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A single-phase full converter with B-2 type of connection has a continuous load current waveform. The thyristor pairs T3, T4 is triggered at ωt =
- a)0
- b)α
- c)π+α
- d)π-α
Correct answer is option 'C'. Can you explain this answer?
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A single-phase full converter with B-2 type of connection has a contin...
First, at α T1-T2 conduct, then after π+α the next pair is excited i.e. T3-T4 and likewise.
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A single-phase full converter with B-2 type of connection has a contin...
Α=120° and T1, T2 is triggered at α=240°.
In a B-2 type of connection, two thyristors are connected in series with the load, and their anodes are connected to the same supply line. The cathodes of the thyristors are connected to opposite ends of the load.
Since the load current waveform is continuous, the firing angle α is greater than the load angle β, which is the angle between the voltage and current waveforms.
Assuming a positive half-cycle of the input voltage, the thyristors T3 and T4 will conduct for the first 60° of the voltage cycle, while T1 and T2 will be off. During this time, the load current will flow through T3 and T4, and the voltage across the load will be positive.
At α=120°, T3 and T4 will turn off, and T1 and T2 will turn on. The load current will now flow through T1 and T2, and the voltage across the load will be negative.
At α=240°, T1 and T2 will turn off, and T3 and T4 will turn on again. The load current will flow through T3 and T4, and the voltage across the load will be positive again.
This cycle will repeat for each half-cycle of the input voltage, resulting in a continuous load current waveform.
It's worth noting that the B-2 connection is not commonly used in practice, as it requires a higher voltage rating for the thyristors and has a lower efficiency compared to other connection types.
In a B-2 type of connection, two thyristors are connected in series with the load, and their anodes are connected to the same supply line. The cathodes of the thyristors are connected to opposite ends of the load.
Since the load current waveform is continuous, the firing angle α is greater than the load angle β, which is the angle between the voltage and current waveforms.
Assuming a positive half-cycle of the input voltage, the thyristors T3 and T4 will conduct for the first 60° of the voltage cycle, while T1 and T2 will be off. During this time, the load current will flow through T3 and T4, and the voltage across the load will be positive.
At α=120°, T3 and T4 will turn off, and T1 and T2 will turn on. The load current will now flow through T1 and T2, and the voltage across the load will be negative.
At α=240°, T1 and T2 will turn off, and T3 and T4 will turn on again. The load current will flow through T3 and T4, and the voltage across the load will be positive again.
This cycle will repeat for each half-cycle of the input voltage, resulting in a continuous load current waveform.
It's worth noting that the B-2 connection is not commonly used in practice, as it requires a higher voltage rating for the thyristors and has a lower efficiency compared to other connection types.
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A single-phase full converter with B-2 type of connection has a continuous load current waveform. The thyristor pairs T3, T4 is triggered at ωt =a)0b)αc)π+αd)π-αCorrect answer is option 'C'. Can you explain this answer?
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