I probably have this diagram for my project. Or problem.


a) only 2 wires out of wall, B & W;

b) switch ON, lights ON

Ceiling electrical box:

One black and one white connects to the ceiling light fixture. enter image description here enter image description here

Project: I need to add a combo switch/outlet

UPDATE after suggestions

Grey= Neutral Is this right? If it is, it seems to me receptacle will be ON at all times, right?

enter image description here

  • How do you plan to switch the lighting outlet if you make the switch control the receptacle? Dec 6 '17 at 12:41
  • 1
    You can't even express the concept that white is a color and neutral is a purpose and they aren't the same thing. You will need that concept to win here. Dec 6 '17 at 19:23

Three options

Run a new /3 cable

You either replace the existing cable with /3 cable (which is now required in new work), in which case your three wires are:

  • always-hot (traditionally black, often white-with-black-tape, sometimes white)
  • switched-hot (ideally red, sometimes black, often white-black-tape, often white)
  • neutral (always white)

And these three go to the lamp/receptacle combo, always-hot to the common bar, switched-hot to the switch side, and neutral to the receptacle side.

Run a whole separate /2 cable

In a perfect world you source that from another location which is not a lighting circuit, but does provide GFCI, AFCI or whatever other protection you need owing to that location. But if you need to, you can get this from the lighting box, if Code allows this, e.g. In bathrooms it might not. In this case your two new wires are:

  • always-hot (black usually)
  • neutral (always white)

And you break the tab off the hot side of the switch/recep combo. These two new wires go to the receptacle side, neutral to the taller pin side. The two existing wires go to the switch side only. You must break off the tab.

Smart switches

Talking something like Insteon.

In this case, rewire so all black wires are always-hot, and all white wires are in fact neutral. Always-hot and neutral go to your smart switch w/ receptacle.

Up in the lamp, the switched-hot to the lamp is now dangling, as it has nowhere to connect to, there are no switched-hot wires up there, you killed them.

So you get a smart-switch module, that connects to always-hot and neutral. It also has a "Load" line which goes to switched-hot, and that's where the lamp's switched-hot wire goes.

Now if you leave the light on, you can turn it off with your phone. From Kazakhstan.

  • Thank you so much for your detailed answer. Option one. /3 goes to the combo box, black to the darker screws on the right side (both connected to each other), red to brass screw (left / upper) and neutral to the silver screw (left lower). What about the light? Forget the original 12/2 cable? and install the light in either the red or black wire? Thank you in advance Dec 6 '17 at 19:53
  • 1
    In the case of 12/3 cable (must be 12/) you replace and eliminate the /2 cable. The light attaches to neutral as before, and its other wire to the switched-hot wire, which it sounds like you have chosen red for. Red is a great choice. Dec 6 '17 at 20:07
  • Would you be kind to check whether the last update is correct (at question)? Also, how can I put the light in line with the receptacle so when the switch is OFF, the plug is also off?. And a very stupid question... If the hot is interrupted by a switch, does that mean all the features downstream will have no power? Dec 7 '17 at 4:43
  • 1
    Yes, that looks right. If the hot is interrupted by a switch, it becomes switched-hot. Any load which draws from switched-hot and neutral will only work with the switch on. So on your switched receptacle, switch red and black, and that will have that effect. Dec 7 '17 at 6:54
  • I will get back tomorrow to that project and will get the new 12/2, break the connection. I tried the 12/3 and either the light was always on, or only the receptacle. I did put the red on the top brass screw, white at the silver screw and the source at the bridged side. Didn't work. Dec 20 '17 at 6:13

Your first diagram shows a switch loop.

Neutral and hot wires arrive at the light box, the neutral stays in that box, while only the hot is sent to the switch box (customary using white with black marking), and returns with the (customary) black wire, which is now a "switched hot".

If you want an outlet at the switch box, you need to bring a neutral from the light box. A customary way would be to wire a 14/3 or 12/3 from the light box to the switch box: - white neutral - black hot - red switched hot (returns from the switch and connects to the light fixture black).

You can then decide wether you want this outlet to be always hot, or if you want the outlet to be switched.

The wire gauge depends on the breaker size. 15A breaker requires 14 gauge or bigger. 20A breaker requires 12 gouger bigger.

  • So do I need to bring a new 12/2 wire pigtailed from the ceiling main wire? In this case, there would be the current B and W from the light fixtures + new wire truly hot and neutral? At the switch box we would have 2 cables (each one black and white). Dec 6 '17 at 9:04
  • 1
    Yes, this is valid. You can bring a second 12/2 romex from the light box to the switch box, to carry the "line" (constant hot + neutral).
    – albator
    Dec 6 '17 at 17:29
  • Yeah use the new 12/2 as the feed for the receptacle and leave the switch leg as-is.
    – ArchonOSX
    Dec 6 '17 at 20:04

This set up worked well. The bridge (wall receptacle/switch) on the right side (two black screws) intact. The plug is ON at all times. The light works as expected. I wanted the switch to control the receptacle but didn't manage to do it.

Ceiling light box had 2 sets of wires: B and W from the wall switch/plug. It does require extra wire. So 12/3 (or 14) needs to be added. 2 wires will NOT work.
Ceiling box: B hot and W were there.


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