1.7.2. Operation of the half-wave rectifier with active - inductive load and limited inductance

Conditions: =0, r_a=0, L_a=0, 0 < L_d < ∞

Figure 1.5

L_d is connected in series with a load for smoothing a rectified current.

Equivalent resistance and inductance of a circuit are

According to second Kirchhoff’s law

where

Characteristic equation is

From here

We should search out solution as i=i_ss+i_fr; where i_fr – a free component, i_ss - a steady state component.

Where

Constant A we could be found from initial conditions:

At

Then

from here

Let's designate

Then

U, I

Figure 1.6

Let's find the energy reserved in inductance L by period

Energy which inductance accumulates

Energy which inductance takes out

Thus, in the time of period

A direct component of the rectified voltage is

where  - a pulse load current duration.

A direct component of a rectified current is

It is possible to determine λ from the condition: at  = λ  i_d =0

If L_d would increase, λ also increases and I_dm goes down, hence, K_r decreases. Thus, it is possible to use L_d as the filter of a load current.

1.7.3. Operation of the half-wave rectifier with resistive-capacitive load

Conditions: α=0, r_a=0, L_d=0, 0 < C <

Figure 1.7

The equations describing electric processes in the circuit are

Let's designate:

λ - a pulse thyristor current duration,

φ – a delay angle of the thyristor switching on is fixed concerning a point of natural switching on.

Let's consider intervals:

1)

The thyristor is on-state:

U_T=0;

Let's find I_T.

2) -

The thyristor is off-state:

.

According to second Kirchhoff’s law

Characteristic equation is

.

Root of this equation is

.

Then a voltage across the condenser is

.

Current through the load is

.

Current of the condenser is

.

Voltage across the thyristor is

.

Constant A we could be defound from the condition:

At

Then

.

Let's define

At

Figure 1.7

According to the diagram of changing the function tg()

limits by

.

For active load:

for the capacitive load: ,

From the condition of the no interrupted process, i.e. periodicity of electromagnetic processes in the circuit, we could find φ:

This transcendental equation can be solved graphically or by numerical methods by computer.

e,u e,i

Fig. 1.8 – The diagram of electric processes in a single-phase half-wave circuit of the rectifier with resistive-capacitive load.

1.8. A single-phase full-wave rectifier with a centre tap

The single-phase full-wave rectifier with a centre tap is in fact two-phase one because the secondary winding of the transformer with a centre tap creates two EMF which are equal by module, but are in opposite directions.

Figure 1.9