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

Conditions: Ld = , r_a=0, L_a=0, 0 < <

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 =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  angle.

The full power of a primary winding of transformer is

,

where P_d – the power of the direct components of the rectified voltage U_d and the rectified current I_d.

The full power of the secondary windings of the transformer is

.

Type power is

.

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

| L_d I_TAV I_TM I₂ I₁ S₁ S₂ S_T U_RM

FWCT | 0 0.5I_d 1.57I_d 0.780I_d 1.11I_d 1.23P_d 1.74P_d 1.48P_d 3.14U_d

|  0.5I_d I_d 0.707I_d I_d 1.11P_d 1.57P_d 1.34P_d 3.14U_d

HW | 0 I_d 3.14I_d 1.570I_d 1.22I_d 2.69P_d 3.49P_d 3.09P_d 3.14U_d

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

S_T 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.

1.8.4. Consideration of a stage of switching of thyristors for a full-wave rectifier with centre tap and active - inductive load with infinite inductance

Conditions: Ld = , r_a=0, L_a=0, 0 < <

As there is inductance L_a connected in series with thyristor, a current i_T can not change by jump. The current of the switching on thyristor will be increased from 0 up to I_d and the current of the switching off thyristor decreases from I_d up to 0, i.e. during the same time two thyristors of same group are on state. These phenomena names as a stage of a current commutation. Duration of the stage of switching terminates as an angle γ.

Figure 1.16

Let’s consider that VS1 is turned on just as a control pulse is applied and this moment should be a moment of a reference mark. VS2 is beginning to turn off.

During an interval γ i_T1 is growing up from 0 to I_d;

i_T2 is growing down from I_d to 0.

According to first Kirchhoff’s law

I_d = i_T1 + i_T2

Since L_d =  a load current have only a direct component; hence, an alternating component of thyristors’ currents flows through phases of the transformer and the thyristors witch are both turned on.

We should designate it as i_C.

Then

i_T1 = i_C; i_T2 = I_d – i_C;

Inductance of a branch of the thyristor

х_а = L_a

According to 2-nd Kirchhoff’s law

i_K = i_T1

Let's discover γ from a condition:

At  = γ i_T1 (γ ) = I_d

According to the theorem of the equivalent generator during the interval γ

Figure 1.17

Let's discover an average rectified voltage

where no-load EMF is

Commutating voltage drop is

From here, we could get an equation of the external characteristic