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Topic: Solid State Relay - Project Enclosure Material? (Read 1 time) previous topic - next topic


I'm building a sous vide cooker using an SSR to control a GFCI outlet that I'm plugging 3 120V 300W (2.5A) heating elements into. The 120V 20A SSR I have is http://www.phidgets.com/products.php?category=9&product_id=3953_0 with the http://www.phidgets.com/products.php?category=9&product_id=3955 heatsink.

With the heatsink, according to Phidget my max load in open air is 15A but unless my math is wrong, I'll only be using 7.5A. The heatsink has a max operating temperature of 80C after which, I assume, it will fail and probably fail closed.

In normal operating conditions the relay will only be closed for short periods of time, other than the initial "burn up" to get the water up to temperature, which could potentially take a while.

So here's the question: I have two enclosures sitting around that I can fit this into: one is plastic and the other is aluminum. Assuming I'd want to keep the whole thing "sealed" (ie. no intentional ventilation to keep dust out) would the plastic enclosure be sufficient or will heat become an issue? I'm concerned with using aluminum, even though it'd be better at heat dissipation, just because it's metal and I've never really done any projects that used more than 9V of load. I'm sure that's just paranoia speaking but fear is a mighty motivator.

I'm leaning towards the plastic enclosure and drilling ventillation "grids" and possibly using some pc fans to keep air circulation but I'd like to avoid that if possible.



I'm leaning towards the plastic enclosure and drilling ventillation "grids"

That would be my choice I would not think there is any need for a fan.
But if you have to go with no ventilation then chose the metal container.


If you use the aluminium enclosure, you could screw the SSR directly to it instead of using the heatsink.
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Jan 05, 2013, 05:03 pm Last Edit: Jan 05, 2013, 05:05 pm by retrolefty Reason: 1
As a 'rule of thumb' a SSR (thyristor based) will have a constant almost 2vac voltage drop across it's output terminals when conduction current, so X the actual current flow will give you the heat dissipation of the device in watts. So 15 watts of heat for a 7.5 amp load. So let that be a planning tool for how hot you think the inside of your 'vented' box might get if it had a 15 watt incandescent lamp turned on inside it.


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