Operation of a full-wave rectifier with a centre tap with an active load

Figure 1.10

A requirements to equipment could be determined at α = 0, this condition fixes a peak load of a transformer and thyristors.

;

From here the rms-voltage of the secondary winding is

and a peak inverse voltage is

U_RM = 2E_2m = πE_d0.

An average rectified current is

or

From here at α = 0

It is obviously, that loading of thyristors and the transformer have maximum at α = 0. It is visible from the waveforms of currents and voltages at the diagrams, therefore requirements to thyristors and the transformer we shall determine proceeded from this condition.

Average current of thytistor is

Current of a secondary winding of the transformer at a = 0 is

Let's define

Let's define full power of primary winding S₁, full power of secondary winding S₂ and type power of the transformer S_T

External characteristic Ud = f (Id)

U_dα = E_dα – (r_a + r_T) I_d,

where r_T – a forward dynamic resistance of the thyristor (it is considered as a constant), E_dα – a external EMF at no-load,

At I_dα = 0 Þ U_dα = E_{dα .}

Figure 1.11

Operating ratio of the transformer by power is

That is greater in 2 times than for a half-wave circuit as there is no permanent magnetizing and К_P→1 at a= 0 since the transformer is not loaded by the higher harmonics of a current i₁.