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Maximum Power Transfer
+−
+
−LV LR
LISR
SV SL
L
SVR
IR
=+
( )2 2
21 1 12 2 2
LL L L L S
SL
L
RPav V I R I VR R
= = =+
2
8 L
SVR
SR LR
( ) ( )( )
22
4
21 02
S SL LS S
S
LLL
L L
R R RdPav RR R
R RV R
R+ − +
= = ⇒ =∂ +
For fixed and ,s sV R∴maximum average power transfer to load LRoccurs when
Maximum Power Transfer
+−
+
−LV
LIS S SZ R Xj= +
SV
SL
L
SVZ
IZ
=+
2 21 1 1cos cos2 2 2L L L L L L L L LPav V I Z Z I Z R I= = =
, 0L L L LZ R jX R= + >
L L LV Z I=
( ) ( )2 2
2 2 21 12 2S S
S S S
L L
L L L
V VR RZ XR XZ R
= =+ + + +
To maximize , necessary to setL Sav LP X X−=
( )2
212L S
Lv
Sa
L
VR
RPR
⇒ =+
( ) ( )( )
22
4
21 02
S SL LS S
S
LLL
L L
R R RdPav RR R
R RV R
R+ − +
= = ⇒ =∂ +
2
m ax8L
Sa v
SVP
R=
LRSR0
SX−
LX
LavP
Theorem : Maximum Power TheoremOptimum Load Impedance
opt sLZ Z=
+−
+
−LV
LIS S SZ R Xj= +
SV
SL
L
SVZ
IZ
=+
( )1 cos2S S S LSPav ZV I Z= +
, 0L L L LZ R jX R= + >
opt SLZ Z=Max. Power Theorem : Conjugate-match condition
( )21 Re2 S S LI Z Z= +
( )21 Re2 SL LZI Z= +
( )2
21 Re2
SLS
S L
ZZ
VZ
Z= +
+
( )( ) ( )1 cos2 S S S SL LZ I I ZZ Z= + +i
( ) ( )2 2
2 21 1Re 22 2 2
S SS S S S
SS S
V VZ Z R
RZ ZPav∴ = + =
+
2
24 L
S
Sav
VR
P= =
Efficiency 1 or 50%2 2
S
L L
L
av av
av av
P PP P
= = =
L SZ Z=
Comments :1. Under conjugate-match condition, 50% of the power delivered by the
source is lost as heat dissipation in RS. Power company neverconjugates their loads!
2. Max. Power Theorem is used extensively in communication circuits to extract maximum power from preceding stages.
∴
Network Functions
j UU U e=
( )( )( )
Y jH jU j
ωωω
LinearElements
No independent Sources
u y
( )( ) cosu t U t Uω= + ( )( ) cosy t Y t Yω= +
j YY Y e=
Definition
is called a Network Function.
+−iV
( )( )( )
o
i
V jH jV j
ωωω
N+
−oV
0oI =
iI
( )( )( )
o
i
V jH jI j
ωωω
N+
−oV
0oI =
+−iV
( )( )( )
o
i
I jH jV j
ωωω
N oI
iI
( )( )( )
o
i
I jH jI j
ωωω
N oI
Typical Application of Max. Power Theorem
+−
1600SZ = Ω
SV
( ) ( )22 22
Re1 1 1610 50 10 0.52 2 1600 16
Lo S
S L
ZP V WZ Z
− = = ≈ = + +
16LZ = Ω
HI-FIAmplifier
Loudspeaker
Input impedance = 16 Ω⇓
10V=
Let average power oP =
delivered to loudspeaker
For maximum power transfer, make 1600 .LZ = Ω
( )21 160010 252 1600 1600oP W = = +
For maximum power transfer, make 1600 .LZ = Ω
( )21 160010 252 1600 1600oP W = = +
Use a transformer :
:1n16 Ω ≡ ( )2 16 1600LZ n= =
2 100 10n n⇒ = ⇒ =
10:1HI-FI
AmplifierLoudspeaker25W 25W
non-energtransformer is 25 Watts of ric powe⇒
is delivered to loudspeaker.
FREQUENCY RESPONSE
C L R
( )( ) ( )
( )( ) ( )2 22 2
11
1 1 1 1
C LRZ j jC CR L R L
ω ωωω ωω ω
− −= +
+ − + −
01 1 Resonant frequencyC L LC
ω ω ωω= ⇒ = ←
0 ω
R( )R jω
2R
2R
−
00ω
0ω
ω
2R
R0ω =
ω = ∞0
0ω ω=( )R jω
( )X jω
( )Z jω
Resistance function ( )R jω→ ( )X jωReactance function ←
( )X jω
FREQUENCY RESPONSE
C L R
( )( ) ( ) ( ) ( )
2 2
2 22 2
11
1 1 1 1
C LRZ jC CR L R L
ω ωωω ωω ω
− = + + − + −
( )( )1
1tan 1
C LZ jR
ω ωω − − − =
01 1 Resonant frequencyC L LC
ω ω ωω= ⇒ = ←
0 ω
R
( )Z jω
2π
2π
−
00ω
0ω
ω
( )Z jω
2R
R0ω =
ω = ∞0
0ω ω=( )R jω
( )X jω
( )Z jω
( )Z jω
FREQUENCY RESPONSE
C L R
( ) ( )1 1Y j j CR Lω ω ω= + −
01 1 Resonant frequencyC L LC
ω ω ωω= ⇒ = ←
0 ω
1R
( )G jω
Cω
0
0ωω
( )B jω
ω = −∞
0
0ω ω=
( )G jω
( )B jω
( )Y jω
1R
ω = ∞
( )G jω ( )B jω Susceptance function ←
Conductance function ↑
1Lω
−
FREQUENCY RESPONSE
C L R
( ) ( ) ( )221 1Y j CR Lω ω ω= + −
( ) 11
tan 1C LY j
R
ω ωω − − =
01 1 Resonant frequencyC L LC
ω ω ωω= ⇒ = ←
0 ω
1R
( )Y jω
2π
2π
−
0 0ω
0ω
ω
( )Y jω
ω = −∞
0
0ω ω=
( )G jω
( )B jω
( )Y jω
Magnitude function←
Phase function←
1R
( )Y jω
ω = ∞
Resonance
( 2) 1 , ( 2) 1Y j Z j= =
CLRRi Li Ci
1Ω 14 H 1 F
v+
−
sitωπ
2π 3
2π 2π
cosSi tω=
1
0
1−
CYLYRYRI LI CI
V+
−01 iI e=
( 2) ( 2) ( 2) 1V j Z j I j= =
( 2) ( 2) ( 2) 1R RI j Y j V j= =
( 2) ( 2) ( 2) 2 90L LI j Y j V j= = ∠ −
( 2) ( 2) ( 2) 2 90C CI j Y j V j= = ∠
2π0
π
32π
2π tω
( ) ,Ci t A
⇓2ω =
01 2LC
ω = =
CI
LI
RI
2
2−
10
2C L RI I I= =
no-gain propertydoes not hold forRLC circuits.
Resonance
1( 1) 10 71.6 , ( 1) 71.610
Y j Z j= ∠ − = ∠
CLRRi Li Ci
1Ω 14 H 1 F
v+
−
sitωπ
2π 3
2π 2π
cosSi tω=
1
0
1−
CYLYRYRI LI CI
V+
−01 iI e=
1( 1) ( 1) ( 1) 71.610
V j Z j I j= = ∠
1( 1) ( 1) ( 1) 71.610R RI j Y j V j= = ∠
4( 1) ( 1) ( 1) 18.410L LI j Y j V j= = ∠ −
1( 1) ( 1) ( 1) 161.610C CI j Y j V j= = ∠
2π
0π
32π
2πtω
( ) ,Ci t A
⇓1ω =
01 2LC
ω = =
CI
LI
RI
10
1 0I = ∠71.618.4−
161.6
Resonance
3( 2) 4 10 , ( 2) 250Y j Z j−= × = Ω
CLRRi Li Ci
250 Ω 14 H 1 F
v+
−
sitωπ
2π 3
2π 2π
cosSi tω=
1
0
1−
CYLYRYRI LI CI
V+
−01 iI e=
( 2) ( 2) ( 2) 250V j Z j I j V= =
( 2) ( 2) ( 2) 1R RI j Y j V j A= =
( 2) ( 2) ( 2) 500 90L LI j Y j V j A= = ∠ −
( 2) ( 2) ( 2) 500 90C CI j Y j V j A= = ∠
2π0
π
32π
2π tω
( ) ,Ci t A
⇓2ω =
01 2LC
ω = =
CI
LI
RI
500
500−
10
500−
500
Instantaneous, Average, Complex Power
+−+
−v
( )( ) cosv t V t Vω= +
Instantaneous Power
N
i
( )( ) cosi t I t Iω= +
( ) ( ) ( ) cos( ) cos( )p t v t i t V I t V t Iω ω= = + +
1 1cos( ) cos(2 )2 2
V I V I V I t V Iω= − + + +
constant Sinusoid of twice the frequency
Average Power
0
1 1( ) cos( )2
T
avP p t dt V I V IT
= −∫
t
( )v t
2T πω
=
t
( )i t
( )p t
avP
0
Power being returned by N
V Z I=j V j Z j IV e Z e I e= i
( )j Z IZ I e +=
V Z I Z V I∴ = + ⇒ = −
INDUCTOR1. 0 in CAPACITORavP =
Average Power 1 cos2avP V I Z=
Remarks:
pa2. ssIf N ive contains only elements, then0 cos 0avP Z≥ ⇒ ≥
90 90 for passive NZ∴ − ≤ ≤
I
IZ
V Z I=
0
Complex Po2
er 1w P V I
( )1 12 2
j V j I j V IP V e I e V I e− −= =i
Active power
Z
1 1cos sin2 2
P V I Z j V I Z = +
ReavP P ImQ PReactive power
1Re Average (Active) power2 in Watts1Im Reactive power2
avP V I
Q V I
= = ∴ = =
Effective (RMS) Value
21 ( )T
RMS oX x t dt
T ∫
Definition : Given any periodic waveform x(t) of period T,
the RMS (root-mean-square) or effective value of x(t)
is defined as
R( )i t
Interpretation
RRMSI
Let WR = average power dissipated in Resistor
( )( ) 2 2
0 0
1 1( ) ( ) ( )T T
R RMSW Ri t i t dt R i t dt R IT T
= = = ∫ ∫
The same average power is dissipated if the resistor is driven
by a dc current source of value IRMS ; hence IRMS is called
the effective value of i(t).
∴
2RMSI
Sinusoidal waveforms: ( ) cos( )x t X t Xω= +
2RMS
XX =
1. Since most instruments measure RMS values;
hence our 60Hz-sinusoidal voltages are rated in
RMS values.
2.
: Line voltage =110 magnitude=Exam 2(11 )le .p 0V V⇒
cos cos2 2av RMS RMS
V IP Z V I Z= =i
Note:
Significance of Complex Power
Most electrical machines are designed to withstand a
maximum voltage magnitude |V| and a maximum current
magnitude |I|. Hence, electrical machines are rated in
maximum in KVA, and not in maximum
average power dissipation Pav.
12
P V I=
12
Power factor cosav V I ZPPF
V I V I=
cosPF Z∴ =
1 for Resistors0 for INDUCTORS and CAPACITORSPF =
Note:
1. For power generation companies, it is important to keep the
PF of the load (customers) be as close to unity as possible.
2. The Watthour meter measures Pav, not |P|.
Example: If a factory dissipates 10KW of power with 50 %
PF, then the power company must generate 20 KVA of
power.
A penalty is usually levied for low PF customers.∴
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