How does a smart switch get its power?

Question

I bought a smart wall switch.

The switch it replaced had only 2 wires (in and out).

My knowledge of electricity might be only limited, but how can this work?

A normal switch just opens a circuit so the current no longer goes through right? So then if there is no longer any current, how does the smart switch gets power?

Answer 1

Smart switches that don't require a neutral connection power themselves by leaking power through the lamp (or other device) being switched. Normally it's such a tiny amount of power that it doesn't cause the lamp to light up, but sometimes a low-wattage LED light won't allow enough leakage current, so some manufacturers have a product that allows more current to leak like the Lutron MLC: (Minimum Load Capacitor). This is recommended for their "no neutral required" switches in the Caseta Advanced Wiring Guide "to help ensure proper operation of the switch with LED, CFL, fluorescent, and ELV lighting loads"

This wiring diagram shows how the LUT-MLC is wired in parallel to the lamp to allow more leakage current:

(image from Lutron MLC wiring instructions)

This diagram also shows the current flow from the Live connection into the smart switch, out the switched hot and through the lamp being powered (and optionally through the MLC), then then back through the lamp's neutral connection.

Answer 2

I've been looking into 'two wire' smart switches which might be compatible with LED lighting, and came across Yoswit - current performance/availability unknown. However, in their documentation they suggest that an LED load of around 5W is the minimum which can provide sufficient leakage current to power the electronics of the smart switch in the 'off' state.

This leakage will be either through the lamp itself (potentially causing a slow ticking as the internal capacitors charge up enough to trigger the LED), through discharge resistors in the lamp's power supply (potentially exceeding their thermal design), or through a parallel 'dropper/leakage' capacitor.

Yoswit claim a standby power for their switch of <1W, but this is still significant compared with a single LED bulb. An array of several LED bulbs is more likely to provide enough leakage to power the switch without illuminating the bulbs.

Answer 3

3 ways:

1. Wire themselves between supply-hot and neutral. They obtain power in the normal and ordinary way.

2. Wire themselves between supply hot and Safety Ground. This is either illegal foreign dreck, or a reputable maker has appealed to Underwriter's Laboratories (UL) for a waiver to do this, because presumably they've shown UL that the device won't degrade grounding protection and electrify all the grounds e.g. if a ground wire breaks. NFPA has told UL to quit doing that.

3. Wire their electronics in series with the incandescent bulb, and then leak power through the bulb to power themselves. This will make the incandescent glow, too dimly to be seen. CFL and LED driver circuits will have a different reaction. They will either

   • glow visibly (because they are more efficient than incandescent)
   • allow the trickle of current to pass, but only if they are designed to do this
   • block the current and prevent the switch from working
   • take damage or burn up, potentially starting a fire

One solution to #3 is that a specially made "resistor" modules (it's not quite a resistor) can be added in parallel with the light(s) to facilitate that leakage current. Never use random electronics parts in mains electrical installation, always use products UL-listed for that use. Several companies make "resistor" modules that will solve this problem, and these are readily available.