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The double three-phase rectifier with balancing reactor

Прочитайте:
  1. Consideration of a stage of switching of thyristors for a full-wave rectifier with centre tap and active - inductive load with infinite inductance
  2. Operation of a full-wave rectifier with a centre tap with an active load
  3. Operation of a full-wave rectifier with centre tap and active - inductive load and limitеd inductance
  4. Operation of a full-wave rectifier with centre tap and active - inductive load with infinite inductance
  5. Operation of single-phase half-wave rectifier with active load
  6. Operation of the half-wave rectifier with active - inductive load and limited inductance
  7. Operation of the half-wave rectifier with center tap and opposite-EMF and active-inductive load
  8. Operation of the half-wave rectifier with center tap with opposite- EMF and active load
  9. Operation of the half-wave rectifier with resistive-capacitive load
  10. Operation of the rectifier with opposite- EMF

Figure 1.37

1.12.1 Conditions: a=0; Ld = ¥; 0 < Lbr <¥.

At this scheme there are 2 secondary windings located on the same core of a transformer and differentially connected. The rectifier has two commutating groups of thyristors: VS1, VS3, VS5 and VS4, VS6, VS2 which are separated by reactor Lbr®¥. Load is arranged between centre tap of balancing reactor and common cathodes of thyristors. There are 2 types of these rectifier regimes:

1) Independent operation of two three-phase rectifiers with center taps;

2) A six-phase regime.

 

Figure 1.38

Regime 1. As Lbr®¥ so switching from the thyristor of one group to the thyristor another group is impossible so both thyristor groups operate as three-phase rectifiers with centre tap independently one from other.

Let there is the highest phase EMF in the anode circuit of the thyristor VS1 of thegroup 1 and VS2 of the group 2.

 

Lengthways of contour O1, VS1, VS2, O2, Lbr according second Kirchhoff’s low

e2a + e2Z = UK .

For intervals of operation of two any thyristors of different groups

eII + eI = UK.

Voltage drop across a half-winding of balancing reactor is .

Then for the contour O1, VS1, Ld, Rd, O:

.

For the contour O2, VS2, Ld, Rd, O:

.

From here

.

Without taking into account losses of the transformer

For intervals of operation of two any thyristors of different groups

As Lbr have limited value through the contour O1, VS1, VS2, O2, Lbr flows the alternating current iK limited by inductance Lbr. The phase lag of the first harmonic iK(1) concerning voltage UK is p¤2. This alternating current iK is imposed on a direct component of a thyristor current of group 1 and groups 2.

Regime 2 At reduction of a load current to value Id/2=I_к m the circuit will be operate as an equivalent six-phase rectifier due to field current of balancing reactor is not provided needed value due to Lbr is reduced and it is possible to consider Lbr asanode inductance. The g angle depends from a load current Id. Simultaneously from one up to four thyristors are conducted.

The waveform of the rectified voltage is under construction in view of g. For no-load regime ed(u) is a top envelope of phase EMF of both groups. So average value of ed(u) is

Thus, the external characteristic will be as

 

 

Figure 1.39

1.12.2. Definition of parameters for a choice of thyristors, calculation of the transformer and the balancing reactor

As Ikm @ 0.5¸1% IdM it is possible to not take into account ik for a choice of thyristors and calculation of the transformer and to consider a current through a thyristor and a secondary winding of the transformer

Hence, calculation of a current through thyristor will be as for the 3-phase rectifier with center tap at Ld = ¥, but under condition of ITM=Id/2

The waveform of an primary current

 

 

Figure 1.40

From here we shall define RMS currents values of primary and secondary windings of the transformer, and also full power of windings and type power of the transformer

Let's define type power of the current-balancing reactor ST br

The average voltage of the current-balancing reactor for half-wave determines Φm

Let's reduce the voltage UKAV to equivalent sinusoid with frequency 50 Hz

where - a form factor.

For a regime 1

From here result is exceeded

Кс > 1 takes into account operating conditions of the core in the real circuit at f=150 Hz.

Usually Кс=2, from here

For the controlled rectifier

Form Uk would depend from α.

If the angle of regulation varies in a range , that

For

Thus with growth α, UKAV and STbr are increased hence, STbr depends from αmax.

Further, as well as at α=0, we shall define

It is possible to use functional association

where

It is necessary to take into account, that as Id would not depend from α, that

where

 
 

 


Figure 1.41

 


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