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Link Posted: 10/13/2004 4:55:53 PM EDT
[#1]

Quoted:
Ok, not trying to hijack the thread but I have a quick question about a statement on pg 1.  My battery is about to give out on my UPS for my computer.  Not fitting into the UPS housing aside, can I use  a deep cycle marine battery to replace it for a longer run time??



Yes. A typical UPS has 10 or 20 amp-hour gel-cells, which might be good for 15 minutes of run-time at full load. A typical size 27 RV/Marine deep-cycle battery has around 110 amp-hours, so your run-time will be increased 5 to 10 times.

You'll need some sort of container to put the batteries in (non-gel batteries are messy, and require occasional watering). An Igloo ice chest works fine.

dablues, with all due respect, you are mistaken about installing diodes in parallel-connected batteries. Nobody (and I mean NOBODY!) does it in the RV/Marine or Alternate-energy industries. Open the battery compartment of any motorhome, yacht or home solar power installation, and you'll find parallel-connected batteries being used with no problems whatsoever.

If your batteries died prematurely, it was almost certainly due to some other cause (i.e., improper voltage setting during  the recharge or maintenance phases, or lack of water).
Link Posted: 10/14/2004 3:00:37 AM EDT
[#2]
While theoretically all you need to do is "jumper" from the original battery connectors to the new battery, I would think some kind of current limiting device might not be a bad idea.  The kind of current you can generate out of the big battery compared to what  the UPS is designed to handle may lead to fried UPS.  It probably could be as simple as an automotive grade 12 volt  wired in series in both leads.

This is a method used in model railroading (and I personally have never done it but have seen it done) to prevent rail switch points from welding themselves to rails in the event of a short or similar  power glitch.  I think the way it works and am more than willing to be edumacated on this, since I am going from memory and not a reference (although www.wiringfordcc.com/track.htm#5-4%20Bulb covers it) is that at low current there is not much resistance in the bulb but as the current increases to "hazardous levels" like welding a wrench with your car battery,  the bulb lights up and the resistance goes way up.  YMMV

Link Posted: 10/14/2004 4:04:35 AM EDT
[#3]
Link Posted: 10/14/2004 9:07:56 AM EDT
[#4]

Quoted:

Although a deep cycle battery has a heck of alot better amp hours, they tend to sacrafice voltage for amperage, current.  A 10 volt alarm should give you maximum use of a deep cycle, however if you are powering 12vdc sensative electronics off the same battery sources it may create issues.  In this case, where I needed to maintain 12vdc, I have gotten better run times out of large cranking batteries than deep cycles.  Otherwise, powering motors, lights, and such the deep cycle is far superior to the cranking.  I guess I'm saying you should know what you plan to run off your system before you plan it.  
Tj




All batteries, not just deep cycles, will show a drop in V as load increases and as the battery's energy depletes.  TJ, I think you had a crappy deep cycle battery to have a vehicle battery out last it in run time.

FYI, marine batteries are not really deep cycle batterys, they are more of a compromise between a cracking battery and a deep cycle.  They are more designed for the purpose your using them for however, you can shop for batteries truely designed for the purpose.

Allot of electronics and radios will use a linear regulator or switching converter to drop the 12V to a lower voltage that the electronics can use.  This helps you in the case of a loaded battery.

One solution is to replace your 12V marine batteries with 14V units.
Another solution is to simply add another battery in parallel.

A third solution is to add a switching converter that can operate in buck and boost modes. The switching supply will maintain a constant 12V no matter where the battery V is at as long as the battery can supply the current (remember Pin=Pout then factor in efficiency).
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