Pe1

Multiphase star rectifier • For larger (>15kW) output power. • Harmonics and fundamental component. • Filter size decreases with the increase of harmonics. • Q phase filter has fundamental component of qf frequency. • Also known as star rectifier.

•

May be considered as q single-phase half-wave rectifier. • K-th diode conducts during the period when the voltage of k-th phase is greater than that of other phases. • The conduction period of each diode is 2π/q. • Primary must be delta connected to compensate the dc component flowing through secondary windings.

Assuming a cosine wave from π/q to 2π/q, the average o/p voltage is. 2 Vdc = 2π / q

π /q

π /q

q V cos ω t d ( ω t ) = V sin ωt m ∫0 m π 0

= Vm

q π sin ..... (1) π q

Output rms voltage is,  2 π /q  2 Vrms =  V cos ω t d ( ω t )  ∫0 m 2 π / q  

1/ 2

 q  ω t sin 2ω t  π / q   q = Vm   + = V   m 4  0   π  2  2π

 π 1 2π    + sin   q  q 2

1/ 2

..... (2)

If the load is purely resistive then peak current through the diode is Im = Vm/R and the rms current through a diode is,  2 Is =   2π

1/ 2

π /q

 1 I s = Im   2π

 2 2 ∫0 I m cos ωt d (ωt )  π 1 2π  + sin q q 2

1/ 2

  

V = m R

(Integrating over the whole period, 2π)  1   2π

π 1 2π  + sin q q 2

1/ 2

  

=

Vs R

Vs is the rms voltage of a transformer secondary

Three phase bridge rectifiers • Full-wave rectifier. • Gives six-pulse ripples on the o/p voltage. • Conduction sequence of diodes are, 12,23,34,45,56,61. • The pair of diodes which are connected between that pair of supply lines having the highest line-toline instantaneous voltage will conduct. • Line to line voltage, Vij = 3V p

•

Average o/p voltage,

π /6 2 Vdc = 3Vm cos ωt d (ωt ) ∫ 0 2π / 6

=

3 3

(Vm is the peak phase voltage)

V =1.654V

m π m Rms o/p voltage, 1/ 2 π / 6 1/ 2  π / 6 6.3  ωt sin 2ωt    2  2 2 Vrms =  3Vm cos ωt d (ωt ) = Vm   +   ∫ 0 2 π / 6 π 2 4    0  

2π 18  π / 6 1 = Vm   + sin 4 6 π  2

1/ 2

  

1/ 2

18 18 3  = Vm  + .  12 4π 2 

3 9 3  = 1.6554Vm = Vm  +   2 4π 

For purely resistive load, peak diode current, I m = 3Vm / R Rms diode current, I r = 2. 2  2π

π /6

∫

0

1/ 2

 I m2 cos2 ωt d (ωt ) 

2π 1 π 1 = I m   + sin 6 π  6 2

1/ 2

  

1/ 2

π /6  2  ωt 1   =   + sin 2ωt   0  π  2 4

= 0.5518 I m

Rms value of transformer secondary current,

 4 I s = 2.  2π

∫

π /6

0

1/ 2

 I m2 cos2 ωt d (ωt ) 

2π  2 π 1 = I m   + sin 6 π  6 2

1 2π 4  π = Im   + sin 6  π  2.6 4

1/ 2

  

1/ 2

  

= 0.7804 I m

Where Im is the peak secondary line current = peak diode current,

I m = 3Vm / R

3-phase bridge rectifier with RL load vab = 2Vab sin ωt for L

iL =

π 2π ≤ ωt ≤ 3 3

di + RiL + E = 2Vab sin ωt dt 2Vab E −( R ) t sin(ωt − θ ) + A1e L − ... (3 - 81) Z R

Where load impedance Z = ( R + (ωL) ) and load impedance angle θ = tan The constant A1 can be found from the initial condition: at ωt = π 3 , i = I . 2

2

1

2

L

I1 =

2Vab  π ) E − ( R ) (π  sin − θ  + A1e L 3ω − Z R 3 

I1 +

E 2Vab  π ) − ( R ) (π  − sin − θ  = A1e L 3ω R Z 3 

 E 2Vab  π  ( R ) (π ) A1 =  I1 + − sin − θ  e L 3ω R Z 3  

1

−

 ωL   .  R 

iL =

 2Vab E sin (ωt −θ ) + I1 + − Z R 

2Vab π  −( R L )t ( R L )(π 3ω) E sin  −θ e e − Z 3 R  

iL =

 2Vab E sin (ωt −θ ) + I1 + − Z R 

 π  2Vab π  ( R L ) 3ω−t  E sin  −θ e − . . . (3 - 82) Z 3 R  

Under steady-state condition,

I1 (1 − e

 R   π 2π  −     L   3ω 3ω 

)=

iL (ωt = 2π / 3) = iL (ωt = π / 3) = I1.

 R   π 2π   R   π 2π  −    2Vab  π   L  3ω − 3ω   E  L 3ω 3ω   sin ( ωt − θ ) − sin − θ e  − 1 − e   Z  3   R    R  π

I1 =

2π 

2Vab Z

π    −  sin ( ωt − θ ) − sin − θ e L  3ω 3ω  E 3  −  R   π 2π  −    R (1 − e L  3ω 3ω  )

2Vab Z

π  −   sin ( ωt − θ ) − sin − θ e  L  3ω  E 3  − for I1 ≥ 0 . . . (3 - 83) R π    −   R (1 − e  L  3ω  )

 R  π 

I1 =

Which after substituting in eq (3-82) and then simplifying, gives

iL =

(

)

 2π − θ − sin π − θ   R  π   sin −  3  2Vab  E 3    L  3ω  ( ) sin ω t − θ + e −    R  π  −  Z    R  L  3ω  ( 1 − e )   for π/3 ≤ ωt ≤ 2π / 3 and iL ≥ 0 ... (3 - 84)

Rms diode current can be found from above equation as,

 2 Ir =   2π

1/ 2

 3 2 i d ( ω t ) ∫π / 3 L  2π

Rms o/p current is the combination of rms current of each diode. I rms = 3I r

Effects of source and load inductance at π vac and vbc are equal and I dc still flows through D1. The current id 1 decreases, resulting and induced voltage across L1 of + vL1 and o/p becomes vL = vac + vL1. At the same time o/p voltage due to vbc is vL = vbc − vL 2 . Result is anode voltage of D1 and D3 are equal and both diodes conducts for some time which is called commutation angle µ .