From: Jonathan Kirwan
Subject: Re: 9V bat ->2AA + switcher ?
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NNTP-Posting-Date: Fri, 06 Dec 2002 01:44:22 GMT
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Date: Fri, 06 Dec 2002 01:44:22 GMT
On 4 Dec 2002 08:27:01 -0800, email@example.com (Steve Sousa)
>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
Yes, those figures are what Energizer shows in a web page table.
Here's what I have from Energizer for their Consumer/OEM data
sheets for their alkaline batteries:
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
All the above applies, assuming the battery can be used down to
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
>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
Okay, I can also calculate those figures at 3mA, from the curves
for 3mA draws:
AA = 3000 mAh / 3 mA = 1000 h, estimated
AAA = 1500 mAh / 3 mA = 500 h, estimated
P3 = 1110 mAh / 3 mA = 370 h, estimated
>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.
Okay. Put in my terms, this means that you've discovered that
things don't work when the 9V battery gets down to 7.2V and that
you'd like to design in something which will take some battery
configuration and deliver a constant 7.5V (a comfort margin of
sorts) until the battery is drained to 0.8V per cell (in order
to get most of the energy out of it.) If that's the case, then
you need 7.5V @ 3mA, or 22.5mW, continuously.
>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.
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
need to use a buck-boost switcher in order to both decrease and
increase the input voltage during operation. That topology is a
bit more complex to manage well with efficiency. You might also
invert the voltage with a simpler topology and work it that way,
If you use either an AAA or an AA battery (or two or three or
four), then you'll use a boost topology (or an inverter, again,
I suppose.) This is less complex than a buck-boost.
You are right about that what matters is the energy. This is
measured in Joules. In electronics, this is volts*amps*time (in
seconds.) A switcher will convert input energy into output work
with some efficiency.
For example, if your input voltage were 24V and your output
requirements were 6V @ 100mA, then the input would have to
supply at least 25mA (100% efficiency) to do that. It's simply:
input current = (output volts/input volts)*output current
at 100% efficiency.
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
3mA. So the life expectancy estimations will get a little more
complex. But even at 4.8V in the 9V battery case, the maximum
draw should be about 5mA or so which is nicely below the knee in
the 9V energy performance, so you'll still get good use.
Let's look first at the estimated current draw, given 100%
efficiency. For the 9V battery, let's estimate the average
voltage over life at 7V; and for a pair of AA and AAA, let's
estimate the average voltage over life at 2.2V combined. Mean
current draw will then be (7.5/7)*3 or 3.2mA on the 9V, and
(7.5/2.2)*3 or 10mA on the AA/AAA battery set. We've already
got an estimate of the mAh for the 9V, but let's look up that
figure for the AA/AAA combination... ah, 3000mAh, just as before
for 3mA draw. Sounds fine.
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
Your power consumption (7.5V @ 3mA) is the same as 22.5mJ/s.
So, assuming 100% efficiency, you should get:
9V: 28kJ/(22.5mJ/s) or about 350 hours.
AA pair: 24kJ/(22.5mJ/s) or about 300 hours.
Assuming 100%. With 80% estimate, these would be reduced to:
9V: 280 hours.
AA pair: 240 hours.
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
so. This difference will cut into your 9V battery, a bit,
making the actual life somewhat less from what you might
otherwise guess. Let's use a rough estimate for the 9V battery
system, using say 10 ohms as the average over the life:
3.2mA^2 * 10 ohms * 280 h * 3600 sec/h = about 100 Joules
So, out of some 28kJ, you'd waste 100J. Nothing much to worry
about. The AA and AAA situation would be:
10mA^2 * .75 ohm * 240 h * 3600 sec/h = about 65 Joules
Approximately the same -- around 0.3%. Nothing to worry about.
There will be many other sources of losses worse than that.
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.
Hope that helps some.