I am looking at something like this LVD 12V and am confused as to the purpose of the two negative terminals. On the BATT side I will have the + coming from the + side of the battery and the - going to a ground (this is the test for the voltage supply). On the LOAD side I will have the + side going to a light that is already grounded. When the switch is allowing current due to the voltage being above 10.7V it should pass that down to the LOAD. So why is there a second - terminal on the LVD?

  • \$\begingroup\$ 1) So that you don't have to connect 3 wires together. 2) Sometimes the disconnect is performed or sensed via the negative connection. \$\endgroup\$ Jul 15, 2016 at 3:45
  • \$\begingroup\$ So from what I am trying to accomplish, it makes sense that I will have that terminal disconnected? \$\endgroup\$ Jul 15, 2016 at 3:46
  • \$\begingroup\$ Who knows. But I wouldn't count on it working properly. \$\endgroup\$ Jul 15, 2016 at 3:49
  • \$\begingroup\$ Maybe I'm missing the whole idea of how the switch works. I couldn't find a diagram for it. I need it to detect when a power source is < 11V so that it will not send power to an automotive bulb in this circumstance. The bulb is in a socket that is grounded, I'm just replacing the positive source with this source based on voltage. \$\endgroup\$ Jul 15, 2016 at 3:56
  • \$\begingroup\$ The battery connects to one side. The load connects to the other. \$\endgroup\$ Jul 15, 2016 at 4:02

1 Answer 1


@SteveG pointed out that the 50 mA operating current at 12V suggests that the unit contains a relay. I agree. This is reinforced by the fact that the 24V version has a 25 mA operating current - ie double voltage but same power. This is typical of relays but would make less sense if it had a solid state switch. This means that the unit is probably but not certainly isolated between input and output when not operated. This would explain the use of two wires for each of input and output. You can check for non operated isolation as mentioned below.

Their spec sheet is inadequate.

If the unit contains a relay it is probably input-output isolated when not operated. (A relay will make a nice comforting click when operated and you can perhaps feel it operate as well if you hold the unit.)

If it uses a 'solid state' switch then it is PROBABLE but not certain that the 2 negatives are hard-joined electrically, and that the extra wire is for convenient connection.
An Ohmmeter should confirm this, or not.
Also, when isolated, check connections and voltages on output leads.

While it is possible to make such a device with semiconductor switches and also with true floating input/output when in isolate mode this would be unusual.
ie usually you get a SPST switch with common grounds
rather than a DPDT switch.

  • \$\begingroup\$ The fact that the LVD12V takes 50mA quiescent current implies that it uses a relay to do the switching, so it would be simple to use a DPDT relay. In this case the two negatives would be separate and only connected together when the battery voltage is above 10V. As Russell says, an ohm meter check between the two negatives will confirm. \$\endgroup\$
    – Steve G
    Jul 15, 2016 at 7:47
  • \$\begingroup\$ @SteveG Good point. I agree. Answer amended. The fact it says 12V/50 mA and 24V/25 mA makes a relay VERY likely. Relays are often a very good choice despite their limitations. Sadly :-). \$\endgroup\$
    – Russell McMahon
    Jul 15, 2016 at 11:56
  • \$\begingroup\$ I have them ordered and will report back with my findings. Thanks for all the help. \$\endgroup\$ Jul 15, 2016 at 13:14

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