Efficient switching for kid's toy?

Discussion in 'The Projects Forum' started by svdsinner, Mar 19, 2013.

  1. svdsinner

    Thread Starter Member

    May 17, 2011
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    I'm working on a "robot unicorn" for my 5yo daughter. I've got most things figured out, but need to do some work to keep battery consumption reasonable. (Significant: In the end, this is a kid's toy. I want to design it well, but it doesn't merit exotic efforts or expenses to attain "perfect". A $4 good-enough circuit trumps a $100 awesomely efficient super-circuit)

    The module I'm struggling with is the "switch" between the battery and the rest of the electronics. The behavior I am trying to achieve is essentially, a bistable multivibrator:

    • Initially off/open. It is important that battery drain is really, really low in this state so the battery isn't wasted while the toy is off.
    • When an "ON" input goes high, voltage should flow freely from "Battery In" (+11.1V) to "Main Power Out" (This will power on the microcontroller that will handle everything from there) This would be a stable state. (Power continues after the "ON" input goes low.)
    • When an "Off" input goes high (>2V) , "Main Power Out" goes off. (This is to allow the micro controller to send the power off command to shut everything down.)
    • Repeat
    • The circuit needs to be designed for reasonably low battery consumption, with the lowest usage in the "Off" state.
    I've found various simple schematics for a BMV, but they fail on some or all of the below:


    • Way too much power consumption for a battery powered device.
    • Not suitable for high side switching
    • Designed for much higher loads than I need (excessive components and complexity)
    Does anyone have a good suggestion for how to design this circuit while avoiding the pitfalls above?


    Bonus question:
    I'm also letting the micro-controller control power to various sub-systems. (E.G. Turn off power to the audio amplifier when no sound effects are being played to prevent wasting the battery) Is there anything better than a simple NPN transistor as a switch? (Better = lower battery drain with similar simplicity)
     
  2. MrChips

    Moderator

    Oct 2, 2009
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    Is there a reason you do not want to use a mechanical ON/OFF switch?
     
  3. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    I've done some tricks with low-dropout regulators. The new ones have an enable input that can be controlled by a micro. When off the LDO draws very close to zero, and it's easy to turn on and off.

    In this project I used a pushbutton as an input that also twigs the power on. You may not need the extra diodes depending on your power switch (just a pushbutton by itself would probably do). Even if you don't need the regulation off the battery you can just use a higher voltage LDO and it will still act as a switch, though a MOSFET would also work for that.

    The biggest drawback I can see is these chips don't come in DIPs, you're stuck with an SOIC surface mount thing. I used an adapter to hand wire mine in.
     
  4. svdsinner

    Thread Starter Member

    May 17, 2011
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    Kids don't turn switches off. If it doesn't shut itself down after a period of inactivity, it will constantly get left on until the batteries are drained.

    The time I spend designing this feature will be made back many times over in time saved not constantly dealing with dead batteries.
     
  5. ErnieM

    AAC Fanatic!

    Apr 24, 2011
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    Oops... generally MOSFETs would have the advantage over transistors in load switching as you're not wasting base current and still get an excellent low voltage drop.
     
  6. John P

    AAC Fanatic!

    Oct 14, 2008
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    I did this for a lower-voltage battery powered device once. It used a P-channel FET on the high side of the battery, with a 1M resistor as a pullup to make sure the gate stayed high. But then there was dual control to allow sinking current away from the gate, thus turning power on. One control was a pushbutton to ground, which was momentary. But once power turned on, a microcontroller started up, and this operated another transistor whose output also kept the FET on. When the user wanted to turn power off, it was done via the processor, which switched off the transistor.
     
  7. MrChips

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    Oct 2, 2009
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    That sounds like a reasonable thing to do.

    If you are using a microcontroller, the first thing I would do is select an ultra-low power mcu. There are many available from which to choose. When in sleep mode these will consume less than 1μA. You don't need external circuitry for power management.

    Use an interrupt input to the mcu to wake up the mcu. The interrupt can come from any switch, pushbutton or keypad.
    Have an internal timer that gets reset when ever there is activity. If the timer times out, the mcu shuts every down and goes into SLEEP mode.
     
    Last edited: Mar 19, 2013
  8. John P

    AAC Fanatic!

    Oct 14, 2008
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    I don't see how the processor can be powered continuously. The battery voltage is too high for it, and if there's a regulator, it would sit there drawing power all the time. That's why I say use a switch to turn the system on, then let the processor handle it. Or are there regulators that draw negligible power if the device they're feeding is in sleep mode?

    Just looking around briefly, I found the LM9036Q, which claims to have "ultra-low quiescent current" but it's still 20uA typical. That seems like a lot.
     
  9. tracecom

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  10. MrChips

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    Let's find out what battery is being used.
    What is the technology, voltage and capacity?
     
  11. thatoneguy

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  12. richard.cs

    Member

    Mar 3, 2012
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    Take a look at RB's thyrister based design: http://www.romanblack.com/self_swi.htm

    Basically push button to turn on. Micro goes into sleep mode (and draws less than the holding current) to turn off.
     
  13. svdsinner

    Thread Starter Member

    May 17, 2011
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    For posterity, while pursuing the LDO idea with the data sheet for an LM7809, I realized that I could eliminate the LDO altogether like below:
    (Simple, and draws almost <.1pV current when off, and only 2mA when On)
    [​IMG]
     
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