Hi

Me and some of my colleagues had a discussion about how 1-phase motion detectors work. They're used for controlling lights etc. Connected like this:

How do they power the electronics inside the detector?

Normally we would use two phase in and get both out to the lights.

Talking about 230v 50hz mains for the records.

 

How do they power the electronics inside the detector?

With an internal power supply.

Normally we would use two phase in and get both out to the lights.

Only one phase needs to be switched. Instead of connecting to "N" it would connect to "L2"

 

With an internal power supply.

And how do you power that with only 1 phase going to the detector? Check my schematic. That's all wiring excluding the electronics inside the motion detector.

It's this detector:
http://translate.google.com/transla...l3exp=1848&l2exp=&l1exp=1845&template=product

Datasheet (page 2 for english): http://elko.no/elko2_nor/frontend/mediabank/17942/Bruksanv4860.pdf

 

Good question. Maybe it uses the load to drop the line voltage at a ver small operating current.

 

The bulbs provide enough current for the internal supply to work.

 

Yeah, but when the switch/triac is closed you have the same mains potential on both side of the switch. Tho I could see how that would work if the lights had to go off for a while in order to power the caps before it could turn on the lights again. But I think the lights can be constant on if someone is in the detection area. Will check out that, but if someone got some ideas. Please tell.

 

It would take only a few volts to operate the electronics. Thus, a small value resistor inline with the load would provide enough voltage without affecting the power delivered too much.

 

Wouldnt that be a seriously big resistor consider the detector can take a resistive load up to 350W?

 

Depends on what you mean by seriously big. If, for example, the electronics required only 1V ( a reasonable assumption) then the resistor would be 350W/230=1.5W.

Or, you could just use a couple diodes to get ~1.5V, no matter what the current is.

 

I might have missunderstood you or? Do you mean a resistor in series with the load? Why do you divide 350W by 230V and get 1.5W as a result? Wouldnt that be current as a result, 1.5A?

 

I assumed that the electronic would work on 1V.

I=350/230 Load current ( same as resistor current in series )

P(resistor)=V*I = 1*350/230 = 350/230. to a good approximation.

But, a couple series diodes would provide a near constant voltage for the electronics. This would not work when the load is switched off. So, the diodes would need to be switched for a bleeder resistor divider when the load is off.

Not saying I know how it works, but this is one possibility.