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Fibaro switch wiring lack of knowledge!


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Ok !

I got this little fella working with my door bell button ( just a standard button ) 

So I get a z wave alert if the buttons pressed .

What I don’t understand is my button is meant to be powered to light it up .

How do I wire it so button is powered and lights up - but also only triggers a closed circuit ( no current ) for the fibaro sensor .

?

Might not be possible. Though they sell an ‘implant’ but I can’t see much difference between that and this universal sensor .

 

https://manuals.fibaro.com/binary-sensor/

 

https://www.fibaro.com/en/products/smart-implant/
 

It’s probably simpler than I think !

Have a similar issue with the front door built in led . Fed off a 12v transformer . Assume I wire live straight through then just use the sensor to close the circuit on the ground .

 

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EC274D91-EB22-42E7-BF4C-DEAA0606B380.jpeg

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Thinking about the illuminated doorbell....

 

Maybe I power it; but stick the sensor on the gnd line and define it's initial state as 'closed' i.e. circuits complete so doorbell lights up.

Then when the button is pushed the circuit is then open ( door bell not illuminated ) and then the sensor will detect it as 'open' ???

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AHHHHHHHHHHHHHHH!!!!

 

So that's how a DC illuminated doorbell works.

Well I never!

 

Can tell I'm a software engineer as this was definitely a hardware problem! ?

 

Right illuminated doorbell works and triggers z wave sensor!

 

Think I'll get my multi meter for door led ( just to convince myself ); also for automatic door unlock. But assume it's the same principle.

 

I learnt something!!!

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Hmmm, I'm still a bit confused.

I've set in1 as 'normally open' . But physically the 'input' isn't open is it?. Am I still sticking 12V through the sensor when I shouldn't be??

Can understand if it was a physically mechanical relay I could see it was open.

Sensor isn't burning out yet!!.

I guess I'm asking how are the input's 'opened' and 'closed' ???. It's not a real physical disconnection is it?

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I am not entirely sure how you have it wired, but an illuminated bell push has the lamp in parallel with the switch contact.  the current that flows (from a door bell transformer) through the lamp is not enough to activate the door bell.  When the button is pressed much more power is drawn activating the bell.

 

To use that same illuminated bell push you would have to feed it with say 12V dc via a resistor.  The resistor value would determine the brightness of the light.  Connect the bell push to an input and in the non pressed state it would read high or "on"  when you press the bell push it would short circuit the lamp and pull the input down to 0V and it would read as low or "off" .  You would therefore have to invert that input in software to activate something when the button is pressed.

 

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2 hours ago, ProDave said:

I am not entirely sure how you have it wired, but an illuminated bell push has the lamp in parallel with the switch contact.  the current that flows (from a door bell transformer) through the lamp is not enough to activate the door bell.  When the button is pressed much more power is drawn activating the bell.

 

To use that same illuminated bell push you would have to feed it with say 12V dc via a resistor.  The resistor value would determine the brightness of the light.  Connect the bell push to an input and in the non pressed state it would read high or "on"  when you press the bell push it would short circuit the lamp and pull the input down to 0V and it would read as low or "off" .  You would therefore have to invert that input in software to activate something when the button is pressed.

 

There's no bell.

Basically it's an illuminated push button.

So it's lit ( supplied with 5v). . When you push it the light goes out. So I *assume* it makes an open circuit ( hence the light going out ). Not sure how the sensor works.

Will draw up my wiring and see if that helps you ( or anyone else ) make sense of it !

Edited by pocster
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The button is open circuit  when the light is on, so if fed from 5V there will be 5V across the bell push. Connect that to an input and it will read as "ON"

 

Push the button and the light is shorted out by the switch. that's why the 5V feed should have some form of current limiting e.g a resistor.  While the button is pressed there will be 0V across it, so the input will read as "OFF" hence the need to invert it in software.

 

Where is this 5V from that is illuminating the light?

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42 minutes ago, ProDave said:

The button is open circuit  when the light is on, so if fed from 5V there will be 5V across the bell push. Connect that to an input and it will read as "ON"

 

Push the button and the light is shorted out by the switch. that's why the 5V feed should have some form of current limiting e.g a resistor.  While the button is pressed there will be 0V across it, so the input will read as "OFF" hence the need to invert it in software.

 

Where is this 5V from that is illuminating the light?

Psu for the doorbell light 

psu for the universal sensor 

 

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Right !

This is how I think I’ve got it wired !! 
I don’t need 2 psu’s . The ‘doorbell’ is in one room ; where I’m fiddling with the sensor another ( long cable inbetween ! ) .

My ‘logic’ is ...

 

The sensors default state is ‘open’ ; so in1 has no current ( still don’t understand how it works internally ) .

So 12v is flowing to illuminate the door bell and not through the sensor (????)

When you press the door bell ; the 12v is now an open circuit the sensor becomes closed .

 

erm - I’m confusing myself ?

 

It works but I’m not confident I’m not sticking 12v into the sensor .

Ran it for a few hours it didn’t get hot ?

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Edited by pocster
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Looking at the door built in leds .

12v transformer . Stick my meter on it get 12v as expected . Swap my probes and get -12v . I’m confused !

Assumed I’d get 0 ; if that were true I’d then use sensor input in2 to control the 0 ( gnd? ) line . But it’s not . Erm - confusion reigns again !!

 

How long before I bust my sensor ! ?

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Oh dear!

The harder I try the more confused I become.

Moved it all into 1 room. Decided that *obviously* it could all run off 1 psu .

But it doesn't work!!!. Sensor power in is off the same psu that's the only difference ( both are rated at 12v 5A output). Sensor is recognised, but when the doorbell is pushed no 'changed' state occurs. Gone back to 2 psu's and it's ok. I just don't get it. Is it something to do with GND????

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You need a resistor in series with the 12V power supply. At the moment when you push the button you short out the power supply. Never a good idea.  It probably has current limiting otherwise it would have already gone bang.  and anything else the power supply is feeding will lose it's power when the button is pressed.

 

And yes reversing youe meter probes will read -12V why do you think it would read 0?

 

What is the maximum voltage input rating of the sensor?  Is that happy with 12V?  If not then I would be using a resistor and zener diode to limit it to 5V

 

EDIT you posted your last reply while I was typing.

 

Yes that explains my point, when the button is pressed you are shorting out the (in that example common single ) power supply so the sensor could not work.

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1 hour ago, ProDave said:

You need a resistor in series with the 12V power supply. At the moment when you push the button you short out the power supply. Never a good idea.  It probably has current limiting otherwise it would have already gone bang.  and anything else the power supply is feeding will lose it's power when the button is pressed.

 

And yes reversing youe meter probes will read -12V why do you think it would read 0?

 

What is the maximum voltage input rating of the sensor?  Is that happy with 12V?  If not then I would be using a resistor and zener diode to limit it to 5V

 

EDIT you posted your last reply while I was typing.

 

Yes that explains my point, when the button is pressed you are shorting out the (in that example common single ) power supply so the sensor could not work.

Why do I think when it’s reversed it should read 0 ? Because I don’t understand ?

If it’s shorting out the psu then that explains why I need 2 ?

So it’s a fudged work around ! . Essentially as the psu has some short protection is the only thing making it ‘work’ .

Sensor max in voltage is 24dc I think .

 

So a resistor on the 12v + psu is all I need ? ( then I can run off 1 psu ? ) .

Erm , by what magic do I compute what resistor value I need ?? ??

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3 hours ago, ProDave said:

1 watt @ 12V is about 80mA

 

Use about a 75 ohm 2 watt resistor in series with the +12V to the bell push.

Simple as that ? . Will stop me shorting the transformer ?

The doorbell is ( naturally ) outside so I couldn’t see the transformer as I pushed it I.e watch it’s led have an epileptic seizure .

 

@ProDave yeah ?

https://www.amazon.co.uk/Sourcingmap-US-SA-AJD-68313-30Pieces-Tolerance-Resistors/dp/B00SWK3DMM/ref=mp_s_a_1_3?dchild=1&keywords=resistor+75+ohm+2+watt&qid=1590422581&quartzVehicle=72-1783&replacementKeywords=resistor+75+ohm+watt&sr=8-3

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56 minutes ago, ProDave said:

That will do, shame you have to buy that many, but anywhere else if you buy 1 the postage will cost you that.

Ordered!

I'll stick one in and let you know what happens. The rest are yours if you like!

 

Appreciated! ?

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