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