TSTE25 Power Electronics. Lecture 3 Tomas Jonsson ICS/ISY

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1 TSTE25 Power Electronics Lecture 3 Tomas Jonsson ICS/ISY

2 Outline Rectifiers Current commutation Rectifiers, cont. Three phase Inrush and short circuit current Exercises 5-5, 5-8, 3-100, 3-101, 3-102

3 Effect of L s on current commutation Current commutation = current path changed from one diode to another Commutation not instantaneous when L s nonzero Stored magnetic energy change Use simplified example Output represented by constant dc current source v s >0

4 Effect of L s on current commutation Current commutation = current path changed from one diode to another Commutation not instantaneous when L s nonzero Stored magnetic energy change Use simplified example Output represented by constant dc current source v s <0

5 Source inductance effects, cont Waveform if L=0 Prior to ωt = 0, v s is negative, current flow through D2 v d = 0, i s = 0

6 [A] TSTE25/Tomas Jonsson Current commutation During commutation (ωt > 0) v s positive, D1 turns on i D1 i D2 i D1 = i s i D2 = I d i s i D1 +i D2 = I d D2 stops conducting when i D2 = wt [deg] Valid for 0 < i s < I d After commutation completed

7 Commutation current Commutation current Temporary current contribution related to energy transfer

8 Current commutation waveforms Large L s used to clearly show effect Time for commutation depend on L s size and current change in L s

9 Current commutation time i s through inductor starts at zero, end at I d when ωt=u v L = di s 2V s sinωt = L s dt = ωl di s s d ωt 2V s sinωtd ωt = ωl s di s 0 < ωt < u Integrate both sides, left is area A u (voltage * angle) u A u = 0 2V s sinωtd ωt = I ddis 2V s 1 cosu = ωl s = ωl s I d 0 Commutation angle can be calculated cosu = 1 ωl si d 2V s

10 Half-wave rectifier output voltage V do = Ideal average voltage of half-wave rectified voltage (effect of the commutation inductance L s neglected) V d0 = 1 2π π 2Vs sin ωt dωt = 2V s 0 cos ωt 2π 0 π = 2 2V s 2π V d0 0.45V s 1.5 v d 1 v s V d

11 Output voltage incl commutation voltage drop V d = V d0 - V d = V d0 A u 2π = 0.45V s - ωl s 2π I d Commutation voltage drop appears as a resistance to the dcside current. R comm = ωl s 2π

12 Commutation conclusions Conduction: Magnetic energy is stored related to the inductance of the conduction path Commutation Transfer of current between two paths: Stored magnetic energy needs to be transfered! Output voltage reduction proportional to I d and L s

13 Exercise 5-5 Consider the basic commutation circuit of Fig. 5-11a with I d = 10 A. a) With V s =120 V at 60 Hz and L s = 0, calculate V d and the average power P d b) With V s =120 V at 60 Hz and L s = 5 mh, calculate u, V d, and P d c) With data as in b) calculate u, V d, and P d with I d = 20 A

14 Effect of L s on current commutation Current commutation = current path changed from one diode to another Commutation not instantaneous when L s nonzero Magnetic energy change Use simplified example Output represented by constant dc current source v s >0

15 Effect of L s on current commutation Current commutation = current path changed from one diode to another Commutation not instantaneous when L s nonzero Magnetic energy change Use simplified example Output represented by constant dc current source v s <0

16 Current commutation in full-bridge Same principle for area A u due to L s

17 Rectifier during current commutation v s negative before t = 0 D3 and D4 conducting i s = -I d v s positive D1 and D2 starts conducting (Short circuit path through D3 and D4) i u are commutation currents Valid for -I d < i s < I d v d = 0 during commutation

18 Current commutation angle i s change is double that of previous example (from -I d to I d ) cosu = 1 2ωL si d 2V s

19 Full-bridge rectifier output voltage V do = average voltage of full wave rectified voltage (effect of the commutation inductance L s neglected) V d0 = 1 π π 2Vs sin ωt dωt = 2V s 0 π cos ωt 0 π = 2 2V s π V d0 0.9V s V d = V d0 - V d = =V d0 A u = 0.9V π s - 2ωL s I π d v d v s V d

20 Exercise 5-8 In the single-phase rectifier circuit shown in Fig. 5-14a, V s = 120 V at 60 Hz, L s = 1 mh, and I d = 10 A. 1. Calculate u, V d, and P d 2. What is the percentage voltage drop in V d due to L s?

21 phase full-bridge rectifier, general view Less ripple on output Handles higher power No current in neutral wire

22 phase full bridge rectifiers, L = 0 One diode in each group is conducting at any time

23 TSTE25/Tomas Diode Jonssonrectifier u a - u b u a + D1 D3 D5 To load 0 u b - D4 D6 D2 u c 6-pulse Graetz rectifier bridge From load

24 TSTE25/Tomas Diode Jonssonrectifier u a - u c u a + D1 D3 D5 To load 0 u b D4 D6 D2 u c - 6-pulse Graetz rectifier bridge From load

25 TSTE25/Tomas Diode Jonssonrectifier u b - u c u a D1 D3 D5 To load 0 u b + D4 D6 D2 u c - 6-pulse Graetz rectifier bridge From load

26 TSTE25/Tomas Diode Jonssonrectifier u b - u a u a - D1 D3 D5 To load 0 u b + D4 D6 D2 u c 6-pulse Graetz rectifier bridge From load

27 TSTE25/Tomas Diode Jonssonrectifier u c - u a u a - D1 D3 D5 To load 0 u b D4 D6 D2 u c + 6-pulse Graetz rectifier bridge From load

28 TSTE25/Tomas Diode Jonssonrectifier u c - u b u a D1 D3 D5 To load 0 u b - D4 D6 D2 u c + 6-pulse Graetz rectifier bridge From load

29 v do = TSTE25/Tomas Jonsson 3-phase full bridge rectifier waveforms Every diode conducts 1/3 of the cycle Output waveform contains 6 segments v d = v Pn v Nn Instantaneous current commutation due to L = 0 1 Τ π 3 v dmax = Τ π 6 න Τ π 6 2V LL = 1. 35V LL 2 V LL cosωtd ωt

30 Principles of AC/DC conversion, 6-pulse bridge TSTE25/Tomas Jonsson I d U a I a a U b I b b U d U c I c c U ac U bc U a U b U c t

31 Exercise In the ideal three-phase rectifier circuit, construct the wave forms of diode D1 and D2 voltages and currents.

32 Input line current 3ph rectifier No 3rd harmonic Compare with single phase PF = RMS current: I s = 2 3 I d Fundamental current: I s1 = 1 π 6I d = 0.78I d DPF = cosϕ 1 = 1.0 PF = P S = V si s1 cosϕ 1 V s I s = I s1 I s cosϕ 1 = 3 π = 0.955

33 Single phase rectifier, input current Fourier analysis gives additional harmonic components Remember calculation uses RMS of I s, I s1 and I d I s1 = 2 π 2I d = 0.9I d I sh = 0 for even harmonics I sh = I s1 h for odd harmonics

34 Source inductance effects, DC current load Source L not 0 Only one current commutation at a time 6 commutations during one line-frequency cycle

35 Current commutation Current commutation phase c -> phase a (D5 -> D1) A u indicates the current commutation voltage drop i c + i a = I d A u only half of area between v a and v c because of two inductances I ddiu A u = ωl s = u = 0 0 2vLL sin(ωt) 2 u van v cn 0 2 d ωt = d ωt = 2v LL(1 cos u ) 2

36 Effect of commutation inductance I ddiu A u = ωl s = ωl s I d = 0 u van v = cn 0 2 u d ωt = 0 2vLL sin(ωt) 2 d ωt = 2v LL(1 cos u ) 2 V d = ωl si d π 3 = 3 π ωl si d V d = V d0 - V d = 1.35V LL 3 π ωl si d

37 Exercise A 3-ph rectifier feeding a constant current load has the following data: V LL = 400 V at 50 Hz, L s = 7 mh. The ac-side rms current I s = 10 A. Calculate a) The dc-side current b) Average dc-side voltage at given current c) The active power d) Diode average current e) Diode rms current

2016-11-09 39 Exercise 3-102 Using results from exercise 3-101, calculate Diode conduction losses (p D1 = u D1 i D1 dt) using the diode BYW29E with V

38 Exercise Using results from exercise 3-101, calculate Diode conduction losses (p D1 = u D1 i D1 dt) using the diode BYW29E with V F0 =0.79 V, r F = ohm (T j =25C) Use the diode on-state model below where i D can be expressed with its average and rms current as calculated above

39 Inrush current LC-circuit fed by voltage step Worst case when input voltage at maximum when applied v d = 2 2V s (single phase) v d = 2 2V LL (three phase) Peek voltage twice the input voltage step DC circuit needs to support twice the peak input voltage! Alternative: limit current, using resistor. Short resistor after start using thyristor

40 V F Ph [H] TSTE25/Tomas Jonsson Short circuit 10.0 [uh] D1 D Vs Is BRK1 #1 #2 Fault [uf] Ud 0.23 D4 D2

41 Short circuit current oscillation

2016-11-09 43 Short circuit current Peak current

42 Short circuit current Peak current መI sc = U d L C Frequency of resonance ω 0 = 1 LC

EN Power Electronics and Machines

EN Power Electronics and Machines EN 206 - Power Electronics and Machines Phase Controlled Rectifiers Suryanarayana Doolla Department of Energy Science and Engineering Indian Institute of Technology, Bombay suryad@iitb.ac.in Prof. Doolla