1.8.1. Operation of a full-wave rectifier with a centre tap with an active load
Conditions: La=0, Ld=0, ra=0, 0 <α < π/2
A peak thyristor current is
The function of load voltage is
The average rectified voltage is
It is a control characteristic
iV1
uV1
iV1, iV2
u, i |
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
URM = 2E2m = πEd0.
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 = 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 ST
External characteristic Ud = f (Id)
Udα = Edα – (ra + rT) Id,
where rT – a forward dynamic resistance of the thyristor (it is considered as a constant), Edα – a external EMF at no-load,
At Idα = 0 Udα = Edα .
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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 КP→1 at = 0 since the transformer is not loaded by the higher harmonics of a current i1.
1.7.2. 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<∞
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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
U2,
i
Id(m)
e2
id
icв
Id
inp
φ
π
0 |
Figure 1.13
Thus, Id does not depend from Ld
Let's analyse
φ |
Figure 1.14
At Ld → 0 and Ld → ∞
Time constant of the load circuit
If Ld → 0; 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 () 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.
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Figure 1.15