How to connect a solid-state relay to an output MOSFET, which has 2 drain pins without the ground which I am refering to the image in the attachment?



enter image description here

  • 1
    \$\begingroup\$ You should link to the datasheet, not the catalog page. Which is the "Output MOSFET which has two drain pins without the ground"? Are you referring to the ones in the image or something else? Please edit your question to clarify. Welcome to EE.SE. \$\endgroup\$
    – Transistor
    Commented Jun 10, 2021 at 18:48
  • \$\begingroup\$ If you send sufficient current through the left diode, the two right terminals will be a closed switch with some small resistance. If there is no current through the diode, the two right pins will behave like an open switch (with a little bit of parasitic capacitance between the terminals) \$\endgroup\$
    – tobalt
    Commented Jun 10, 2021 at 19:54
  • \$\begingroup\$ Where would you expect to find and/or how would you use a ground on the secondary? \$\endgroup\$
    – winny
    Commented Jun 10, 2021 at 19:59
  • \$\begingroup\$ In that case, I can reverse the connections on the output side or? \$\endgroup\$
    – HaK
    Commented Jun 10, 2021 at 21:23

1 Answer 1


As you can see in the image you posted, the output MOSFETs have their sources connected together and to the negative side of the photodiode. When the LED (left side) shines light onto this photo diode, the gates of the MOSFETs become positively biased compared to the source, opening a channel between the output terminals.


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