What is the output when the alarm is tripped? (modulated signal, +30VDC, etc?)
1179724
According to the specifications the output interface is a relay contact rated at 1 amp max and 36 volts.
1179725
There is no output voltage when tripped. There is a relay contact that you can configure for normally open or normally closed, then just run the voltage of your choice through the relay. The unit works very well. (check relay contact ratings on spec sheet)
1179726
I have not yet fitted up my sensor, but I believe operation is as follows -: The transmitter sends out a continuous signal at 1.92 khz, which is picked up by the receiver. When this beam is interrupted, the relay in the receiver switches states - if the jumper located just above the relay is set to 'NO', then the supply voltage will appear at terminals COM & OUT. I plan on using both sensors on 12 volts DC, so the output would power a 12 volt lamp or buzzer, 1 amp max. If the jumper is set to 'NC',there would be 12 volts at the output, which would drop to zero when the beam is broken. Hope this is helpful, Morris N. I have not yet fitted up my sensor, but I believe operation is as follows -: The transmitter sends out a continuous signal at 1.92 khz, which is picked up by the receiver. When this beam is interrupted, the relay in the receiver switches states - if the jumper located just above the relay is set to 'NO', then the supply voltage will appear at terminals COM & OUT. I plan on using both sensors on 12 volts DC, so the output would power a 12 volt lamp or buzzer, 1 amp max. If the jumper is set to 'NC',there would be 12 volts at the output, which would drop to zero when the beam is... » Read More I have not yet fitted up my sensor, but I believe operation is as follows -: The transmitter sends out a continuous signal at 1.92 khz, which is picked up by the receiver. When this beam is interrupted, the relay in the receiver switches states - if the jumper located just above the relay is set to 'NO', then the supply voltage will appear at terminals COM & OUT. I plan on using both sensors on 12 volts DC, so the output would power a 12 volt lamp or buzzer, 1 amp max. If the jumper is set to 'NC',there would be 12 volts at the output, which would drop to zero when the beam is broken. Hope this is helpful, Morris N. « Show Less
1670891
Mine read just at 12V when received so I didn't charge any more before installing in alarm system. It is working fine.
2215208
I have the same issue with one of my switches, it is ON all the time, have seen similar comments, it appears to be a common failure with these units. With that said I view it as an inconvenience not a safety issue. The output is only 12V and as long as each individual output is protected (as described) then each one should trip if it gets shorted or overloaded. However to be safe I would unplug the whole unit anytime you mess with the output wiring.
2135711
No this will not sound alarm if bulb burns out. This is monitoring the incoming voltage & sounds alarm when power is lost.
228
Could be several possibilities. Check to see if your battery is older or not sufficiently holding a charge and or that all the connections and cables are clean and tight. You don't mention whether your alarm goes off constantly however even if it did you would have to try a different sensitivity setting and you may still have to refer back to the above assuming the unit is not defective in some way. Disconnectthe alarm and see if your battery still goes dead. if it does then you have a short or another problem or possibly something i mentioned. Good luck.
246000
Austin,I used this unit integrated into an alarm unit for a friend, but as I remember it is has a dc power jack on it for the 12 volt output. It also includes a dc power cable with female connectors on each end. One end plugs into the power jack, the other end will plug into whatever you to operate. You will need a matching male dc power plug, which can be purchased at Radio Shack etc. Hope this helps,Jack
899413
This is a car or house alarm system siren. I bought it to go outside my house in addition to the inside siren to make sure my neighbors hear my alarm going off. It hooks up to the bell connection on the wiring panel, along with and on top of the inside siren wires. It is powered by the system which I believe is 12vdc. You could use this siren for other purposes using 12v but do not use a regular speaker in the place of this siren for an alarm system. I am told it will fry your alarm system. I hope this helps to answer your question.
850705
I can't remember exactly, but the alarm was driving me CRAZY! I finally went through the instructions again and figured it out. Sorry I can't be of more help.
1762630
No, the "low voltage" alarm would go off. You need panels that will output 23 to 26 volts to power this inverter.
334913
If I unplug the unit, the alarm goes activates. If I plug it back in, the alarm goes off. Hope this helps.
1806605
Honestly its a cheap product. It never worked for me even when I tested it straight to the battery... But to answer ur question it simply taps into the positive and negative on the alarm system
2229574
Happened to me at 1 am a few times! It is alarm. I don't have it handy or I could give more details. The instruction sheet should help you out though. Good luck :)
2162813
D R D I/m sorry. I did not buy this item for myself. I know the person I bought it for likes the alarm but I do not know how they have it hooked up. Sorry .
894230
I have not found anything. Just assume it's a 5V input, and output is inverted (High input = relay OFF)
1360092
The only time I got an alarm is when I had the leads inserted into the wrong hole. Double-check that the probe wires are correct: if you are trying to measure resistance, make sure that you don't have one plugged into the current port, or something like that.
344164
all that you need is a 110v outlet to plug it into. the siren beeps after 30 seconds when the alarm is set and immediately when the alarm is unarmed. it goes off immediately when a door or window is opened with alarm armed.
1904726
sorry I'm not that savy with what your saying. Should I put a volt meter on the 2 lines and physically trip the sensor itself and look for a voltage spike or a contunity test doing the same thing?