> I did some more math, but seemed to have misplaced the paper right
> now. But the thought was to use AA sized NiMH batteries.
>
> Intuitively, I'm guessing my idea won't work. I'm hoping someone will
> point out where this plan will run in to difficulty:
>
> Am I correct in thinking that connecting the 10V output of the solar
> panel on the backpack to eight AA's in series (9.6V) should to charge
> them up?
>
> And if I connect leads across five of those batteries I'll have six
> volts, which looking at my Nokia phone charger and my WristPDA wall
> warts, is what they put out. So it could charge them, and I'm guessing
> any mini-USB chargeable device?
Sounds reasonable to me. It will effectively be a sort of voltage divider.
> And if I put twelve of them in series I'll have 14.4V which will be
> enough to run an inverter. Then the trick would be to switch which
> eight I was charging so that all twelve of them end up being charged
> from time to time.
Correct me if I'm wrong, but this sounds like:
Input +
Input -
| |
[+ 4 batteries -]---[+ 4 batteries -]---[+ 4 batteries -]
| |
Output + Output -
And then at some point switching to:
Input + Input -
| |
[ + 4 batteries -]---[+ 4 batteries -]---[+ 4 batteries -]
| |
Output + Output -
Here's my guess at what would happen:
Initially, all batteries are charged. The load gets connected, and all
the batteries start draining. Since the solar cell is in parallel with
one of the batteries, its current gets folded into the current rating
of that battery. The 4 batteries that are not connected to the solar
cell would be charged by being in series (through the load) with the 8
that are charged. So far so good.
I think the place that this would run into a problem is if you
disconnected the load, but left the solar cell in place. Now the
batteries that are on the charger (parallel with the solar cell) would
charge up, but the ones that are not would remain discharged. Now when
the load is next connected, there would be a voltage across the
semi-discharged cells. The batteries would draw current to charge.
Since the batteries would draw whatever they could get away with, this
might lead to heating and potential damage.
I may be wrong about this, but I think the system with all batteries
charged can be treated as three voltage sources (two in parallel and
one in series with them) and a load (the rest of the circuit). If one
set of batteries is discharged, then it becomes part of the load
instead of a voltage source. The main places I see this theory falling
down are A) my ignorance of modern battery chemistry current draw
while charging and B) the fact that it's been a long time since
Electrical Engineering 101. Is there a professional in the house?
> Questions for Abe:
> If this does work, would you be able to eliminate either your voltage
> regulator or DC DC converter?
In theory, yes, I could tap the pack at various points and get the
voltages I need. In practice, it would be off by a bit, as each sell
doesn't generate 1V increments. I could get 4.8 and 6, but getting 5V
on the nose out of 1.2V packs isn't doable.
Also, the output voltage drops as the pack discharges. My 12V boost
converter gets me 12V from 7.2 down to 5 (or 6, I forget). Below
about 5.35 volts, the 5V regulator doesn't reliably output 5V. I have
the "low battery" trip point set at 6V as a safety margin.
> What is your wearable setup comprised of? I'm interested to see your
> average power consumption and how long your wearable lasts.
Numbers coming Real Soon Now(tm), I swear. :-)
My motherboard claims to draw 2A, but I have a scan converter and a
USB keyboard drawing power off the same source. The 5V 5A regulator
still gets really hot.
These are some sites I made while working on the wearable, not really
complete documentation yet. Beware, reckless self-aggrandizement
ahead:
http://www.aculei.net/~ams/cgi-bin/wiki.cgi/WearableTechSpec
http://www.aculei.net/~ams/cgi-bin/wiki.cgi/WearableComputer
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