From: Jonathan Kirwan
Subject: Re: 9V bat ->2AA + switcher ?
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NNTP-Posting-Date: Sat, 07 Dec 2002 01:50:38 GMT
Organization: AT&T Broadband
Date: Sat, 07 Dec 2002 01:50:38 GMT
On 6 Dec 2002 16:24:36 -0800, email@example.com (Steve Sousa)
>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??
On the data sheets, look at the curves in the chart which is
titled, "Constant Current Discharge." I selected the 3mA line
and simply looked upwards until I found the black line intercept
(black for 0.8V discharge level.) If you multiply the hours
shown there by the (mA) rating line you used, then you get the
mA-h, right? So if I read out 370 hours for 3mA, then the mA-h
result is 370 * 3 or 1110 mA-h. Which is what I showed above.
(The rated value for the 9V is actually 595mA-h, I just rounded
it in the chart. The estimations were simply taken from that
curve, though. However, if you eyeball the 25mA line on that
chart and the black line, you'll see it's just what they say it
is there. So the rated value is a "typical" value, not a worst
>> >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. Same place that I did, then. I just did the simple
multiplications to get the mA-h figure.
>> 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.
Yes, but isn't that just what I said?? If you use an external
battery with a switcher, the switcher will "see" the battery
voltage go down from 9V to 4.8V, once each of the 6 cells in the
9V battery drop to their 0.8V level: 6 * 0.8V = 4.8V. When you
look on the 9V battery charts and they talk about discharging to
0.8V per cell, they are talking about 4.8V. The 9V battery
still uses the same chemistry as the AA and AAA, so the voltage
is still 1.5V per cell. Consequently, 9V/1.5V = 6 cells.
>> input current = (output volts/input volts)*output current
>> at 100% efficiency.
>or input current = (output volts/input volts)*output current *1/eff
Yes, I figured I didn't need to extend the equation with the
>> 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.
Well, you can at least get a kind of bracket around it, yes?
Obviously, if you are only getting 0.8V per cell at the end
point, then the current draw will be at a maximum exactly when
it is more difficult (the battery's internal resistance is at a
maximum.) That will set your worst case. The best case will be
with a fresh battery with nearly 1.55V per cell (call it 1.5V.)
So, in the case of the 9V battery, you have two situations to
bracket it: 9V and 4.8V. You know you need 3mA at 7.5V on the
output. So, the current draw will be between (7.5/9)*3 mA and
(7.5/4.8)*3 mA, which is about 2.5mA to 4.7mA.
If you look closely at those discharge curves, you'll see that
there is roughly one early slope, a middle slope, and an ending
slope. The way I eyeball the curves for the 9V battery, I'd
select the "Radio" example as the lighter load case and estimate
a median voltage from it of about 7V or maybe a little higher.
That means that your median current will be about 3.2mA.
Nicely, that range from 2.5mA to 4.7mA, with a median of about
3.2mA, is on the good side black line slope found in the
constant current discharge chart, where the first knee appears
at about 6mA. That means you aren't overtaxing the 9V battery
with your needs. Good thing.
>> 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?...
Hmm. Let's double check my numbers, then. Surprises me, too.
We agree on this, yes?
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
So, let's take those estimated mA-h figures @ 3mA. We now need
to have an average voltage over the discharge cycle. Being
slightly conservative, looking at the E91, E92, and 522 curves,
I get something about like:
However, 1.15V * 6 is 6.9V. That's close to that 7.0V figure I
got by eyeballing the 9V curve, so let's just stick with 1.15V
per cell, on average. This means:
Type Volt | Name MeanV mC-hr/sec Total Energy
AA 1.5V | E91 1.15V 3000 mA-h 12,420 J
AAA 1.5V | E92 1.15V 1500 mA-h 6,210 J
9V 9.0V | 522 6.90V 1110 mA-h 27,570 J
That's just one way of doing it. I suppose you could do a
piece-wise integration under the curve, after typing in data, to
get some kind of 'better' estimate.
There's one big error I can think of in this method. The curves
show the voltage at a constant load. In actual fact, the
switcher will present a varying load, pulling more current
towards the end-of-life. So the voltage decline will be more
rapid than they show, suggesting that perhaps the average
voltage figure I used might be a bit high. But I suspect that
since all their chemistries are the same, that if you adjust the
1.15V to 1.05V, for example, you'll have to do that for all
three. So although the total energy will be smaller, it will be
proportionately smaller for all three.
In other words, the comparison probably remains valid, even if
the accuracy might not be as good as it could be.
So it really does look like the 9V stack packs more energy than
two AA batteries. Frankly, that surprises me, too. But unless
you can spot my error, that's where I'm stuck.
>> 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?
I just looked at the data sheets, under the "Internal
Resistance" curves. Look at the E91 and 522 data sheets for it.
>> 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...
Well, 2 AAs have the energy of 4 AAAs. So yes, they have more.
>> Hope that helps some.
>It does help a lot, thank you.
No problem. I only hope I didn't miss some important detail.