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Door bell enhancement with web service

Discussion in 'General Electronics Discussion' started by MarkNL, Dec 22, 2015.

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  1. MarkNL

    MarkNL

    4
    0
    Dec 22, 2015
    Hi Folks,

    Over the holidays I have decided to enhance our current doorbell such that it will also ring the families mobile phones when someone rings. We live in a multistory house where the doorbell is often not heard. I will be using an ESP8266 (wifi arduino) and the pushover app on mobile phones. I have everything working with the ESP8266 and the Pushover service signaled by one of the 3.3v INPUT digital pins.

    My question to all of you is how I can tap into the doorbell configuration to get a DC 3.3v high signal (or a 0v low signal) from the door bell circuit. The doorbell consists of 3 components: A 15v AC/AC transformer from 220v mains, a 15v AC bell, and a doorbell. Note that the doorbell works with 15v AC and not DC. It is a simple loop circuit activated by the doorbell.

    I have tried adding in a bridge rectifier with a voltage divider to signal 3.2v. It works great with the power supply alone but the bell will not ring if I add it into the loop.


    BTW: The Mrs likes the current bell and the doorbell button and I have been told that these are not to change.

    Any other ideas of how to build a circuit?

    Thanks,

    Mark
     
  2. Anon_LG

    Anon_LG

    453
    117
    Jun 24, 2014
    Does your doorbell have a small led indicator light that is activated when the switch is pressed? If so, utilise this to form an optocoupler, rather than messing about with overcomplicated circuitry.

    If no indicator light is present, utilise an N channel enhancement MOSFET on the rectifier output, FET's have extremely high gate impedances (10^15Ω<) and will not disrupt any circuitry of his sensitivity.

    Provide us with a picture of the exterior and the internal circuitry, a nice clear photo.

    When this project is underway, ensure that you create a project log, it sounds interesting.
     
  3. MarkNL

    MarkNL

    4
    0
    Dec 22, 2015
    Hi Lavaguava,

    Thanks for the quick response. No LED. This is an old school doorbell configuration from the 70s.

    I am not sure that I follow how the N-channel mosfet would be hooked up. Are you saying that the + side of the rectifier output would go to the Source and the - side to the Gate and the Drain going to a voltage divider?

    Thanks,

    Mark
     
  4. gorgon

    gorgon

    603
    24
    Jun 6, 2011
    I suppose you have a pilot light in the doorbell(button) that is turned off when you push the button? If that is correct there will always be a small voltage over the AC Bell. What you can do is to put the bridge rectifier directly over the AC Bell, and you will then see small rectified voltage (or none), and the full rectified 15V AC voltage when the button is pushed. Add a capacitor(1uF/35v) at the + and - terminals of the bridge to make DC voltage, and use a resistor divider to get the correct voltage to trigger the radio circuit.
    A total resistance of the divider of around 20k should be ok. You'll then get around 20mW power dissipated in the 2 resistors, when the bell is used. Values of 22k and 1k5 should give you a trigger voltage for a transistor as described below of around 1.3V.
    When rectifying 15VAC you'll get around 21V DC over your capacitor.

    I would have used a NPN bipolar transistor to trigger the radio circuit, you will then need around 1-1.5V out from the resistor divider, via a 10k resistor into the base of the transistor. You can then use the collector of the transistor on the active low input on your radio circuit, it might be possible that you need a pullup resistor to the 3V3 line of your radio circuit. Remember also to connect the emitter of the transistor and the 0V of the radio circuit to the - of the bridge rectifier.
     
    Last edited: Dec 22, 2015
    Anon_LG likes this.
  5. Anon_LG

    Anon_LG

    453
    117
    Jun 24, 2014
    The gate of the MOSFET should be connected to the positive rectified output. Of course this does require an external 3.3 (or whatever input is required) supply. I was riding on the LED optocoupler idea though.

    What are you going to be powering your esp module off? Connect this to the drain of your nchannel, and the gate to the rectified positive ouput. The rectified output will only be present on the gate when the switch is pressed. Use the arduino to poll/interrupt sense a level above, say 5 volts (or something near the peak of the rectified AC) and do your thing with the ESP. You may want a voltage divider to reduce the peak voltage of your rectified output.

    The above are just ideas, read gorgon's suggestion as well.
     
  6. MarkNL

    MarkNL

    4
    0
    Dec 22, 2015
    Thanks everyone for all the help.

    I am not sure why I did not think about this earlier but Gorgon's suggestion to place the bridge rectifier directly in parallel with the bell did the job. While the power supply voltage is 15v, it seems that the measured voltage when actually ringing the doorbell is around 7.5v. I was not expecting this and have had to readjust the voltage divider circuit to get it back to 3.2 volts.

    I am a bit nervous that the voltage value after the voltage divider may go higher than 3.2 if the voltage of the goes above 7.5 to the bell.

    Any comments on the voltage drop from 15v to 7.5v? I assume that it has to do with Ohm's law and the load of bell pulling the voltage down. Any explanations for this? Are there any risks of the voltage being higher and blowing my ESP?

    Mark
     
  7. gorgon

    gorgon

    603
    24
    Jun 6, 2011
    You don't say if the 7.5V is AC or DC?

    You may have a full 15V AC over the bridge rectifier, if the Bell should break and be open circuit. To protect agaist that you can use a 3V3 zener diode over the input resistor, to GND(-). You should also add a series resistor like 10-22k before the input, if you connect directly to the control circuit, and not via the transistor. This serial resistor will reduce the possible current into the input's overvoltage protection, to 2-1mA even if the zener is not working and the input voltage is rectified 15V AC.
     
  8. MarkNL

    MarkNL

    4
    0
    Dec 22, 2015
    I do not have any zener diodes handy so i ordered some 3.3 and 5v units from china that should show up in the 2nd week of January. Being an impatient person, I also tried using an LM1117 3.3v regulator on the other side of the bridge rectifier which seemingly delivers the desired result on a multimeter. My assumption being that the LM1117 would be some circuitry clamping any spikes from getting through.

    HOWEVER, when I view what is happening on my scope, it seems that there is a very short burst of energy (<1ms) of around 10v when I take my finger off of the door bell button. How would a Raspberry PI, ESP8266 or Arduino digital input react to such a short spike?

    My assumption is that this is a result of some of the inductive energy being released from the bell. Would a standard 3.3 or 5v zener diode protect against this better than the internal circuitry of an LM1117?

    How does one get around this? a relay or opto isolator?

    Thanks,

    Mark
     
  9. gorgon

    gorgon

    603
    24
    Jun 6, 2011
    It all depends on where in your circuit you see the spike.
    Is it after the bridge rectifier?
    Do you use a capacitor after the rectifier, and if so what size?
    A zener diode will normally take care of the spikes, and if you add a capacitor over the zener it will also help.

    Generally high voltage spikes is not good for the electronics, and depending on the energy in the spike, you may burn the input protection and damage the chip.
     
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