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From: "petrus bitbyter"
Subject: Re: 9V bat ->2AA + switcher ?
X-Newsreader: Microsoft Outlook Express 5.50.4807.1700
Date: Wed, 04 Dec 2002 21:13:29 GMT
NNTP-Posting-Date: Wed, 04 Dec 2002 22:13:29 MET
Organization: chello broadband
You'r confused because you mix up power, energy and battery capacity. As you
power = voltage * current
energy = voltage * current * time
batt. cap. = current * time
Comparing capacity of batteries makes sense only when the batteries are of
the same voltage. In this case you can either compare a series of six 1.5V
batteries with the 9V battery or one 1.5V battery with one cell of the 9V
battery. (A 9V battery consists of six small 1.5V cells with roughly the
capacity off half a AAA.)
The energy that these batteries deliver, is capacity * voltage [mWh]. As the
voltage degrades over time, the energy cannot be calculated easily. It also
depends on the lowest voltage the load requires to function correctly.
What you realy want is a 7.5V power source capable to deliver 3mA for as
long as possible. The easiest way is to take a series of six 1.5V batteries.
Six AAA batteries will run about twice as long as a single 9V one.
Using a switcher is a little bit more complicated. Achieving 80% efficiency
at such a low power will not be easy. As your load requires about 22mW, your
switcher has to do his job with less then 6mW. I guess 50% will be more
realistic. It's no use to go for small batteries and a switcher to get a
longer runtime. Only two or three fat ones may do the job.
I think you'd better look for a box for six penlites (AA).
"Steve Sousa" schreef in bericht
> I've been trying to figure if it would be worth to change a power
> supply from a 9V bat to a couple of AAA bats + a switching regulator,
> but i'm getting confused, I don't know if i could actually get a
> longer running time.
> The data i have is:
> Type Vol. Cap(mAh)
> AA 1.5 2850
> AAA 1.5 1250
> P3 9.0 625
> The load takes about 3mA. The discharge curve to .8 Volts per cell
> gives the following running times:
> AA 1000 h
> AAA 500 h
> P3 370 h
> BUT the current cut-out for the 9V bat is at 7.2 which gives a running
> time of about 170 h. The current configuration is not regulated, but
> when regulated, it should be at about 7.5 volts.
> Lets say the switcher has an efficiency of 80%. How would using a
> single cell or 2 or three in series influence the running time?
> My first guess is that if i use a single bat at 80% i would get (for
> AAA) 1000mAh, but if i put two in series, with the same switching
> efficiency i still get that figure because i can only extract the same
> 1000mAh but at a higher voltage. Another aproach would be to start
> with 1250mAh at double the voltage i get half the capacity?? I don't
> think that is correct, because what matters is the total amount of
> energy used??? It's certainly very simple, but i'm confused, i don't
> know how to calculate this, the switcher confuses me.
> Any help apreciated.
> Steve Sousa
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