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Doorbell transformer to 7805?

Discussion in 'Electronic Basics' started by Jeff Dege, Jan 29, 2006.

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  1. Jeff Dege

    Jeff Dege Guest

    I've built a 418MHz RF receiver into a Radio Shack project box together
    with a couple of relays. When I push one of the button on the
    transmitter, one of the relays closes.

    My intent is to wire this into the doorbell circuit, so I can ring the
    doorbell remotely, when no one is there. I can then use this to
    desensitize the dog to the doorbell - teach him to not get so excited
    every time he hears it ring.

    I have a clip for a 9V battery inside the box, but I also included a set
    of binding posts so that I could power it from an external source. I'd
    thought that I might be able to power it from the doorbell circuit. But
    the truth is I don't know beans about what sort of power doorbell circuits

    I know they run at a low voltage. That's clear from the wiring used. And
    I can see the step-down transformer.

    I also know that the solenoids in the doorbell itself have to be powered
    by a DC current. Applying AC to a solenoid accomplishes nothing.

    But I don't know whether the current is rectified at the transformer or
    inside the doorbell. If the former, I might be able to connect it
    straight to the 7805 voltage regulator inside the box. If the latter, I
    need a rectifier circuit.

    So - can anyone tell me exactly what sort of current I can expect to find
    in a doorbell circuit? And what I would need to condition it before
    feeding it to a 7805 voltage regulator?

    I am myself persuaded, on the basis of extensive study of the historical
    evidence, that... the severity of each of the contractions - 1920-21;
    1929-33, and 1937-38 - is directly attributable to acts of commission
    and omission by the Reserve authorities and would not have occurred
    under earlier monetary and banking arrangements.
    - Milton Friedman
  2. ehsjr

    ehsjr Guest

    It's AC, likely 16 volts. AC will operate a solenoid
    designed for AC.

    Use the circuit below.

    110v 16v ----
    ---|| +-----+---[R1]---+-In|7809|Out-+-> To
    || ----- | | | ---- | 7805
    ||---|~ +|---+ |+ [C2] | [C3] Vin
    || | BR | [C1] | | |
    ||---|~ -|---+ | +------+------+
    || ----- | | |
    ---|| +-----+-----------------+---------> To Gnd

    BR = 50v (or higher) bridge rectifier
    R1 = 100 ohm 2 watt
    C1 = 1000 uF 35 volt
    C2 = .33 uF
    C3 = .1 uF

    The purpose of R1 is to reduce the heat in the
    7809, but you should still use a small heat sink
    on it. If we knew the current that your receiver
    draws when it energizes the relay, R1 could be
    better selected. I guessed a max of 100 mA and
    assumed a 16V doorbell xformer.

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