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I need to control the shutdown of a load that operates with a voltage of 25V and a maximum RMS current of about 90 mA.

This load is working almost always and rarely needs to be turned off.

For which, I think I can use a solid state relay for AC loads, similar to this (NC Normally closed):

enter image description here

The doubt, or problem I have, is that I cannot find an NC type SSR with zero crossing detection.

On the other hand, SSRs with NO type, there are several options with zero crossing.

My question is.

Could there be a problem using an SSR like in the image to control the mentioned load without zero crossing detection?

Any comment or suggestion is welcome.

Regards

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  • \$\begingroup\$ your question cannot have a no answer because the load is unknown \$\endgroup\$
    – jsotola
    Commented Jun 24 at 1:19
  • \$\begingroup\$ Thanks for answering. The load is an LED type lamp, which obviously has electronic components inside for its operation, therefore it isn't a linear load. \$\endgroup\$ Commented Jun 24 at 15:13

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This type of SSRs with back-to back MOSFETs can switch both DC and AC. This type of relay doesn't rely on zero crossing. When the relay is off, both MOSFETs are off. One of the MOSFET body diodes is forward biased, the other one is reverse-biased. That's enough to prevent current from flowing.

MOSFET body diode solid state relay

somewhat related answer: LCA717 SSR connected to a digital pin

¹ Be careful, in general not all SSRs can switch both DC and AC.

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  • \$\begingroup\$ Thanks for answering. Do these protection diodes exist in all models of solid state relays like the image I posted in the OP? Since in the sdataheet of the device that I want to use, nothing is mentioned about said diodes, nor does it represent them in the equivalent schematic. The load is an LED type lamp, which obviously has electronic components inside for its operation, therefore it is not a linear load. \$\endgroup\$ Commented Jun 24 at 12:17
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    \$\begingroup\$ MOSFET body diodes exist in all power MOSFETs. Body diodes are parasitic diodes. If body diodes aren't drawn on the schematics (for brevity), then they're implied. If the power MOSFETs didn't have body diodes, then an SSR with one MOSFET (not two) would be able to switch AC. More here. \$\endgroup\$ Commented Jun 24 at 13:17
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90mA at 25VAC isn't a whole lot of power; I suspect the reason you aren't finding zero-crossing detecting relays at your power level is because there aren't a whole lot of situations where suddenly switching on a few dozen mA would cause a problem for the source. In other words, if your circuit can handle it, it shouldn't be a problem.

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Thyristor type zero-crossing devices wait until the voltage across them is less than some relatively low voltage like 5V before they turn on. They turn off when the current drops to zero.

One of the purposes is to reduce EMI so that the device does not switch on with a very high voltage across it, resulting in a big current spike. They don't generally force complete cycles, only complete half-cycles.

With a depletion MOSFET SSR when you apply current the MOSFETs will turn off following a substantial and variable delay, they will do that regardless of the current. If you have an inductive load, there could be a voltage spike that would require something like a bipolar TVS to deal with so that it does not damage the SSR. It might cause too much EMI.

When you remove the current, again after a delay, usually less, the MOSFETs will switch on, regardless of the voltage across them. If you have a load that is capacitive (like a switching power supply) or behaves more like a tungsten lamp or motor you'll get a spike in current that could damage the SSR, and might cause undesirable EMI.

With such a low current, a combination of TVS, RC snubber and maybe a series resistance or NTC current limiter could deal with all combinations (or not, perhaps), but the details would depend on the load and what trade-offs you are prepared to make.

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  • \$\begingroup\$ Thanks for answering. The load is an LED type lamp, which obviously has electronic components inside for its operation, therefore it is not a linear load. \$\endgroup\$ Commented Jun 24 at 12:17

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