2-Batteries

This is how I finally set up my glider--with two main battery switches.
But if you want to run this way without bouncing the logger, the two
switches should be on for the whole flight. You don't want to run one
battery down like a fuel tank, then switch when it is empty. If you do,
you will get a big current spike when you switch them both on. At best,
this wastes energy from the new battery by dumping it into the dead
one. You can also burn-out your switch, weld the contacts, or blow a
fuse.

What you want to do is run two identical batteries, bought at the same
time from the same production batch. Then always charge and run them
together. They should age together, and share the load properly this
way. You only want to use the two switches in opposite positions to
separate the batteries for testing, or if one battery fails.

wrote:
By all means a second battery should be installed in our electrically
driven modern sailplanes. After many years of quickly flipping my
3-position battery switch, and trying not to have my logger to
momentarily dropout, I have concluded that is best to use 2 single-pole
battery switches. That way one can have either or both batteries
connected at the same time.

I saw the light when Jim Hendrix brought his sailplane to Caddo Mills
for Wing Deturbulator flight testing, and it was wired like that. You
will hear much more about that amazing new invention at the coming SSA
Convention.

Thermally,

Dick Johnson

Being the incurably curious type, I decided to test this theory. I
took two known good 7Ah 12V SLA batteries and discharged one to 8 volts
(resting) with a 12V light bulb. The other battery I topped off with a
charger to 13.6 volts. I connected the two using less than two feet of
18 gauge wire and a ordinary toggle switch. Using a 60 Mhz bandwith
oscilloscope and a hall-effect type current probe I looked at the
resulting waveform when I closed the switch; a nice square edged rise
to about 3 amps, tapering down to 2.5 amps in a few seconds. Because I
didn't know what the frequency response of this current probe was, I
inserted a precision .001 ohm current shunt in line (very high
frequency response) and used the scope to watch the voltage drop across
it. The results were identical; no current spike, no inrush of current
- just a nice square edged waveform rising to about 3 amps. This simply
isn't going to weld contacts, burn out switches or blow (properly
sized) fuses. As for "wasting energy" by dumping from the good battery
into the dead battery when switching over - just do the math. Even if
the two batteries were connected for as much as 5 seconds while
switching from one to the other (two switch or "make before break"
switch arrangement), you will be using less than one thousanth of the
good battery's capacity to charge the "dead" battery.

RF

Doug Haluza wrote:
This is how I finally set up my glider--with two main battery switches.
But if you want to run this way without bouncing the logger, the two
switches should be on for the whole flight. You don't want to run one
battery down like a fuel tank, then switch when it is empty. If you do,
you will get a big current spike when you switch them both on. At best,
this wastes energy from the new battery by dumping it into the dead
one. You can also burn-out your switch, weld the contacts, or blow a
fuse.

What you want to do is run two identical batteries, bought at the same
time from the same production batch. Then always charge and run them
together. They should age together, and share the load properly this
way. You only want to use the two switches in opposite positions to
separate the batteries for testing, or if one battery fails.

wrote:
By all means a second battery should be installed in our electrically
driven modern sailplanes. After many years of quickly flipping my
3-position battery switch, and trying not to have my logger to
momentarily dropout, I have concluded that is best to use 2 single-pole
battery switches. That way one can have either or both batteries
connected at the same time.

I saw the light when Jim Hendrix brought his sailplane to Caddo Mills
for Wing Deturbulator flight testing, and it was wired like that. You
will hear much more about that amazing new invention at the coming SSA
Convention.

Thermally,

Dick Johnson

On Dec 29, 6:25 pm, "Tinwings" wrote:
Being the incurably curious type, I decided to test this theory.
-- snip --
...you will be using less than one thousanth of the good battery's
capacity to charge the "dead" battery.

Holy cow!! Blasphemy!!!!

We can't have people posting empirical evidence complete with a
complete description of the experiment messing up our continuing
propagation of "old wives' tales" in this group.

Seriously, thanks for doing the experiment and reporting the results
here. Perhaps now we can close out the power switch discussion once
and for all. Naaaah, I'm sure it will come back.

-Tom

Excellent! Beat to the punch... I was reading this thead and thinking
I'd just go measure this to quiet concerns about inrush currents.
I've.been operating my glider with 2 x 12 Ah batteries with seperate
spst master switches on each battery. I'll run one battery at a time
but both will be on for a short while when switching batteries. I've
never had a problem with switches and currents and never expected that
I would.

This really should not need an experment to prove it... (but its fun
and who would believe anything on ras without it?). While AGM batteries
have very low internal resistance when fully charged, like all lead
acid batteries the internal resistance increases as they discharge due
to the reduction in electrolyte conductivity (the sulfuric acid is
turning into water as the sulfate ions form lead sulfate on the
plates). Internal resistance might go from ~0.002 ohm fully charged to
maybe an 0.1 to an ohm or so if you really discharge things (your
mileage will vary widely). You won't find the internal resistance specs
anywhere except at full charge, but you can infer them from the
standard manufacturers discharge curves (usually voltage vs. log time
at various discharge currents). That fall off in voltage measured in
those discharge curves is actually telling a lot about the internal
resistance increasing as the battery discharges. The internal
resistance increasing inside the battery causes the external measured
voltage under load to decrease - this dominates the voltage measured
under load much more than the cell chemical potential decreasing due to
the weaker electrolyte concentration. The battery voltage is not
getting lower as much as it is getting harder to pull whatever is in
the battery out. And for our switch story, it is luckilly also harder
to push charge back into the flat battery.

So while a fully charged lead acid battery and especially (for their
size) AGM batteries can sink huge currents into a short circuit the
other battery just does not look like a short circuit if it is
discharged, and if it is not very discharged then the small voltage
differences (fractions of a volt) don't generate a large current even
with fairly small total internal resistance of both batteries.

I run my glider with two 12 Ah batteries and two seperate master
switches because I like to be able to see what is going on and I like
the redundancy and to be able to control things. I don't like the diode
idea since in normal operation you can't see if you have a weak battery
(like when you make a mistake and don't fully charge one battery - I'm
more worried about operator error (me) like that than by one of the
batteries actually being sick. I just think I'm more likely to spot a
mistake like that with the batteries not wired in parallel with diodes.

BTW in the test described below the current starting at 3 amps and
dropping to 2.5 amps is casued by surface charge -- surface chemsiry
effects of the electrolyte in the very porous surface of the plates. It
it also the reason why a "12 volt" lead acid battery measure up over 13
volts, if you burn of the surface charge you'll find the true open
circuit cell voltage is around 12.5 volts (depends slightly on battery
chemsitry and temperature). Surface charge is why a damaged old battery
can sometimes charge up over 12 volts but rapidly fall and why just
pulling a battery off charge and measuring it's open circuit voltage
without either waiting (many hours) or deliberately drawing current to
burn off the surface charge is next to useless. But it is always
amusing watching people do this at the gliderport... (see the surface
charge effect slides in the link below, they show the surface charge
being discharged and the battery voltage allowed to relax back again
prior to a proper open circuit voltage measurement to estimate the
state of charge - a measurement that even if properly made tells you
*nothing* about the actual battery capacity).

I was asked to give a talk at one of our PASCO seminars on glider
batteries. I'm not sure how useful the slides will be by themselves but
they are available here
http://www.pacificsoaring.org/articl...SCO%202006.pdf

Anyhow again thanks for the measurements


Darryl Ramm
DG-303 6DX


Tinwings wrote:
Being the incurably curious type, I decided to test this theory. I
took two known good 7Ah 12V SLA batteries and discharged one to 8 volts
(resting) with a 12V light bulb. The other battery I topped off with a
charger to 13.6 volts. I connected the two using less than two feet of
18 gauge wire and a ordinary toggle switch. Using a 60 Mhz bandwith
oscilloscope and a hall-effect type current probe I looked at the
resulting waveform when I closed the switch; a nice square edged rise
to about 3 amps, tapering down to 2.5 amps in a few seconds. Because I
didn't know what the frequency response of this current probe was, I
inserted a precision .001 ohm current shunt in line (very high
frequency response) and used the scope to watch the voltage drop across
it. The results were identical; no current spike, no inrush of current
- just a nice square edged waveform rising to about 3 amps. This simply
isn't going to weld contacts, burn out switches or blow (properly
sized) fuses. As for "wasting energy" by dumping from the good battery
into the dead battery when switching over - just do the math. Even if
the two batteries were connected for as much as 5 seconds while
switching from one to the other (two switch or "make before break"
switch arrangement), you will be using less than one thousanth of the
good battery's capacity to charge the "dead" battery.

RF

Doug Haluza wrote:
This is how I finally set up my glider--with two main battery switches.
But if you want to run this way without bouncing the logger, the two
switches should be on for the whole flight. You don't want to run one
battery down like a fuel tank, then switch when it is empty. If you do,
you will get a big current spike when you switch them both on. At best,
this wastes energy from the new battery by dumping it into the dead
one. You can also burn-out your switch, weld the contacts, or blow a
fuse.

What you want to do is run two identical batteries, bought at the same
time from the same production batch. Then always charge and run them
together. They should age together, and share the load properly this
way. You only want to use the two switches in opposite positions to
separate the batteries for testing, or if one battery fails.

wrote:
By all means a second battery should be installed in our electrically
driven modern sailplanes. After many years of quickly flipping my
3-position battery switch, and trying not to have my logger to
momentarily dropout, I have concluded that is best to use 2 single-pole
battery switches. That way one can have either or both batteries
connected at the same time.

I saw the light when Jim Hendrix brought his sailplane to Caddo Mills
for Wing Deturbulator flight testing, and it was wired like that. You
will hear much more about that amazing new invention at the coming SSA
Convention.

Thermally,

Dick Johnson

One reason to run the batteries in parallel is that the slower the rate
at which you discharge a battery, the more total power you can get from
it. I.e. if you discharge a battery at 1 amp it will not deliver as
many amp hours than if you discarge it at 0.5 amps. This is somewhat
offset by the need ( or good idea) to isolate the batteries with low
loss zener diodes. The 0.25 volt drop of the zeners means that in
effect you can only discharge you batteries to 11.25 rather than to 11
volts - which sacrifices some capacity. You have to get out the data
sheets for the diodes and batteries to see if this is a good tradeoff.

I believe that using one battery switch rather than multiple spst
switches is a bad idea because the single switch becomes a single point
of failure. Having 15 batteries is of no value if the switch breaks.

Tinwings wrote:
Being the incurably curious type, I decided to test this theory. I
took two known good 7Ah 12V SLA batteries and discharged one to 8 volts
(resting) with a 12V light bulb. The other battery I topped off with a
charger to 13.6 volts. I connected the two using less than two feet of
18 gauge wire and a ordinary toggle switch. Using a 60 Mhz bandwith
oscilloscope and a hall-effect type current probe I looked at the
resulting waveform when I closed the switch; a nice square edged rise
to about 3 amps, tapering down to 2.5 amps in a few seconds. Because I
didn't know what the frequency response of this current probe was, I
inserted a precision .001 ohm current shunt in line (very high
frequency response) and used the scope to watch the voltage drop across
it. The results were identical; no current spike, no inrush of current
- just a nice square edged waveform rising to about 3 amps. This simply
isn't going to weld contacts, burn out switches or blow (properly
sized) fuses. As for "wasting energy" by dumping from the good battery
into the dead battery when switching over - just do the math. Even if
the two batteries were connected for as much as 5 seconds while
switching from one to the other (two switch or "make before break"
switch arrangement), you will be using less than one thousanth of the
good battery's capacity to charge the "dead" battery.

If the batteries are both "good", but not charged equally, then the last
part is true. If a battery happens to go "bad" and won't take a charge
then the situation is different. Then the "bad" battery will just suck
down the extra power from the good battery and that power will not be
recovered.

Gary Emerson wrote:
Tinwings wrote:

Being the incurably curious type, I decided to test this theory. I
took two known good 7Ah 12V SLA batteries and discharged one to 8 volts
(resting) with a 12V light bulb. The other battery I topped off with a
charger to 13.6 volts. I connected the two using less than two feet of
18 gauge wire and a ordinary toggle switch. Using a 60 Mhz bandwith
oscilloscope and a hall-effect type current probe I looked at the
resulting waveform when I closed the switch; a nice square edged rise
to about 3 amps, tapering down to 2.5 amps in a few seconds. Because I
didn't know what the frequency response of this current probe was, I
inserted a precision .001 ohm current shunt in line (very high
frequency response) and used the scope to watch the voltage drop across
it. The results were identical; no current spike, no inrush of current
- just a nice square edged waveform rising to about 3 amps. This simply
isn't going to weld contacts, burn out switches or blow (properly
sized) fuses. As for "wasting energy" by dumping from the good battery
into the dead battery when switching over - just do the math. Even if
the two batteries were connected for as much as 5 seconds while
switching from one to the other (two switch or "make before break"
switch arrangement), you will be using less than one thousanth of the
good battery's capacity to charge the "dead" battery.


If the batteries are both "good", but not charged equally, then the last
part is true. If a battery happens to go "bad" and won't take a charge
then the situation is different. Then the "bad" battery will just suck
down the extra power from the good battery and that power will not be
recovered.
Sorry, didn't read your post fully. You were only talking about the
very short period when the batteries are still connected while switching
from A to B. The nice thing about the diode setup is you'll never have
to fiddle with a battery switch again.

The previous poster was making measuremnts that disproved the argument
that there is a loss while temporarilly having both masters on while
switching batteries.Whichever way you slice that it is just not an
issue that causes any significant power loss, It is totally
insignificant, you lose much more power though ohmic loses in fuses
and circuit breakers.

Of course that relies on the operator not to accidently leave two
master switches both on all the time (without taking the diode
approach).

Darryl

Gary Emerson wrote:
Tinwings wrote:
Being the incurably curious type, I decided to test this theory. I
took two known good 7Ah 12V SLA batteries and discharged one to 8 volts
(resting) with a 12V light bulb. The other battery I topped off with a
charger to 13.6 volts. I connected the two using less than two feet of
18 gauge wire and a ordinary toggle switch. Using a 60 Mhz bandwith
oscilloscope and a hall-effect type current probe I looked at the
resulting waveform when I closed the switch; a nice square edged rise
to about 3 amps, tapering down to 2.5 amps in a few seconds. Because I
didn't know what the frequency response of this current probe was, I
inserted a precision .001 ohm current shunt in line (very high
frequency response) and used the scope to watch the voltage drop across
it. The results were identical; no current spike, no inrush of current
- just a nice square edged waveform rising to about 3 amps. This simply
isn't going to weld contacts, burn out switches or blow (properly
sized) fuses. As for "wasting energy" by dumping from the good battery
into the dead battery when switching over - just do the math. Even if
the two batteries were connected for as much as 5 seconds while
switching from one to the other (two switch or "make before break"
switch arrangement), you will be using less than one thousanth of the
good battery's capacity to charge the "dead" battery.

If the batteries are both "good", but not charged equally, then the last
part is true. If a battery happens to go "bad" and won't take a charge
then the situation is different. Then the "bad" battery will just suck
down the extra power from the good battery and that power will not be
recovered.