Introduction to Rectifiers, Exercises of Analog Electronics

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MIT OpenCourseWare http://ocw.mit.edu 6.334 Power Electronics Spring 2007 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.

Chapter 2

Introduction to Rectifiers

Read Chapter 3 of “Principles of Power Electronics” (KSV) by J. G. Kassakian, M. F. Schlecht, and G. C. Verghese, Addison-Wesley, 1991. Start with simple half-wave rectifier (full-bridge rectifier directly follows). Ld

• − Id D1 Vx
• id + ω t VsSin(ωt) D2 Vx^ Vo π 2 π − − VsSin(ωt) D1 ON D2 ON Figure 2.1: Simple Half-wave Rectifier In P.S.S.: < vo > = < vx > vs = (2.1) π 10

12 CHAPTER 2. INTRODUCTION TO RECTIFIERS

i Lc D VsSin(ωt) D2 i2^ Id Figure 2.3: Special Current

• D 1 off no longer valid.
• But just after turn on i 1 still = 0. Therefore, D 1 switches from D 2 and D 2 to D 1. are both on d Lc uring a commutation period, where current i D

VsSin( ω t) D2 Vx_ (^) Id i Figure 2.4: Commutation Period D 2 will stay on as long as i 2 > 0 (i 1 < Id). Analyze: di 1 1 = Vs sin(ωt) dt Lc ∫ (^) ωt Vs i 1 (t) = sin(ωt)d(ωt) 0 ωLc Vs (^0) = ωLc cos(Φ)|ωt Vs = [1 − cos(ωt)] (2.4) ωLc

2.1. LOAD REGULATION 13

i u Id ω t Figure 2.5: Analyze Waveform Commutation ends at ωt = u, when i 1 = Id. Commutation Period: Vs ωLcId Id = ωLc [1 − cos u] ⇒ cos u = 1 − Vs

As compared to the case of no commutating inductance, we lose a piece of output voltage during commutation. We can calculate the average output voltage in P.S.S. from < Vx >: 1 ∫ (^) π < Vx > = Vs sin(Φ)dΦ 2 π (^) u Vs = [cos(u) + 1] 2 π ωLcId f rom bef ore cos(u) = 1 − Vs XcId = 1 − Vs Vs ωLcId < Vx > = [1 − ] (2.6) π Vs So average output voltage drops with:

1. Increased current

2.1. LOAD REGULATION 15

Id

Id 2 π^ slope 2 π ω Lc ω Lc 2Vs ω Lc −

− π Vs − π Vs Figure 2.7: DC-Side Thevenin Model All due to non-zero commutation time because of ac-side reactance. occurs in most rectifier types (full-wave, multi-phase, thyristor, etc.). rectifier has similar problem (similar analysis). Read Chapter 4 of KSV. This effect Full-bridge

• D D D D VsSin( ω t) Lc 2Vs π Full−Bridge Vs Vx Id (^) π 1/2−Bridge Id Vs 2Vs −^ ω^ Lc^ ω^ Lc Figure 2.8: Full-Bridge Rectifier