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 = = 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

= 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. =

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

= 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

= 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) 22 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

= 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.

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