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Need help with Hacked Wireless Outlet -- Earth Ground

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Telemachus

New Member
Question first, and then longer explanation:

How do I eliminate the high voltage differential between a rectifier circuit and earth ground?

Now the wordy part:

I have embedded a cheap etekcity wireless outlet (http://www.etekcity.com/product/100068.html) into a bug fan to be able to control it with a reed switch connected to a door, so that when the door opens, the fan turns on. The idea was that the rf outlet gives me a way to turn it on via remote control, as well as automatically when the door is open. The outlet was much cheaper than creating my own wireless interface with a uC and relay.

Attached is a picture of the PCB of the etekcity outlet. Blue wire on bottom right is neutral, left brown is Line in, and middle brown is line out.

Best I can tell, these outlets first drop the voltage to 24 volts through a coil on the opposite side, and then DC through a rectifier (U1). A voltage regulator (U2) drops the voltage down to 5V for the uC and RF unit. The uC on the top left controls the transistor (q1) through Pin 3 output with a current limiting resistor (R6). The relay is a 24V relay.

My idea was to use a hard wired Reed switch to either connect 5V VCC to the current limiting resistor that goes into Q1, or to connect the emitter to collector on Q1. Jumper wires on either of these seems to work well to control turn the relay on using a reed switch.

I embedded the circuit into the fan, wired it all up, and thought it would work great, since the exposed wires to the reed switch are 24V max.

However, one of the wires to the reed switch came into contact with the chasis of the fan, which is connected to earth ground, and popped the GFI. A little probing around confirmed that just about ANY spot on the PCB post rectifier (VCC, GND) in relation to earth ground, shows 110V (I'm not sure how much current, but enough to pop the GFI).

How do I eliminate the high voltage differential between a rectifier circuit and earth ground? I'd like to make sure that these wires to the reed switch are only capable of low voltage potential.

Thanks in advance!
 

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Les Jones

Well-Known Member
Most Helpful Member
I think one of these wireless remote controls may suit your need better. The relays in them are only rated at 120 volts so if you live in a country with 240 volt mains they would not be suitable. They require an external 12 volt DC supply. (It does not need to be a regulated supply as the relay coils are the only thing fed directly from the 12 volts. There is a 5 volt regulator for the electronics.) You could modify one channel so that the relay was driven directly from the reed switch on the door. You would then wire these relay contacts in parallel with the one used with the remote control. (I have traced out the schematic of these units so let me know if you want a copy.)

Les.
 

kinarfi

Well-Known Member
Post a photo of the other side also, please
Jeff
 

Telemachus

New Member
You don't, it's all live to the mains - and you shouldn't be doing something so dangerous if you don't understand it.
Of course, you are right. Please help my ignorance if you are able.

What you are teaching me, is that despite a rectifier or voltage regulator circuit, the entire circuit remains at high voltage compared to Earth ground?

Attached is the other side. The little connector on the left is something I added hoping to connect the reed switch.
 

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Telemachus

New Member
I think one of these wireless remote controls may suit your need better. The relays in them are only rated at 120 volts so if you live in a country with 240 volt mains they would not be suitable. They require an external 12 volt DC supply. (It does not need to be a regulated supply as the relay coils are the only thing fed directly from the 12 volts. There is a 5 volt regulator for the electronics.) You could modify one channel so that the relay was driven directly from the reed switch on the door. You would then wire these relay contacts in parallel with the one used with the remote control. (I have traced out the schematic of these units so let me know if you want a copy.)

Les.
Les thanks for the idea. Then i would just use a separate DC wall wart to run the relay/reed circuit?

I really thought that these outlets would be ideal since they have a low voltage circuit already set up.

Maybe i could add an opto isolator to completely remove the Reed switch circuit from the rest?
 

schmitt trigger

Well-Known Member
The low voltage circuit is indeed low voltage, but since it may be running from a transformerless supply, its "common" terminal is actually floating more than 100 volt above "true ground".

The cheapest way I can think is to hack the circuit, remove the transformerless supply components, and replace it with a proper, isolated supply. Either you purchase the transformer or purchase a wall wart.

EDIT; please post photos of the other side of the board, too.
 

Les Jones

Well-Known Member
Most Helpful Member
This is how I would suggest using the module. This picture is the underside of the board.

Image2B.jpg

Cut the track at point X You will need to solder a wire to the solder pads just above and to the left of the arrow head.
This is how you will connect things up.
Telemachus.png

Here is the schematic.
4 channel remote.jpg
I've not yet got around to drawing it with "Eagle PCB"
The channel A button on the remote will toggle the fan on and off. (Or A will switch it on and B will switch it off depending on the position of the jumper.)

Les
 

Telemachus

New Member
Okay, so what if I use a 5/12 volt wall wart on the Reed switch circuit, and use that with a current limiting resistet to trigger an opto isolator that will connect 5V VCC in the outlet circuit to the current limiting resistor going to Q1?
 
Last edited:

Les Jones

Well-Known Member
Most Helpful Member
Telemachus,
Bear in mind that if the reed switch on the door is the normal type used for alarms the contacts will be closed when the door is closed and open when the door is open.

Les.
 
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