of transformer secondary and primary windings;
𝑘
— transformation ratio;
w1, w2 — number of turns of primary and
secondary windings of the transformer;
S2Н, S1Н — rated power ratings of transformer
secondary and primary windings;
S2Н = 3∙U2Н∙I2Н, S1Н = 3∙U1Н∙I1Н — rated
power ratings of transformer secondary and primary
windings;
Рd0 = Ud0 IdH∙— notional rated power at the
rectifier output.
3 RESULTS AND DISCUSSION
We will consider the operation of 6Y/Δ when the
connection between mains and phase B is open, i.e.,
when the key K in figure 1, a is open.
In this case, the circuit will switch to an
incomplete phase mode. Phase voltages and currents
В and bΔ are equal to zero. Only the phases A, С, aΔ
and сΔ are involved.
The vector diagrams of the phase and line
voltages of the secondary 6Y/Δ open phase
transformer are shown in Figure 2, b. From this
figure it can be seen that the phase voltages aD and
сD (U’a and U’с), and linear (U’aс and U’сb) are
reduced cos π/3 times, and modulo become equal to
each other:
𝑈′
𝑎
𝑈′
𝑎𝑐
𝑈′
𝑐
𝑈′
𝑐𝑏
𝑈′
𝑏𝑐
𝑈
2𝐿
∙
𝑐𝑜𝑠
𝜋
3
√
3
2
𝑈
2𝑃𝐻
. (22)
Thus, the six-pulse rectifier circuit of the inverter
in question is converted into a two-pulse bridge
circuit when the B phase circuit is interrupted. As a
single-phase secondary voltage, two phase voltages
of equal magnitude and direction are connected in
parallel (u’a and -u’с) or linear (u’aс and u’bс)
voltages, of those phases which are not in phase
mode. In the case of a disconnection in phase A,
these will be u’ab and u’cb, and in the phase С these
will be u’aс and u’ab.
Since the break occurs in phase B, there will be
voltages applied to the rectifier u’ac and u’bc, the
amplitudes of which are
√
3
2
less than the
amplitudes of the voltages uac and ubc of the full-
phase mode (Figure 2, b).
At the interval where the highest potential at the
point aΔ and bΔ, and the smallest at the point сΔ
(Figure 3, a), the current flows as follows. The first
half of the load current flows out of the point aΔ,
and the second from the point bΔ. Running through
VD1 and VD3 respectively, these currents are
connected at the common cathode of the rectifier,
and then through the load circuit and VD2. They
then branch out again at the point сΔ, passing
through the phases aΔ and сΔ. Voltage is applied to
the load u’ac(bc) (Figure 2, b), and the rectifier
operates VD1, VD3 and VD2 (Figure 3, a).
At the interval where the highest potential at the
point сΔ, and the lowest – at the point aΔ and bΔ.
(Figure3, b), the load current flows through the
circuit: point cΔ, VD5, load. The current then splits -
the first half flows through the VD4 circuit, point
aΔ, phase aΔ, point сΔ, and the second by the circuit
VD6, point bΔ, phase cΔ, point сΔ. Voltage u’ca(сb)
is applied to the load (Figure 2, b), and VD5, VD4
and VD6 operate in the rectifier (Figure 3, b).
Figure 3: Instantaneous 6Y/Δ operation diagrams in an
open phase mode.
Since 6Y/Δ operates as a two-pulse rectifier in
partial phase mode, its average rectified voltage can
be found from the expression (16). Therefore,
according to (22), (1) and (2), we obtain:
𝑈
′
𝑑0
2
√
2
𝜋
∙𝑈
′
2
2
√
2
𝜋
∙𝑈
2𝑃𝐻
∙
√
3
2
√
6
𝜋
∙𝑈
2𝑃𝐻
1
√
3
∙𝑈
𝑑0
. (23)
It follows from (23) that the rectified voltage
6Y/Δ is reduced by a factor of 1.732 when there is
an open circuit in any phase of the primary winding.
If the inverter operates according to Algorithm 1
(Zd = const), then the rectified current is reduced by
the same amount and the power delivered to the load
is reduced by a factor of 3:
𝐼
√
∙𝐼
, (24)
𝑃
𝑈
∙𝐼
√
∙𝑈
∙
√
∙𝐼
∙𝑃
. (25)
If the inverter operates according to Algorithm 2
(Pd = const), then the rectified current 6Y/Δ in
partial phase mode must increase as much as the
voltage has decreased, i.e., 1.732 times