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Subject: Battery Faq patch (seeing as you asked :)
From: Andrew Jamieson (ajamiesn_at_optusnet.com.au)
Date: 2002-09-08


I've been holding this one in, 'cos I'm soon to make a tool that will allow
me to give a full graph of AJB current draw over any arbitrary time period,
and then I'll contribute a bit more on the battery / current draw side.

But for now:

Q12 talks about what is better; switching the unit off, or leaving it on
pause. Now the amount of energy stored in a battery is usually expressed in
mAh, or the amount of mA that a battery can provide over the period of one
hour (energy can be expressed by the equation I^2 . t ). Technically this
scales, so that a 900mAh battery can provide 900mA for one hour (when fully
charged), or 450mA for 2 hours, or 100mA for 9 hours, etc. This isn't quite
so true in practice, but that's a whole branch of science unto itself :)

Anyway, when you leave your unit on for 15 mins, you are using 94mA * 0.25 h
= 23.5mAh of energy from your battery. If you turn it off and on, then you
are spinning up the disk once for say 10 secs (lets keep it in round figures
for now), and then back to idle, so that's: 800mA * (10/60 mins, / 60 to
get hours = 0.0027 hours) = 2.22 mAh of energy. Thus, it is much better to
turn your unit off, than leave it in pause. I'm sure this will all change
once we have a deep sleep mode :)

If its needed / desired, I can volunteer to provide some support to the
battery faq. I've worked designing battery operated devices for ~ 10 years,
amongst other things, and have lived though most idiosyncrasies of battery
performance / operation.

Keep up the good work,

A

----- Original Message -----
From: "Björn Stenberg" <bjorn_at_haxx.se>
To: <rockbox_at_cool.haxx.se>
Sent: Sunday, September 08, 2002 11:05 PM
Subject: [Sven.Radke_at_gmx.de: battery faq]

> Sven Radke had some comments regarding the Battery FAQ:
>
> ----- Forwarded message from "Radke, Sven" <Sven.Radke_at_gmx.de> -----
>
> Date: Sun, 8 Sep 2002 02:55:10 +0200
> From: "Radke, Sven" <Sven.Radke_at_gmx.de>
> To: Björn Stenberg <bjorn_at_haxx.se>
> Subject: battery faq
>
> > A good circuit for powering from a car battery or other source
> with a higher voltage would be a 600 mA fuse and a 9V regulator like
> a 7809 of your favourite manufacturer in series.
>
> it's that if u use just a 7809 the voltage is about 8 volts, I
> experienced it 2 times (1 a while ago but I remembered when the 2nd
> spit out just 8 volts)
>
> the 7809 needs a lil buffering the energy, I made this witch adding a
> 470uf/25V condensator to input and output
>
> also it is important that is has a little load to work correctly
>
> I made this with a LED what also tells me if the thing is working
>
> I'm a pro tv service technican btw
>
> cheers,
> Sven
>
> ----- End forwarded message -----
>
> FAQ patches are welcome.
>
> --
> Björn



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