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

Figure 1.12

The given conditions take place for low-power rectifiers with the inductive filter.

According to second law for electric circuits

Then a current i_d we could present as the sum of free component i_fr and steady-state component i_ss

;

where amplitude of the forced component

;

a phase displacement of the steady-state component

.

The current waveforms of the thyristors VS1 and VS2 are identical at quasistate duty

i_T1=i_T2,

hence, at and value of the currents of thyristors are equal (current flowing through inductance can not vary by jump).

From here we should find the constant A

Thus, currents i_d, i_T and i₂ at interval are

Let's construct the graphs of the currents waveforms

Figure 1.13

Thus, I_d does not depend from L_d

Let's analyse

Figure 1.14

At L_d → 0 and L_d → ∞

Time constant of the load circuit

If L_d → 0; Þ τ_d → 0; φ→ 0.

Then

If L_d → ∞ then

and τ_d → ∞; therefore i_for → 0;

The current curve i_d have only i_fr, but so far as i_d (0)= i_d (p) и τ_d = ∞,

therefore i_d=I_d.

Let’s find a free component of load current

Thus, the current i_d is ideally smoothed.

i_d(υ) and u_d(υ) is not depended from R_{d .} If to determine u_d(υ) as a voltage drop across R-L-load - then u_d(υ) is not depended from L. If one across R-load - then u_d(υ) depends from L and

u_d (υ) = i_d (υ) R_d.

If Х_d ≥ 5 R_d it is possible to consider, that Х_d = ∞ and i_d is ideally smoothed.

Figure 1.15

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