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Knightrous
Site Admin
Joined: 15 Jun 2004
Posts: 8511
Location: NSW
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I didn't want to create a new thread, so I'm posting my NiCD based question here.
I recieved my 40 CP2400SCR's today (Thanks Glen) and I'm just thinking about battery configuration. I had an idea, but I wanted to get some confusion settled before trying it. My idea is to make 8 x 6v packs from 40 cells. My reason for this is that I can then make 12v/18v/24v/30v/36v pack combinations by just putting 6v packs in series to make the required voltage. This would save me remaking packs when I want to change voltage.
The thing I'm questioning come back to the discussion of resistance of batteries. As Jeff F said, if you put packs in parralel, you 1/2 the resistance, but if you put them in series, you double the resistance (Correct me anyone). My question is, if I was to make a 24v pack from 4 x 6v packs, ignoring wire and plug resistances, would I have the same resistance as if it was a single 24v pack? I don't think it will be different, but I have this annoying feeling that something isn't right.
Comments, corrections and suggestion welcome and wanted! _________________ https://www.halfdonethings.com/
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Tue Feb 08, 2005 8:01 pm |
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Spockie-Tech
Site Admin
Joined: 31 May 2004
Posts: 3160
Location: Melbourne, Australia
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You're both right..
a 24v pack does have more resistance than a 12v pack, but it also has more volts to push current through that resistance.. Its resistance relative to voltage you have to think about..
[lecture mode]
[
In a 24v system, to get say 480watts of power through, you need a current flow of 20 amps (watts = volts * amps).
If you put .09 ohms of resistance in the way of that 20 amp flow (20 cells * .0045 ohms per cell - not including braid/connectors etc), then that resistance will cause a 1.8v drop (Volts = I(current) * R(esistance)).
1.8v lost * 20 amps of power is 36 watts of heat being generated in your batteries.. coincidentally, since we have 20 cells (as well as 20 amps), then thats 1.8 watts of heat per cell..
Leaving an actual 444 watts being delivered to the load and 36 watts keeping our batteries toasty warm.
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Now, at 12v, to get 480 watts of power, we need a current flow of 40 amps..
now our 10 cell pack has half the resistance (.045 ohms) of the 24 pack since we only have half the number of cells.. ,
but
we are pushing twice the current through it to get the same power.
40 amps flowing through .045 ohms gives you, yes, a 1.8v drop again. (V=I*R)
but now the current flowing through those cells (and causing that voltage drop) is double the 24v pack, so the power loss is 1.8v * 40amps = 72 watts ! Double the power loss that we had before ! Leaving only 408 watts getting through to your load..
And to make matters worse, not only are our batteries radiating double the heat they were before, but that heat is now all coming from half the amount of cells (10 instead of 20), so now instead of each cell dissipating 1.8watts of heat, they are now dissipating 7.2 watts each, or 4x as much ! MMmm. marshmallow packs anyone ?
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So to sum up.. yes, a 24v pack has double the resistance of a 12v pack, but
if
it is being called upon to deliver the same amount of power (watts not amps), then it will run much cooler and more efficienctly than a 12v pack will.
Of course if you hook up a 12v pack and a 24v pack to the *same motor*, then the 24v pack will be delivering twice as much current as the 12v pack would (and 4x the power), so you have to redo the numbers).
but if Power requirements are equal, then its better (more efficient, and cooler) to do it with a higher voltage and lower amps than a lower voltage and higher amps..
Thats why main power transmission lines run at 60Kv and more, then get converted to 240v only when it gets close to you..
[/lecture mode]
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Clear as mud now ? _________________ Great minds discuss ideas. Average minds discuss events. Small minds discuss people
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Tue Feb 08, 2005 10:53 pm |
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ffej
Experienced Roboteer
Joined: 22 Jun 2004
Posts: 595
Location: Kurrajong, NSW
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No, thats not really it, its true that the charger does overcharge them a little bit, but no where near that (batteries would be getting cooked all over the joint if it were). Most NiCd's and NiMh's (pretty much all batteries actually) require 140% of the total in to get 100% of power out, due to inefficencies during charging, Eg . . . a 2400mAh pack requires 2400 / 100 * 140 = 3360mAh in order to get 2400mAh from a pack that was close to dead flat.
Also, if the pack isnt even warm after charging, chances are its not fully charged, assuming your charging at 1C or greater. _________________ Jeff Ferrara
fb@ffej.net
ffej.net
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Sat Feb 12, 2005 6:58 pm |
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