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Operation of a full-wave rectifier with centre tap and active - inductive load with infinite inductance

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Conditions: Ld = ¥, ra=0, La=0, 0 <a <

 

Figure 1.16

 

The peak current of a thyristor is ;

The average current of thyristor is

;

The peak-inverse voltage is

;

The average rectified voltage is

It is a control characteristic.

Requirements to equipment could be determined at α = 0, this condition defines a most loading of the transformer and thyristors.

Then

.

Thus

.

Let's define a secondary winding voltage

From this equation is defined a peak-inverse voltage

Hence, a powers of the transformer windings we could define at a=0.

Let's determine an effective value of the current of a secondary winding of transformer and rms-current of a thyristor

I2 does not depend from a control angle.

Let's determine the an effective value of the current of a primary winding of transformer

I1 also does not depend from a angle.

The full power of a primary winding of transformer is

,

where Pd – the power of the direct components of the rectified voltage Ud and the rectified current Id.

The full power of the secondary windings of the transformer is

.

Type power is

.

Let's compare single-phase circuits: with centre tap (FWCT) a nd half-wave (HW) by the thyristors and the transformer loading:

| Ld ITAV ITM I2 I1 S1 S2 ST URM

FWCT| 0 0.5Id 1.57Id 0.780Id 1.11Id 1.23Pd 1.74Pd 1.48Pd 3.14Ud

| ¥ 0.5Id Id 0.707Id Id 1.11Pd 1.57Pd 1.34Pd 3.14Ud

HW | 0 Id 3.14Id 1.570Id 1.22Id 2.69Pd 3.49Pd 3.09Pd 3.14Ud

 

Apparently from the table active load is heavier than active-inductive load as thyristors as the transformer.

ST is less twice for full-wave rectifiers than for half-wave ones since there is no flux of the forced magnetization through a transformer core so current loading of thyristors is less in 2 times, but also number of thyristors is in 2 times more than for half-wave schemes.

 


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