question about battery capacity, current and joules

Discussion in 'General Electronics Chat' started by toffee_pie, Jan 30, 2015.

  1. toffee_pie

    Thread Starter Active Member

    Oct 31, 2009
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    Guys,

    I have a question on current and Joules.

    Take a circuit that for example uses up 42J over 18 seconds, that would need 6.4mA of current to obtain 42Joules.

    How do I find out the actual device current over that period? is it 115mA ? (6.4mA x 18)
     
  2. BobTPH

    Active Member

    Jun 5, 2013
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    42 J over 18 seconds is 2.33 W. The current will depend on the voltage. P = I * V.

    Bob
     
  3. toffee_pie

    Thread Starter Active Member

    Oct 31, 2009
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    assuming 3.6V then it would be 23mA?
     
  4. ronv

    AAC Fanatic!

    Nov 12, 2008
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    1 Joule = 1 watt/second.
     
  5. wayneh

    Expert

    Sep 9, 2010
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    A Joule is 1 watt-second. To calculate wattage, you need to know voltage, which you have not specified. Watts are volts•amps.
     
  6. toffee_pie

    Thread Starter Active Member

    Oct 31, 2009
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    I know, I want to translate it to actual device current over that time period.,

    I am finding it hard to explain, but take something that uses various amounts of energy, how to calculate actual device current required over the entire period?

    ie [ 34J @ 12S , 65J @ 23S, 82J @ 23S).

    assume voltage is 3.6 for the sake of it.
     
  7. BobTPH

    Active Member

    Jun 5, 2013
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    As I said before, 42 J over 18 sec is 2.33W.

    To get 2.33 W at 3.6V the current is 2.33/3.6 = 647mA

    Bob
     
  8. BobTPH

    Active Member

    Jun 5, 2013
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    No you don't. If you expend 42 Joules over 18 seconds that is enough to calculate the power, as I did.

    Bob
     
  9. toffee_pie

    Thread Starter Active Member

    Oct 31, 2009
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    ok thanks .:)
     
  10. ronv

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    34J/sec X 12 sec = 408J or 408 watts per second. 408/3.6 volts = 113 amp seconds. Or for a battery .031 amp hours.

    I think. :D
     
  11. toffee_pie

    Thread Starter Active Member

    Oct 31, 2009
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    I have been thinking about this again, how do i figure out mAh capacity needed from executions of energy (in Joules or Watts, doesn't matter right now (as Lesley Nielsen might say)).

    example, [5.6W and 4.2W and 7.2W] what capacity would this take up in mAh? , the total used is not 5.6+4.2+7.2 as the time intervals would be varying.

    I know how to convert Wh to Ah but the above are Joules, not Wh

    for an accurate answer do i need to integrate over time period? (integral 5.6W + integral 4.2W + integral 7.2W)
     
  12. ronv

    AAC Fanatic!

    Nov 12, 2008
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    Only the joule has time specified for you. 1 Joule is 1 watt for 1 second.
    Maybe it would make more sense to you if you look at it from the opposite direction.
    Lets say you have a 1 amp hour battery and it is a 1 volt battery. It can supply 1 amp at 1 volt for 1 hour. Hence 1 amp hour. Or you can say it can produce 1 watt for 1 hour. -- 1 amp X 1 volt X 1 hour.. The battery could deliver 3600 joules. 1 amp at 1 volt for 3600 seconds.
    Or 2 amps at 1 volt for 1800 seconds.
     
  13. wayneh

    Expert

    Sep 9, 2010
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    Yup, agreed. I was thinking amperage - as the OP asked about - but wrote wattage.

    Gonna let Ron slide on "watt/sec"? (I did!)
     
  14. WBahn

    Moderator

    Mar 31, 2012
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    You do? I sure don't. How many Ah is 300 kWh? That's about my typical electricity usage each month.

    1 Wh is a unit of energy.

    1 Ah is a unit of charge.

    What you are missing is a unit of energy per unit charge, or something with units of energy/charge, namely joules/coulomb (J/C).

    It just so happens that 1 V is defined as 1 J/C.
     
  15. ronv

    AAC Fanatic!

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    QUOTE]
    I think I understand what your asking. :D
    I would just use ah for all the different time periods, then add them up.

    1 amp for 1 hour = 1 ah +2amps for 1 hour = 2 ah for a total of 3 ah used.[/QUOTE]
     
  16. WBahn

    Moderator

    Mar 31, 2012
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    This is like asking how to determine how far you traveled on your vacation if you drove at 50 mph, 30 mph, and 75 mph?

    Do you see what's missing?

    Even assuming that you are delivering each of these powers from the same voltage (probably the case), the amount of time that you are at each power level is critical to determining the total charge delivered. Even if we assume that each power was delivered for the same amount of time, it makes all the difference if the total time was one minute or if it was one day.
     
  17. toffee_pie

    Thread Starter Active Member

    Oct 31, 2009
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    [/QUOTE]


    that does not work as easy as that i am afraid, I have done work previously and the capacity (I stress capacity) needed from a battery does not equate to all the individual currents added up., hence my question.
     
  18. toffee_pie

    Thread Starter Active Member

    Oct 31, 2009
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    yeah, so i need to integrate the power across the particular time interval perhaps? simply adding up individual values wont work.

    assume 5W is done across a time span of 22ms and 14W is done over 4 seconds, i would use integration over said time periods and get an average of both ? (3.6V stays constant)
     
  19. WBahn

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    Mar 31, 2012
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    Yes. Though you don't need the average, but the total charge so with an average you would still need to multiply by the total time. With 22ms for one and 4s for the other, the calculation will be dominated almost entirely by the 4s one.

    Leverage your units.

    Needed Capacity = [(5W)(22ms) + (14W)(4s)]/(3.6V)
    Needed Capacity = {[(5)(22m) + (14)(4)]/(3.6)}[Ws/V)][(1VA)/(1W)][(1h)/(3600s)]
    Needed Capacity = {[(5)(22m) + (14)(4)]/[(3.6)(3600)]} Ah
    Needed Capacity = 4.33 mAh
     
  20. WBahn

    Moderator

    Mar 31, 2012
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    that does not work as easy as that i am afraid, I have done work previously and the capacity (I stress capacity) needed from a battery does not equate to all the individual currents added up., hence my question.[/QUOTE]

    Capacity is a highly nonlinear function, so the capacity itself is a function of the current draw (and the temperature and the duty cycle and the history of the battery and ....)
     
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