Stop power loss on relay in standby mode

Discussion in 'Digital Circuit Design' started by irozak, Apr 14, 2018.

  1. irozak

    Thread Starter New Member

    Apr 3, 2015
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    My project is battery powered (1.3Ah), has a mercury switch which activates an alarm for a few seconds then goes back to standby. This timer switch is perfect for the job but it still draws 2.4mA in standby even after I removed the led's. I'm guessing the culprit is a capacitor soaking up the power. I would like the battery to last months between charges but this current loss would kill it in a couple of weeks.
    Is there any way to achieve zero drain in standby mode?

    timer relay_irozak.jpg


    Mods Note:
    Please don't upload the colorful image with *.png, and use it as *.jpg.
     
  2. crutschow

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    Mar 14, 2008
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    Capacitors don't conduct DC so it has to be something else taking the current.
    Do have a schematic or other info on the timer board?
     
  3. irozak

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    Apr 3, 2015
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    No nothing. It's just an ebay special.
     
  4. crutschow

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    So you have no directions on how to use it or connect it or what it does?
     
  5. AnalogKid

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    Link to the ebay page ? - DUH.
     
  6. ebp

    Well-Known Member

    Feb 8, 2018
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    Photo of the flip side? Since there are two pots I'm guessing a pair of (probably CMOS - current too low for bipolar 555s) one-shots that suck current through the timing resistors. It might be one bipolar 555 with pot for each timing resistor & jumpers config as monostable or astable.
     
  7. Hymie

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    Mar 30, 2018
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    I offer the following two circuits which use zero power in standby.

    I have used the first circuit; capacitor C1 needs to be a sufficiently large value such that the charging current through the relay coil causes the relay to energise and close the contacts. From memory I used a 1,000uF capacitor, resistor R1 is required to discharge the capacitor ready for the circuit’s next trigger signal.

    If the relay contacts are required to be closed for longer than a momentary action, then the effective capacitance value is multiplied by the dc gain of the transistor Q1 (shown in the second circuit) – and values adjusted by trial & error to achieve the required on time for the relay.

    Err – the transistor Q1 should be a PNP type and not as shown (NPN).

    Zero power standby_Hymie.jpg
     
    Last edited: Apr 15, 2018
  8. Hymie

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    Should have added to my above post – that the relay in the circuit will only energise while the tilt switch is closed.
     
  9. irozak

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    Apr 3, 2015
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    Thanks Hymie. I have got the idea (I think) but will need more assistance if I had to build it.
    The tilt switch would be a momentary action, then the relay should activate for 15 seconds.

    Here is a bit more info on the board I have. It's a shame because it is perfect except for the standby loss.
    https://www.ebay.com.au/itm/DC-5-12...hash=item2a87da4102:m:maRhG677VAMXF7H8c158aIA
    Also attached pic of the chip on the back of the board.

    Thanks for your help.
     
  10. ebp

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    I can't recall ever seeing any component with such a long string of characters in the marking. It appears to be a microcontroller. What a micro doing next to nothing needs with 2.4 mA is beyond my ken. Perhaps the pots are voltage dividers across the supply and read by analog inputs. 5k pots used that way would account for 2 mA with 5 V supply; 2.4 mA isn't a big stretch because end to end tolerance of even good brand-name pots is generally pretty sloppy.
     
  11. irozak

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    Apr 3, 2015
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    Searched for the chip specs to no avail.
    I have used Ton to get the 15 seconds. Toff is wound back so it doesn't cycle. Maybe I'll have a play with the pots and see if it makes any difference.
     
  12. crutschow

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    Mar 14, 2008
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    Below is the LTspice simulation of a 555 circuit that should do what you want.

    In the quiescent state, transistors Q1 and Q2 are off, completely removing the power to the 555, so the circuit draws only the leakage current of the transistors.
    When triggered, the 555 is initially powered from the input trigger, causing the 555 output to go high. This turns on Q1 and Q2, bootstrapping the 555 power through D1 so it remains on when the input trigger is removed.
    After ≈15s the 555 output pulse goes low, shutting Q1 and Q2 back off and removing the 555 power.

    If you just need a pulse output and isolation is not required, then for no more than a 200mA load, you could use the "Out" pulse and eliminate the relay and D2.

    upload_2018-4-15_19-34-13.png
     
    Last edited: Apr 15, 2018
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  13. irozak

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    Apr 3, 2015
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    Thank you very much crutschow, This is very helpfull to a novice like me and I will try to build it.
    The load is 250mA so I guess I will need the relay.
    Please confirm Vcc is 12v in, and TRIG is 12v power from my (momentary) mercury switch.
    Can you please explain the purpose of cap C3?
     
  14. Hymie

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    Mar 30, 2018
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    In crutschow’s circuit Vcc is the output of your mercury switch.
    The Trigg input to the 555 timer IC (pin 2) in unconnected.
    Capacitor C1 is the timing capacitor.
    Capacitor C4 ensures the timer IC triggers at switch on (power up) from the mercury switch.
    Capacitor C3 is providing smoothing of the supply source.
    Note that the permanent 12V supply is connected to the emitter of transistor Q2.
    The transistor Q2 is capable of 600mA, so no need for the relay (or diode D2).
     
    Last edited: Apr 16, 2018
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  15. joeyd999

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  16. irozak

    Thread Starter New Member

    Apr 3, 2015
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    No relay! This is getting simpler by the minute. Thanks
     
  17. crutschow

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    No. It won't work with the Trig input open.
    The "Trig" label means it's connected to the "Trig" label near C1.

    Here's the circuit modified to show the wire connection:

    upload_2018-4-16_19-29-24.png
     
    Last edited: Apr 16, 2018
  18. irozak

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    Apr 3, 2015
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    OK got it thank you.
    Would you also agree that the relay is not required for 250mA load?
     
  19. crutschow

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    Yes, 250mA should be fine.
    What is the nature of the load?
    If it's inductive, it needs a diode across to suppress the inductive transient.
     
  20. irozak

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    Apr 3, 2015
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