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Battery powered device question

Discussion in 'Power Electronics' started by dxpwny, Apr 11, 2015.

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  1. dxpwny

    dxpwny

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    Jul 9, 2011
    I have a small sealed device that has a USB interface and runs off an internal rechargable battery when not connected to a USB port.

    I'd like to add an external 9V battery to extend it's life before a recharge is needed. I attached a 9 V battery to a 78L05, and then the 5 V output to the USB connector. Made it run nearly twice as long (it uses only about 27 mA).

    Only afterwards did I take the time to find out that the device will run with as little as 3.8 V applied. So, by using the 78L05, I was 'wasting' some of the batteries power - the 78L05 would 'shut off' once the battery voltage dropped below about 5V.

    So, I'm wondering what would be the most efficient method of attaching a common 9V battery to the USB connector and allow the device to be able to run as long as possible.

    I thought of connecting the 9V battery to a LM317 set to output about 3.8V. To make things as efficient as possible, I'd have to make the adjustment resistors a high value to lessen the 'wasted' current ... right ?

    Also thought of just using a resistor / zener diode with voltage as close to 3.8 V as I can find. In that case I'd want to use a low value resistor to keep its voltage drop as low as possible ... right ?

    Which approach would likely be more efficient ... or is there another possibility I am not seeing ?
     
  2. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    Using a regulator will WASTE just as much energy, not matter how you configure it.
    Just get 3 x AAA cells and connect it directly to the device, making sure you have the polarity correct. You can add a diode in one lead if you are not sure about the polarity and then remove it. The device may need more than 3.8v as it has an internal regulator and may need 5v.
     
  3. davenn

    davenn Moderator

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    Sep 5, 2009
    no, that's wrong ... the resistor values set the voltage not the current
    and the result would be just as wasteful and useless as the 7805

    again, wrong, The zener sets the voltage and there is some current limiting caused by the resistor
    the problem with the zener method is that the as the load current varies, the voltage from the zener/resistor
    combo will also vary

    With your very low current required around 27mA and as long as that doesn't vary significantly, you will
    probably get away with this method

    cheers
    Dave
     
  4. KTW

    KTW

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    Feb 22, 2015
  5. dxpwny

    dxpwny

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    Jul 9, 2011

    I applied 3.8 volts and it fully charged.

    Also should have added that the smaller the better.

    3 AAAs would only give me about 3.6 to 4.5 V and about 800 mAH. I found 9V batteries that can give me more then that.
     
  6. davenn

    davenn Moderator

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    a 9V battery may give more voltage but its current capability is so much less only ~ 300 mA
     
  7. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    "a 9V battery may give more voltage but its current capability is so much less only ~ 300 mA"

    It's not the mA but the mAHr that is important.
     
  8. davenn

    davenn Moderator

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    Sep 5, 2009
    both

    but initially, if the device requires 1A and the battery can only supply 350mA, then it is irrelevant
    what the mA/h rating is ... it aint going to work!!
     
  9. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    "but initially, if the device requires 1A "

    We are not talking about device requirement. We are talking about the ability of one battery charging another battery. Keep to the requirement.

    "300 mA" means nothing. It is an instantaneous value. We are talking about capacity - mA-Hr.
     
  10. davenn

    davenn Moderator

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    Sep 5, 2009
    total garbage ... he's looking for external power NOT external charging
     
  11. Colin Mitchell

    Colin Mitchell

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    Aug 31, 2014
    Sorry, you are right.
    I was put off with his remark:
    "I applied 3.8 volts and it was fully charged."

    The only problem is this:
    Any external battery is just going to constantly charge the internal battery and the charging circuit will consume some of the energy.
    You are not going to be able to do this successfully.
     
    davenn likes this.
  12. davenn

    davenn Moderator

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    all cool :)

    it is tricky when things are not initially explained and then its easy for the topic to wander off-course and all sorts of misunderstandings to occur :rolleyes:

    as in this case
     
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