Need Rx/Tx with ultra low quiescent current

Discussion in 'The Projects Forum' started by mpboden, Feb 4, 2013.

  1. mpboden

    Thread Starter New Member

    Jan 19, 2013
    4
    0
    Hi,

    This is my first time posting here, and I'm a newbie to electronics. Thanks in advance for helping me out.

    I'm building a wireless circuit to remotely trigger my doorbell. (The purpose is to aid in training our dog that goes nuts every time it rings. This will allow me to ring it from anywhere in the house randomly.) I've actually already completed the project, and it works as intended. However, I've discovered after the fact that the circuit is drawing 9mA while idle, and this is simply too high to allow the 9v battery source to last very long. This was not something I had considered during the design. Anyway, the circuit is powered with a 9v battery, which is regulated to 5v with a LM317. I'm also using the following Rx/Tx pair:

    https://www.adafruit.com/products/1096
    https://www.adafruit.com/products/1095

    So, as the title states, I'm looking for something that will draw ultra low quiescent current. At it stands, 9mA is too large for the long term draw on a 9volt battery. A little research has already suggested I replace the LM317 with a MCP1702, which has a 2uA quiescent current. But this will only lower the quiescent current ~3.5mA, which leaves ~5.5mA still drawn by the RF receiver when idle. Thus, I'd like to find an alternative that could lower that down into the uA range as well. Any ideas? Is this even possible?

    On the other hand, perhaps I'm working too hard at this and I should simply add an AC-to-DC converter and use the 16vac that feeds the doorbell? That would eliminate the 9v battery altogether.

    Thanks again for any suggestions.
     
  2. THE_RB

    AAC Fanatic!

    Feb 11, 2008
    5,435
    1,305
    Or power it from the doorbell switch so it only receives power while someone is pressing the button?
     
  3. mpboden

    Thread Starter New Member

    Jan 19, 2013
    4
    0
    Thanks for the response, but unless I'm misunderstanding something, that's not going to work. The purpose of this circuit is so that the bell can be dinged without someone having to be at the button at the front door.
     
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