:Hi to newsgroup : :I have a problem that needs some assistance. :I have a 220V 50 HZ industrial relay at 20 amps. :The arming inductor is rated at 6 watts . I need to :use a varistor as surge arrester but i don't know to :calculate the energy that will be produced and the :varistor ratings . As i understand it must be rated for :275 VAC rms.But how is the current capability related :and the energy (joules) that it is rated .Is there a rule :of thumb for such calculations ?. :Any help would be apprecciated . : :
I assume by "arming inductor" you mean the relay coil. If the coil is rated at
6W this means a coil current of less than 30mA so it is hardly a large relay. At best it might come into the category of "contactor".
Your main problem is defining what you are trying to do by suppressing the back emf from the relay coil. In most cases the reason is to prevent arcing at a series contact (or switch) which is used to activate the relay in question. A varistor across the relay coil will not be sufficient in itself since the voltage required to maintain an arc at the series contact will be be lower than the clamp voltage of the varistor, therefore considerable arcing at the contact can occur before the clamp voltage of the varistor is exceeded. In most cases a C-R arc suppression combination across the series contact is necessary.
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