On Tue, Dec 7, 2010 at 12:35 AM, Sean Breheny <shb7@cornell.edu> wrote:

> A linear regulator tends to draw a constant current from the source
> regardless of the source voltage. This is what Olin saw in the graph.
> The values on the low end where the current falls off as the voltage
> drops are likely the "dropout" region of the linear regulator, where
> the output voltage is no longer being regulated but rather is
> following the input voltage minus an offset.
>

I thought that graph was *not* from an iPhone, but some Android one? What
also Olin said is that solarwind would need to measure his iPhone and draw
that graphs to go any further from this point.

Tamas




>
> Sean
>
>
> On Mon, Dec 6, 2010 at 6:13 PM, V G <x.solarwind.x@gmail.com> wrote:
> > On Mon, Dec 6, 2010 at 1:52 PM, Olin Lathrop <olin_piclist@embedinc.com
> >wrote:
> >
> >>
> >> I took Martin's data and made a plot of it, which is attached.
> >>
> >> This says the phone has just a linear regulator between the USB power
> and
> >> the LiIon battery.  If there was a switcher, the current would not hav=
e
> >> been
> >> so completely flat from 4.8V on.  If the iPhone does this then just 4
> NiMH
> >> batteries driving the USB power will work, but leave a significant
> fraction
> >> of the NiMH battery capacity unused.
> >>
> >
> > Olin,
> >
> > Why would you say that? I suspect that there is some sort of a switchin=
g
> > power supply inside the iPhone from various random articles I read on t=
he
> > Internet. Of course, they could be wrong.
> >
> > However, a threshold voltage of 1.15 volts/cell would not allow the pac=
k
> to
> > drop below 4.6 volts. At that voltage, the pack would be still be drain=
ed
> > above 90% (if not 95%?) according to the chart here:
> >
> > http://www.eneloop.info/home/performance-details/discharge-current.html
> >
> > I think that's a reasonable voltage to cut off at.
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> >
>
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