TSTE25 Power Electronics Lecture 3 Tomas Jonsson ICS/ISY
2016-11-09 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
2016-11-09 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
2016-11-09 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
2016-11-09 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
[A] TSTE25/Tomas Jonsson Current commutation During commutation (ωt > 0) v s positive, D1 turns on 12 10 8 2016-11-09 6 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 6 4 2 0 0 5 10 15 20 25 30 wt [deg] Valid for 0 < i s < I d After commutation completed
Commutation current 2016-11-09 7 Commutation current Temporary current contribution related to energy transfer
2016-11-09 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
2016-11-09 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
2016-11-09 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 0.5 0-0.5-1 -1.5 0 30 60 90 120 150 180 210 240 270 300 330 360
2016-11-09 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
2016-11-09 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
2016-11-09 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
2016-11-09 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
2016-11-09 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
2016-11-09 16 Current commutation in full-bridge Same principle for area A u due to L s
2016-11-09 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
2016-11-09 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
2016-11-09 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 1.5 1 0.5 v d v s V d 0-0.5-1 -1.5 0 30 60 90 120 150 180 210 240 270 300 330 360
2016-11-09 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?
2016-11-09 21 3-phase full-bridge rectifier, general view Less ripple on output Handles higher power No current in neutral wire
2016-11-09 22 3-phase full bridge rectifiers, L = 0 One diode in each group is conducting at any time
TSTE25/Tomas Diode Jonssonrectifier 2016-11-09 23 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
TSTE25/Tomas Diode Jonssonrectifier 2016-11-09 24 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
TSTE25/Tomas Diode Jonssonrectifier 2016-11-09 25 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
TSTE25/Tomas Diode Jonssonrectifier 2016-11-09 26 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
TSTE25/Tomas Diode Jonssonrectifier 2016-11-09 27 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
TSTE25/Tomas Diode Jonssonrectifier 2016-11-09 28 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
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 2016-11-09 29
Principles of AC/DC conversion, 6-pulse bridge TSTE25/Tomas Jonsson 2016-11-09 30 I d U a I a a 1 3 5 U b I b b U d U c I c c 4 6 2 U ac U bc U a U b U c 1 6 1 2 3 2 3 4 5 4 5 6 t
2016-11-09 31 Exercise 3-100 In the ideal three-phase rectifier circuit, construct the wave forms of diode D1 and D2 voltages and currents.
Input line current 3ph rectifier No 3rd harmonic Compare with single phase PF = 0.9 2016-11-09 32 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
2016-11-09 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
2016-11-09 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
Current commutation Current commutation phase c -> phase a (D5 -> D1) A u indicates the current commutation voltage drop 2016-11-09 36 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
2016-11-09 37 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
2016-11-09 38 Exercise 3-101 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 F0 =0.79 V, r F = 0.013 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
2016-11-09 40 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
V F Ph 0.0001 [H] TSTE25/Tomas Jonsson 2016-11-09 41 Short circuit 10.0 [uh] D1 D3 0.0 50.0 Vs Is BRK1 #1 #2 Fault 100.0 [uf] Ud 0.23 D4 D2
2016-11-09 42 Short circuit current oscillation
2016-11-09 43 Short circuit current Peak current መI sc = U d L C Frequency of resonance ω 0 = 1 LC
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