I just completed a battery system using lithium batteries for a customer.  =
The batteries are 3.7V and 6.1 Amp-hours(Ah).  I needed outputs at 12V, 150=
mA and 5V, 600mA.  To accomplish this I paralleled 2 batteries because the =
end goal was to supply the two outputs for at least 8 hours.  =


>From what has been discussed and what you have said I think you're on the r=
ight track.  For my application one of the main considerations was exactly =
what Jinx posted, power =3D power on each side of the conversion.  With 2 c=
onversions happening they both added together and had to be something the b=
attery could cope with.    =


If you are going to use batteries remember that the Ah rating is based on w=
hat's called C/5, which is the maximum discharge rate that will get that po=
wer rating.  Take the Ah (or mAh in the case of most AA rechargables) and d=
ivide by 5 to find this value.  Discharging the battery above that value wi=
ll result in it being flat faster.  I'm sourcing my batteries from Ultralif=
e and they have more information on their website at www.ultralifebatteries=
.com.

As a side note I used Microchip's MCP1650 series for the converters.  It wa=
s very easy because everything is given to you including the pcb layout.  A=
ll I needed to do was ensure I had enough battery capacity.  Best of luck w=
ith your design.

Regards,
Andy



> Date: Fri, 27 Jun 2008 01:04:49 -0700
> From: joecolquitt@clear.net.nz
> Subject: Re: [EE] Trying to understand step-up regulators
> To: piclist@mit.edu
> =

> > this is that a 5 V battery is more "powerful", so where the hell is tha=
t =

> > extra energy going to come from?
> =

> You'd might think of it in wattage or VA terms. Power =3D Power on
> both sides of the equation because of conservation (ignoring losses)
> =

> 1 amp supplied from a 1.5V battery (1.5W) has the same energy
> as 0.3 amps from a 5V battery
>  =

> > So is that how it all works? If you want to use a single AA battery to =

> > put 25 mA through a 200 ohm resistor then:
> >     * 87 mA has to come from the AA battery
> >     * The regulator steps up to 5 V and gives out 25 mA
> =

> 25mA into 200 ohms dissipates P =3D I*I*R =3D 0.125W
> =

> So ideally you need 0.125W from the1.5V battery
> =

> P =3D I*V, 0.125 =3D 0.83A, 83mA. But efficiency is 95% -> 87mA
> =

> > Let's say I have two AA batteries, each of which is 500 mA hr. If I
> > put them in series to form a 3 V battery, then what's the capacity of
> > this 3 V battery? Is it 500 mA hr at 3 V? (The energy conservation
> > principle is telling me that yes it is 500 mA hr at 3 V, but I just want
> > to be sure)
> =

> Batteries in series add voltage, batteries in parallel add capacity
> =

> -- =

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