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nuckolls.bob(at)cox.net Guest
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Posted: Thu Mar 27, 2008 6:11 pm Post subject: Just for grins |
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Last week when I put the UltraLast cells on the hand-dandy-
battery-runner-downer I had a single loose Maxell AA cell
laying on the bench. I threw it on the tester at 1/6th
the benchmark loads I've used to compare various brands of
batteries. My benchmark testing is done at 300 mA, this
cell was loaded to 50 mA.
The differences in apparent capacity are pretty profound.
See:
http://www.aeroelectric.com/Pictures/Curves/300_vs_50mA_discharge_of_AA_cell.jpg
A reduction of load to 50 mA allowed me to get twice
the stored energy from the cells chemistry.
Bob . . .
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( . . . a long habit of not thinking )
( a thing wrong, gives it a superficial )
( appearance of being right . . . )
( )
( -Thomas Paine 1776- )
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echristley(at)nc.rr.com Guest
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Posted: Fri Mar 28, 2008 6:32 am Post subject: Just for grins |
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Robert L. Nuckolls, III wrote:
the energy from the system if I wire the batteries in parallel? Could
you run a test with 150mA load? That would simulate dual batteries
loaded in parallel vs a single battery.
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nuckolls.bob(at)cox.net Guest
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Posted: Fri Mar 28, 2008 1:40 pm Post subject: Just for grins |
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At 10:27 AM 3/28/2008 -0400, you wrote:
Quote: |
<echristley(at)nc.rr.com>
Robert L. Nuckolls, III wrote:
>
>See:
>
>http://www.aeroelectric.com/Pictures/Curves/300_vs_50mA_discharge_of_AA_cell.jpg
>A reduction of load to 50 mA allowed me to get twice
>the stored energy from the cells chemistry.
When designing the battery configuration, I will get more than twice the
energy from the system if I wire the batteries in parallel? Could you run
a test with 150mA load? That would simulate dual batteries loaded in
parallel vs a single battery.
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EXCELLENT question. Actually, doubling the batteries up in
parallel will get you MORE than twice the energy. The reasoning
here is that energy lost in the battery's internal resistance
is a Watt=I(squared)*R function.
Went to the bench an measured one of those same Maxell cells
open circuit at 1610 mV. I put a 25 ohm load across the cell
and it dropped to 1580 for a difference of 30 millivolts.
The load was 1580mV/25ohms = 63mA . Therefor the internal
resistance of this cell is 30mV/63mA or about 475 milliohms.
Okay, 475 milliohms loaded to 300 mA will drop 140 millivolts
and toss off .140v x .3A or 42 milliwats of heat. Now, let
us connect two cells in parallel. Each cell now sees 1/2
the total load or .15A x .475 ohms = 71 millivolts drop
and .071 x .15 = 10 milliwatts of heating. Considering
the two cells in parallel they now each toss off 10 milliwatts
for a TOTAL of 20 milliwatts . . . where the single cell
was tossing off twice that amount with the same load.
Paralleling cells has the effect of paralleling internal
resistance (making it go down) while doubling the chemical
mass (electrical capacity). Given that current in each
internal resistance falls by 1/2, the power dissipated in
each falls to 1/4th the original thus allowing both cells
to deliver more of their content while doubling their total
potential energy as a team. Lowering effective internal
resistance is always a good thing for return on investment
in electro-chemical energy.
Bob . . .
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