6.334 Power Electronics Spring 2007

Transcription

1 MIT OpenCourseWare Power Electronics Spring 007 For information about citing these materials or our Terms of Use, visit:

2 Chapter 3 Power Factor and Measures of Distortion ead Chapter 3 of Principles of Power Electronics (KSV) by J. G. Kassakian, M. F. Schlecht, and G. C. Verghese, Addison-Wesley, 99. Look at the AC side. Definitions and Identities Two functions X and Y are orthogonal over [a, b] if: b X(t)Y (t)dt = 0 (3.) a Now: π sin(mωt) sin(nωt + φ)dωt = 0, if n = m sinusoids of different frequencies 0 are orthogonal. 6

3 7 π 0 sin(ωt) cos(ωt)dωt = 0 sine and cosine are orthogonal. In general: π sin(ωt) sin(ωt + φ) = cos φ (3.) π 0 These definitions will be useful for calculating power, etc. Suppose we plug a resistor into the wall. wire Fuse i + VsSin(ωt) V L Figure 3.: esistor P = < V i > = V MS i MS = i (3.3) MS The fuse is rated for a specific MS current. Above that, it will blow so that dissipation in wire does not start a fire. Neglecting wire, for 5V AC,MS, 5A MS fuse, we get.7kw max from wall. Suppose instead we plug an inductor into the wall. Neglecting wire :

4 8 CHAPTE 3. POWE FACTO AND MEASUES OF DISTOTION wire Fuse i + VsSin(ωt) V L Figure 3.: Inductor V s i = ωl cos(ωt) (3.4) < P > = V (t)i(t)d(ωt) π V s = sin(ωt) cos(ωt)d(ωt) πωl = 0 (of course) (3.5) Mathematically, it is because V and i are orthogonal. While we draw no real power, we still draw current. π i MS = i (ωt)d(ωt) π 0 V s = (3.6) 60Hz,L 0mH i MS 5A (3.7) So we still will blow the fuse (to protect the wall wiring), even though we do not

5 9 draw any real power at the output! (some power dissipated in wire ). 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 > eal Power P.F. = = (3.8) V MS i MS Apparent Power For a resistor < P >= V MS i MS P.F. = best utilization. For a inductor < P >= 0 P.F. = 0 worst utilization. Consider a rectifier drawing some current waveform, i(t) + ectifier VsSin( ωt) V(t) Figure 3.3: ectifier Express i(t) as a Fourier series: i(t) = i n sin(nωt + φ n ) Sum of weighted shifted sinusoids (3.9) n=0 Note: i MS = i + i + + i + n < P > = V (t)i(t)d(ωt) π π

6 0 CHAPTE 3. POWE FACTO AND MEASUES OF DISTOTION = V s sin(ωt) i n sin(nωt + φ n ) π π n = V s i n sin(ωt) sin(nωt + φ n ) (3.0) n=0 π By orthogonality all terms except fundamental drop out. < P > = V s i sin(ωt) sin(ωt + φ ) π V s i = cos φ = V s,ms i,ms cos φ (3.) So the only current that contributes to real power is the fundamental component in phase with the voltage. We can break down into two factors: V MS i,ms P.F. = cos φ V MS i MS i,ms = cos φ (3.) i MS i,ms P.F. = ( ) cos φ i MS = k d (distortion factor) k θ (displacement factor) (3.3) k d, distortion factor ( ) tells us how much the utilization of the source is reduced because of harmonic currents that do not contribute to power.

7 k θ, displacement factor ( ) 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).. n= i THD = n i (3.4) This measure the MS of the harmonics normalized to the MS of the fundamental (square root of the power ratio). Distortion factor and THD are related: n= i THD = n i i MS i,ms = i,ms THD = i i MS i MS,MS = + THD i,ms i MS = i + THD,MS k d = (3.5) + THD Example: V = V s sin(ωt)

8 CHAPTE 3. POWE FACTO AND MEASUES OF DISTOTION ( ) 4 i i pk n = πn i(t) = square wave i 0 = i ave = i pk THD = % i pk 4 π k d = i pk = π P.F. = 0.63 (3.6) i(t) Ipk π π ω t Figure 3.4: Example (Passive) Power Factor Compensation (KSV: Section 3.4.) Lets focus on the displacement factor component of power factor. For simplicity, lets assume a linear load (e.g. -L) so that voltages and currents are sinusoidal. For sinusoidal V and i: φ is the power factor angle: < P > P.F. = = cos φ (3.7) V MS i MS Leading φ < 0 Capacitive Lagging φ > 0 Inductive

9 3 eal power: P = V MS I MS cos φ (3.8) Define reactive power as:. Q = V MS I MS sin φ (3.9) Q S P Figure 3.5: eactive Power In vector form S = P + jq. In phaser form V, i S =< V I > units Apparent Power S = S = V MS I MS V A Average Power e{s} = P = V MS I MS cos φ W eactive Power Im{S} = Q = V MS I MS sin φ V A We can use these results to help adjust the displacement factor of a system. (make Q net 0).

10 4 CHAPTE 3. POWE FACTO AND MEASUES OF DISTOTION i VsCos(ω t) L Im +(ω L) θ S ω L i* v e i Figure 3.6: -L Load Suppose we have an -L load (e.g. an induction machine): V s ωl i(t) = ω L + cos(ωt arctan( )). since S = V I ωl voltage-current phase φ = arctan( ) ωl P.F. = cos(arctan( )) = + ω L < (3.0) factor. We can add some additional reactive load to balance out and give net unity power S = V MS I M S = V s (3.) ω L + P = S cos φ = V MS I MS cos φ

11 5 V s = (ω L + ) jq = js sin φ = jv MS I MS sin φ ωlv s = j (ω L + ) (3.) (3.3) So we have real and reactive power. Suppose we add a capacitor in parallel: i VsSin(ωt) C Figure 3.7: Capacitor Z c Z c V phase i phase i S P = jωc = e j π ωc = ωce j π (3.4) = 90 π = V s ωc sin(ωt + ) (3.5) = V MS I MS = V s ωc (3.6) = 0 (3.7)

12 6 CHAPTE 3. POWE FACTO AND MEASUES OF DISTOTION Q = j V s ωc (3.8) So by placing the capacitor in parallel: VsCos(ω t) P, Q L Q C Figure 3.8: Parallel Capacitor make jq and jq cancel: Q + Q S = P + jq + jq = 0 ωlv s j (ω L + ) j V ωc = 0 s L C = (3.9) ω L + Example: ω = 377AD/sec (ωhz) = Ω L =.7mH C =.3mF

13 7 If we know our load, we can add reactive elements to compensate so that no displacement factor reduction of line utilization occurs. eal, reactive power definitions are useful to help us do this. This does not help with distortion factor.