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I just noticed this SSR that was for DCV instead of all the others that are normally for AC.

enter image description here

enter image description here

Although I can not find the datasheet for it, it does seem compatible with Arduino. eBay auction site and also here is where I found it.

However, I am just wanting to double check 2 things here:

  1. My thoughts on how to go about wiring it:

    enter image description here

    I have 2 scenarios here. One that runs 12V out and another that does ground. Is this SSR capable of doing the 0vdc out since I am seeing the output is 5V-220V DC? I'm using the 0V DC to "push" a button (pushing = grounding it) so that the object it's connected to turns on or off.

  2. I gather that when I send 5V DC to the SSR (via Digital Pin 1 or 2) it will open up the line for the 12V/0V to flow out the negative (-) side? Not sure if these are NO or NC? And as a side note, the diagram in the picture and the label on the SSR itself seems to be backwords? It says +/- and +/- while the diagram says -/+ and -/+?

Addition:

This image shows that they have some resisters in place: enter image description here enter image description here

Can anyone tell by those images above what type of resister they are using? Seems to me they are dropping the voltage by that on the arduino board side?

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  • \$\begingroup\$ While this will probably work fine, you're likely to solve the same problem cheaper with either an optocoupler (like a 4N35 or whatever you can find cheaply) or with a simple MOSFET transistor and a gate bleed resistor (if galvanic isolation is not needed.) \$\endgroup\$
    – Jon Watte
    Commented Apr 4, 2014 at 4:11
  • \$\begingroup\$ Your second circuit, switching the ground side of a circuit, is incorrect. The "+" output terminal must always be more positive than the "-" terminal, so you want the "-" terminal grounded, and the "+" terminal going to the switched circuit. \$\endgroup\$
    – Peter Bennett
    Commented Apr 4, 2014 at 6:11
  • \$\begingroup\$ I think the diagrams are rotated about the Y axis, not the X axis. Then the pins and polarity symbols would line up properly. \$\endgroup\$
    – user95251
    Commented Dec 24, 2015 at 2:05

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Re your thoughts on wiring it, I think your second diagram has the output polarity reversed. The (-) should go to the power supply and the (+) to the load.

If I misinterpreted what you meant by "IN" and "OUT", then the other diagram is reversed- (+) should go to the power supply and (-) to the load.

The diagrams are not backwards. This one:

enter image description here

Is looking at the bottom of the SSR (pins pointing at your face) as if you rotated the relay 180° in the top photo about the X-axis (label would not be visible).

This one:

enter image description here

Is looking down through the top of the relay, with the relay in the top photo rotated about the Z axis 180° (writing on the label would be upside-down).

The output will be normally open (no voltage applied to the input). When sufficient voltage/current (at least 6mA at 3V) is applied, it will close. Note that the "off" leakage specification is quite high (1.5mA).

The SSR incorporates 1500V isolation so you can use it either as a high-side switch or a low-side switch, as you have shown (except for the polarity).

Edit: The schematic you linked is wrong-- the transistors drive the input of the SSR with series and shunt 10K base resistors. The collectors of the transistors go to the SSR control inputs. They reversed collector and base on each transistor.

enter image description here

The other two 1K resistors are for indicator LEDs that are in parallel with the SSR inputs. The SSRs can accept 3~32V in but the LEDs need a series current-limiting resistor each.

The transistor drive circuit is only required if your drive circuit cannot supply at least 3V with 6mA current. Not a problem for most 5V logic.

The LED indicators can be omitted if you don't need visual indication (plus they draw additional current from your drive circuit- a few mA more, so the requirement is more like 10mA).

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  • \$\begingroup\$ Can you enlighten me on what Note that the "off" leakage specification is quite high (1.5mA). means? \$\endgroup\$
    – StealthRT
    Commented Apr 4, 2014 at 3:19
  • \$\begingroup\$ @StealthRT: When the input is fully off, the output will pass 1.5mA regardless. \$\endgroup\$
    – Ignacio Vazquez-Abrams
    Commented Apr 4, 2014 at 3:23
  • \$\begingroup\$ @StealthRT It means that when the SSR is off (no power to the control input) the output will "leak" as much as 1.5mA. For example, if you're using it to control power to an LED, you will see light from the LED unless you shunt the LED with a resistor. \$\endgroup\$
    – Spehro Pefhany
    Commented Apr 4, 2014 at 3:23
  • \$\begingroup\$ What resistor value would I need in order to bring that 1.5 to 0? I don't need voltage going to the product when it should be "off" completely. \$\endgroup\$
    – StealthRT
    Commented Apr 4, 2014 at 17:59
  • \$\begingroup\$ @StealthRT Unfortunately, the only correct answer is 0\$\Omega\$. Anything higher and some current will go to the load. An LED is a bit easier because it won't conduct appreciable current below a volt or two depending on type. \$\endgroup\$
    – Spehro Pefhany
    Commented Apr 4, 2014 at 18:02

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