I’d like to detect whether a circuit (in this case, a recessed light) has power without actually tapping into the power line. I’ve read about Hall Effect sensors, but am not sure that’s the best path.
Does anyone have any recommendations?
@normelton, is the power AC or DC? What is the voltage?
Take a look over here. The suggested circuit acts like a typical non-contact AC voltage probe. You could easily use it as-is since the Photon can handle the 5V input on a digital pin.
AC power, 120V.
After some more googling, I stumbled on this:
It seems to be everyone on one ready-made shield (a good thing for a noob like me).
@normelton, CE makes great stuff! However, it might be overkill here as you indicated you want to know if the lights are on or off, not how much current they are consuming. If your lights take very little current, the sensor may be sensitive enough. Also note that you will need to pass a SINGLE wire through the toroid coil since having 2 AC wires (live and neutral) will cancel each other out! What kind of current are you looking at?
Yep, and it would also prevent me from using my relay shield. I’m going to lower a motorized window shade any time the light above our shower is turned on. The relay + motor piece is fairly straight forward, but I’m less confident on the control part.
The light is a 10w LED fixture, so very very low amperage.
@normelton, perhaps the approach needs to be rethought. What if you used a wifi connected bulb like Lifx and got its status from the web? Never tried it but there are lots of smart members who might help!
Well I do have a backup plan. Change the light switch to a single-throw double-pole switch, and hook the second pole up to a pin on the photon.
Are you afraid the current is so low that it might be hard to detect?
@normelton, I’m pretty sure what no matter where you live, that goes against the electrical code! It’s also incredibly dangerous and a sure way to fry a Photon. Another approach is to drive an 120 VAC relay and use the contacts to detect the switch closure (complete isolation from the live AC side). Yet another possible way is to use a light sensor to detect the light in the shower. Yet another approach is to use a magnetic reed switch to detect when the shower door is opened, indicating shower use.
You need to figure out where you expect to put your blind controller, how you will power it and THEN how will you detect the light in the shower. How you power your controller and blinds will impact your wiring and your approach IMO.
Agreed, lots of possibilities. I’m no guru on the electrical code (hence my hesitation to mess with the actual house wiring), but a STDP switch would isolate the low-voltage Photon side from the actual light circuit. Basically two independent light switches operated by a single throw, one circuit providing house power to the light, the other circuit closing the low-voltage loop to the Photon. But, regardless, it’s not the best solution.
I need to crawl up in the attic and see what the wiring possibilities are.
Thanks again for your help, the brainstorming will continue!
@normelton, two independent switches in the same electrical box with 102VAC! A bigger issue is how will you power the blind/photon and how far is the controller box from the shower?
I have used hall switches to detect current flow in the 5-10 amp range very reliably over long periods of time.
Have you considered a simple CdS (cadmium sulfide) light sensor? They can be found in night lights sold by Walmart for about a buck a piece.
@butchalline, hall effect sensors are great but need to be either inserted in-circuit (ACS7xx device for example) or via a toroid, requiring a single wire be inserted through the toroid for monitoring its current. The OP didn’t want any intrusive. Great idea on the light sensor!
A light sensor is actually my “most favored” idea at the moment. One came with my Photon starter kit. I’d like to install it in the recessed can that is above the ceiling. The question is whether the recessed fixture emits enough light backwards, into the recessed can, to be detected by the sensor. I need to pull one down and play with it to find out.
@normelton, you don’t need a lot of light, especially if you hook up the detector to an analog input pin. Then you can measure “off” vs “on” levels and adjust your code accordingly.
That’s the plan, I’ll let you know what I find!