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Solid State Relay G3MB

This relay's output says "2 A at 100 to 240 VAC". If I wanted to connect something like DC12v 2A to this relay output,

  1. It is not going work?
  2. It might work, but it will damage the relay circuit?

I know this is a very stupid question, I googled but couldn't find a page explaining the differences between VAC and VDC output relay.

Hope somebody can clarify the differences. Thanks in advance.

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3 Answers 3

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NO you cannot. Nada, no way Jose. 決して、今日ではなく、決して

This is a Triac controlled SSR and these latch ON with DC.

Read the fine print in the 1st page summary.

"The G3MB-202PEG-4-DC20MA crosses directly to the Motorola M0C2A-60 series power triac."

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  • \$\begingroup\$ youtube.com/watch?v=qGQdbUiTd54 So its bidirectional. If it's bidirectional, why not DC? DC only goes to one direction. Not trying to outsmart you, but some sites say a triac can be used on DC or AC. \$\endgroup\$
    – kukrt
    Sep 7, 2018 at 3:23
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    \$\begingroup\$ You can use a triac to turn on a DC circuit, but it won't turn off until the current drops to zero - this happens 100 or 120 times a second for AC, but never for DC. \$\endgroup\$ Sep 7, 2018 at 4:04
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    \$\begingroup\$ That is what I called a DC Latch. The holding current must be extinguished from load side to turn off Triac. ( < 15 mA in typ spec.) More correct to say an SCR is a DC latch and a Triac is a bidirectional or bipolar DC latch. \$\endgroup\$ Sep 7, 2018 at 4:17
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First be sure this relay is built with MOSFETs or SCRs. If built with triacs they will NOT work with DC. Worse case is you have to buy a compatable or conventional relay, else:

As long as you do NOT exceed the voltage or current rating of the output, it is a "Don't care" scenario. With an AC rating you could change polarity of the DC voltage and it would still be "Don't care".

Be aware that you maybe running the relay at its current limit, so if it gets hot put it on a large heatsink. Since it expects a much higher voltage it may have some internal resistance that will cause a voltage drop under full load. If less than a 1 volt drop I would just live with it. If the drop is several volts than you need a source greater than 12 volts to make up for the loss.

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  • \$\begingroup\$ "Normal" mechanical relays can be used to switch AC or DC with no problems (with suitable regard for ratings). However, solid state relays designed to switch AC normally use Triacs as the switching element - Triacs only switch off when the current falls to Zero, so they cannot be used to control a DC circuit. \$\endgroup\$ Sep 7, 2018 at 3:15
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Voltage and current inside a relay works in a slightly different way.

First of all, current will create heat inside the relay. Depending on the conductor size, you will likely have a maximum current that can pass through the relay without burning it up. In your case: 2A. Therefore, no matter what the voltage is, you will need to be bellow 2A on your load.

For the voltage, you need to picture how a relay is made inside your head. A metal piece is moving back and forth creating or cutting a contact between two port. That metal piece is controlled by a electric coil that clamp the contact on. When the contact is not clamped, there is a distance in between the contact and the metal piece. While air is a nice dielectric (insulator) there is a fixed voltage that it can withstand before it begins to be a conductor. Think of lightning, electric ark, welding. Therefore, if the voltage is above the rated voltage, you will most likely create an ark inside your relay and this is not a desirable thing at all.

In the case of a solid state relay (your case) the switching is done with a semiconductor. Therefore, the dielectric will be different and there won't be any moving parts inside it. But, exceeding the voltage may break down the device or cause it to ark in surprising ways.

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  • \$\begingroup\$ This relay I am concerned about is an SSR type, which I believe there is no moving parts inside. \$\endgroup\$
    – kukrt
    Sep 7, 2018 at 2:49
  • \$\begingroup\$ My bad, you are quite right. The same problem might however occur inside the switching semiconductor. The dielectric will be different than air in that specific context. I will edit my answer. \$\endgroup\$ Sep 7, 2018 at 2:52

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