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Approx. Ohm's law for AC?

Discussion in 'Electronic Design' started by Robert Moore, Oct 12, 2005.

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  1. Robert Moore

    Robert Moore Guest

    Calculations aside, what is a good rule of thumb for estimating
    battery life in a circuit where the vast majority of current is
    consumed as AC?

    As a specific example, 4 x AAA cells rated at 1100 mAH's outputting a
    2Vpp sinewave at 60Hz. How would one adapt a DC result from the basic
    ohm's law equation to estimate the appropriate resistive load for a
    50 hour battery life?

    Given the low frequency, can I simply upgrade the battery life by a
    factor of RMS, or is there more to it.

    The drive circuit current requirements can be ignored for this

    Thanks very much for any advice.

    Robert Moore
  2. Phil Allison

    Phil Allison Guest

    "Robert Moore"
    ** ????????

    ** You need to know the power in the load, the percentage efficiency of the
    60 Hz drive amplifier and the battery power capacity. Battery capacity
    depends on the acceptable end point voltage. You do not say the type of
    cells but from the mAH rating I take it they are NiMH - so the end point is
    about 1.1 volts.

    The rating suggests 4 cells each with 1.2 volts nominal delivering 110 mA
    for 10 hours ( 5.3 Watt hours)

    If discharged over 50 hours, a current of 22mA is likely = 105 mW ( at
    4.8 volts)

    If the drive amp is say 70 % efficient ( class B at peak level) then the AC
    power can be 73 mW.

    A 2 volt p-p sine wave is 0.71 volts rms.

    The load can consume 73 mW so solving for R

    R = V squared / Power

    = 6.8 ohms.


    The assumption about the drive amp may be quite optimistic, it may have
    significant DC idle current and/or not be 70 % efficient when delivering 0.7
    volt rms.

    This will dramatically affect the available power.

    ........ Phil
  3. Where can I get some of these 60Hz AC AAA cells? And what happens if
    you mix them with the 50Hz batteries?

    Of course, you are using some sort of inverter. You will need to know
    the switching topology, output waveform, etc. in order to calculate the
    power consumption of a resistive load.
    The "RMS factor" depends on the inverter's output waveform. Vp * sqrt(2)
    only works for sine waves.
  4. Phil Allison

    Phil Allison Guest

    "Paul Hovnanian P.E."

    ** Yawn .......

    ** The OP said " 2Vpp sinewave".

    ** Err - the OP did say "sinewave".

    You in need of new glasses ?

    .............. Phil
  5. Pooh Bear

    Pooh Bear Guest

    The battery actually supplies *DC* regardless.

    The load circuit may indeed generate AC.
    You can't. Not from that info. You mentioned a voltage. Voltages don't
    take current.

    Now... if you had a load across that 2V sinewave it would be possible to
    determine the current drawn from the battery and calculate batery life
    from that.

  6. Phil Allison

    Phil Allison Guest

    The Graham Stevenson Idiot
    ** The OP is asking for a resistance value to go with a specified AC

    ** That is JUST what the OP is asking for.

    .......... Phil
  7. Ted Edwards

    Ted Edwards Guest

    I believe 1.0V is more widely accepted.

  8. Phil Allison

    Phil Allison Guest

    "Ted Edwards"

    ** Depends if you ask the cell makers or electronic equipment makers.

    .......... Phil
  9. Robert Moore

    Robert Moore Guest

    Thank you everyone for the replies.

    Here are a few more details if anyone is inclined to comment further.

    The cells are AAA alkaline, not NiMH.

    Starting voltage is around 6.25. Cut-off is 5V, being the lower Vss
    limit of the amp which is a CA 3130.

    Phil's product of 6.8 ohms seems very low to me. Can we confirm this?

    I get 909 ohms for DC. 2V/(1.100AH/50Hrs). Is AC really that different
    in practice?

    Robert Moore
  10. ehsjr

    ehsjr Guest

    WattHours in - WattHourslost = Watthours out.
    I don't know how you can avoid calculations, even
    if you ignore losses. You still need to figure the
    watthours available from the supply, and the watthours
    consumed in the load.

  11. Phil Allison

    Phil Allison Guest

    "Robert Moore"

    ** Have you checked maker's data to verify the 1100mAH figure applies to a
    1.25 volt per cell cut-off point ?

    Non rechargeable cells have large operating voltage ranges.

    ** So this op-amp is your 2vp-p, 60 Hz source ?

    It has little current output ability - you need to reconsider.

    ** Sorry if simple calcs have no meaning to you.

    ** Huh ??

    Should be 90.9 ohms ?

    ** You do not have 2 volts DC - you have 0.71 volts rms, sine wave.

    The relative power into a given load is 8 times less.

    ......... Phil
  12. The Phantom

    The Phantom Guest

    Phil is giving you the right info. Explanation below.
    This is the wrong result, but you didn't even do this division right. Looks to me like
    1.1AH/50Hrs is .022 amps; then 2v/.022 amps gives 90.9 ohms. But besides that error, the
    2VAC is a peak-to-peak value. The RMS value of the sine wave is .707 volts, and that's
    the number you need to use. And there's more to it than that, anyway.
    You have to do this on an energy basis. I am going to do the calculations with some
    false assumptions. You say the cut-off voltage is 5V, but this is for 4 cells in series,
    so the cut-off per cell is 1.25 volts. If you only take the cells down to 1.25 volts, you
    won't get the full 1.1 Ah out of them.


    But, let's assume that you do. Then let's use as the average cell voltage during
    discharge as (6.25 + 5.00)/2 = 1.40625, or about 1.4 volts. (The real discharge curve is
    non-linear, but for this first crack at it, we'll assume it's linear with time.) If we
    assume that we'll get exactly 1.1 Ah during the 50 hour discharge, that means the current
    delivered by the cells for 50 hours would be 1.1/50 = .022 amps if the output voltage per
    cell were a constant 1.4 volts. So, if the 4 cells in series can put out .022 amps at 5.6
    volts, that is a power of .1232 watts that the 4 cells can deliver for 50 hours.

    This is the key to solving this problem; you have to calculate the *power* the cells can
    deliver as DC and then figure out what resistive load will consume this power as AC (with
    conversion efficiency taken into account).

    The DC provided by the 4 cells has to get converted to AC somehow, and that process will
    happen with less than 100% efficiency. Phil picked 70% efficiency, so let's go with that
    number. Thus, you will be able to provide .7 * .1232 watts, or .08624 watts to your load
    for 50 hours. The AC voltage is 2 volts, peak-to-peak, which is .707 volts RMS. Now
    we're finally at a point where simple AC ohm's law can be used. Power =
    volts^2/resistance (P = E^2/R), so .08624 = .707^2/R, or R = .5/.08624 = 5.798 ohms.

    The reason I get a slightly different result than Phil is because you told us that the
    cells are alkaline, which have a higher voltage during discharge than NiMH cells do.

    But, I will say that you'll get substantially less than full capacity from these cells
    if your endpoint voltage is 1.25 volts per cell. See the curves in the PDF linked above.
  13. Deefoo

    Deefoo Guest

    discharge curve is

    (6.25 + 5.00)/2 = 1.40625 ????


  14. Phil Allison

    Phil Allison Guest


    ** Yawn .....

    ........ Phil
  15. The Phantom

    The Phantom Guest

    Ya got me! Of course, I meant to further divide by 4 to get the per cell voltage. :)
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