Operation of a full-wave rectifier with centre tap and active - inductive load and limitеd inductance
Conditions: La=0, α=0, ra=0, 0<Ld<∞
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 id we could present as the sum of free component ifr and steady-state component iss
;
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
iT1=iT2,
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 id, iT and i2 at interval are

Let's construct the graphs of the currents waveforms


Figure 1.13
Thus, Id does not depend from Ld
Let's analyse

Figure 1.14
At Ld → 0 and Ld → ∞
Time constant of the load circuit

If Ld → 0; Þ τd → 0; φ→ 0.
Then

If Ld → ∞ then

and τd → ∞; therefore ifor → 0;
The current curve id have only ifr, but so far as id (0)= id (p) и τd = ∞,
therefore id=Id.
Let’s find a free component of load current


Thus, the current id is ideally smoothed.
id(υ) and ud(υ) is not depended from Rd . If to determine ud(υ) as a voltage drop across R-L-load - then ud(υ) is not depended from L. If one across R-load - then ud(υ) depends from L and
ud (υ) = id (υ) Rd.
If Хd ≥ 5 Rd it is possible to consider, that Хd = ∞ and id is ideally smoothed.
Figure 1.15
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