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I would like to switch a 10A AC load with a solid state relay (SSR) and I am wondering if a higher current SSR (e.g. 40A) would be more efficient and produce less heat than a lower current rated SSR (e.g. 25A).

Is the SSR's loss resistive (like a MOSFET) and improved with a higher rated SSR, or is it based on a forward voltage drop (like a transistor) meaning the wasted power would be unrelated to the SSR's current rating?

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    \$\begingroup\$ I tried to test this once. I used triacs (without heatsinks) rated for 4A to those rated for 40A. Heating was almost similar for all triacs. Even the 40A triac didn't seem like a good option to operate a 10A load due to heating. If you choose to add a heat sink, then you can probably use it. However you should be able to find electro-mechanical relays which might be cheaper and smaller than (triac + heat sink) assembly. At least that's what I did because I didn't need any sort of dimming control on that high load. \$\endgroup\$ – Whiskeyjack Nov 20 '15 at 6:18
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Most Solid State Relays use either a triac or inverse-paralell-connected SCRs as the switch element.

The forward voltage drop when the switch is turned ON is about 1 Volt (average). This voltage drop is relatively constant, no matter what the current rating of the switch element is.

So, no. I don't think that using a SSR that is much larger than is actually needed will give significantly lower losses.

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