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6.334 Power Electronics
Spring 2007
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Chapter 3
Power Factor and Measures of
Distortion
Read Chapter 3 of “Principles of Power Electronics” (KSV) by J. G. Kassakian, M.
F. Schlecht, and G. C. Verghese, Addison-Wesley, 1991. Look at the AC side.
Definitions and Identities
Two functions X and Y are orthogonal over [a, b] if:
Z b
a
X(t)Y (t)dt = 0
(3.1)
Now:
•
R 2π
0
sin(mωt) sin(nωt + φ)dωt = 0, if n =
6 m ⇒ sinusoids of different frequencies
are orthogonal.
16
17
•
R 2π
0
sin(ωt) cos(ωt)dωt = 0 ⇒ sine and cosine are orthogonal.
In general:
1 Z 2π
1
sin(ωt) sin(ωt + φ) = cos φ
2π 0
2
(3.2)
These definitions will be useful for calculating power, etc.
Suppose we plug a resistor into the wall.
Rwire
i
Fuse
+
V
VsSin(ωt)
RL
−
Figure 3.1: Resistor
P = <Vi>
= VRM S iRM S
= i2RM S R
(3.3)
The fuse is rated for a specific RMS current. Above that, it will blow so that
dissipation in Rwire does not start a fire. Neglecting Rwire , for 115VAC,RM S , 15ARM S
fuse, we get ∼ 1.7kW max from wall.
Suppose instead we plug an inductor into the wall.
Neglecting Rwire :
18
CHAPTER 3. POWER FACTOR AND MEASURES OF DISTORTION
Rwire
i
Fuse
+
V
VsSin(ωt)
L
−
Figure 3.2: Inductor
i=−
Vs
cos(ωt)
ωL
(3.4)
1 Z
<P > =
V (t)i(t)d(ωt)
2π
Vs2 Z
sin(ωt) cos(ωt)d(ωt)
= −
2πωL
= 0 (of course)
(3.5)
Mathematically, it is because V and i are orthogonal. While we draw no real
power, we still draw current.
iRM S
s
1 Z 2π 2
=
i (ωt)d(ωt)
2π 0
Vs
= √
2ωL
@115V, 60Hz, L ≤ 20mH → iRM S ≥ 15A
(3.6)
(3.7)
So we still will blow the fuse (to protect the wall wiring), even though we do not
19
draw any real power at the output! (some power dissipated in Rwire ). In this case we
are not utilizing the source well.
Power Factor
To provide a measure of the utilization of the source we define Power Factor.
.
P.F. =
<P >
Real Power
=
VRM S iRM S
Apparent Power
(3.8)
For a resistor < P >= VRM S iRM S → P.F. = 1 best utilization. For a inductor
< P >= 0 → P.F. = 0 worst utilization.
Consider a rectifier drawing some current waveform,
i(t)
+
VsSin(ωt)
Rectifier
V(t)
−
Figure 3.3: Rectifier
Express i(t) as a Fourier series:
i(t) =
∞
X
in sin(nωt + φn ) Sum of weighted shifted sinusoids (3.9)
n=0
Note: iRM S =
s
1 2 1 2
1
i1 + i2 + · · · + i2n + · · ·
2
2
2
1 Z
<P > =
V (t)i(t)d(ωt)
2π 2π
20
CHAPTER 3. POWER FACTOR AND MEASURES OF DISTORTION
X
1 Z
Vs sin(ωt)
in sin(nωt + φn )
2π 2π
n
Z
∞
X
1
=
Vs in sin(ωt) sin(nωt + φn )
n=0 2 2π
=
(3.10)
By orthogonality all terms except fundamental drop out.
1Z
<P > =
Vs i1 sin(ωt) sin(ωt + φ1 )
2 2π
Vs i1
=
cos φ1
2
= Vs,RM S i1,RM S cos φ1
(3.11)
So the only current that contributes to real power is the fundamental component
in phase with the voltage.
VRM S i1,RM S
cos φ1
VRM S iRM S
i1,RM S
=
cos φ1
iRM S
P.F. =
(3.12)
We can break down into two factors:
P.F. = (
i1,RM S
) ·
cos φ1
iRM S
= kd (distortion factor) · kθ (displacement factor)
(3.13)
• kd , distortion factor (≤ 1) tells us how much the utilization of the source is
reduced because of harmonic currents that do not contribute to power.
21
• kθ , displacement factor (≤ 1) tells us how much utilization is reduced due to
phase shift between the voltage and fundamental current.
Total Harmonic Distortion (THD)
Consider another measure of distortion: Total Harmonic Distortion (THD).
v
uP
i2n
. u
6
T HD = t n=1
2
i1
(3.14)
This measure the RMS of the harmonics normalized to the RMS of the funda­
mental (square root of the power ratio). Distortion factor and THD are related:
T HD
v
uP
u n=1
i2n
6
t
=
2
i1
v
u
u i2RM S − i21,RM S
= t
2
i1,RM S
T HD2 =
i2RM S
i21,RM S
iRM S
i1,RM S
i2RM S
−1
i21,RM S
= 1 + T HD2
=
kd =
√
s
Example:
V
1 + T HD2
= Vs sin(ωt)
1
1 + T HD2
(3.15)
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CHAPTER 3. POWER FACTOR AND MEASURES OF DISTORTION

³ ´

ipk

 in = 4
πn
2
i(t) = square wave 

1
 i =i
0
ave = 2 ipk
T HD = 121%
kd =
=
ipk
2
·
4
π
ipk
√
2
·
√1
2
2
π
P.F. = 0.63
(3.16)
i(t)
Ipk
π
ωt
2π
Figure 3.4: Example
(Passive) Power Factor Compensation (KSV: Section 3.4.1)
Lets focus on the displacement factor component of power factor. For simplicity,
lets assume a linear load (e.g. R-L) so that voltages and currents are sinusoidal.
For sinusoidal V and i:
P.F. =
φ is the power factor angle:
• Leading φ < 0 Capacitive
• Lagging φ > 0 Inductive
<P >
= cos φ
VRM S iRM S
(3.17)
23
Real power:
P = VRM S IRM S cos φ
(3.18)
.
Q = VRM S IRM S sin φ
(3.19)
Define reactive power as:
Q
S
P
Figure 3.5: Reactive Power
~ = P + jQ. In phaser form V~ ,~i → S
~ =< V I ∗ >
In vector form S
units
Apparent Power
~ k= VRM S IRM S
S =k S
VA
Average Power Re{S} = P = VRM S IRM S cos φ W
Reactive Power Im{S} = Q = VRM S IRM S sin φ V AR
We can use these results to help adjust the displacement factor of a system. (make
Qnet → 0).
24
CHAPTER 3. POWER FACTOR AND MEASURES OF DISTORTION
i
2
2
R +(ω L)
ωL
θ
L
VsCos(ω t)
R
Im
S
i*
R
v
Re
i
Figure 3.6: R-L Load
Suppose we have an R-L load (e.g. an induction machine):
i(t) = √
Vs
ω 2 L2
+
R2
cos(ωt − arctan(
ωL
))
R
.
since S = V I ∗
ωL
)
R
ωL
P.F. = cos(arctan(
))
R
R
<1
= √ 2
R + ω 2 L2
voltage-current phase φ = arctan(
(3.20)
We can add some additional reactive load to balance out and give net unity power
factor.
S = VRM S IRM S
=
Vs2
√
2 ω 2 L2 + R 2
P = S cos φ
= VRM S IRM S cos φ
(3.21)
25
=
Vs2 R
2(ω 2 L2 + R2 )
(3.22)
jQ = jS sin φ
= jVRM S IRM S sin φ
= j
ωLVs2
2(ω 2 L2 + R2 )
(3.23)
So we have real and reactive power.
Suppose we add a capacitor in parallel:
i’
C
VsSin(ωt)
Figure 3.7: Capacitor
1
jωC
1 −j π
=
e 2
ωC
Zc =
π
1
= ωCej 2
Zc
(3.24)
Vphase − iphase
= −90◦
i′
= Vs ωC sin(ωt +
π
)
2
(3.25)
S ′
= VRM S IRM S
=
1 2
V ωC
2 s
P′ = 0
(3.26)
(3.27)
26
CHAPTER 3. POWER FACTOR AND MEASURES OF DISTORTION
1
Q′ = −j Vs2 ωC
2
(3.28)
So by placing the capacitor in parallel:
P, Q
L
Q’
VsCos(ω t)
C
R
Figure 3.8: Parallel Capacitor
S = P + jQ + jQ′
make jQ and jQ′ cancel: Q + Q′ = 0
j
ωLVs2
1
− j Vs2 ωC = 0
2
2
2
2(ω L + R )
2
C =
Example:
ω = 377RAD/sec (ωHZ)
R = 1Ω
L = 2.7mH
⇒ C = 1.32mF
L
+ R2
ω 2 L2
(3.29)
27
If we know our load, we can add reactive elements to compensate so that no dis­
placement factor reduction of line utilization occurs. Real, reactive power definitions
are useful to help us do this. This does not help with distortion factor.