door buzzer with timer

Thread Starter

ck87651

Joined Jun 9, 2015
3
Greetings from a first time poster and electronics novice. I have been searching and reading up on solutions to my project for the past hour or so and think I may have some direction, but I'd like to get some advice before I end up moving in a direction that may not be the best.

My project: We have a security system at our office building and each desk has a button wired in to the system that will unlock the front door for 8 seconds. The security company wired in a buzzer that will activate as long as the door is unlatched. Even if someone opens the door the buzzer stays on, which is confusing for the customers and annoying for everyone else.

Here is what I have to work with: When the button is pressed, a 12v circuit is activated for 8 seconds that drives the buzzer directly. I'm not able to access/modify anything on the security system, all I have is the 12v circuit to work with and it is separate from the door lock mechanism so I can do what I want with it and not affect the locking mechanism.

Here is what I'd like to accomplish: The buzzer activates until the door opens. Once the door opens, the buzzer stops for at least 8 seconds until the security system closes the latch on the front door and disconnects the 12v power to the buzzer.

Here is what I'm thinking: Installing a magnetic door switch between the 12v source and the buzzer and integrating a 555 one shot circuit inline that will keep the power from the buzzer for at least 8 seconds. Is this something that sounds reasonable? If so, are there any suggestions on what I should know before going into this? Any suggestions on how to integrate the 555 timer? Any help or direction would be greatly appreciated.
 

AnalogKid

Joined Aug 1, 2013
10,986
Yes, a 555 will do what you want, and there probably are at least three different ways to wire it up. But your timeout period equals the signal time, so it looks like your goal is to silence the timer for the remainder of the 8-second activation period, no matter how much time is left. Therefore...

Another approach is to use the 12 V power and the door switch as the set and reset inputs to a flipflop. In this way, you have no timing elements. The security system provides that with its timed 12V power cycle. This way, if the 12 V timing changes (security system firmware upgrade, etc.) you don't care. The buzzer sounds until the door opens, then stays silent until the 12 V goes away and comes back, whenever that is.

Flipflop powered by 12 V, comes up in the Set state that drives the buzzer. Door switch drives the Reset input. Done. This can be done with two CMOS gates and a driver transistor, or just two transistors and have the buzzer driven directly by one of them. What is the make/part number/current requirement for the buzzer? The two-transistor method is old school - lotsa cred.

ak
 

AnalogKid

Joined Aug 1, 2013
10,986
The attached schematic has three versions. Top right is the 2-transistor version.

Top right is essentially the same thing but uses a darlington transistor array. The 2003/2004 is a great part for industrial-strength logic. Each transistor is rated for 50 V and 500 mA, and has a snubbing diode built-in. I included a schematic of one of the sections. Note that the unused section does not have to have it's input tied to anything because of the internal resistors to GND.

Lower center is the CMOS gate version.

ak
 

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Thread Starter

ck87651

Joined Jun 9, 2015
3
Wow thanks for the great reply. I really like the flipflop idea, makes a lot of sense and I'm glad you mentioned it. To answer your question, I'm using a Philmore B812 Miniature DC Buzzer, 12V. I'm leaning toward the two transistor (top right) schematic because it looks the easiest to build and most likely that our local electronics shop has that stuff in stock.

Does the buzzer we have change any of the design?
 

AnalogKid

Joined Aug 1, 2013
10,986
Several places show the typical current at 15 mA at 12 VDC. This is a pretty low value, so all discrete transistors in all schematics can be changed be a 2N4401, 2N222, 2N3906, etc. The capacitor guarantees that the system always powers up in the alarm state.

ak
 

Thread Starter

ck87651

Joined Jun 9, 2015
3
Well it looks like the project has taken a bit of a turn. Manger didn't like the idea of the buzzer staying on until the door is opened, he wants the buzzer on for a second or so then off. Any design ideas for this? I'm still searching for solutions but mostly learning the basics along the way (reading schematics, building a circuit with breadboard) so I'd appreciate any ideas you may have. Thanks again ak
 

AnalogKid

Joined Aug 1, 2013
10,986
You could go back to the 555, but where's the fun in that? Two ways come to mind.

First is basically 1/2 of the flipflop, with the power-on capacitor now acting as a timer. The left schematic shows this. R1 and C1 set the turn-off delay. Three time constants equals 95%, and 95% of 12 V is 0.6V. The threshold voltage for a MOSFET is not as precisely defined as the cutoff voltage for a bipolar transistor, but Q1 should be off by then. R2 discharges the cap after the +12 V is removed. Again, it has to be large enough so that the R1-R2 node is not high enough to partly turn on Q1 when the cap is fully charged. You can play with these values to get the performance you need.

Second, the right hand schematic is the ultra-minimalist approach. The buzzer will start out loud then trail off rather than end crisply. The component values are more speculative than for the left schematic. Just a thought.

Don't worry about the 1% resistor values, those are what's in my library. Nearby 5% values are fine for this.

ak
 

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In reality,it sounds like someone should have used a real "door prop alarm" which also integrates to the lock strike. They don't work like you want.

It sounds like you want an "Open the door stupid!" until they do.
 
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