Lasted edited by Andrew Munsey, updated on June 15, 2016 at 1:18 am.
Virgin 12-volt lead-acid batteries increase in voltage under load for a while after an initial rapid drop, before they gradually drop in voltage. Natural occurrence could be mistaken for "over unity" if a person is not aware of it.
Jan. 30, 2005
This is probably something that people seasoned in batteries know like the back of their hand. I did not.
About a month ago I was running a control experiment on two new 12-V NAPA tractor batteries. I just hooked them up to a light bulb and recorded volts-amps over time.
(ran Dec. 31, 2004 - Jan. 1, 2005)
The first bulb was rated 12V, 10W
The "before" voltage was 12.61v.
After I connected the bulb, the voltage dropped rapidly down to 12.23v (0.865 amps) over two minutes but then it began to climb gradually up to 12.34v (0.863 amps) over the next three hours. It may have gone up for longer than that. I don't know because I retired to bed. When I woke in the morning, the voltage was down to 12.16v.
(ran Jan. 1-2, 2005)
On the second brand-new NAPA 12-V tractor battery, I plugged in a 14-V, 200mA bulb, and saw the same phenomenon.
The before voltage was 12.66v. When the bulb was plugged in, the voltage dropped quickly to 12.46v (0.207 amps) after 11 minutes. The voltage then climbed from there, peaking at 12.52v (0.207 amps) in 2.5 hours, when it began gradually dropping.
The point is that the phenomenon that I've seen quite often during a Directory:Bedini SG:Replications:PES:Sterling Allan:Data of battery voltage climbing after an initial drop is 'not unique to the Bedini_SG system', but is a characteristic of lead-acid batteries.
Something to bear in mind as you tinker with batteries.
It has nothing to do with radiant energy.
Sent: Monday, January 31, 2005 6:57 AM
Subject: [Bedini_SG] Re: virgin battery voltage rises under load
Good post, and very true. The reason the voltage drops when the
battery is being discharged is due to internal resistance i.e., the
voltage across the batteries internal resistance works against the
battery's polarity. The opposite happens when the battery is being
charged i.e., the voltage across the batteries internal resistance
works with the batteries polarity. I posted a formula on this some
time ago on how to calculate the voltage changes.
Also, as the battery warms up due to current, the internal resistance
will decrease and so the voltage across the battery will increase
while under load. This depends on the type of materials. The
resistance of some materials increase and some decrease with an
increase in temperature. So it's possible that some batteries may
even decrease in voltage when warmed up.
I'm sure there are other minor forces involved though.
Sent: Monday, January 31, 2005 2:21 AM
Subject: [sebcar] Re: virgin battery voltage rises under load
This phenomenom can occur generally with all cells depending on
the load placed on them, their initial temperature and their amp/hour
capacity. If the load is not too high but not too low (say from C to
C/5 depending on the cell technology) their voltage will initially
drop based on the current drawn and level off, then exhibit internal
increase in temperature which, depending on their initial temperature
may decrease their internal resistance and show a slight increase in
voltage on the output as they warm up. This increase indicates a
slight increase in efficiency as less power is lost in thermal
emissions. The overall capacity may even increase, but this merely
indicates that the cell is behaving more efficiently and loosing less
energy internally in heat. The contrary is also true, if the cell is
too hot it's unloaded/loaded voltage will drop and if cooled it will
rise. There are graphs by the various manufacturers stating this
clearly (Power transfer Vs Temp).
There is also a phenomenon called "decrystalization". This can
happen in brand new or inactive old cells. This can be seen placing a
light load (C/20) on a cell and watching the voltage drop. The
voltage will not drop evenly and jump up and down while discharging.
This is because in new cells or very old inactive cells (NiCd, NimH)
there are clusters of crystalized electrolyte "dipoles" which
constantly change the cell's internal resistance. In new cells they
need to be "equalized" and broken down and this is accomplished by
charging and discharging the cell for a few times after which the
cell will have reached full stated capacity. There are also
techniques used to decrease even more the internal resistance of
cells (in NiCds) which are dangerous and can destroy or explode a
cell but have shown that they are effective under heavy discharge
conditions (20-40C, called "pushing").