Operation of a full-wave rectifier with centre tap and active - inductive load with infinite inductance
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
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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
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From this equation is defined a peak-inverse voltage
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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
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I2 does not depend from a control angle.
Let's determine the an effective value of the current of a primary winding of transformer
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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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