A3
There are several situations, where the pocket amp could benefit from a higher voltage
power supply - when driving high impedance headphones, when the amplifier is being
fed from a high gain equalizer or when the listener just wants more volume. With very
high impedance headphones (600 ohms or more), the amp may not be able to develop
sufficient voltage across the load for maximum power transfer. If the amp is fed from an
equalizer or tone control with a high boost, the output of the pocket amp could be driven
into clipping. (Refer to the net http://sandro0713.spaces.live.com/blog/)
3. Power consumed in Bluetooth.
The bluetooth has low power consumption longevity for battery-powered devices.
During data transfer the maximum current drain is 30mA. However during pauses or at
lower data rates will be lower.
References
[1] Fundamentals of Electric Circuits, Charles K. Alexander, Matthew N. O. Sadiku, Tsing
Hua University Publishing
[2] http://www.google.com
[3] http://sandro0713.spaces.live.com/blog/
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