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Battery pack with indicator

Discussion in 'Power Electronics' started by CoDi, Feb 10, 2014.

  1. CoDi

    CoDi

    24
    3
    Dec 30, 2013
    I am thinking of creating an external battery pack for a camera which internal circuit works at 7.2v.

    When shopping for cells I found this:

    http://www.tmart.com/2pcs-Ultra-Fire-18650-3.7V-5000mAH-Lithium-Battery-Yellow_p191451.html

    Two of this in series would offer 7.4v and 5000mAh, so it would give juice for quite a while to power the camera. Anybody has experience with these cells? Do these cells need circuit protection?

    Also, I would like to add a custom battery fuel gauge indicator similar the one the macbooks pro use:

    http://icdn6.digitaltrends.com/image/battery-indicator-macbook-pro-650x0.png

    But I don't know where to start on creating something like that because I can only find already made ones that are too big or bulky.

    Thanks
     
  2. gregfox

    gregfox

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    Mar 25, 2013
    I've used the Ultra fire cells before, and if you get 75% of what they claim, you're golden!
     
  3. CoDi

    CoDi

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    Dec 30, 2013
    Thanks gregfox. That is exactly what I was fearing, too good to be true. So what brands are more reliable for Li-Ion cells?

    As usual the big challenge is space, which is approximately 100mmx26mmx26mm. It could roughly fit two 26500 or C size cells placed one on top of the other perfectly or 8 AA cells.

    I am trying to get at least 7.4v (it would be better if I could get to 12v) and at least 3000mAh.

    I have been browsing so many battery websites and I can not seem to find the perfect fit. I found this but I suspect the quality might be poor as well:

    http://efestpower.com/Product/1652072426.html
     
    Last edited: Feb 13, 2014
  4. gregfox

    gregfox

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    3
    Mar 25, 2013
    Battery

    HI,
    Take a look at Hobby King. I find their batteries to be closer to their claims. They have many sizes, you may find one to fit.
     
  5. gregfox

    gregfox

    149
    3
    Mar 25, 2013
    I data log the batteries into a constant current source for testing.
     
  6. CoDi

    CoDi

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    3
    Dec 30, 2013
    Thank you so much for the Hobby King recommendation! They have a great selection and great prices.

    Unfortunately today I realized I made a mistake, the camera needs 12v instead of 7.4v. Since the space is limited this changes everything dramatically.

    I could use a step up from 7.4v to 12v but it would reduce the capacity.

    So again I am hitting another wall... Now I need to fit ideally 4x Li-Ion cells to get 14.4v and regulate it to 12v with at least 2000mAh. Would a step down from 14.4v to 12v give me more amperage?
     
  7. gregfox

    gregfox

    149
    3
    Mar 25, 2013
    Amperage is determined by the battery capacity, so stepping down will not give you more amperage. I'm sure you can find a suitable battery at 14.4 (remember fully charged it will be 16.8V).
    You could step it down to 12V, and I have used the attached circuit to do just what you need.
     

    Attached Files:

  8. CoDi

    CoDi

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    3
    Dec 30, 2013
    Thank you so much gregfox!

    So if I step up from 7.4v to 12v then there will be a loss in capacity? right? Is there a math to calculate the amperes that get lost? by doing that is it going to generate heat?

    And if the input in the camera is 12v-20v, do I still need to step down or regulate the voltage to 12v? would this process generate heat as well?
     
  9. BobK

    BobK

    7,578
    1,633
    Jan 5, 2010
    Stepping up from 7.4V to 12V will require 12/7.4 x the current from the batteries, plus the converter is only operating at 80 90% efficiency, so divide by 0.8 again to get the answer. It will require about twice the current from the batteries as you are going to use.

    Bob
     
  10. gregfox

    gregfox

    149
    3
    Mar 25, 2013
    Hope it works out, you may have to play with it a bit.
     
  11. CoDi

    CoDi

    24
    3
    Dec 30, 2013
    Thanks guys!

    I do not know if I am doing something wrong but if I follow the math:

    12v / 7.4v x 3000mAh = 4864.86

    then 4864.86 / 0.8 = 6081 mAh???

    Am I doing it wrong? Or is it instead:

    7.4v / 12v x 3000mAh = 1850

    1850 / 0.8 = 2312.5 mAh

    The result is the loss or the final amperage?
     
  12. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,158
    2,676
    Jan 21, 2010
    OK, stop right there.

    you have divided 12 by 7.4 and that will give you the raw ratio of currents if you step up from 7.4 to 12V (you would multiply by that.

    But then you multiply by a battery capacity. That's wrong for one or more of 3 reasons:

    1) the capacity in mAh will fall when the voltage is stepped up, so a 3000mAh 7.4V battery, can deliver 3000 / (12 / 7.4) mAh when stepped up to 12V. But that's a charge, not a current.

    2) perhaps you meant current drawn from the battery. Presumably that means you meant the load draws 3000mA at 12V. In that case, the calculation is correct, but the units should be mA, not mAh.

    3) perhaps you were looking a the current from the battery to see what current you could get at 12V. If this is the case most of the rest of your calculations were backwards.

    I'm going to assume what you meant was (2) because the rest of the calculations seem to support that.

    So, for a 3A load at 12V, you have a 4.9A load at 7.4V

    It looks like you're expecting around 80% efficiency from your boost converter.

    And yes, that means the draw from the 7.4V battery to supply 12V at 3A is around 6A.

    Be very careful with the difference between mA and mAh -- it's similar to the difference between speed (mA) and distance (mAh).
     
  13. CoDi

    CoDi

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    Dec 30, 2013
    Thanks steve for the explanation.

    I apologize for the confusion but this is exactly what I needed:

    Two 3.7v 3000mAh Li-Ion cells in series obtain 7.4v 3000mAh and stepped up to 12v will run at 1875mAh. I hope I am correct now :)
     
  14. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,158
    2,676
    Jan 21, 2010
    No, you're exactly wrong.

    nothing can run at some mAh.

    It's like saying that I have a really fast car it can go as fast as 25 kilometers.

    According to the previous calculations it might have an effective capacity of 1500mAh at 12V (not 1875mAh)

    Given that, if the device draws 250mA, the battery will last approximately 1500/250 = 6 hours.
     
  15. BobK

    BobK

    7,578
    1,633
    Jan 5, 2010
    Although he is using the wrong terminology, the effective capacity of the batteries does scale the same way as the current draw. I.e. if I am drawing twice the current needed to operate the device, I can basically derate my battery capacity by a factor of 2. So 3000 mAH batteries will act like 1500 mAH batteries when boosted by 12/7.4 at 80% efficiency.

    Bob
     
  16. (*steve*)

    (*steve*) ¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd Moderator

    25,158
    2,676
    Jan 21, 2010
    But the next thing he'll need is a 1500mAh boost converter... And so we'll be asking the question again.
     
  17. gregfox

    gregfox

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    Mar 25, 2013
    Hi CoDi, I a bit confused.
    Is it your intention to use a ~7.5V battery and boost it up to 12V or use a ~14.8V battery and step it down to 12V?
     
  18. BobK

    BobK

    7,578
    1,633
    Jan 5, 2010
    Why not use an 11.1 V (3 cells) and call it even? It would likely work quite reasonably with a 12V device.
     
  19. BobK

    BobK

    7,578
    1,633
    Jan 5, 2010
    Agree, the correct units should be used.

    I cringe every time I see a newspaper article about a new solar installation that says it will produce 10 Megawatts per year. The always seem to get power / energy units wrong.

    Bob
     
  20. CoDi

    CoDi

    24
    3
    Dec 30, 2013
    Thanks guys for all your help, and so sorry I keep using the wrong terminology. I also forgot to multiply 1875 by 0.8 to obtain the 1500mAh.

    gregfox, I am just looking for all the options. Either 7.4v battery pack stepped up to 12v or a 14.8v stepped down to 12v. Whatever is more efficient, takes less space and lasts longer.

    Maybe I should have been more clear from the beginning. What I am trying to create is a pistol grip that would provide external power to the camera and serve as a LANC controller at the same time. This is the camera:

    http://www.blackmagicdesign.com/products/blackmagicpocketcinemacamera

    It uses rechargeable EN-EL20 7.4v Lithium Ion batteries. Even though the camera uses these 7.4v batteries, the external DC input is rated 12v-20v. And I believe the camera draws between 800mAh to 1000mAh, I have to confirm (I hope I am not saying something wrong again). That is why I am trying to reach 12v.

    My knowledge in electronics is very basic as you can tell, I am more of an enthusiast designer. I am doing this for fun and to see if it would be possible. So the big challenge is to figure out if it would all fit inside of the grip.

    [​IMG]
    [​IMG]
    [​IMG]
     
    Last edited: Feb 15, 2014
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