Two battery questions - K2 portable operations

[Doug Forman]

I'm assembling a portable kit in which I can fit two small sealed
lead-acid batteries.

For my specific size requirements, I have the option of putting (2) 12v
(3.0Ah) batteries in parallel, or (2) 6v (3.0Ah) batteries in series to
obtain the 12vdc I need.  (the specific batteries I am considering are
the PowerSonic PS-1230 - Mouser # 547-PS-1230 or the PowerSonic PS-630 -
Mouser # 547-PS-630)

It seems to me that with [ 2 x 6v (3.0Ah) = 12v (3.0Ah) ] while [ 2 x
12v (3.0Ah) = 12v (6.0Ah) ].  Maybe the simplicity of this makes me
question this assumption, or maybe it's been too long since I did these
kinds of calcs  ;-)

Could someone please confirm or deny my thoughts here?

Also, I'm trying to calculate (roughly) the useful battery capacity
based on considering the minimum current draw from a rig, and the
maximum current draw from the same rig.  For example, given the battery
configurations above, and considering the K2 approximate minimum current
draw of 150mA and approximate maximum current draw of 2A (at 10w), what
is the potential operating time ranges available from the batteries?

Thanks in advance & 73,

Doug  N7BNT

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[N2EY]

In a message dated 10/16/04 2:15:31 AM Eastern Daylight Time,
[hidden email] writes:


> It seems to me that with [ 2 x 6v (3.0Ah) = 12v (3.0Ah) ] while [ 2 x
> 12v (3.0Ah) = 12v (6.0Ah) ].  


That's correct. You'll have twice the capacity. But for the same size and
weight, it should be possible to find two 6V 6AH batteries to put in series.

>
>
> Also, I'm trying to calculate (roughly) the useful battery capacity
> based on considering the minimum current draw from a rig, and the
> maximum current draw from the same rig.  For example, given the battery
> configurations above, and considering the K2 approximate minimum current
> draw of 150mA and approximate maximum current draw of 2A (at 10w), what
> is the potential operating time ranges available from the batteries?

That depends entirely on how much time (percentagewise) is spent at each
condition.

Here's how to make the calculation:

First off, I'll assume CW. Analysis of typical plain-language Morse Code has
shown that the key-down time in a typical transmission is 44%. (derived from
the standard word "PARIS") In the case of a QSK transmitter like the K2, it
means that *while you are transmitting*, the rig draws 2A 44% of the time and 150
mA 56% of the time. The "key down" time is only 44% of the transmit time.

But how much of the time are you transmitting?

Unless you're sending bulletins, a lot of messages, or calling CQ a lot, you
probably spend at least 60% of the time listening and 40% transmitting. So the
actual keydown time is 40% times 44% (.4 x .44 = .176) = 17.6%. To allow for
tuneup and such, let's say 20%.

So the K2, when operated in CW under the conditions assumed, draws 150 mA 80%
of the time and 2 A  20% of the time. So in a typical hour of operation, the
K2 will use up:

.15 x .8 = .12 AH receiving
2.0 x .2 = .4   AH transmitting

.4 + .12 = .52 AH total

Using a 6 AH battery, you could get up to about 12 hours operating time as
described above.

However, such calculations are very approximate. For one thing, they assume
that your battery is fully charged at the start and can deliver name-plate
capacity. Also assumes that your keydown time is as calculated. Less keydown time
produces a lot more operating time. For example, if you listen more and send
less, the operating time increases dramatically. Look at what happens if the
operating is such that the rig then draws 150 mA 90% of the time and 2 A  10% of
the time. So in a typical hour of operation, the K2 will use up:

.15 x .9 = .135 AH receiving
2.0 x .1 = .2   AH transmitting

.2 + .135 = .335 AH total

Using a 6 AH battery, you could get up to about 18 hours operating time at
the lower duty cycle.

Note also that battery capacity isn't linear. Lead-acid battery capacity is
specified at a certain current demand level. At lower levels it increases, and
at higer levels it decreases. If a given lead-acid battery can deliver 2 A for
8 hours, (16 AH at the "8 hour rating"), do not assume the same battery can
deliver 8 A for 2 hours - you'll be disappointed, because it will run out of
juice before then. OTOH, at 0.2 A it will probably last longer than 80 hours,
indicating a capacity greater than 16 AH.

Last time I used my K2 on FD (all CW), a single "7 AH" battery lasted more
than 12 hours and was still going strong.

73 de Jim, N2EY



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[Ed Worst]

Hi Doug,

You're right: 2 x 12 is better than 2 x 6 (because you get twice the current
capacity from the two 12 packs).  But a caution... you can't just wire the
battery packs in parallel.  If one battery has even a slightly different
voltage than the other, the higher voltage battery will discharge into the
lower voltage battery which could be a disaster.  I've seen circuits for
putting batteries in parallel (they're called 'battery isolators'); there
might even be one in the handbook or maybe someone on the reflector can
point you to one.  Otherwise a Google search would certainly find it for
you.  It could be as simple as a pair of Schottky Barrier Diodes (they have
a lower voltage drop than normal silicon diodes).  I seem to remember seeing
a circuit in QST lately (meaning 'in the past 5 years or so').  An article
search on arrl.net might find something.

As far as calculating battery life, it seems straightforward, but you have
to make a lot of assumptions to do it.  For example:  if you assume you
listen (0.15A) 90% of the time, and transmit (2A) 10% of the time, your
average current drain is (0.15 + 2 x 0.1)A = 0.35A per hour.  At 0.35A/hr, a
3AH battery would last about 8.5hr.  A pair of batteries would last about
17hr.  The big assumption in that example is the 10% transmit time.  If
you're using CW, your average current draw is much less than the "key-down"
current because you're only drawing 2A during the key-down times, and 0.15A
the rest of the time.  If you're on SSB, you have a similar situation, only
with voice peaks instead of key-down.  So that 17hr is a really conservative
estimate, and could be off by a factor of 2.  In addition to the
calculations, you should do a "real-world test".  I went through the same
calculation scenario with my little K1, concerned that a 2.5A gel cel
wouldn't be enough.  The actual usage however showed that I came home from
my camping trip with a lot of capacity to spare, plus an extra gel cel that
never got used.

Hope this helps.

73,
ed - k9ew


----- Original Message -----
From: "Doug Forman" <[hidden email]>
To: <[hidden email]>
Sent: Saturday, October 16, 2004 1:14 AM
Subject: [Elecraft] Two battery questions - K2 portable operations

> I'm assembling a portable kit in which I can fit two small sealed
> lead-acid batteries.
>
> For my specific size requirements, I have the option of putting (2) 12v
> (3.0Ah) batteries in parallel, or (2) 6v (3.0Ah) batteries in series to
> obtain the 12vdc I need.  (the specific batteries I am considering are
> the PowerSonic PS-1230 - Mouser # 547-PS-1230 or the PowerSonic PS-630 -
> Mouser # 547-PS-630)
>
> It seems to me that with [ 2 x 6v (3.0Ah) = 12v (3.0Ah) ] while [ 2 x
> 12v (3.0Ah) = 12v (6.0Ah) ].  Maybe the simplicity of this makes me
> question this assumption, or maybe it's been too long since I did these
> kinds of calcs  ;-)
>
> Could someone please confirm or deny my thoughts here?
>
> Also, I'm trying to calculate (roughly) the useful battery capacity
> based on considering the minimum current draw from a rig, and the
> maximum current draw from the same rig.  For example, given the battery
> configurations above, and considering the K2 approximate minimum current
> draw of 150mA and approximate maximum current draw of 2A (at 10w), what
> is the potential operating time ranges available from the batteries?
>
> Thanks in advance & 73,
>
> Doug  N7BNT
>
> _______________________________________________
> Elecraft mailing list
> Post to: [hidden email]
> You must be a subscriber to post to the list.
> Subscriber Info (Addr. Change, sub, unsub etc.):

http://mailman.qth.net/mailman/listinfo/elecraft
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> Elecraft web page: http://www.elecraft.com
>


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[G3VVT]

In practice it is possible to use 12V batteries in parallel without  any
problems apart from a possible one of charging where one battery could hog  the
charge current at the disadvantage of the other.
 
Had number of 48V solar powered communication systems at my place  of work to
maintain for many years that had banks of batteries in  parallel. One of
5400AH capacity used 4 series banks of 6 V cells in  parallel and another of
7500AH capacity had 2 series banks of extremely large 2V  batteries in parallel.
These get over the problem of current sharing when  charging by applying a time
limited equalise charge on a daily basis when  the normal charging cycle was
complete (can be 120A on a clear day with the  larger system!). The only
concession to separating the banks were that isolation  switches fitted to each
separate bank of batteries for maintenance  purposes.
 
The problem that does happen in the end is that as the batteries reach the  
end of their life they can either go high internal impedance which effectively  
removes that particular battery out of the chain and drops the system  
capacity or internally S/C on individual cells which can drag down the whole  
battery bank voltage. However this is normally after much usage. The better you  
maintain the batteries by careful charging and not over discharging, the less  
likely this will happen. Telecom sites normally have a low voltage  disconnect
(LVD) to protect the batteries from being over discharged. With a K2  or
similar it would be a matter of monitoring the battery voltage to ensure that  it
does not fall below 1.75V per cell or 10.5V on a 12V battery.
 
Even with placing batteries in parallel not being a problem, it would  
probably be better to go for 2 X 6V batteries in series particularly if a 6AH  
version could be obtained with the ease in charging with this  configuration.
 
If there is any interest by anybody using 12V battery power, I have made up  
a simple, small LVD circuit with a relay, transistor, zener diode(s) and a few
 other bits and pieces on strip board that removes the battery from  circuit
when the 12V battery voltage falls to 10.5V. Have fitted them to the two  
repeaters I maintain here in the UK to protect the back up  batteries from being
over discharged under power failure. Have the scanned  schematic available for
e-mail if anybody wants to build such an  item.
 
Bob, G3VVT
K2 #4168
IRLP node 5140
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[Doug Forman]

Thanks to everyone who responded publicly and privately regarding my K2
portable battery questions.  Very helpful information was shared and I
certainly learned more about using SLA batteries generally, and in
combination with each other too.

Here are some of the suggestions gleaned from multiple emails:

1.  In a perfect world, it's not advisable to connect two identical SLA
batteries in parallel  (since if one battery is lower than the other,
the high one will discharge into the lower one in an attempt to charge
it and equalize voltages.)  Instead, use a switch, or a battery isolator.

2.  Don't let the battery voltage drop below approx 10.5v on a 12v
battery to prevent damage to the battery.

3.  Keep batteries charged using a trickle charger when not in use (I
use an A&A Engineering "Smart Battery Charger")

4.  Consder the relative weight of the battery for portable ops.

5.  Don't attempt to charge a SLA battery directly from a solar panel
without a charge controller.  (found a 10/2001 QST article featuring the
SunLogic Micro M+ Charge Controller that looks appealing!)

6.  Some ARES folk like 17Ah SLA's for portable ops, commenting that
they "are a great size for portable backpack operation".
(www.eham.net/articles/2228)

Thanks to everyone for responding!

73,

Doug - N7BNT






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