Usually such devices are not self powered, and a gander at the
HWell website states an input power, which, to me, means that the
device is not self powered.
You need to find out the current draw of your bell/ringer and get
a power supply that will power it and the photo sensor. The photo
sensor uses 50 mA. That is a negligible quantity. Look at the name
plate of the bell/ringer and find out what the current draw is,
multiply that number by 2, and find a power supply that will
provide the current and the proper voltage while making sure that
the available current is *more* than the sum of the sensor and bell
together. Getting more power than you need is not a bad thing.
This leads to the next thought in this process. How handy are you?
There are different types of power supplies, and I'd recommend getting
a cheap wall-wart type for this project as long as you know how to
use the meter and how to determine the polarity of the power supply.
If there is any thought of expanding this system, then I'd ask you
to go and look for a quality power supply. Of course, this leads
to more complications...such as have you ever spliced into a 120vac
cord and do you know how to properly connect such a cord to a
I meant I think he device powers itself *and* the alarm..a power source did
come with the device
Yes I do: I have installed panels and sub panels..light switch boxes and
outlets; fans and lights from it.
I made up my own 4 outlet receptacle for a long extension cord I use when I
play in my band at functions. Yes I can splice and dice.
Yes I have some knowledge.
I guess from now on when I ask a question I need to put my resume, basic
though it is :-)
Cool, then you won't have to worry about getting a power supply.
Good to know, and this is something you should have stated from the get
On your desk/workbench you may:
Hook up the power supply to the sensor.
Face the sensor at a reflective surface, or the beam reflector that
should have come with the unit, making sure that there is a couple
of feet between the two.
Energize the power supply.
Set your meter for Ohm/continuity and place the red lead on the COM
terminal, and the black lead on the NC terminal. Does it show zero
Ohms or continuity? While holding the test leads on the terminals
with one hand, move your other hand in front of the sensor and block
the beam for 1 second, then remove your hand from in front of the
beam. Does the Ohms/continuity change?
Move the black lead to the NO terminal and place your hand in front
of the beam as before. The meter readings should be exactly opposite
from the previous readings.
The combination that does not show continuity when your hand is not
blocking the beam, but does show continuity when your hand is blocking
the beam, is the combination that you want to use.
Power goes in on the COM, and goes out on the NO or NC to the bell.
Power Supply (+) ====> COM
NC OR NO ====> (+)Bell(-) ====> Power Supply (-)
Some might have you "sink" the bell, but I won't go there...
Please let us know which state is normally closed with this device.
Powered or unpowered. It might be a useful bit of information for
someone in the future.
Are you using a controller?
I'm assuming that you are using a power supply, a photo sensor, and a
The power supply will have a positive (+) and a negative (-) output.
The bell, as you stated, has a (+) and a (-).
The photo sensor should have a (+) and a (-) , COM, NC, and NO.
Wire from the power supply (-) to the photo sensor (-), and to the
Wire from the power supply (+) to the photo sensor (+), and the photo
Wire from the photo sensor (NO) to the bell (+).
Getting ready to test the system ..the dc power supply has a plus, minus,
then a M italic type symbol with an I on the top as if it was a whisk broom
with a handles and someone was sweeping.. It is a short circuit protected
power supply so I assume it's a ground of some sort...just want to be sure..
Ok so I called up a company inthe us for help..they said that the alarm has
to be powered, so connect the - from the power supply to the com on the
reciever, then from NC on reciever to alarm - . The + from power supply goes
direct to + on alarm.
Before I did all this I continuity tested the connection and it was
closed..This meant that with the power on it would open the circuit, which
it did when I powered up. So when beam is broken the circuit closes and
alarm goes off. which it did when I connected the alarm.
I used same power supply for all three and worked fine.
So thanks to all for the responses
I guess I'm done
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