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Why when I apply a firing angle alfa with a thyristor controlled regulator the current is zero in the red line drawn in the figure? I think that the thyristor opens the circuit, but why it does so? How can I understand when the thyristor stops the conduction and then restart it? I know that the thyristor stop the conduction when there is a zero cross point of the voltage, but in that point the voltage is not zero. enter image description here

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  • \$\begingroup\$ Does the current through an inductor remain in phase with the voltage? \$\endgroup\$ – a concerned citizen May 4 at 10:18
  • \$\begingroup\$ @aconcernedcitizen no it is out of phase pf pi/2 without thyristor, but with the thyristor with alfa higher than pi/2 and lower of pi, it is out of phase of alfa. \$\endgroup\$ – Samuele Benito Di Gioia May 4 at 10:20
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With or without thyristor, the same differential equations apply: you have a voltage across an inductor, therefore there will be a time when the voltage and current will not be both above zero. The conduction will stop when the current tends to drop below zero and the voltage is below zero (i.e. reverse polarized). The field collapses, since there is no voltage to sustain it. Unless the second thyristor acts, in which case it will be as if there are no switches (ideal case, no dead-time), or the waveform will have a dead-zone. Here is how the angle α=[1/4,2/4,3/4]*π affects the waveform for a single thyristor, (black trace) and two anti-parallel thyristors (red trace, overlapping on the positive side with the black trace), when driven with almost no dead-time:

thy

In all cases for the single thyristor, the current stops because the voltage is negative, therefore it is blocked and the current simply decays to zero and stays there. For the two thyristor case, the current starts ramping up (or down, reverse polarized) as soon as the second thyristor enters the conduction mode. If there was only the 2nd thyristor there, the waveform would have stopped ramping up and stayed at zero (the reverse of the black trace).

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    \$\begingroup\$ @aconcernedzitizen thank you very much! It is all clear now! \$\endgroup\$ – Samuele Benito Di Gioia May 4 at 11:24

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