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ARCHIVED FORUM -- April 2007 to March 2012
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This is the first Archived Forum which was active between 17th April 2007 and 1st March February 2012

 

Latest post 01-27-2009 8:15 PM by Calvin. 9 replies.
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  • 01-24-2009 3:13 PM

    • Calvin
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    A8 power consumption

    Well was using a pair of A8 earphones with my ipod touch for a few months and noticed that the battery lasted nowhere near the time Apple claimed it should.  Anyway, then one of the ear's sound wasn't working (owing to the age of the headphones and my habbit of knotting the wires when not in use, I'm going to get the soldering iron out at some point and fix the jack).  Well I went and got a pair of £30 Sony headphones and noticed that I could turn the volume setting way down and hence the battery lasted an awful lot longer.

    The questions is if anyone can explain why the B&O earphones need so much more power?  Is it just that the sound drivers which give a better sound are made of some weird polymer that's heavier and needs more of a magnetic field to move?  As an aside, I would say that once they're fixed. I'd far rather be using the A8's than the Sony ones, the sound quality and design more than outweigh any other concerns.

  • 01-24-2009 3:32 PM In reply to

    • lausvi
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    Re: A8 power consumption

    My experience is very different. I use my iPod and A8's mainly during the nights and I never need to raise the volume - with some tracks even the most quiet level is almost too much. Also I remember seeing a note that the A8's are very easy to drive (compared to usual headphones) and so they should cause lower power consumption. No idea what's going on in your case. Geeked

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  • 01-24-2009 4:34 PM In reply to

    • Puncher
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    Re: A8 power consumption

    A8's are nominally 32 Ohms - pretty standard stuff.

    Generally speaking, you aren't learning much if your lips are moving.

  • 01-24-2009 4:48 PM In reply to

    • Calvin
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    Re: A8 power consumption

    Oh right, very interesting....   As the fault with mine is clearly that a break has occurred where the jack and wire connects inside the plastic casing, maybe this was connected to the higher power consumption.  I suppose if the wire had part broken, it could have been causing a minor short or something.  

  • 01-26-2009 3:17 PM In reply to

    • Calvin
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    Re: A8 power consumption

    Just got back from a few drinks with a friend who also owns an ipod touch and some A8s.  I raised the point and he agreed with me and said that the B&O headphones burnt through his battery in half the time.  He also said that Phillips stuff was the same and that it was maybe a shared drivers issue.  On the upside, he pointed out that he normally prefers his B&O ones though.

  • 01-26-2009 4:12 PM In reply to

    • Puncher
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    Re: A8 power consumption

    Calvin:

    Just got back from a few drinks with a friend who also owns an ipod touch and some A8s.  I raised the point and he agreed with me and said that the B&O headphones burnt through his battery in half the time.  He also said that Phillips stuff was the same and that it was maybe a shared drivers issue.  On the upside, he pointed out that he normally prefers his B&O ones though.

     

    My apologies, I've re-checked and you're quite correct, the A8's are significantly lower impedance than the standard Apple earbuds as follows -

    Apple     32 Ohm

    A8         19 Ohm

    Form1    32 Ohm

    This will explain your runtime differencesSmile

     

     

    Generally speaking, you aren't learning much if your lips are moving.

  • 01-27-2009 6:14 AM In reply to

    Re: A8 power consumption

    Shouldn't a lower impedance mean that you need to turn up the volume less and therefore the battery will last longer, or have I misunderstood something?

     

    Simon

     

     

  • 01-27-2009 8:07 AM In reply to

    • Dave
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    Re: A8 power consumption

    The lower the impedance, the higher level of energy required, Someone tell me if i'm incorrect

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  • 01-27-2009 8:45 AM In reply to

    • Puncher
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    Re: A8 power consumption

    The impedance related bit is easy, the complicated bit is what actually happens within the iPod.

    Here we go with a whole string of assumptions, all of which we'll assume are true for a first pass aproximation of the problem.

    If the sensitivity (dB/mW) of the earphones in question is the same then they require the same input power to produce the same output SPL.

    Because of the impedance difference this means that the speakers are operating at different voltages and currents, for example -

    @20mW input, the 19 Ohm A8's are operating at an (rms) voltage of 0.616V and a current of 32.4mA, while the 32 Ohm Apple earbuds have 0.8V and 0.025mA respectively.

    The A8's are operating at a higher current (albeit a lower voltage) than the Earbuds. What happens next depends upon the iPod internals. Assuming the iPod doesn't have clever voltage supply control schemes i.e. the headphone amplifier voltage rail is a fixed value, then it is all down to the topology of the headphone amplifier.

    If the amp. is a linear type, then the current difference above will result in a reduction in battery runtime when running with A8's. The major power hungry components in the iPod are the HD (if present), the backlight and the audio out, therefore although the A8 current is ~ 30% higher than the Earbud the runtime reduction will be less than this when including the fixed drain of the other components.

    If however the amp. is a switching amplifier, then the efficiency will be very high (it is packets of energy being switched into the load rather than linear voltage regulation) and the difference in runtime due to the earphone impedance will be negligable .

    So it seems the answer is some where between hardly noticeable and less than 30% when listening at the same volume.

    That's that cleared up thenLaughing

    Generally speaking, you aren't learning much if your lips are moving.

  • 01-27-2009 8:15 PM In reply to

    • Calvin
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    Re: A8 power consumption

    I would also add that the optimum efficiency of any audio system is when the impedance of the speakers matches that of the output.  For those unfamiliar with the technicalities, impedance is different to resistance.  You have a resistance along with an inductance, as the magnetic fields themselves create an electromagnetic field.  A quick Google search will answer all questions probably although it helps to understand i based (complex) maths if you want to play with the numbers

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