low power draw indicator

Discussion in 'General Electronics Chat' started by bug13, Apr 20, 2015.

  1. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    Hi guys

    I am looking for an indicator that draws little or no power. (for my battery power device) Any idea? The indicator will have two state, to show the device is operating, or to show the device is power off.

    Thanks guys!!
     
  2. dl324

    Distinguished Member

    Mar 30, 2015
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    How about an LED to indicate power on/off? You could operate it at a low current to minimize power.
     
  3. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    Hi dl324,

    Thank you for your replay. That could work. But in my case, the indicator also needs to be easily visible from distance during the day.. For that reason, I can't run a LED at low current.
     
  4. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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  5. ScottWang

    Moderator

    Aug 23, 2012
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    You could using TLC555 cmos type that it draw almost less than 1mA, and set a high current and less duty cycle as flashing light, if you set it right then the battery will get alive for a long long time.
     
  6. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Yes a flipdot might be the best way, but really it depends on how much energy you need to flip it, and how often you would od that.
    The smallest from those needs 250mA for 1.5ms, which is 0.375mJ.
     
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  7. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    My indicator needs to change state 2-4 time at the most in a day, so on average 250mA for 1.5ms is not much. I think on average work out to be ~20nA in a day.

    Using a TLC555 drawing 1mA is still too much for my application.
     
  8. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    You could use a cmos variant of a 555, but the charging current of the timing cap will still be there. Can you post your circuit?
    Also note the flipdot needs at least 12V to flip, so you might need some charge pump to get that.
     
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  9. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    My circuit is just a low power PIR sensor from Panasonic, which is powered by a digital pin. A pic and two push button. And a solid state relay. Plus the usually passive stuff.

    The one I'm interested in is the 30 series, I read it requires min 4.5V. And others are 7.5v, am I missing something here? (You say 12v)

    Thanks for pointing out the charge pump, I didn't think of that. Lucky I have some spare pins available to make a charge pump.
     
  10. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    You´re right with the 4.5V, I must have mixed it up somehow.
     
  11. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Can't make out from the flipdot info whether the pulse polarity needs to change to flip it back :confused:.
     
  12. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Yes it does, look at the page which describes the principle, you need to magnetize that core in the other direction, than it holds this state.
     
  13. dl324

    Distinguished Member

    Mar 30, 2015
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    It would be helpful if you provided as much detail as possible when seeking assistance...

    Low power and visibility from a distance during daylight are typically conflicting requirements.

    Check out the LM3909 LED flasher. It will operate from 1.2V to several hundred volts and uses very little current over battery lifetime. You just add a timing cap which is also used to boost voltage to drive the LED.
     
  14. Alec_t

    AAC Fanatic!

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    What is your battery voltage?
     
  15. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    Here's a possible driver for a Flipdot, using a 6V supply :
    FlipdotDriver.gif
     
  16. wayneh

    Expert

    Sep 9, 2010
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    I wonder if you could somehow use light-focusing, fiber optic plastic like the stuff that is used in gun sights. This would take advantage of ambient light to produce an intense dot. Of course the challenge is how to toggle the appearance from one state to another.
     
  17. crutschow

    Expert

    Mar 14, 2008
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    How much average current can you tolerate for the indicator?
    What is the battery and what does the rest of the circuit draw?
     
  18. ian field

    Distinguished Member

    Oct 27, 2012
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    You probably can't get much better than an LED, regular types are typically 20mA but you can get ultra-efficient 3mm types that at least give an indication at only 2mA.

    If you really need a micropower solution, try a couple of CMOS NAND gates wired as an astable - make the mark space ratio very unequal so the LED only blinks breifly during its on period - even with 50/50, the LED would only be drawing current half the time.
     
  19. #12

    Expert

    Nov 30, 2010
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    This does not seem important now, but the TLC555 idles at 250 ua Maximum under room temperature conditions.
     
  20. bug13

    Thread Starter Well-Known Member

    Feb 13, 2012
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    Sorry about that, I should have include more information. But all of you inputs are appreciated!

    I am using lithium battery, so it's about 3.3V normally.

    It sounds like it's not something I can buy off the shelf, but it's a really cool idea, I would like to look into it.

    I am hoping on my average current for the indicator is under 50uA. my circuit draw is under 10uA at sleep. The circuit is sleeping all the time. For a designer, my circuit doesn't power off, the two states that the indicator needs to show to an end user is 1) device is sleeping but interrupt enable.(aka operating); 2)device is sleeping but interrupt is disable. (aka for a user, power off).

    The indicator will change state between 2-4 tims / day at the most. My interrupt will be trigger maybe up to 10 times /day. But most likely will only be triggered 2-4 times.

    I actually haven't look into the battery current draw, can I assume you mean battery leakage current? This is the batteries I will be using:
    http://data.energizer.com/PDFs/l92.pdf

    I still think the flip dot indicator will draw less current on average / day, as my device will likely operating ~12 hours/day. (Indicator needs to show the device is operating). But i the indicator only needs to change state 2-4 time, so on average, the flip dot indicator draws virtually no power. Am I correct?
     
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