Chapter 1 Introduction and Analysis Methods 1.1 Switching Power Electronics Read Chapter 1 of “Principles of Power Electronics” (KSV) by J. G. Kassakian, M. F. Schlecht, and G. C. Verghese, Addison-Wesley, 1991. Linear Regulator + Vx − io + Vin + Vo − − Figure 1.1: Linear Regulator Control vx such that Vo = Vo,REF : Simple, accurate, high-bandwidth, but Pdiss = < vx io > > 0 (1.1) Pout Pin Vo io = Vin io Vo = Vin η = (1.2) @ Vin = 15v, Vo = 5v → η = 33% (1.3) For eﬃciency we will consider switching power converters: q(t) 1 q(t) Vin + − 0 + Vo DT T v(t) − = dVin Figure 1.2: Considering Switching Power Convertor Add ﬁltering: NOTE: Only lossless elements. L, C (energy storage). Use semiconductors as switches. Switches: Block V, carry I, but NOT at the same time! + Vin + + Vo − Vx − v(t) − Vx ~ dVin + + Vin + Vo − − − Figure 1.3: Add Filtering 1.2 1.2.1 Analysis Techniques Methods of Assumed States Semiconductor switches are typically not fully controllable. Let’s consider how to analyze a switching circuit in time domain: Simple Rectiﬁer Example: (trivial but fundamental) + Vd − id VsSin(ωt) + Vo − Figure 1.4: Simple Rectiﬁer Diodes: Uncontrolled • Cannot sustain positive voltage (will turn on) • Cannot sustain negative Current (will turn oﬀ) id Vd Figure 1.5: Diode The method of assumed states allows us to ﬁgure out which un/semi-controlled switches are on as a function of time. 1. Assume a state (on/oﬀ) for all un/semi-controlled switches. 2. Calculate voltages and currents in the system (linear circuit theory). 3. See if any switch conditions are violated (e.g., “on” diode has negative current and “oﬀ” diode has positive voltage.) 4. If no violations, then done, else if violation assume a new set of states go back to step 1. + Vd id VsSin(ωt) − + Vo − Figure 1.6: Simple Rectiﬁer • If Vs sin(ωt) > 0 and we assume diode oﬀ: vd > 0, since this is not possible diode must be on during this condition. • If Vs sin(ωt) < 0 and we assume diode on: id < 0, since this is not possible diode must be oﬀ during this condition. Vs sin(ωt) > 0 → diode on: + Vo VsSin(ωt) − Figure 1.7: Simple Rectiﬁer with Diode On Vs sin(ωt) < 0 → diode oﬀ: + Vo VsSin(ωt) − Figure 1.8: Simple Rectiﬁer with Diode Oﬀ Vo Vs = π ωt π i=(VsSin(ωt))/R 2π VsSin(ωt) Figure 1.9: Rectiﬁer Waveform Very simple example but principle works in general. 1.2.2 Periodic Steady State In power electronics we are often interested in the periodic steady state. In periodic steady state the system returns to the same point at the end of cycle (beginning matches end), so things are operating cyclicly. In periodic steady state (P.S.S.): V = L di dt di > dt di = L< > dt = = 0 →< V > = 0 dt (1.4) Therefore, in P.S.S.: • Inductor < VL >= 0 → average didtL = 0 • Capacitor < VC >= 0 → average dVdtC = 0 The P.S.S. conditions are useful for analysis. Consider adding a ﬁlter to smooth the ripple current in our simple rectiﬁer: + Vd − + VL − + + VsSin(ωt) Vx Vo − − Figure 1.10: Simple Rectiﬁer with Filter If we assume diode is always on in P.S.S., then: + VL − + Vo VsSin(ωt) − Figure 1.11: Simple Rectiﬁer with Filter and Diode On Vs sin(ωt) − VL − vo = 0 < Vs sin(ωt) > = 0 in P.S.S. < VL > = 0 in P.S.S. vo = 0 (1.5) If diode were always on < Vo >= 0 and io must be < 0 part of the time. We know diode must turn oﬀ during part of cycle by the method of assumed states. What happens: Vx ωt io π 2π VsSin(ωt) Figure 1.12: Rectiﬁer with Filter Waveform Negative voltage for part of cycle drives i → 0. Exact analysis in KSV, Section 3.2.2. Good for review of time-domain analysis. Main point: Method of assumed states and P.S.S. condition are useful tools to determine system behavior. Now, in P.S.S. < Vx >=< Vo >, since < VL >= 0. < Vx > is pos neg 1 2 sin, so we lose some voltage as compared to a pos 1 2 1 2 sin plus some sin. Solution: Free-wheeling diode → Half-wave Rectiﬁer. + Vd D1 VsSin(ωt) D2 - + + VL io + Vx Vo - - Figure 1.13: Simple Rectiﬁer with Free Wheeling Diode D2 clamps so that Vx never goes negative. io “free-wheels”. Using method of assumed states: • D1 conducts when Vs sin ωt > 0. • D2 conducts when Vs sin ωt < 0. Io = V0 R Vx ωt π 2π VsSin(ωt) Figure 1.14: Rectiﬁer with Free Wheeling Diode Waveform < VL > = 0 in P.S.S. < Vo > = < Vx > 1 Zπ Vs sin(φ)dφ = 2π 0 = Vs π (1.6) NOTE: This circuit is rarely used in line applications today for several reasons, but the analysis technique is the key point. Full-wave rectiﬁer is more common. NOTE: For analyzing output current, output voltage, etc., we can do an equivalentsource replacement. Linear circuit with sum of fourier sources. io Veq(t) + Veq(t) Vo - t Figure 1.15: Linear Circuit with Sum of Fourier Sources ∞ X Bn cos(nωt + φn ) (1.7) X vo (ω) ⇒ Vo = |H(nω)|Bn cos(nωt + φn + < H(nω)) vx (ω) n (1.8) Veq = n=0 If H(ω) = Main point: We can replace diﬃcult to handle part of circuit with an equivalent voltage source, then use linear circuit theory to analyze from there. Summary of analysis thechniques: • Method of assumed states • Periodic Steady State • Equivalent source replacement

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