Welcome @lhagva99, to the community!
Well, to put it simply, the thyristors work like a switch through which the current of 12A to be switched flows. If this switch gets too hot because your heat sink is too small, this is caused by too much power loss. This power loss can effectively be seen as the product of the current flowing (12A in your case) and the voltage drop across the thyristor. Since you cannot reduce the current, you would have to reduce the voltage drop. However, the BTW69 has only approx. 1V (at 125°C even less), smaller values can currently only be achieved with MOSFETs for semiconductors, but this would change your circuit considerably and make it more expensive and is therefore probably out of the question.
In the case of a heating application, i.e. a resistive consumer, it could make sense to use an additional relay that is switched on approx. 200...500ms after the thyristors are switched on and whose relay contacts short-circuit the thyristors. Of course, you should then proceed in reverse when switching off and only switch off the thyristors after switching off the relay. This has the additional advantage that the relay contacts have no problem with arcing and therefore only have to be designed for the current.
BTW: one of the two fuses is redundant, since only one will blow out in the event of an overload anyway.
Hope that helps?
Regards
/Peter
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