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Rectifiers in Power Electronics

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Rectifiers in Power Electronics
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  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 VxD1 id+ + ω t VoVsSin( ω t) D2Vx π 2 π −− VsSin( ω t) D1 ON D2 ON Figure   2.1:   Simple   Half-wave   Rectifier   In   P.S.S.:   < v o   >   =   < v x   >   v s   =   (2.1)   π   10   Rectifiers in Power Electronics  v s If    L d   Big   →   i d   ≃   I  d   =   πR   (2.2)   If    LR   d   ≫   2   ωπ   ⇒   we   can   approximate   load   as   a   constant   current.   2.1   Load   Regulation   . Now   consider   adding   some   ac-side   inductance   L c   (reactance   X  c   =   ωL c ).   •   Common   situation:   Transformer   leakage   or   line   inductance,   machine   winding   inductance,   etc.   •   L c   is   typically   ≪   L d   (filter   inductance)   as   it   is   a   parasitic   element.   Lc Ld D1 + RVsSin( ω t) D2 Vx − Figure   2.2:   Adding   Some   AC-Side   Inductance   Assume   L d   ∼∞   (so   ripple   current   is   small).   Therefore,   we   can   approximate   load   as   a   “special”   current   source.   v x “Special”   since   < v L   > =   0   in   P.S.S.   I  d   = < >   (2.3)   ⇒   R   Assume   we   start   with   D 2   conducting,   D 1   off    ( V     sin( ωt )   <   0).   What   happens   when   V     sin( ωt )   crosses   zero?    i1 Lc D1 VsSin( ω t) D2i2 Id Figure   2.3:   Special   Current   •   D 1   off    no   longer   valid.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   i1 D1 + VsSin( ω t) D2 Vx Id _ i2 Figure   2.4:   Commutation   Period   D 2   will   stay   on   as   long   as   i 2   >   0 ( i 1   < I  d ).   Analyze:   di 1   1   =   V   s   sin( ωt ) dt   L c       ωt   V   s i 1 ( t )   =   sin( ωt ) d ( ωt )   0   ωL c   V   s   0   =   ωL c   cos(Φ) | ωt   V   s   =   [1   −   cos( ωt )]   (2.4)   ωL c    i1 u Id t ω Figure   2.5:   Analyze   Waveform   Commutation   ends   at   ωt   =   u ,   when   i 1   =   I  d .   Commutation   Period:   V   s   ωL c I  d I  d   =   ωL c   [1   −   cos   u ]   ⇒   cos   u   = 1   −   V   s   (2.5)   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   < V   x   > :   1       π   < V   x   >   =   V   s   sin(Φ) d Φ   2 π   u   V   s   =   [cos( u )   +   1]   2 π   ωL c I  d from   before   cos( u ) = 1   −   V   s   X  c I  d   =   1   −   V   s   V   s   ωL c I  d < V   x   >   =   [1   −   ]   (2.6)   π V   s   So   average   output   voltage   drops   with:   1.   Increased   current    π u VsSin(wt) Vx ω t π +u2 π 2i1 Id ω t u +u 2 +u D1 D2 π π π 2 π D1+D2 Figure   2.6:   Commutation   Period   2.   Increased   frequency   3.   Decreased   source   voltage   We   get   the   “Ideal”   no   L c   case   at   no   load.   We   can   make   a   dc-side   thevenin   model   for   such   a   system   as   shown   in   Figure   2.7.   No   actual   dissipation   in   box:   “resistance”   appears   because   output   voltage   drops   when   current   increases.   This   Load   Regulation   is   a   major   consideration   in   most   rectifier   systems.   •   Voltage   changes   with   load.   •   Max   output   power   limitation  
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