Operation with Inductive Load in Cycloconverter
- In practice the load is never purely resistive. In case of the inductive loads, due to the energy stored by the load inductance, the conducting SCR continues to conduct in the next half cycle as explained in the AC to DC converters.
- Therefore as shown in the figure / table a portion of negative half cycle will appear in the positive half cycle of the load voltage waveform and a portion of positive half cycle will appear in the negative half cycle of the load voltage waveform as shown in figure / table.
- For a divided by three output frequency the sequence of triggering the thyristors is as given in the figure / table.
|
Interval |
α to π + α |
Π + α to 2π + α |
2π + α to 3π + β |
3π + β to 4π + α |
4π + α to 5π + α |
5π + α to 6π α |
|
Conducting SCR |
S1 |
S3 |
S1 |
NONE |
S4 |
S5 |
Operation with Inductive Load in Cycloconverter :
The sequence of operation is as follows :
- SCR1, is turned on at instant (ωt = α), this will make the load voltage positive.
- This SCR will conduct throughout the first positive half cycle and continues to conduct in the negative half cycle of the ac supply due to the stored energy by the load inductance.
- At instant (ωt = π + α), SCR3, is triggered. This will turn off the conducting thyristor S1 due to the line commutation.
- The load voltage becomes positive. SCR3, will conduct upto the instant (ωt = 2π + α) where it is turned off due to the “line commutation” when SCR1, is turned on.
- The load voltage will be positive in the third half cycle of the supply and SCR1, conducts from (2π + α) to (3π + α) .
- At (ωt = 3π + α), the SCR2, is triggered , in order to develop a negative load voltage. But at this instant SCR1 is already in conduction & so SCR2, cannot be turned on. SCR2, cannot turn on as long as SCR1, is in conduction as they are connected in antiparallel manner. SCR1, will turn off “naturally” when all the load stored energy is returned back.
- If the trigger pulses are not being applied continuously to then none of the will conduct during the rest of the fourth half cycle and the load voltage remains zero as shown in figure / table.
- This will happen again in the beginning of the positive half cycle of the output where an effort is made to turn on SCR1, when SCR2, is still conducting. In this way all the SCRs may remain off during a part of the half cycle, the duration of which depends on the load power factor and firing angle α.
