TSTE19 Power Electronics. Lecture3 Tomas Jonsson ICS/ISY

Transcription

1 TSTE19 Power Electronics Lecture3 Tomas Jonsson ICS/ISY

2 Outline Rectifiers Current commutation Rectifiers, cont. Three phase

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 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 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 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 = 0 [A] 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 = 2 sinω = =ω ω 0<ω< 2 sinω ω =ω Integrate both sides, left is area A u (voltage * angle) = 2 sinω ω = 2 1 cos =ω =ω Commutation angle can be calculated cos =1 ω 2

10 Half-wave rectifier output voltage V do = Ideal average voltage of half-wave rectified voltage (effect of the commutation inductance L s neglected) = sin 1.5 = cos = v d 1 v s V d

11 Output voltage incl commutation voltage drop V d = V d0 - DV d = V d0 = 0.45V s - Commutation voltage drop appears as a resistance to the dcside current. R comm =

12 Commutation conclusions Conduction: Magnetic energy is stored related to the inductance of the conduction path Commutation Transfer of current between two paths: ÞStored 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 i b) calculate u, V d, and P d with I d = 20 A

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

15 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

16 Current commutation angle i s change is double that of previous example (from -I d to I d ) cos =1 2ω 2

17 Full-bridge rectifier output voltage V do = average voltage of full wave rectified voltage (effect of the commutation inductance L s neglected) = 2 sin = cos = v d 1 v s V d

18 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?

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

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

21 Diode rectifier 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

22 Diode rectifier 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

23 Diode rectifier 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

24 Diode rectifier 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

25 Diode rectifier 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

26 Diode rectifier 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

27 3-phase full bridge rectifier waveforms Every diode conducts 1/3 of the cycle Output waveform contains 6 segments = Instantaneous current commutation due to L = 0 = 1 π 3 = 2 2 cosω ω =

28 Principles of AC/DC conversion, 6-pulse bridge 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 wt

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

30 Input line current 3ph rectifier No 3rd harmonic Compare with single phase PF = : = 2 3 : = 1 π 6 = 0.78 = cosϕ = 1.0 = = cosϕ = cosϕ = 3 π = 0.955

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

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

33 Transfer of current from valve 1 to valve 3 U = 0.5 ( U +U ) d a b I I U a X c U b X c I d I 1 t I 3 u t

34 Current commutation Current commutation phase c -> phase a (D5 -> D1) A u indicates the current commutation voltage drop = A u only half of area between v a and v c because of two inductances = ω = = () ω = ω = ( )

35 Effect of commutation inductance =ω =ω = = ω = () ω = ( ) D = ω 3 = 3 ω = - D = 1.35 ω

36 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 maximum ac-side rms current I s = 10 A. Calculate a) Max dc-side current b) Average dc-side voltage at max current c) Max active power d) Diode average current e) Diode rms current

37 Exercise Using results from exercise 3-101, calculate Diode conduction losses (p D1 = u D1 id 1 ) using the diode BYW29E with V 0 =0.79V, R s = ohm (T j =25C) Use the diode on-state model below wherei D can be expressed with its average and rms current as calculated above

38 Inrush current LC-circuit fed by voltage step Worst case when input voltage at maximum when applied =2 2 (single phase) =2 2 (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

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