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Burglar Alarm to Door Buzzer Using a Monostable Circuit

Former Member

At work we are currently using what I suspect is a burglar alarm as a door buzzer. It's inappropriately loud to have buzzing the entire time the door is ajar (from barely open onward), as we're a motel and it's frankly obnoxious. However in order to hear people come in from the basement, it's what the folks in charge decided to use. I would very much like to create a monostable circuit so it buzzes for a small amount of time then turns off until the door is shut again, re-arming the buzzer. The buzzer works using a momentary switch that is pressed when the door is shut and released when the door is open. This switch is connected to what looks like a transformer core, and the device also plugs into our AC outlet.

 

I have looked at circuits online, but even using a few simulators I was unable to produce results like I wanted. Can I have some advice? I'm new to electrical engineering.

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  • For the curious, I drew a diagram of what I saw at work. This circuit is connected in a manner that fits my drawing, more or less (I don't know about the transformer connections exactly) and behaves the same as the circuit at work.

     

    image

    Here is my current circuit, which I believe behaves as intended: a ballpark 1 second pulse to the relay when the door is shut after opening it. Having this happen when the door is initially opened would be perfect but this is good. I am curious whether or not the placement and bias of my diode is correct, because my simulator doesn't show me the expected feedback. The transformer section is less than important as the buzzer already functions when connected to the mains outlet.

     

    image

Reply

  • For the curious, I drew a diagram of what I saw at work. This circuit is connected in a manner that fits my drawing, more or less (I don't know about the transformer connections exactly) and behaves the same as the circuit at work.

     

    image

    Here is my current circuit, which I believe behaves as intended: a ballpark 1 second pulse to the relay when the door is shut after opening it. Having this happen when the door is initially opened would be perfect but this is good. I am curious whether or not the placement and bias of my diode is correct, because my simulator doesn't show me the expected feedback. The transformer section is less than important as the buzzer already functions when connected to the mains outlet.

     

    image

Children
  • in reply to Former Member

    Yes, your schematic looks good to me. I think you will find that your button on the door is open when the door is closed and closed when the door is opened so it will work just as planned. Buzzer should ring for about a half a second because of the R1 C1 values and then go quiet until the door is closed and reopened. Also be aware that the C3 value of 3.3 uF is not critical, a 2uF to a 5uF will still work the same. Your protection diode is drawn correctly and unless your simulator is quite sophisticated it may not recognize its purpose but it is important. When you wire this up be very careful with the mains voltage. It might be better to put your relay switch in the seconday just like the door switch is presently placed so that you do not have to bring mains voltage to the circuit board.

    John