best way to drive an AC inductive load

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The max peak current rating of the diode which is the spec you need to look at is 400mA for the 1N4148. The relay coil current is ~180mA which is also the current you'll get with the back EMF when the driver opens so a 1N4148 will be fine. But the relay release time will increase which could be a problem and that can be helped by using a 4V zener in series with the diode to allow the back EMF voltage rise up to about 5V.

Also a while back you asked about hot vs neutral switching. Here in the U.S. we have 2 types of 2 prong plugs, one polarized and one not. If a non-polarized plug is used then you have no idea which is wire is hot which is usually not a problem. If it is a concern then a polarized 2 prong plug or a 3 prong grounded plug, which end up also being polarized, could be use. No matter what type of plug you use you are still relying on the electrician to have wired the outlet correctly. So bottom line for safety is to consider both sides as possible hots.

Ray
 
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Perhaps it is better to use a MOV or Schotky diode with faster switching speed
 

Perhaps it is better to use a MOV or Schotky diode with faster switching speed
in which case would I need fast switching? Is it to guarantee longer life to the relay or for some other reason?

thanks

Davide

---------- Post added at 14:28 ---------- Previous post was at 14:25 ----------

Thanks Ray,
now that you mention this, I actually realized that I might be able to drive the relay directly from the output of a 555 without the need of a transistor in between. The 555 should be able to drain/source up to 200mA so the 185mA should be Ok, right? Or is there anything I am missing? Obviously the flyback diode would stay where it is.

 

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