电力转换技术：script_pct_english

Universit?t Bremen Otto-Hahn-Allee, NW1 Institut für elektrische Antriebe, 28359 BREMEN Leistungselektronik und Bauelemente Tel.: 0421-218-4436 Prof. Dr.-Ing. B. Orlik Fax: 0421-218-4318 Power Converter Technology Version 1.5 April 2006 IALB – Institute of Electrical Drives, Power Electronics and Components - 1 - 1 One–Quadrant Converter As an example of a one-quadrant converter a buck converter (also known as step-down converter) is presented in the picture below. The name of the step-down converter results from the arithmetic mean at the output side which is less than the arithmetic mean at the input side. Figure 1-1 shows a typical step-down converter with an idealized freewheeling diode and an idealized switch as a replacement for real power-semiconductor devices. U D Figure 1-1: One–quadrant converter U L : Voltage at the load-side U D : Direct voltage of the source S closed: DL Uu = : t on S opened: u L = 0: t off Destination: D L Uu ≤≤0 43421 0 0 1 0 f T tt t U tt ttU u offon on D offon offonD L = + = + ?+? = (T 0 = Period of oscillation) Note: Figure 1-2: Current and voltage at an idealized switch P VS = i S ? u S P VS : Power losses at the switch; at a real switch (IGBT, MOSFET, etc.) the power losses are P VS > 0. This means less efficiency and the need of bigger heat sinks. Requirement: Power losses P VS at the switches have to be as low as possible! IALB – Institute of Electrical Drives, Power Electronics and Components - 2 - Switching Performance: Parasitic capacities cause a high current at the moment of turning on the switch by a high voltage change. This causes a current peak at the switch (which might be dangerous for the switch). Parasitic capacities may result from turning a freewheeling diode off, too! Parasitic inductances at the load side (or the inductive load itself) impel the current after turning the voltage of the source off. This is a result of the magnetic energy which is still stored in the magnetic field of the inductances. Opening the switch without any freewheeling diode in the circuit will lead to an electric arc which can destroy any semiconductor switch (this effect can be observed by turning a usual light switch off). Because of this effect a freewheeling diode is used in any circuit containing strong inductive components. Turn on: capacitive parasitic effects Turn off: inductive parasitic effects Off On Figure 1-3: Parasitic effects at the switch t t UL UZ uSpule iL100% = U/RZ 63% τ SoffenSgeschlossen t1 t2 iL1 iL0 Figure 1-4: Current and voltage at a quasi-static state IALB – Institute of Electrical Drives, Power Electronics and Components - 3 - The displayed characteristics in Figure 1-4 are only for steady-state condition (quasi-static state). The inductance at the load side works as an energy storage after opening the switch. The current can only flow across the freewheeling diode. The negative field voltage across the inductance stays at a constant level until the magnetization of the inductance is not longer able to supply the necessary current: i L = i D . Calculation of i L : S closed: LindR uuu =+ DLL Uu dt di LRi ==+? (I) R u dt di R L i L L =+ ; R L =τconstant Time ; Possible highest current level R u i L Lstat = Solution of der homogeneous differential equation: iRL di dt L L ?+ =0 ??=?iR L di dt L L ?= ∫∫ R L dt di i L L ln lnc R L ti L1 ?= ice L R L t =? ? 1 ice Logen t ,hom =? ? 1 τ R L =τwith Special solution thru variation of constants: ckt 1 = () ikte L t =? ? () τ IALB – Institute of Electrical Drives, Power Electronics and Components - 4 - di dt kt e ke L tt =?? ? ? ? ? ? ??+? ? ? ? () 1 τ ττ Insert this expression in I provides: D ttt UekekLekR = ? ? ? ? ? ? ? ? ?????+?? ?? ? ? τττ τ 1 D ttt Ue L eRkeLk = = ? ? ? ? ? ? ? ? ?????+?? ??? ? 444344421 0 τττ τ τ t D e L U k = ? ττ τ t D t D e R U e L U k =?= R U ee R U i D tt D specialL =?=? ? ττ , General solution: R U eciii D t specialLogeneousLL +?=+= ? τ 1,hom, Insert initial values: R U c R U ectii DD L +=+?==== ? 1 0 10 0)0( τ R U c D ?=? 1 Solution (for 0 1 ≤≤tt, S closed): ? ? ? ? ? ? ? ? ?=?=? ?? ττ t D t DD L e R U e R U R U i 1 for i L0 0≠ 00 1)( L t L D L iei R U ti + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?= ? τ IALB – Institute of Electrical Drives, Power Electronics and Components - 5 - S opened: 0=+ indL UU iRL di dt L L ?+ =0 iR L di dt L L ?=? ?= ∫∫ R L dt di i L L ?+ = R L tc i L ln ln 1 ice L t = ? 1 τ Calculation of c 1 : τ 0 10 )0( ? ==== ec R U tii D L R U c D = 1 respectively if the final value R U D is not reached: cit i LL11 1==( ) Solution (S opened): general: τ t D L e R U i ? = respectively for tttt 112 ≤ ≤ + : it ie LL tt ()= ? ? 1 1 τ IALB – Institute of Electrical Drives, Power Electronics and Components - 6 - Time average of the current i L : it t L ()== 1 i L1 = 00 1 L t L D iei R U + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? τ it t t L ()=+ = 12 i L0 = ie L t 1 2 ? τ i L1 = ie L t 0 2 τ ie L t 0 2 τ 00 1 1 L t L D iei R U + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?= ? τ ττ 11 0 1 t L t D eie R U ?? + ? ? ? ? ? ? ? ? ?= ? ? ? ? ? ? ? ? ?=? ??? τττ 112 1 0 t D tt L e R U eei ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ?= ∫∫ + ??? dteidteie R U T i tt t tt L t t L t D L 21 1 11 1 0 0 0 1 1 τττ ; Ttt 012 = + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ?+= ??? 111 1 211 101 0 τττ τττ t L t L t DD eieie R U t R U T ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?+ ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ?+= ?? 111 1 211 001 0 τττ τττ t L t L t DD eieie R U t R U T ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?+ ? ? ? ? ? ? ? ? ?+= ?? τττ ττ 121 01 0 1 1 tt L t DD eeie R U t R U T R u R U tt t t R U T i LDD L = + == 21 1 1 0 1 IALB – Institute of Electrical Drives, Power Electronics and Components - 7 - 2 Four-Quadrant-DC Chopper with Separated Voltage Sources The following diagram shows the equivalent circuit diagram of a four-quadrant-DC chopper converter. U D U D 2 Figure 2-1: four-quadrant-d.c. chopper converter Basically there are two different cases: Case 1: S 1 closed, S 2 open ? 2 D L U u = Case 2: S 1 open, S 2 closed ? 2 D L U u ?= Case 1: S 1 closed, S 2 open Case 2: S 1 open, S 2 closed D D Figure 2-2 L D InductorR u U UU ==+ 2 L D InductorR u U UU =?=+ 2 IALB – Institute of Electrical Drives, Power Electronics and Components - 8 - 00 1 1 2 2 2 2 L t L D D t L DL L DL L iei R U R U eCi R U dt di R L i U dt di LRi + ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? =+= =+ =+ ? ? τ τ differential equations general solution 00 1 1 2 2 2 2 L t L D D t L DL L DL L iei R U R U eCi R U dt di R L i U dt di LRi + ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ?? =?= ?=+ ?=+ ? ? τ τ Initial value (IV): i L0 = 0 ? ? ? ? ? ? ? ? ? ?= ? τ t D L e R U i 1 2 new IV: R U i D L 2 0 ?= ? ? ? ? ? ? ? ? ?=? ? τ t D L e R U i 2 1 when t → ∞ when t → ∞ when t → ∞ new IV: R U i D L 2 0 = ? ? ? ? ? ? ? ? ??=? ? τ t D L e R U i 2 1 new IV: R U i D L 2 0 = Diagram 2.3 shows the current and voltage characteristics. t 1 : time in which S 1 is closed t 2 : time in which S 2 is closed Generally the average voltage u L is as following: 22 D L D U u U ≤<? () ? ? ? ? ? ? ? ? ?= + ? = 12 22 0 1 21 21 T tU tt ttU u DD L where T 0 = t 1 + t 2 IALB – Institute of Electrical Drives, Power Electronics and Components - 9 - D D Figure 2-3: Current and voltage characteristic in quasi-static operations The disadvantage of four-quadrant-d.c. current converter is that two voltage sources are needed, though only one of them is in operation. IALB – Institute of Electrical Drives, Power Electronics and Components - 10 - 3 Four-Quadrant-DC Chopper Figure 3.1 shows the equivalent circuit diagram of a four- quadrant-d.c. chopper. D Figure 3-1: Four-Quadrant-dc chopper The four switches are always cross interlocked. Case 1: S 1 and S 4 closed, S 2 and S 3 open V 1 = U D and V 2 = 0 ? u L = V 1 – V 2 = U D Case 2: S 2 and S 3 closed, S 1 and S 4 open V 1 = 0 and V 2 = U D ? u L = V 1 – V 2 = – U D Case 1 Case 2 00 1 L t L D L DL L DL L L LInductorR iei R U i R U dt di R L i Uu dt di LRi uUU + ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ?= =+ ==+ =+ ? τ Differential equation of the load Solution 00 1 L t L D L DL L DL L L LInductorR iei R U i R U dt di R L i Uu dt di LRi uUU + ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ??= ?=+ ?==+ =+ ? τ IALB – Institute of Electrical Drives, Power Electronics and Components - 11 - Initial Value (IV): i L0 = 0 ? ? ? ? ? ? ? ? ?=? ? τ t D L e R U i 1 New IV: R U i D L ?= 0 ? ? ? ? ? ? ? ? ?=? ? τ t D L e R U i 2 12 when t → ∞ when t → ∞ when t → ∞ New IV: R U i D L = 0 ? ? ? ? ? ? ? ? ??=? ? τ t D L e R U i 2 12 New IV: R U i D L = 0 The average of voltage load is as following: D L D UuU ≤≤? ? ? ? ? ? ? ? ? ?= + ? = 12 0 1 21 21 T t U tt tt UU DD L Figure 3.2 shows the current- and voltage characteristic. D D Figure 3-2: Current- and voltage characteristic IALB – Institute of Electrical Drives, Power Electronics and Components - 12 - 4 Fourier Series Expansion ∑∑ ∞ = ∞ = ++=Φ 11 0 sincos 2 )( k k k k tkctkc A t νν ν π = 2 T A T tdt T 0 0 1 = ∫ Φ() Time average c T tktdt k T = ∫ 2 0 Φ()cos ν { s T tktdt k T = ∫ 2 0 Φ()sinν Angular frequency of the k-th order harmonic component Figure 4-1: Fourier–Analysis IALB – Institute of Electrical Drives, Power Electronics and Components - 13 - Analysing the curve of U L in U L1 and U L2, a line function without sinusoidal function part is developed. ? ? ? ? ? ? ++++= ...3cos 3 3sin 2cos 2 2sin cos 1 sin 2 2 )( 1111 1 α α α α α αα π α D L U U ? ? ? ? ? ? ++++?= ...3cos 3 3sin 2cos 2 2sin cos 1 sin 2 2 )( 1111 2 β β β β β ββ π β D L U U since, ? ? ? =+ += πβα πβα 11 ? ? ? +? ? ? ? ? +? ? +? ?( + ? ?= ...)(3cos 3 )(3sin )(2cos 2 )(2sin )cos( 1 )sin 2 2 )( 1 111 2 πα απ πα απ πα απαπ π α D L U U ? ? ? +? ? ? ? + ? +?+ ? ?= ...)3cos( 3 3sin 2cos 2 2sin )cos( 1 sin 2 2 )( 1111 2 α α α α α ααπ π α D L U U UU U LL L () () ()ααα=+ 12 ? ? ? ? ? ? ++++ ? = ...3cos 3 3sin2 2cos 2 2sin2 cos 1 sin2 2 22 )( 1111 α α α α α απα π α D L U U since 2 and 2 0 1 111 0 T t t T t πααπα =?== IALB – Institute of Electrical Drives, Power Electronics and Components - 14 - ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? + ? ?= ...23cos 3 3sin + 22cos 2 2sin 2cossin 4 2 2 2 )( 0 0 0 0 00 1 ? 0 1 0 T t T t T t T t T t T tUT t U tU D U D L L π π π π ππ π ππ π 4434421 ν νπ π ν ? ? ? ? ? ? ? ? ?= 0 1 sin 4 ? T t u U D L value)(r.m.s. Value Square -Mean -RootComplex : ~ ,...6,4,2 when 0 2 sin since ,...7,5,3,1 2 :case Special 0 1 == ? ? ? ? ? ? =?= ν π νν T t Generally, Ν∈ ? ? ? ? ? ? ? ? ?= ν 1 2 ? :componentDirect 0 1 0 T t UU DL 0 0 0 1 :ffrequency operating of multiples are sfrequencie Hamornic f T f ?=?= νν ν R–L–Load: ~ ~ ()ZRjLZ R L LL =+ ? = +ωω 22 $ $ () i u RfL L L ν ν πν = + 2 0 2 2 Lower harmonic current is obtained by higher operating frequency f 0. R T L<< ? ? ? ? ? ? 2 0 π $ $ i u fL L L ν ν πν ≈ 2 0 for high f 0 and with i U R i L L L 0 0 == $ (Direct component of current.) IALB – Institute of Electrical Drives, Power Electronics and Components - 15 - 5 Pulse Pattern Generation Figure 5.1 shows the block diagram of gating of a four-quadrant-controller. SET Two Point Element Figure 5-1: Block diagram of a gate control The simplest approximation of a non-linear two-point-element is the comparator shown below. Figure 5-2: Comparator with offset U 2 = 0 applies to an ideal two-point-element. The next two illustrations 5.3 and 5.4 show the pulse-pattern-generation with two different reference voltages. In opposite to saw-tooth shaped reference voltage, with triangular reference voltage two sampling information are available in one period. IALB – Institute of Electrical Drives, Power Electronics and Components - 16 - D D D D Figure 5-3: Pulse pattern generation with saw-tooth-shaped reference voltage set D D D D Figure 5-4: Pulse pattern generation with triangular-shaped reference voltage Period T 0 holds: t 1 + t 2 = T 0 = constant IALB – Institute of Electrical Drives, Power Electronics and Components - 17 - Moreover the frequency holds: () () 2010 1 0 0 0 K≈ =<< Lset Lset Uf f T fUf 5.1 Two-Step Control of 4-Quadrant-Actuator In two-step control a pulse pattern is developed through the comparison between set and actual value. This control is only reliable with a energy store in the load circuit. So for the inductive loads the load current is applied as controlled variable in two-point-control. The following example shows the two-point current control. The set value i L set is given as sinusoidal. ()ti Lset ωsin= Figure 5.5 shows the block diagram of a two-point closed-loop control: set s Two-point-Element with hysteresis Figure 5-5: PT 1 -element with two-point-controller The next illustration shows the current response and the pulse pattern caused by the control. IALB – Institute of Electrical Drives, Power Electronics and Components - 18 - set set D set D Figure 5-6: Current response and pulse pattern Differential equation of R-L-load ( ) DL Uu ±= : { 1 00 0 L0D 1 1 I , Uofation Standardiz V U u RI U I i dt I i d R L R u i dt di R L uRi dt di L D L V L D L Lset L Lset T L L L LL L ±==+ ? ? ? ? ? ? ? ? ↓ =+ =+ 321 The switching time is as following: t T V i I V i I t T V i I V i I Lsoll L Lsoll L Lsoll L Lsoll L 1 1 1 0 1 0 2 1 1 0 1 0 = ?+ ?? = ++ ?? ln ln ε ε ε ε IALB – Institute of Electrical Drives, Power Electronics and Components - 19 - The average load voltage is as below: D L U tt tt u 21 21 + ? = ? ? ? ? ? ? ? ? ?+ ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ++ ? ? ? ? ? ? ? ? +? ? ? ? ? ? ? ? ? ++ ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ?+ ? ? ? ? ? ? ? ? +? = + ? = εε εε εε εε 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 21 21 ln ln L Lset L Lset L Lset L Lset L Lset L Lset L Lset L Lset D L I i V I i V I i V I i V I i V I i V I i V I i V tt tt U u The following illustration presents the response characteristic of two-point-control. S set Figure 5-7: Response characteristic The mean gain V m is calculated as below: V VV m = ? ≈ 11 11 ε The frequency holds: IALB – Institute of Electrical Drives, Power Electronics and Components - 20 - ? ? ? ? ? ? ? ? ?+ ? ? ? ? ? ? ? ? ?? ? ? ? ? ? ? ? ? ++ ? ? ? ? ? ? ? ? +? = + = εε εε 0 1 0 1 0 1 0 1 121 ln 111 L Lset L Lset L Lset L Lset I i V I i V I i V I i V Ttt f set Figure 5-8: Frequency characteristic curve ()() ()() () () ε ε ε ε εε εε ? + ?? = ? + ? = ?? ++ ? = ? ? ? ? ? ? ? ? === 1 1 1 2 1 2 1 1 11 11 1 0 max0 ln2 1 ln 1 ln 1 0 V V T V V T VV VV T I i fff L Lset IALB – Institute of Electrical Drives, Power Electronics and Components - 21 - 6 One–Quadrant–Converter with Thyristor and Turn-off Arm D Figure 6-1: One-quadrant-d.c. chopper This circuit can be used to actively turn off a Thyristor (high voltage device). With the advancing development of interruptible electronic, high voltage valve devices (IGBT, IGCT, GTO), this circuit is not used in new development. It is explained here for its historic value and to reinforce the understanding of power electronic circuits. T 1 and T 2 are alternate: T 1 switched on, T 2 switched off T 1 switched off, T 2 switched on IALB – Institute of Electrical Drives, Power Electronics and Components - 22 - 2U D U D T 1 off, T 2 on T 2 off T 1 on U D -U D U D -U D Because of D2 C charges the wave ends here Figure 6-2: Current- and voltage characteristic of a one-quadrant d.c. chopper before I: ? T 1 on, T 2 off ? C is charged ( DC uu ?= ) ? 0; 1 == TDL uuu I: ? T 2 is triggered. D Figure 6-3: Current and voltage when T 1 off und T 2 on ? 0=?? LCD uuu DDDCDL uuuuuu ?=??=?=? 2)( ? i L is constant, since current is stored by the inductor. ? C charges linearly, since i L is constant. IALB – Institute of Electrical Drives, Power Electronics and Components - 23 - ? u C idt C itt i CL L CL ==??+= ∫ 11 0 ; i ? DCT uuu ?== 1 II: ? u C = 0 ? DL uu = ? The inductor L L drives current i L through C and T 2 . At the same time C charges with reversal sign. III: ? blocked through flow 0 2 TDiuuu FLLDC ??=?= IV: ? T 1 is triggered. ? DC uu = ? 0 da , 1 == TDL uuu ? C discharges through L 2 , D 2 ? Freewheeling diode D F is blocked ? Load current i L flows through T 1 V: ? Inductor L 2 continues the ?discharging current“ ? Diode D 2 prevents ?ring-back“ to IV: ideal oscillating circuit is without resistance R it C ()==00 DC utu == )0( uu LC2 0+= L di dt Q C C +=0 d dt Li C i CC ?+?= && 1 0 Statement: ie C t = λ IALB – Institute of Electrical Drives, Power Electronics and Components - 24 - ?=? =? &&& ie i e C t C t λλ 2 ??? +?? =Le C e tt λλ λλ2 1 0 λλ 2 12 11 =? ? ?=±? ? CL j CL , ?= ? ? ? ? ? ? ? ?? ie C j CL t 1 ↓ is a solution of the differential equation i LC tj LC t C = ? ? ? ? ? ? ? ? +? ? ? ? ? ? ? ? ? cos sin 11 equation aldifferenti theof solutions thealso are 1 sin ; 1 cos 21 ? ? ? ? ? ? ? ? ? ? = ? ? ? ? ? ? ? ? ? ? = t CL it CL i CC tindependenlinear with 1 sin 1 cos 2121, CCgeneralC i,it CL kt CL ki ? ? ?+? ? ?= Initial value: it k C ()==0 1 ut u C idt k C LC tLC C kLC CZC t () cos== = =?? ? ? ? ? ? ? ? ? ?? =???? ==∫ 0 111 0 2 02 ?=??ku C L Z2 Solution: ? ? ? ? ? ? ? ? ? ? ???= t CLL C uti DC 1 sin)( IALB – Institute of Electrical Drives, Power Electronics and Components - 25 - 7 Four–Quadrant–DC Chopper with IGBT’s This circuit is typically used to supply a DC machine or a single-phase AC load. D Figure 7-1: Four-quadrant- d.c. chopper With S 1 and S 4 as switches and D2 and D3 as freewheeling diodes current and voltage of a d.c. motor can be tuned by the chopper operation. There are always two transistors together switched on or off: S 1 and S 4 together and S 2 and S 3 together. The two transistor couples are switched on alternatively. However, It must be considered that real transistors are still conductive after being switched off. If the other transistors are triggered without time delay, a short circuit through S1 and S2 or S3 and S4 will occur. To prevent the short circuit a time delay need to be introduced (dead-time). This time delay is typically a delay of a couple of microseconds of the On-Pulse. The four quadrants (in case of a DC machine) of operation are: drive forward (I and U positive), brake forward (I positive, U negative), drive reverse (I and U negative), brake reverse (I negative, U positive). Thus, the machine can be completely controlled. dead Figure 7-2: Short circuit prevention with time delay t dead IALB – Institute of Electrical Drives, Power Electronics and Components - 26 - i L > 0: S 1 and S 4 ?longer on“ than S 2 and S 3 set set set set Figure 7-3: Current- and voltage characteristic when i L > 0 Offset: ?Vu tt T sign i Z tot off L1 0 =? ? ? () () 0 11 1 T tUUtVV offDDdeadset ????=? Offset: 0 1 T tt UV offdead Z ? ?=? IALB – Institute of Electrical Drives, Power Electronics and Components - 27 - i L < 0: S 2 and S 3 ?longer on“ than S 1 und S 4 t t t t dead t dead t off U g2 U g2set U g1 U g1set V 2 V 2set t t i L i L = i D1 i L = i S2 i L = i D1 V=V -V 22Set2 t dead Figure 7-4: Current- and voltage characteristic when i L < 0 IALB – Institute of Electrical Drives, Power Electronics and Components - 28 - () 0 22 1 T tutuVV offDdeadDset ??+??=? 0 1 2 T tt uV offdead D ? ??=? Summary: )( :Offset 0 1 L offdead D isign T tt uV ? ? ?=? but i L is not allowed to be intermittent! Figure 7-5: Characteristic of i L and ?V As it is shown in figure 7.5, current i L through the Offset ?V and the Offset itself are in phase. We add i L (U D ) with i L (?V), then we get a pulsating current which is not desired. As dead-time needs to be longer than off-time (safety margin), an offset is usually present. IALB – Institute of Electrical Drives, Power Electronics and Components - 29 - 8 D.C. Converter A d.c. converter is to convert d.c. current energy with constant voltage into one with continuous varying voltage. There are two different types: ? direct d.c. converter (direct power transmission) ? indirect d.c. converter (galvanic isolation / transformer coupling) 8.1 Direct D.C. Converter 8.1.1 Step-down Converter Figure 8-1 shows the electrical equivalent circuit diagram of a step-down converter. D Figure 8-1: Step-down converter It is defined: U 1 = 1 3 U D U G = 2 3 U D Moreover the current- and voltage characteristics of intermittent and non-intermittent current are shown in Figure 8-2 and Figure 8-3 respectively. IALB – Institute of Electrical Drives, Power Electronics and Components - 30 - D Figure 8-2: Intermittent flow of step-down converter D Figure 8-3: Non-intermittent load current For the mesh loop it holds: – u L + u 1 + U G = 0 ? u 1 = u L – U G Case 1: switch S ist closed (? u L = U D ) u L Uu G L= U 1 T on T off T on u 1 = u L - U G u L 1 3 U D U D - 2 3 U D IALB – Institute of Electrical Drives, Power Electronics and Components - 31 - U 1 = U D – 2 3 U D = 1 3 U D Moreover it holds: u 1 = L di dt L where U 1 = 1 3 U D ? i L = 1 L ∫ u 1 dt = 11 3L U D t + C 1 the gradient m is as below: m = 11 3L U D Case 2: switch S is open (? U L = 0 ) u 1 = 0 – 2 3 U D = – 3 2 U D ? i L = 1 L ∫ u 1 dt = – 3 21 L U D t + C 2 the gradient m is as below: m = – 12 3L U D The intermittent flow holds that the current at d.c. side becomes zero at certain time inside of every period of system voltage. This can be approached through a changed pulse-delay-relation of switch. The period time T 0 holds: T 0 = T closed + T open In the case shown in Figure 8.2 T open is longer than it in non-intermittent operation, because switch S should be open till the magnetic field of inductor completely decays. Then u 1 = 0 , thus u L = U G . In the case u L = U G current i L equals zero. The average voltage u L in non-intermittent case holds: D open L U T T u 0 = IALB – Institute of Electrical Drives, Power Electronics and Components - 32 - 8.1.2 Step-up Converter Figure 8-4 shows the electrical equivalent current diagram of a Step-up converter and in figure 8-5 the accessory current- and voltage characteristic is shown. D Figure 8-4: Step-up converter D Figure 8-5: Current- and voltage characteristic of a Step-up converter When switch S is open, the mesh loop holds: U D = u 1 + U G ? u 1 = U D – U G Switch S is closed: U D = u 1 = L di dt 1 The average voltage u L is as below: G open L U T T u 0 = IALB – Institute of Electrical Drives, Power Electronics and Components - 33 - 8.2 Indirect D.C. Converter Indirect d.c. converter contains an intermediate circuit with a transformer. It is applied, when a large difference between input and output voltage should be reached with the help of the transimission ratio of a transformer or galvanic separation is necessary. 8.2.1 Single-ended Flyback Converter Figure 8-6 shows the equivalent current diagram of a single-ended flyback converter. Figure 8-6: Single-ended flyback converter In such a converter alternating current is driven by the transistor in inverter and the diode in rectifier. The transformer is as energy store applied, i.e. current at the primary side magnetizes the core and current at secondary side demagnetises it. In this circuit the IALB – Institute of Electrical Drives, Power Electronics and Components - 34 - 9 Pulse Controlled Inverter in Three-phase Bridge Connection D Half-bridge 1 Delta Star Figure 9-1: Pulse-controlled inverter; Load as star- and delta switched Interlinked voltage: uVV 12 1 2 = ? uVV 23 2 3 =? uVV uu 31 3 1 12 23 = ? =? ? Z 1 on: upper transistor on, lower transistor off Z 1 off: upper transistor off, lower transistor on In this way every four-quadrant-operation is possible. [see also four-quadrant-d.c. converter]. There are different methods to trigger the IGBT’s. Three of them are shown here. IALB – Institute of Electrical Drives, Power Electronics and Components - 35 - 9.1 Full Block Control (180°-Conduction) In full block control the states are switched from state 2 to 7. uuu 12 1 2 =? uuu 23 2 3 =? ()uuu 31 3 1 =? There are only 2 linear independent equations! Secondary condition: symmetric a.c. loads hold: uuu u uu 123 3 12 0+ + =? = ? ? () () () ?= + =? =? + uuu uu uu 11223 22312 32312 1 3 2 1 3 1 3 2 u u Switching states Potential Interlinked Voltage Phase voltage V 1 V 2 V 3 u 12 u 23 u 31 u 1 u 2 u 3 1 All lower S on 0 0 0 0 0 0 0 0 0 2 Z 1 on; Z 2 , Z 3 off U D 0 0 U D 0 – U D 2 / 3 U D – 1 / 3 U D – 1 / 3 U D 3 Z 1 , Z 2 on; Z 3 off U D U D 0 0 U D – U D 1 / 3 U D 1 / 3 U D – 2 / 3 U D 4 Z 2 on; Z 1 , Z 3 off 0 U D 0 – U D U D 0 – 1 / 3 U D 2 / 3 U D – 1 / 3 U D 5 Z 2 , Z 3 on; Z 1 off 0 U D U D – U D 0 U D – 2 / 3 U D 1 / 3 U D 1 / 3 U D 6 Z 3 on; Z 1 , Z 2 off 0 0 U D 0 – U D U D – 1 / 3 U D – 1 / 3 U D 2 / 3 U D 7 Z 3 , Z 1 on; Z 2 off U D 0 U D U D – U D 0 1 / 3 U D – 2 / 3 U D 1 / 3 U D 8 Z 1 , Z 2 , Z 3 on U D U D U D 0 0 0 0 0 0 Table 9-1: Full block control (2→7) In the following diagram the values in this table are shown graphically. IALB – Institute of Electrical Drives, Power Electronics and Components - 36 - D D D D D D D D D random sines random sines Figure 9-2: Full block control (180°-Conduction) IALB – Institute of Electrical Drives, Power Electronics and Components - 37 - D D D D D D D D D random sines random sines Figure 9-3: Full block control (180°-Conduction) IALB – Institute of Electrical Drives, Power Electronics and Components - 38 - U D U D -U D U D 2 / 3 U D 1 / 3 U D _1 / 3 U D - 2 / 3 U D -U D Figure 9-4: 180°-Conduction Output range of full block control: Interlinked voltage: DDDBlock UUUU ?≈? ? =°??= 1,1 32 30cos 4 ? 12 ππ Phase voltage: DD Block Block UU U U ?≈?== 63,0 2 3 ? ? 12 1 π The maximal output range is equivalent to the amplitude of 1 st . fundamental-wave of Fourier series expansion ( please refer to Fourier series expansion). IALB – Institute of Electrical Drives, Power Electronics and Components - 39 - 9.2 Sine-Triangle Modulation In full block (180°-Conduction) control strong harmonic components appear (please refer to Fourier series expansion). We strive for a harmonic free characteristic of random sine waves. The application of inserted auxiliary signal is a solution: set set set D set set set Figure 9-5: Sine-triangle-modulation In opposite to block control the states 1 and 8 are added. D set D set D Auxiliary signal Figure 9-6: Sine-triangle-modulation Synchronization: not necessary, if Sinesu ff H ?< 15 1 Auxiliary signal: Clock-pulse rate per pro 10 to possibly 20 times higher than the frequency of sinusoidal signals. Output range of sinus-triangle-modulation: Interlinked voltage: DSD UU ?= 2 3 ? 12 IALB – Institute of Electrical Drives, Power Electronics and Components - 40 - Phase voltage: 2 ?? 2 11 D SDZSD U UUU =?=? ?< $$ UU SD Block11 but: with small harmonic components 9.3 Vector Modulation (Space Vector Modulation) Figure 9-7: Vector Modulation Ut U t 1 () $ cos( )=? +ω? Ut U t 2 2 3 () $ cos=? +? ? ? ? ? ? ?ω? π Ut U t 3 4 3 () $ cos=? +? ? ? ? ? ? ?ω? π ? ? ? ? ? ? ? ? ?+?+?=? ππ 3 4 3 3 2 21 )()()( 3 2 )( jj etUetUtUtu complex vector ut U e jt () $ () =? +ω? Phase voltage replaced by interlinked voltage: ut u u j u()=? +? +?? ? ? ? ? ? ? ? 2 3 1 2 1 2 3 12 23 23 ↑ Space vector which describes the dynamic behaviour of the machine. IALB – Institute of Electrical Drives, Power Electronics and Components - 41 - ()ut uuuuuujuuuu()=?? + ?? +? + +? + ? ? +? + ? ? ? ? ? ? 2 9 2 1 2 1 2 1 2 3 2 2 12 23 23 12 23 12 23 12 23 12 ut u u ju()=? + + ? ? ? ? ? ? ? 2 32 3 2 12 23 23 uue u u ue jj =? = + ? ? ? ? ? ? +?? αα 2 32 3 4 12 23 2 23 2 α= ? + arctan 3 2 2 23 12 23 u u u Switching states: u 12 u 23 u α =∠u 1 0 0 0 - 2 U D 0 2 / 3 U D 0° 3 0 U D 2 / 3 U D 60° 4 -U D U D 2 / 3 U D 120° 5 -U D 0 2 / 3 U D 180° 6 0 -U D 2 / 3 U D 240° 7 U D -U D 2 / 3 U D 300° 8 0 0 0 - Table 9-2: Switching states u is the complex space vector of the phase voltages IALB – Institute of Electrical Drives, Power Electronics and Components - 42 - set set Figure 9-8: Vector modulation To realize u set it must be switched between 2 and 3. (In addition still 1 or 8): 0 000 ?+?+?= T T U T T U T T u Zero L L R R set R: right; L: left where : 0 ZeroLR TTTT ++= the period of inverter operation cycle ),( 0 Dset R Uuf T T = ),( 0 Dset L Uuf T T = 000 1 T T T T T T RLZero ??= )( ).60sin(120sinsin 00 R R set L L U T T u U T T ? ° = ° = αα αsin 120sin 0 ? ?° = L set L U u T T αsin 2 3 3 2 0 ???= D set L U u T T αsin3 0 ??= D set L U u T T )60sin(3 0 α?°??= D set R U u T T 000 1 T T T T T T RL off ??= IALB – Institute of Electrical Drives, Power Electronics and Components - 43 - °??= 30cos 2 3 max D VM set Uu Output range of vector modulation: Phase voltage: 3 ? max 1 D U U VM = Interlinked voltage: D UU VM = max 12 ? D UUU SDVM ?=>? 87,0 ?? maxmax 1212 The switching sequence of half-bridge branch Z 1 , Z 2 and Z 3 are so selected that a minimum pulse frequency of pulse-width-modulation inverter arises. It is realized by switching the states 1 and 8 alternatively and the state 2 to 7 should be so switched that the required state can be reached with only one reversal of terminal connections of a half-bridge branch. 9.4 Comparison of Output Ranges Interlinked voltage Phase voltage Full block control (180°-conduction) DBlock UU ? ? = π 32 ? 12 DBlock UU ?= 1,1 ? 12 DBlock UU ?= π 2 ? 1 DBlock UU ?= 63,0 ? 1 Sinus-triangle-modulation (Sinusoidal-delta-connected modulation) DSD UU ?= 2 3 ? 12 DSD UU ?= 87,0 ? 12 2 ? 1 D SD U U = Vector modulation DVM UU = 12 ? 3 ? 1 D VM U U = DVM UU ?= 58,0 ? 1 Tabelle 9-3: Comparison of output ranges IALB – Institute of Electrical Drives, Power Electronics and Components - 44 - 10 One-pulse Electronic Power Converter Figure 10-1: One-pulse electronic power converter α: ignition angle of thyristor 1. case α = 0°: ui VAK d =? ≠00 ut L di dt Ri U t S d ds () sin=? +?= ? ?2 ω ~ ()ZR L=+? 22 ω T L R = Current i d is composed of stationary current stationary d i and dynamic current dynamic d i : dynamicstationary ddd iii += Stationary current: )sin( ~ 2 ?ω ?? ? = t Z U i S d stationary ?=arctan X R X L=ω Dynamic current (transient current): T t S d e Z U i dynamisch ? ?? ? = ?sin ~ 2 Total current: 0 ifor 0 0 =?==+= = vdSVAKtddd uUuiii dynamicstationary i U Z te d S t T =?? ?+ ? ? ? ? ? ? ? ? 2 ~ sin( ) sinω? ? IALB – Institute of Electrical Drives, Power Electronics and Components - 45 - ωt ωt ωt i d I d,stationary I d,dynamic ? Θ u S u S u vd u VAK u d Figure 10-2: Current- and voltage characteristic for α = 0° angle flowCurrent :Θ ωti d =?=Θ 0 sin( ) sinΘ? + ? = ? ??e t T 0 IALB – Institute of Electrical Drives, Power Electronics and Components - 46 - X R e t T <? << ? 11 ?≈+Θ π ? UUtdt dS =?? ∫ 1 2 2 0 π ωωsin( ) Θ UU dS = ? ?? 1 2 2cosΘ π : of Average dd iI Iidt dd = ∫ 1 2 0 π ω Θ I u R U R d Sd = ? ??= 1 2 2cosΘ π 2. case α ≠ 0°: ? ? ? ? ? ? ? ? ? ? ? ? ? ? ??+???=+= ? ? ? ? ? ? ? ? ? ? ? ? ? ? T t S ddd et Z u iii dynamicstationary ω α α??ω )sin()sin( ~ 2 () ( ) UUtdt U t U dS S S =?? = ?? ? ?= ?? ?+ ? + + ∫ 1 2 2 2 2 2 2 π ωω π ω αθα π α α α θα sin( ) cos( ) cos cos Θ = Iidt U R dd d == + ∫ 1 2π ω α αΘ IALB – Institute of Electrical Drives, Power Electronics and Components - 47 - Figure 10-3: Current- and voltage characteristic for α = 90° U vak U vd IALB – Institute of Electrical Drives, Power Electronics and Components - 48 - Approximation for α=0° : L → 0, R ≠ 0 ? ?=0° Figure 10-4: Current- and voltage characteristic for L → 0, R → ?0 i U R t d S =??2sin()ω Approximation for a=0 : R → 0, L ≠ 0 => ? = 90 ° Figure 10-5: Current- and voltage characteristic for R → 0, L ≠ 0 Thyristor does not quench! i U L t d S =???21 ω ω(cos() IALB – Institute of Electrical Drives, Power Electronics and Components - 49 - Through the energy stored in inductor the current flow angle Θ is enlarged. Without delay Θ becomes bigger than π; a time delay of triggering diminishes the current flow angle. The operation serves as current smoothing. IALB – Institute of Electrical Drives, Power Electronics and Components - 50 - 10.1 One-pulse Electronic Power Converter with Counter-E.M.F. Figure 10-6: One-pulse electronic power converter with counter-E.M.F. iuRiU dS dG ≠=?+0: ?= ? i uU R d sg iuuU d S VAK G ==+0: ? = ?uuU VAK S g Figure 10-7: Current- and voltage characteristic The negative waves of u S can not be used as energy transmission. The operation serves for charging accumulators or d.c. motors excitation. Because of the high ripple content of d.c., however, it is only suitable for small output. IALB – Institute of Electrical Drives, Power Electronics and Components - 51 - 11 Double-Pulse Centre-Tap Connection Diagram 11.1 shows a two-pulse star connection with centre tap. Figure 11-1: Double-pulse centre-tap connection Generally : u S = ()2U t S sin ω when L CD → ∞, Ii dd = The next diagram shows the current- and voltage characteristic: IALB – Institute of Electrical Drives, Power Electronics and Components - 52 - ideal with Figure 11-2: Current- and voltage characteristic The following simplifying assumptions together with ()uUt sS =??2cosω hold: L CD → ∞ commutation time x K = 0 for Integration a zero shift of π 2 Thus the ideal direct voltage U di for ignition angle α = 0° holds: ()U q Utdt di S q q = ? ∫ 1 2 2 π ωω π π cos IALB – Institute of Electrical Drives, Power Electronics and Components - 53 - ? U q q UkU di S u S = ? ? ? ? ? ? ? ? ? ? ? ? ? ? = 2 π π sin For the commutation number q=2 the constant of voltage k u holds: k u =≈ 2 209 π , Definition: commutation number q The commutation number q arises from the number of the commutation process during one network period that inside of one group of with each other commutating valves. Current I S1 holds: Iidt I q SS q q d 1 2 2 1 2 == ? ∫ π ω π π where I d = i S for [? ππ qq , ] thus II I q II k SS ddd i 12 2 ==== IALB – Institute of Electrical Drives, Power Electronics and Components - 54 - 12 Phase-Commutated Converter in Three-pulse Star Connection Diagram 12.1 shows the three-pulse star connection. Figure 12-1: Three-pulse star connection The next diagram shows the current- and voltage characteristic: Figure 12-2: Current- and voltage characteristic ? ? ? ? ? ? ?= conducting V3; conducting V2; conducting V1;0 31 211 SS SSVAK uu uuu IALB – Institute of Electrical Drives, Power Electronics and Components - 55 - The valve voltage u VAK1 is calculated as below: u VAK1 = u S1 – u vd with u q q U di S =? ? ? ? ? ? ? ?? π π 2sin ↓=q 3 =?>117, UU SS => The mean value of direct voltag is larger than the r.m.s. value. Gernerally the ideal direct voltage u di is calculated according to: ()u q Utdt di q q S = ? ∫ 1 2 2 π ωω π π cos In spezial case of three-pulse star connection with q = 3 the ideal direct voltage u di holds: () () () uUtd uUtdtU di S di SS = == ? ? ∫ ∫ 1 2 3 2 3 2 2 3 2 3 3 3 3 3 π ωω π ωω π π π π π π cos cos sin Phase current i S equals: i S = I d 3 IALB – Institute of Electrical Drives, Power Electronics and Components - 56 - 13 Three-pulse Star Connection (Controlled Converter Operations) The following illustration shows the electrical equivalent circuit diagram of a three- pulse star connection: Figure 13-1: Three-pulse star connection The controllable semiconductor converter valves, e.g. thyristors, transfer current from one converter arm to the next (commutation) only after follow valve being triggered. The transmission can be delayed. The trigger delay angle α is defined as time interval in which the natural triggering instant (α = 0°) is postponed. The average direct voltage with integration limits of trigger delay angle α can be calculated as following: () ()u q Utdt q U q di q q SS α π α π α π ωω π π α== ? ? ? ? ? ? ?+ + ∫ 1 2 22cos sin cos The arising ideal direct voltage U diα with trigger delay angle α is described as: ()uu di diα α= cos , As the aforesaid that the average direct voltage of line-commutated converter is changed by the cosine function of trigger delay angle α. When the trigger delay angle α = 0° the average direct voltage u diα reaches its maximum. The following illustration shows the current- and voltage response of the three-pulse star connection with α = 30°. IALB – Institute of Electrical Drives, Power Electronics and Components - 57 - Figure 13-2: Voltage characteristic for α = 30° If the trigger delay angle α, as it is shown in figure 13.3, increases to α = 60° the loss of average direct voltage u diα becomes obvious according to u diα = u di cos(α). Figure 13-3: Current- and voltage characteristic with α = 60° IALB – Institute of Electrical Drives, Power Electronics and Components - 58 - Figure 13-4: Valve voltage u VAK1 with α = 60° When α = 90° the average direct voltage becomes zero. When α is greater than 90° the average direct voltage becomes negative and goes on increasing with the increase of trigger delay angle with negative sign. Figure 13.5 shows the voltage characteristic with α = 90°. Figure 13-5: Spannungsverl?ufe bei α = 90° When α = 150° the voltage shape is as it is shown in figure13.6. IALB – Institute of Electrical Drives, Power Electronics and Components - 59 - Figure 13-6: Voltage characteristic with α = 150° With trigger delay angle from α = 90° to 150° and negative average direct voltage it is called inverter operation. The energy flow in this operation is opposite to it is in rectifier operation (α = 0° to 90°), i.e. in inverter operation energy traces from d.c. load over semiconductor converter back into three-phase system. Conclusion: Trigger delay angle α Average direct voltage u diα Operating mode of semiconductor converter 0° – 90° u diα > 0 Rectifier operation 90° – 150° u diα < 0 Inverter operation Theoretically commutation, the transmission from the current conducting valve arm to the next following arm, can take place when trigger delay angle α is in the range from 0° to 180°, since in this range the correct polarity and the higher potential of following phase compared with the previous phase are saved as well. When trigger delay angle grows to more than 180° the commutation voltage (the voltage difference between commutating phases) will change its sign, i.e. the potential of the following phase is lower than the conducting phase. This range is forbidden in natural commutation because it leads short circuit in commutation circuit. IALB – Institute of Electrical Drives, Power Electronics and Components - 60 - However, to prevent the danger in inverter operation, trigger delay angle must not increase up to 180°. A safety clearance to intersection point of phase voltages which is called distinction angle must be kept (Please refer to Fig. 13-7). ? α max = 150° Figure 13-7: Commutation limit IALB – Institute of Electrical Drives, Power Electronics and Components - 61 - 13.1 Three-pulse Star Connection in Pulsating Operation The three-pulse star connection contains a pure ohmic load and no smoothing reactor L CD and for trigger delay angle α = 60° the current- and voltage characteristic is shown in figure 13.8. Figure 13-8: Three-pulse star connection with pure ohmic load and without smoothing reactor Current i d pulsates according to the trigger delay angle α. In general case, no pulsating current i d if α ≤ 30°; Curent i d pulsates when 30° ≤ α < 150°; no current i d if α > 150. As it is shown in Fig. 13.9 the load of 3-pulse star connection is supplemented by a small inductance L L, i.e. the pulsating limit of current i d depends additionally on the inductance L L . Figure 13-9: Pulsating limit depending on α and L L IALB – Institute of Electrical Drives, Power Electronics and Components - 62 - 13.2 Three-pulse Star Connection with Counter-e.m.f. U B The following illustration shows the equivalent circuit diagram: Figure 13-10: Three-pulse star connection with counter-e.m.f. New reference axis Figure 13-11: Current- and voltage characteristic IALB – Institute of Electrical Drives, Power Electronics and Components - 63 - In case of X CD >> R : () uUt SS2 2 3 =?+cos ω π α () uuUX di dt U SLBCD d B2 =+= + ω () () di dt uU X d SB CD ω =? 2 1 ()() ( ) [ ] ()i X uUdt X Ut Udt d CD SB CD SB =?= ?+? ∫∫ 11 2 3 2 ωω π αωcos ( ) [] ?= ? +? +i X Ut UtC d CD SB 1 2 3 sin ω π αω Determination of the constant C: Pulsating condition ()it d ω= =00 () ( ) () it X UC C X U d CD S CD S ωα π α π == ? += ?=? ? 0 1 2 3 0 1 2 3 sin sin The solution is as below: ()( ) [] i X Ut Ut X d CD S B CD =?+??? 1 2 33 sin sinω π αα π ω The mean of current I d is calculated as Iidt d d t L = = ∫ ω ω 0 Θ ()() (){} ?= ? ? ++ ? ? ? ? ? ? ? ? ? ? I U X U X d S CD LL B CD L 3 2 2 333 2 2 π π αα π α π cos cos sin ΘΘΘ IALB – Institute of Electrical Drives, Power Electronics and Components - 64 - 14 Commutation Figure 14-1: Block diagram The equivalent circuit diagram of the block diagram above is shown below: Figure 14-2: Equivalent circuit diagram After simplification finally we get the equivalent circuit diagram shown in Fig. 14-3. Figure 14-3: Simplified equivalent circuit diagram The current transition from a valve arm to the next one can not be abrupt because reactance X k always exists in the commutation circuit. The firing of the following valve V 2 takes place while V 1 is still conducting since because of X k sudden current change is impossible. → two-phase short circuit with e.m.f. u S2 – u S1 It should hold if V 1 conducts, V 3 closed, V 2 should be open. IALB – Institute of Electrical Drives, Power Electronics and Components - 65 - Mesh loop when V 2 is triggered: uuRiX di dt Ri X di dt SS KS K S KS K S 21 1 1 2 2 0?+ + ? ? = ω ω with i S2 = i K and i S1 = I d – i K follows: { u u RI Ri X dI dt X di dt Ri X di dt SS KdKKK d K K KK K K 21 0 0?+ ? + ? ? ? = = ωω ω Thus we have the differential equation which describes the characteristic of commutation current : ? 22 21 Ri X di dt uuRI KK K K SS Kd +=?+ ω Moreover if the ohmic resistance in commutation circuit is neglected (R k →0) and we supposed that the commutation inductance is at the same value, then the commutation equation can be simplified to: uuu X di dt SSK K K 21 2?== ω with R K << X K (4.1) The commutation voltage u k concerns a sinusoidal a.c. voltage created in a polyphase system by the potential difference between two commutating phases. In a three phase system u K is: uu KS = 3 where ( )uUt SS = 2sinω (4.2) Insert equation. 4.2 in equation. 4.1: ()32 2UtX di d t SK K sin ω ω = The short-circuit current i K is composed of: i K = i K ’ + i K ’’ , IALB – Institute of Electrical Drives, Power Electronics and Components - 66 - where i K ’ is quasi-steady current and i K ’’ transient current. With the initial condition i K = 0 for ωt = α and R K << X K the expression of the short- circuit current i K is as () ( )[]i X Ut K K S =? 32 2 cos cosαω and i K ’, i K ’’ is as below: ()i X Ut K K S 'cos=? 23 2 ω ()i X U K K S '' cos= 23 2 α The following diagram shows the current characteristic. Figure 14-4: Current characteristic i K , i K ’, i K ’’ and I d The mesh loop for commutation branch 1 holds (when R K << X K ): IALB – Institute of Electrical Drives, Power Electronics and Components - 67 - () 0 0 0 * 1 * 1 *1 1 =+?? =+ ? +? =++? vd K KS vd Kd KS vd S KS u td di Xu u td iId Xu u td di Xu ω ω ω Branch 2 holds ?+ += ?+ += uX di dt u uX di dt u SK S vd SK K vd 2 2 2 0 0 ω ω * * Adding the two mesh loop equations we get () ?+ +?+ += ??+ = ?= + uX di dt uuX di dt u uu u uuu SK K vd S K K vd SS vd vd S S 12 12 12 0 20 1 2 ωω ** * * The voltage u vd * presents the voltage u vd during the commutating time. Attention: All commutation equations are only vaild during the commutating phase! The following diagram shows the corresponding current- and voltage characteristic. IALB – Institute of Electrical Drives, Power Electronics and Components - 68 - Figure 14-5: Voltage and approximated current characteristic The commutation time t u , the period in which two changing over converter arms are conducting at the same time , is called overlap. The overlap is mostly specified by overlap angle u. During the commutation u vd reaches the average of the two phase voltages that drive commutating current i k . After the commutation uvd jumps to be the potential of phase voltages. The commutation ends, if V 1 extincts. ? i S1 = 0 und i K = I d With the overlap angle u it holds, i K (ωt = α + u) = I d ? () ( )[]I X Uu d K S =?+ 32 2 cos cosαα ()()cos cosαα+= ? ? ?? u XI U kd S 2 32 IALB – Institute of Electrical Drives, Power Electronics and Components - 69 - The overlap time u ω reaches its maximum when α = 0° and when α = 90° it reaches its minimum. Figure 14.6 shows the characteristic of the commutating current i K with α ≤ 150°. Commutating moment Figure 14-6: Commutating current If α > 150° the commutating current is not enough to force the d.c. in extinguishing valve to zero. The following valve can not carry on. This behaviour is also called ?pull out“. Figure 14.7 shows the behaviour of commutating current and phase current. Figure 14-7:Commutating current and phase current with α ≥ 150° IALB – Institute of Electrical Drives, Power Electronics and Components - 70 - 15 Bridge Connection 15.1 Two-pulse Bridge: ?Gr?tz“-Bridge Figure 15-1: Zweipulsige Brückenschaltung Since both the positive and negative parts of i s are constituents of I d, the transformer works to capacity. Bridge connection is used to rectify alternating current. As shown above the bridge connection is with two pulses. There are however bridge connections with higher pulse numbers that run smoother current. Bridge connection with six pulses is mostly often applied. 15.2 Six-Pulse Bridge Connection Figure 15-2: Six-pulse bridge connection IALB – Institute of Electrical Drives, Power Electronics and Components - 71 - uAB vd =? α = 0° ?Natural field commutation: Figure 15-3: Current- and voltage characteristic with α = 0° ( i R is alternating current at the primary side of the transformer.) UU d i =? ?2117, UU dd iiα α=?cos Iii R = ? 14 IALB – Institute of Electrical Drives, Power Electronics and Components - 72 - α = 30°: Figure 15-4: Current- and voltage characteristic for α > 0° IALB – Institute of Electrical Drives, Power Electronics and Components - 73 - 15.3 Bridge Power Rectifier D Figure 15-5: Bridge Power Rectifier ND UU ??= 23 max D Figure 15-6: Current- and voltage characteristic with negligible system inductance iii R = ? 14 - 74 - 16 Literature ? K. Heumann, Grundlagen der Leistungselektronik, Teubner Studienbücher, 1991 ? Michel, Leistungselektronik, Springer-Lehrbuch, 1992 ? Watzinger, Netzgeführte Stromrichter mit Gleichstromausgang, Siemens, 1972 ? Lappe u. a., Leistungselektronik, Technik Berlin–München, 1994 17 Authors Initial version: Petra Rose and Maik Buttelmann wrote the script ?Power Electronics II“ according to Prof. Orlik’s lecture in 1996. Version 1.1: Joachim Spilgies, April 1997 Version 1.2: Dipl.-Ing. Oliver Harling, Juli 2000 Version 1.3: Dipl.-Ing. Oliver Harling, April 2002 Version 1.4: Dipl.-Ing. Timo Christ, September 2005 Version 1.5: Dipl.-Ing. Timo Christ, April 2006