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					          Chapter 2
   AC to DC CONVERSION
        (RECTIFIER)
• Single-phase, half wave rectifier
   –   Uncontrolled
   –   R load
   –   R-L load
   –   R-C load
   –   Controlled
   –   Free wheeling diode


• Single-phase, full wave rectifier
   –   R load
   –   R-L load,
   –   Controlled R, R-L load
   –   continuous and discontinuous current mode


• Three-phase rectifier
   – uncontrolled
   – controlled

                  Power Electronics and            1
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                  Dr. Zainal Salam, 2002
              Rectifiers
• DEFINITION: Converting AC (from
  mains or other AC source) to DC power by
  using power diodes or by controlling the
  firing angles of thyristors/controllable
  switches.

• Basic block diagram



AC input                                DC output



• Input can be single or multi-phase (e.g. 3-
  phase).
• Output can be made fixed or variable

• Applications: DC welder, DC motor drive,
  Battery charger,DC power supply, HVDC

               Power Electronics and                2
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               Dr. Zainal Salam, 2002
Single-phase, half-wave with
          R-load

        +
                                             +
        vs
                                             vo
        _
                                             _




                    vs


                                                  ωt
   vo
               io




                                                   ωt



   Output voltage (average),
                          π
   Vo = Vavg = ∫ Vm sin(ωt )dωt
                          0
               Vm
             =    = 0.318Vm
               π
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                    Dr. Zainal Salam, 2002
                RMS voltage
   Output voltage (RMS)
                  π
                1                         Vm
                  ∫ (Vm sin(ω t ) ) dω t = 2
                                   2
   Vo , RMS   =
                π 0
   Output current (DC),
        Vo 0.318Vm
   Io =   =
        R     R


• DC voltage is fixed at 0.318 or 31.8% of
  the peak value

• RMS voltage is reduced from 0.707
  (normal sinusoidal RMS) to 0.5 or 50%
  of peak value.

• Half wave is not practical because of
  high distortion supply current. The
  supply current contains DC component
  that may saturate the input transformer
                   Power Electronics and       4
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                   Dr. Zainal Salam, 2002
 Half-wave with R-L load
                   i

                                                 +
                                                 vR    +
  +                                              _
  vTNB                                                 vo
  _                                              +
                                                 vL    _
                                                 _

vs = vR + v L
                             di (t )
Vm sin( ω t ) = i (t ) R + L
                              dt
This is a first order differenti al equation.
Solution is in the form of :
i ( t ) = i f ( t ) + in ( t )

where : i f , in are known as " forced" and
" natural" response, respective ly.
From diagram, forced response is :

            Vm 
i f (t ) =      ⋅ sin( ω t − θ )
            Z 
where :
                                                       ωL 
Z=        R 2 + (ω L ) 2 and              θ = tan −1 
                                                         
                                                      R 
                       Power Electronics and                  5
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                       Dr. Zainal Salam, 2002
                    R-L load
Natural response is when source = 0,
             di (t )
i (t ) R + L         =0
              dt
which results in :

in (t ) = Aet τ ; τ = L R

Hence
                              V 
i (t ) = i f (t ) + in (t ) =  m  ⋅ sin(ωt − θ ) + Ae −t τ
                               Z 
A can be solved by realising inductor current
is zero before the diode starts conducting, i.e :

         Vm 
i (0) =   ⋅ sin(0 − θ ) + Ae −0 τ
         Z 
          Vm              Vm 
⇒ A =   ⋅ sin( −θ ) =   ⋅ sin(θ )
          Z               Z 
Therefore the current is given as,
         V 
          Z 
                [
i (t ) =  m  ⋅ sin(ωt − θ ) + sin(θ )e −t τ   ]
or
           V 
            Z 
                    [
i (ωt ) =  m  ⋅ sin(ωt − θ ) + sin(θ )e −ωt ωτ      ]
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                R-L waveform
          vs,

                    io


                     β

     vo




           vR




     vL




0                             2π                        ωt
                π                             3π   4π


    Note :
    vL is negative because the current is
    decreasing, i.e :
                           di
                    vL = L
                           dt
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            Extinction angle
Note that the diode remains in forward biased
longer than π radians (although the source is
negative during that duration)

The point when current reaches zero is when
diode turns OFF. This point is known as the
extinction angle, β .

          V 
           Z 
                [
i ( β ) =  m  ⋅ sin( β − θ ) + sin(θ )e − β   ωτ
                                                     ]= 0
which reduces to :

sin( β − θ ) + sin(θ )e − β   ωτ
                                   =0
β can only be solved numerically.
Therefore, the diode conducts between 0 and β

To summarise the rectfier with R - L load,

           Vm 
           Z 
           
                    [
                  ⋅ sin(ωt − θ ) + sin(θ )e −ωt ωτ     ]
i (ωt ) = for 0 ≤ ωt ≤ β
          0
          otherwise
          

                    Power Electronics and                   8
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                    Dr. Zainal Salam, 2002
 RMS current, Power Factor
The average (DC) current is :
                          β
      1 2π              1
Io =
     2π 0∫ i(ωt )dωt = 2π ∫ i(ωt )dωt
                          0

The RMS current is :
                              β
         1 2π 2             1 2
I RMS =
        2π 0∫ i (ωt )dωt = 2π ∫ i (ωt )dωt
                              0

Power absorbed by the load is :
Po = ( I RMS )2 ⋅ R

Power Factor is computed from definition :
       P
pf =
       S
where P is the real power supplied by the source,
which equal to the power absorbed by the load.
S is the apparent power supplied by the
source, i.e

S = (Vs , RMS ).( I RMS )

              P
⇒ pf =
       (Vs,RMS ).(I RMS )
                    Power Electronics and      9
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Half wave rectifier, R-C Load

          +                 iD                      +
          vs                                        vo
          _                                         _




     vs
     Vm



           π /2   π               2π 3π /2     3π        4π


   Vmax                     vo
   Vmin                                                       ∆Vo
                      iD

                                    α    θ



     Vm sin(ωt )      when diode is ON
vo =  −(ωt −θ ) / ωRC
     Vθ e              when diode is OFF
vθ = Vm sin θ

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               Operation
• Let C initially           • After ωt=π/2, C
  uncharged. Circuit          discharges into
  is energised at             load (R).
  ωt=0
                            • The source
• Diode becomes               becomes less than
  forward biased as           the output voltage
  the source become
  positive                  • Diode reverse
                              biased; isolating
• When diode is ON            the load from
  the output is the           source.
  same as source
  voltage. C charges        • The output voltage
  until Vm                    decays
                              exponentially.




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        Estimation               of θ
The slope of the functions are :
d (Vm sin ωt )
               = Vm cos ωt
    d (ωt )
and
  (
d Vm sin θ ⋅ e −(ωt −θ ) / ωRC         )
             d (ωt )
              − 1  ⋅ −(ωt −θ ) / ωRC
= Vm sin θ ⋅             e
              ωRC 
At ωt = θ , the slopes are equal,

                      − 1  ⋅ e −(θ −θ ) / ωRC
Vm cosθ = Vm sin θ ⋅       
                       ωRC 
  Vm cosθ        1
⇒            =−
  Vm sin θ ⋅    ωRC

  1      1
     =
tan θ − ωRC

θ = tan −1 (− ωRC ) = − tan −1 (ωRC ) + π
              Power Electronics and           12
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              Dr. Zainal Salam, 2002
            Estimation of α

For practical circuits, ωRC is large, then :
                       π      π
θ = -tan(∞ ) + π = −     +π =
                       2      2
and Vm sin θ = Vm

At ωt = 2π + α ,

Vm sin( 2π + α ) = (Vm sin θ )e −( 2π +α −θ ) ωRC
or

sin(α − (sin θ )e −( 2π +α −θ ) ωRC = 0
This equation must be solved numerically for α




                   Power Electronics and            13
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                   Dr. Zainal Salam, 2002
              Ripple Voltage
Max output voltage is Vmax .

Min output voltage occurs at ωt = 2π + α

Reffering to diagram, the ripple is :
∆Vo = Vmax − Vmin

= Vm − Vm sin( 2π + α ) = Vm − Vm sin α

If Vθ = Vm and θ = π 2, and C is large such that
DC output voltage is constant, then α ≈ π 2.

The output voltage evaluated at ωt = 2π + α is :
                        2π +π 2−π 2                   2π 
                      −                             −     
                            ωRC                       ωRC 
vo (2π + α ) = Vm   e                        = Vm   e

The ripple voltage is approximated as :
                    2π                  2π  
                  −                    −     
                    ωRC                 ωRC  
∆Vo ≈ Vm − Vm   e            = Vm 1 − e          
                                                 
                                                 

                    Power Electronics and                        14
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          Voltage ripple-cont’d
    Approximation of exponent term yields:

     −2π ω RC       2π
    e         ≈ 1−
                   ω RC
    Substituting,

              2π  Vm
    ∆Vo ≈ Vm       =
              ω RC  fRC

    • The output voltage ripple is reduced
    by increasing C.
    •As C is increased, the conduction interval
    for diode decreases.
    •Therefore, reduction in output voltage
    ripple results in larger peak diode current.


•    EXAMPLE:
     The half wave rectifier has 120V RMS source at
     60Hz. R=500 Ohm and C=100uF. Determine (a)
     the expression for output voltage, (b) voltage ripple.
                     Power Electronics and               15
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         Controlled half-wave
                  ig                            ia
                                         vo
         ia


 +                         +              α
                                                          ωt
 vs                        vo                        vs
 _                         _

                                ig




                                          α                ωt
Average voltage :

      1 π                   Vm
Vo =    ∫ Vm sin (ωt )dωt = 2π [1 + cos α ]
     2π α

RMS volatge

              π                      2
          1
VRMS   =    ∫ [Vm sin (ωt )] dωt
         2π α

  Vm π
   2
                            Vm     α sin (2α )
=    ∫ [1 − cos(2ω t ] dωt = 2 1 − π + 2π
  4π α

                       Power Electronics and               16
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  Controlled h/w, R-L load
                             i

                                                          +
                                                          vR
          +                                                    +
                                                          _
          vs                                                   vo
          _                                               +
                                                          vL   _
                                                          _


                    vs


                         π                   2π                ωt

               vo

                             io

                         π         β        2π                      ωt
      α

                                                                            −ωt
                                 V 
i (ωt ) = i f (ωt ) + in (ωt ) =  m  ⋅ sin (ωt − θ ) +                 Ae ωτ
                                  Z 
Initial condition : i (α ) = 0,
                                                      −α
             V 
i (α ) = 0 =  m  ⋅ sin (α − θ ) +                 Aeωτ
              Z 
                               −α
         V               
A = −  m  ⋅ sin (α − θ ) e ωτ
         Z               
                                 Power Electronics and                      17
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                                 Dr. Zainal Salam, 2002
          Extinction angle
Substituting for A and simplifying,
           V                                   −(α −ωt ) 
           m  ⋅ sin (ωt − θ ) − sin (α − θ )e ωτ 
            
           Z                                          
                                                           
          
i (ωt ) = for α ≤ ωt ≤ β
          0 otherwise
          
          
          
Extinction angle, β is defined when i = 0,
                                                 (α − β 
             V 
i( β ) = 0 =  m  sin ( β − θ ) − sin ( β − θ )e ωτ 
              Z                                       
                                                        
which can only be solved numerically.
Angle ( β − θ ) is called the conduction angel.
i.e the diode conducts for γ degrees.




                   Power Electronics and                 18
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                   Dr. Zainal Salam, 2002
       RMS voltage and current
             voltage
       Average     :
                β
           1                Vm
      Vo = ∫Vm sin(ωt )dωt = [cosα − cosβ ]
          2π α              2π

             current
       Average                           RMScurrent
                β                             β
            1                            1 2
       Io = ∫ i(ωt )dω          I RMS =    ∫ i (ωt )dω
           2π α                         2π α

                      by
       Thepowerabsorbed theloadis :

       P = I RMS2 ⋅ R
        o

•   EXAMPLES
•   1. Design a circuit to produce an average voltage of 40V
    across a 100 ohm load from a 120V RMS, 60Hz supply.
    Determine the power factor absorbed by the resistance.

•   2.    A half wave rectifier has a source of 120V RMS at
    60Hz. R=20 ohm, L=0.04H, and the delay angle is 45
    degrees. Determine: (a) the expression for i(ωt), (b)
    average current, (c) the power absorbed by the load.
                     Power Electronics and                 19
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       Freewheeling diode (FWD)
     • Note that for single-phase, half wave
       rectifier with R-L load, the load (output)
       current is NOT continuos.

     • A FWD (sometimes known as
       commutation diode) can be placed as
       shown below to make it continuos
                                                 io


                                                         +
                                                         vR
           +                                                      +
                                                         _
           vs                                                     vo
           _                                             +
                                                         vL       _
                                                         _

                io                                           io

                                                                  vo= 0
+                    vo= vs                                                    +
                                     +
vs                                                                             vo
                                     vo                 io
_
                                                                               _
                                     _


          D1 is on, D2 is off                     D2 is on, D1 is off

                              Power Electronics and                       20
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                              Dr. Zainal Salam, 2002
          Operation of FWD
• Note that both D1 and D2 cannot be turned
  on at the same time.
• For a positive cycle voltage source,
   – D1 is on, D2 is off
   – The equivalent circuit is shown in Figure (b)
   – The voltage across the R-L load is the same as
     the source voltage.


• For a negative cycle voltage source,
   – D1 is off, D2 is on
   – The equivalent circuit is shown in Figure (c)
   – The voltage across the R-L load is zero.
   – However, the inductor contains energy from
     positive cycle. The load current still circulates
     through the R-L path.
   – But in contrast with the normal half wave
     rectifier, the circuit in Figure (c) does not
     consist of supply voltage in its loop.
   – Hence the “negative part” of vo as shown in the
     normal half-wave disappear.


                   Power Electronics and                 21
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          FWD- Continuous load
                current

 • The inclusion of FWD results in continuos
   load current, as shown below.

 • Note also the output voltage has no
   negative part.




 output       vo


                         io
                         iD1                     ωt
Diode
current
                        iD2

          0        π            2π          3π    4π


                   Power Electronics and              22
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                   Dr. Zainal Salam, 2002
Full wave rectifier with R load




                             iD1




                                                         io
              is
    +                                                    +
    vs                                                   vo
    _                                                    _




                    Bridge circuit
         is                       iD1

                             +
                             vs1
    +                        _           − vo   +
    vs
    _                       +                       io
                            vs2
                            _

                                   iD2


          Center-tapped circuit



                   Power Electronics and                      23
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           Notes on full-wave
• Center-tapped rectifier requires center-tap
  transformer. Bridge does not.

• Center tap requires only two diodes,
  compared to four for bridge. Hence, per
  half-cycle only one diode volt-drop is
  experienced. Conduction losses is half of
  bridge.

• However, the diodes ratings for center-
  tapped is twice than bridge.

   For both circuits,

   vo =   {V Vsinsintωt
           −
            m
                m
                   ω        0 ≤ ωt ≤ π
                            π ≤ ωt ≤ 2π

   DC voltage :
       1π                   2Vm
   Vo = ∫ Vm sin (ωt )dωt =     = 0.637Vm
       π0                    π
                    Power Electronics and       24
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                    Dr. Zainal Salam, 2002
                 Bridge waveforms

           vs
           Vm


                                  2π            3π   4π
                   π
           vo
           Vm




vD1
vD2
           -Vm
vD3 vD4

           -Vm
      io

 iD1 iD2

iD3
iD4
      is




                       Power Electronics and              25
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                       Dr. Zainal Salam, 2002
Center-tapped waveforms
           vs
           Vm


                              2π         3π   4π
                π
           vo
           Vm




vD1

 -2Vm
vD2


-2Vm
  io

 iD1

iD2

      is




                Power Electronics and              26
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                Dr. Zainal Salam, 2002
    Full wave bridge, R-L load
                                               io




                               iD1
            is                                      +
                                                    vR
     +                                                     +
     vs
                                                    _
     _                                                     vo
                                                    +
                                                           _
                                                    vL
                                                    _




iD1 , iD2

                      π                   2π         3π         4π
 iD3 ,iD4


                 vo
 output                              io




                          vs                              is
 supply




                      Power Electronics and                          27
                       Drives (Version 2),
                      Dr. Zainal Salam, 2002
    R-L load analysis:
approximation with large L
Using Fourier Series, output voltage
is described as :
                      ∞
vo (ωt ) = Vo +      ∑ Vn cos(nωt + π )
                  n = 2, 4...
where
     2Vm
Vo =
      π
     2V     1     1 
Vn = m       −     
      π  n − 1 n + 1
The DC and harmonic currents are :
     Vo                        Vn     Vn
Io =                      In =    =
     R                         Z n R + jnωL

Note that as n increases, voltage
amplitude for the nth harmonic decreases.
This makes I n decreases rapidly for
increasing n. Only a few terms sufficient
to approximate output.

                  Power Electronics and       28
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                  Dr. Zainal Salam, 2002
              R-L load analysis
IfωLis large enough, it is possible to drop all the
harmonic terms, i.e. :
               Vo 2Vm
i(ωt ) ≈ I o =   =    ,               for ωL >> R,
               R   R
The approximation with large L is shown below.

  iD1 , iD2                   exact                  approx.

                     π                2π        3π             4π
   iD3 ,iD4


                vo
   output                             io


    2Vm/R
                         is
   supply




                     Power Electronics and                          29
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                     Dr. Zainal Salam, 2002
               Examples
                     (
I RMS = I o 2 + ∑ I n, RMS 2 = I o)
Power delivered to the load : Po = I RMS 2 R



• EXAMPLE: Given a bridge rectifier has an
  AC source Vm=100V at 50Hz, and R-L
  load with R=10ohm, L=10mH

   – a) determine the average current in the load
   – b) determine the first two higher order
     harmonics of the load current
   – c) determine the power absorbed by the load




               Power Electronics and                30
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               Dr. Zainal Salam, 2002
Controlled full wave, R load




                              iD1




                                             io
               is
   +                                         +
   vs                                        vo
   _                                         _




     1π                  V
 Vo = ∫ Vm sin (ωt )dωt = m [1 + cos α ]
     πα                   π
                    π                   2
          1
 VRMS   =
          πα∫ [Vm sin (ωt )] dωt

                    1 α sin (2α )
        = Vm         −   +
                    2 2π   4π
 The power absorbed by the R load is :

      VRMS 2
 Po =
        R
                    Power Electronics and         31
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                    Dr. Zainal Salam, 2002
     Controlled, R-L load             io




                          iD1
              is                            +
                                            vR
+                                                +
vs
                                            _
_                                                vo
                                            +
                                                 _
                                            vL
                                            _



     io
          α           π β       π+α   2π

     vo



                   Discontinuous mode
                            π+α
     io
      α               π     β         2π

     vo



                   Continuous mode

                   Power Electronics and              32
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                   Dr. Zainal Salam, 2002
           Discontinuous mode
Analysis similar to controlled half wave with
R - L load :

          V 
           Z 
                 [
i (ωt ) =  m  ⋅ sin(ωt − θ ) − sin(α − θ )e −(ωt −α ) ωτ   ]
for α ≤ ωt ≤ β
 Z = R 2 + (ωL) 2
               −1  ωL    L
  and θ = tan   ; τ =
                   R     R
For discontinous mode, need to ensure :
β < (α + π )

Note that β is the extinction angle and
must be solved numerically with condition :

io ( β ) = 0

The boundary between continous and
discontinous current mode is when β in
the output current expression is (π + α ).

For continous operation current at
ωt = (π + α ) must be greater than zero.
                     Power Electronics and              33
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                     Dr. Zainal Salam, 2002
          Continuous mode
i (π + α ) ≥ 0
sin(π + α − θ ) − sin(π + α − θ )e −(π +α −α ) ωτ ≥ 0

Using Trigonometry identity :

sin(π + α − θ ) = sin(θ − α ),

            [
sin(θ − α ) 1 − e −(π    ωτ )
                                ] ≥ 0,
Solving for α

         −1  ωL 
α = tan  
         R 
Thus for continuous current mode,
       −1  ωL 
α ≤ tan  
           R 
Average (DC) output voltage is given as :

     1 α +π                 2Vm
Vo =
     π α ∫ Vm sin (ωt )dωt = π cos α
                     Power Electronics and        34
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                     Dr. Zainal Salam, 2002
      Single-phase diode groups
                           D1
                                                io

                           D3                  vp
          +
          vs                                                 +
          _                                                  vo
                           D4                                _


                           D2              vn
                                                vo =vp −vn


•   In the top group (D1, D3), the cathodes (-) of the two
    diodes are at a common potential. Therefore, the
    diode with its anode (+) at the highest potential will
    conduct (carry) id.

•   For example, when vs is ( +), D1 conducts id and D3
    reverses (by taking loop around vs, D1 and D3).
    When vs is (-), D3 conducts, D1 reverses.

•   In the bottom group, the anodes of the two diodes
    are at common potential. Therefore the diode with
    its cathode at the lowest potential conducts id.

•   For example, when vs (+), D2 carry id. D4 reverses.
    When vs is (-), D4 carry id. D2 reverses.

                      Power Electronics and                       35
                       Drives (Version 2),
                      Dr. Zainal Salam, 2002
       Three-phase rectifiers
                                            D1
    + van           -                                  io
                                            D3

    + vbn       -                           D5
n                                                  vpn
                                                            +
    + vcn       -                                           vo
                                            D2              _

                                            D6     vnn           vo =vp −vn

                                            D4

                van           vbn          vcn
      Vm




                        vp
     Vm


                        vn




                             vo =vp - vn


            0            π                 2π     3π        4π
                         Power Electronics and                        36
                          Drives (Version 2),
                         Dr. Zainal Salam, 2002
      Three-phase waveforms
• Top group: diode with its anode at the
  highest potential will conduct. The other
  two will be reversed.

• Bottom group: diode with the its cathode at
  the lowest potential will conduct. The other
  two will be reversed.

• For example, if D1 (of the top group)
  conducts, vp is connected to van.. If D6 (of the
  bottom group) conducts, vn connects to vbn .
  All other diodes are off.

• The resulting output waveform is given as:
  vo=vp-vn

• For peak of the output voltage is equal to
  the peak of the line to line voltage vab .


                 Power Electronics and         37
                  Drives (Version 2),
                 Dr. Zainal Salam, 2002
 Three-phase, average voltage
    vo
           vo

                  π/3
                                         Vm, L-L


     0
     π/3   2π/3


Considers only one of the six segments. Obtain
its average over 60 degrees or π 3 radians.

Average voltage :
           2π 3
      1
Vo =
     π 3π3 ∫ Vm,L− L sin(ωt )dωt
     3Vm, L − L
   =            [cos(ωt )]ππ33
                          2
        π
     3Vm, L − L
   =            = 0.955Vm, L − L
        π
Note that the output DC voltage component of
a three - phase rectifier is much higher than of a
single - phase.
                        Power Electronics and      38
                         Drives (Version 2),
                        Dr. Zainal Salam, 2002
         Controlled, three-phase
                                         D1

    + van        -                                   io
                                         D3

     + vbn           -
                                         D5         vpn
n
                                                                +
    + vcn        -                                              vo
                                    D2                          _

                                    D6              vnn

                                    D4




             α           van                  vbn         vcn
     Vm


    vo




                         Power Electronics and                       39
                          Drives (Version 2),
                         Dr. Zainal Salam, 2002
    Output voltage of controlled
       three phase rectifier
     From the previous Figure, let α be the
     delay angle of the SCR.

     Average voltage can be computed as :
                2π 3+α
           1
     Vo =       ∫ Vm, L− L sin(ωt )dωt
          π 3 π 3+α

           3Vm, L − L 
         =             ⋅ cos α
           π          



•   EXAMPLE: A three-phase controlled rectifier has
    an input voltage of 415V RMS at 50Hz. The load
    R=10 ohm. Determine the delay angle required to
    produce current of 50A.




                   Power Electronics and          40
                    Drives (Version 2),
                   Dr. Zainal Salam, 2002

				
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