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Member's Forum Doorbell Circuit Help
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  • monostable
  • reed_switches
  • bell
Related

Doorbell Circuit Help

pathfider1998
pathfider1998 over 9 years ago

Hello, I am quite new here, so don't be too harsh.

If this is in the wrong community, tell me where to move this post to so I can get appropriate help.

-------------------------------------------

 

This specific problem has been bugging me for a long time.

*UPDATED*

 

So I have this little old doorbell which uses 3V battery and a generic Normally-open switch of any type to operate.

The doorbell rings when two wires coming from the unit is shorted (connected together).

Minimal duration for the activation is about 0.1ms.

When the switch is closed and held, it rings continuously (ding-dong, 0.5 second delay, ding-dong, delay again...)

 

I was using reed switch (normally-closed) to on a sliding automatic door so the bell rings when the door is opened.

The problem is, the door closes quite slowly because of the safety function (it has been installed so the door doesn't trap hands or other objects), so the doorbell rings 5~6 times constantly while the door closes.

I do not want to annoy people nor I want to drain the battery too fast.

 

I needed something to pulse the transistor or a relay for a brief moment (so the bell rings) and stays off until the door is closed.

So I searched for the monostable circuit that uses 3V or less as a power source. It returned none. (Or, they didn't work as desired)

I tried using some generic 555 timers, but they drew too much power (4.5V) and was not qualified.

I want to avoid using Arduino or ATTiny (or LSxxxx) because they cost a lot here.

 

UPDATE: Here's the explanation:

1. The reed switch is connected parallel to the power in (Vcc) of the control circuit we are trying to build.

So when the magnet is taken away and the reed switch closes, the control circuit kicks in, and when the door closes and reed switch opens, only the control circuit shuts off.

2. The pulse should be sent only once no matter how slow the door closes or where the door is at. There must be no second ringer.

 

It doesn't need to be tiny, just small enough to fit inside the credit card when built on a breadboard (ignore the heights).

Could anyone help?

 

p.s. E14 really needs some sort of "Without Code", "Newcomer Friendly", "Not-a-Coder" or "Logic Gates Only" group.

Somewhere where people like me can ask about the most cost-effective/fool-proof way to make things happen without programming PICs or using ready-made solutions.

Or for people without access to advanced components like ATTiny or LS74xx.

 

Or just make it easier for people to look for communities without having to scroll down 940 miles just to find the end of the list.

(Or maybe I am just silly. Tell me about communities or groups I can join too)

 

Here's the circuit board. Maybe you guys can understand this better.

image

image

Plus: Pin 1 of IC marked, and components are highlighted.

image

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Top Replies

  • dougw
    dougw over 9 years ago +2
    It sounds like you want a directional switch that sends a single pulse if the door opens, but when it closes there is no pulse, just a reset. There are many ways to achieve this, even without chips ..…
  • gadget.iom
    gadget.iom over 9 years ago +1
    Hi Phil Welcome to the Element14 community. Does your reed switch go open or closed circuit when the door is opened?
  • D_Hersey
    D_Hersey over 9 years ago +1
    I think that a CMOS 555, such as lmc555 will work at the low potentials you require.
  • gadget.iom
    0 gadget.iom over 9 years ago

    Hi Phil

     

    Welcome to the Element14 community. image

    Does your reed switch go open or closed circuit when the door is opened?

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  • pathfider1998
    0 pathfider1998 over 9 years ago in reply to gadget.iom

    It needs to go closed circuit (so it turns on the "single bleep" circuit until the door closes).

    It only needs to send ONE pulse to whatever electronically controlled switch I got so it rings the bell, then STOP sending the pulses so it stops ringing.

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  • gadget.iom
    0 gadget.iom over 9 years ago

    Ok. Do you have a reed switch fitted already?

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  • pathfider1998
    0 pathfider1998 over 9 years ago in reply to gadget.iom

    No, I had it removed for now.

    The switch was affixed to the door with a tape so not a big deal.

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  • D_Hersey
    0 D_Hersey over 9 years ago

    I think that a CMOS 555, such as lmc555 will work at the low potentials you require.

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  • D_Hersey
    0 D_Hersey over 9 years ago

    As would a 74HC221

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  • pathfider1998
    0 pathfider1998 over 9 years ago in reply to D_Hersey

    I have to use things I have; I can't order any more components.

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  • shabaz
    0 shabaz over 9 years ago

    One approach (I've not tried it so it would be a bit of an experiment, and isn't great, but it is quick) would be to replace the magnet with a smaller one (you can even get them down to 2-3mm diameter). It would require precise positioning but if this is an automatic door then perhaps it already moves precisely enough). Perhaps from broken earphones..

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  • pathfider1998
    0 pathfider1998 over 9 years ago in reply to shabaz

    Oh. I forgot to mention:

     

    It should not ring twice until the door is fully closed.

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  • clem57
    0 clem57 over 9 years ago

    Simple answer: When the door is open the circuit is open because the wires not touching and no ringing until the door closes to complete the circuit and touch the wires. Cheap and no parts to buy!

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