# Solid state relay voltage current rating

I am looking to use a solid state relay like this one . This relay has a peak load voltage of 350v and continuous dc current rating of 110mA.

My question - if I wanted to control 12v @ 1A DC through this SSR, will it be able to handle it?

350v @ 110mA = 38.5W and 12v @ 1A = 12W, since 12W is less than 38.5W it should work, correct?

• You can't multiply those values. Because current limit you can't. Although 100ma seems too small, maybe it's something different.
– user76844
Jul 7, 2018 at 21:06

My question - if I wanted to control 12v @ 1A DC through this SSR, will it be able to handle it?

No, The limit is 110mA for DC, you can't go beyond 110mA DC without risking damage to the part.

There are other conditions that you can go over the 110mA limit, and the bottom graph shows you why, the switch is anywhere from 10Ω to 16Ω

It appears to me that the switch would be useful for isolation and a switch for triacs or relays or but not for any substantial load.

The manufacturer sets the current limit based on factors other than the maximum power dissipation. Based on your calculation, if the voltage was only 1 volt, then the current could be 38 amperes which is obviously not feasible. In any case, this particular device incorporates a current limiting feature. According to the data sheet, the maximum current at which this limited occurs is 270 ma. Thus you cannot put 1 ampere through it, irrespective of any power considerations. Your best bet is to get a relay rated to handle your current requirement.

• thanks for the info, I had that suspicion even though a friend of mine suggested that with the power calculation I would be ok to run 1A through it at a lower voltage. Jul 7, 2018 at 21:50

350v @ 110 mA = 38.5 W and 12 V @ 1 A = 12 W, since 12 W is less than 38.5 W ...

That is the calculation for the power in the load not in the SSR.

The power dissipated in the SSR is given by $P_{SSR} = V_{SSR}I_{SSR}$. Obviously when $I = 0$ then $P = 0$. When the SSR is on the power can be calculated by $P = I^2 R$ and from the datasheet we can see that $R_{typ} = 18\ \Omega$ so $P = 0.1^2 \cdot 18 = 0.18\ \text W$.

Back to your misunderstanding: the internal wiring and semiconductor cross-sectional area, etc., determine the maximum current that can flow without exceeding the maximum current density (current per unit cross-sectional area) of the device. Operating beyond these limits will cause failure of some part of the device due to thermal effects. Your proposal was to exceed the current ratings by a factor of almost ten. This would be unlikely to last long.