From: email@example.com (Steve Sousa)
Subject: Re: 9V bat ->2AA + switcher ?
Date: 6 Dec 2002 16:24:36 -0800
NNTP-Posting-Date: 7 Dec 2002 00:24:37 GMT
I'd like to thank everyone for helping me out. It easy once you see
the solution. Everyone explained the concept i needed, Mr. Vlad also
mentioned one of the key issues: getting the most energy for $ in
I have some points i would apreciate if Mr. Kirwan could explain.
(that's why i top-posted).
Thanks again to the others.
Now the questions:
Jonathan Kirwan wrote in message news:...
> Yes, those figures are what Energizer shows in a web page table.
That's were i got them.
> Type Volt | Name Rated at 25mA Est. curve at 3mA
> AA 1.5V | E91 2850 mA-h 3000 mA-h
> AAA 1.5V | E92 1250 mA-h 1500 mA-h
> 9V 9.0V | 522 600 mA-h 1110 mA-h
> 0.8V per cell. (Past 6mA, the 9V alkaline runs on a different
> slope and you get less total energy from it, which is why you
> see such a difference in the total for the 9V, between 3mA and
> 25mA draws.)
Were did you get that "est. curve"? There are no capacity/load curves
on the datasheets i got??
> >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
I got those from the curves on the ds, ok.
> Okay, I can also calculate those figures at 3mA, from the curves
> for 3mA draws:
> If you use a 9V battery and if that input is going to vary
> between 9V and 4.8V while providing 7.5V on the output, you'll
This was not the case i wanted to analise, the comparison should be
between the case of an unregulated 9V supply, that "expires" at 7.2V
Vs. some external bats plus switcher, running those bats to 0.8 V.
> input current = (output volts/input volts)*output current
> at 100% efficiency.
or input current = (output volts/input volts)*output current *1/eff
> The current draw on the battery, in any of these switcher
> topologies, will not be constant even with a constant load. As
> the battery voltage decreases the switcher will need to pull
> more current in order to maintain the 7.5V output at a constant
Yes, i wanted to get an estimate by calculating the area under that
constant power curve, but it would not be really representative of
this case. There were no other helpfull curves.
> What's the available energy?
> 9V energy = 7V * (1110mAh / 1000) * 3600 sec/h = 28 kiloJoules
> AA pair = 2.2V * (3000mAh / 1000) * 3600 sec/h = 24 kiloJoules
It seems ilogical that a pair of AA would have less available energy
than a 9V bat, if we consider their volume...Assuming that the 9V bat
is made out of six AAAA (1.5V, 625mAh) You can easily find some
relationship between their rated capacities. Like Mr Bitbyter said: "A
9V battery consists of six small 1.5V cells with roughly the capacity
off half a AAA", this gives us a PP3=3*AAA=1.52*AA, so 2*AA should be
roughly 1.31*PP3 bat. I think it is reasonable to expect a larger mass
of bats to have a larger amount of energy available?...
> Now, let's factor in the battery system's internal resistance.
> The internal resistance of the 9V is always more than 2 ohms,
> however, rising to over 10 ohms near end of life. In the case
> of an AA battery, this varies from say 0.15 ohms to an ohm or
Could you give me a pointer to those figures please?
> So the decision is more about the complexity of the converter
> and its efficiency at your load current and what you think is
> important for the battery type. For example, AA's may be a
> little more widely available around the world than 9V. Or the
> 9V may have a less expensive connector for your needs.
In this case i was hoping to reduce the expense with batteries. You
can get 4 AA or AAA for the price of one PP3, so if at least the AA
worked, i could get more economical running times. In any case, I
never expected the 9V bat to have so much available energy, maybe
because i'm used to using rechargeables, that have a much lower
capacity. I still think 2*AA's must have more energy than 3*AAA's...
> Hope that helps some.
It does help a lot, thank you.