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Forum Burglar Alarm to Door Buzzer Using a Monostable Circuit
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  • monostable
Related

Burglar Alarm to Door Buzzer Using a Monostable Circuit

Former Member
Former Member over 10 years ago

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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  • jw0752
    jw0752 over 10 years ago +1
    Hi Simon, I have set up a mock up of an improved circuit which allows the door switch to close and remain closed. The components that I have selected for C1 and R1 give a time of approximately 5 seconds…
  • D_Hersey
    D_Hersey over 10 years ago

    So, um, the buzzer works from 12VAC, ya' figure? 

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  • Former Member
    Former Member over 10 years ago in reply to D_Hersey

    I really have no idea.

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  • jw0752
    jw0752 over 10 years ago

    Hi Simon, Here is a simple monostable 555 timer circuit that I have used for similar applications. R-1 should be between 10K and 14 Meg and C-1 should be between 100pF and 1000uF. The formula for calculating the time that the output from pin 3 will be high is: Time( in seconds) = 1.10 * R-1 (in Ohms) * C-1 (in Farads). The power to the circuit should not exceed 15 volts DC. If you are going to power a buzzer with the unit you will have to use a relay or other means to boost the power as the output of the 555 timer is limited to about 200 mA.

    image

    John

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  • Former Member
    Former Member over 10 years ago in reply to jw0752

    I've seen that circuit, but been unable to simulate it before purchasing the needed parts.

     

    I would also need to invert the momentary switch output and add to it it to get something that will work with my input and output volatage and current needs.

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  • jw0752
    jw0752 over 10 years ago in reply to Former Member

    Hi Simon, From your description of the circuit and the application of the switch you should not have to invert the input. While your switch is a NC switch the fact that the door is normally closed pushing on it makes it actually a NO switch. You may have to put a capacitor in series with the switch to just give the trigger a pulse when the switch closes instead of a continuous closure, though I am not sure about that. A simulation or a rough prototype would let you experiment. The circuit itself is a standard off the shelf design and will work as I have used it successfully several times in my projects. You will probably have to rectify and regulate the voltage from the transformer to 12 volts for the timer circuit but you could also use the power from the transformer to continue to power the buzzer with a small relay, controlled by the 555,  to turn it on and off. The project you propose is very doable and I agree with you it would be much more professional than a continuous buzzer.

    John

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  • D_Hersey
    D_Hersey over 10 years ago

    Another way to go would be to detect the sounding of the buzzer would be to attach the series combination of a resistor, a half-wave bridge rectifier and the B-E junction of an NPN (E attaches to BOS) transistor.  The collector is connected to TOS through a resistor.  This node goes low when the buzzer sounds, except during the zero-crossings.  An opto-isolator could be used instead here, BTW, shunt the input LED with a reversed signal diode.  So, these zero crossings occur at a rate 2x the line fq.  We can count-and-compare for our duty pulse, ultimately connecting to a relay as John describes.

     

    John's way works:  Using the 555 in the datasheet way is considered unstylish in some design circles, but that's no nevermind.

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  • jw0752
    jw0752 over 10 years ago

    Hi Simon, I have set up a mock up of an improved circuit which allows the door switch to close and remain closed. The components that I have selected for C1 and R1 give a time of approximately 5 seconds. This circuit only turns on an LED but it could just as easily turn on a low current relay which could turn your buzzer on and off. This circuit will only ring the buzzer once for 5 seconds each time the door is opened no matter how long it is held open. It will reactivate only if the door is closed and reopened.

    John

     

    image

     

    image

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  • Former Member
    Former Member over 10 years ago in reply to jw0752

    Thanks, I appreciate it. I'll fiddle with your design and see what I can simulate as far as relays go, that way I can give the schematic to the maintenance guy to get his OK on safety.

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  • jw0752
    jw0752 over 10 years ago in reply to Former Member

    Hi Simon, You probably already know this but it is important to put a diode accross the relay coil reverse biased to the power so that any inductive voltage spikes from the relay coil can not damage your other components. Good luck on your project.

    John

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  • Former Member
    Former Member over 10 years ago in reply to jw0752

    No I didn't know that. I'll have to look up what that means.

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