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Author Topic: Battery bank selection  (Read 1204 times)
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max
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« on: November 10, 2012, 11:35:53 11:35 »

Hi,

I have three 12v batteries 200ah each with independent chargers and a 1000va inverter,
i need to connect the batteries turn by turn when they discharged to feed the inverter.
Please the advise about the battery selection circuit.

Regards
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bobcat1
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« Reply #1 on: November 10, 2012, 11:55:33 11:55 »

Hi

Use LTC4412 in parallel (see data sheet for more info) to automatically select the battery for charging or discharging

All the best

Bobi
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xenix
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« Reply #2 on: November 13, 2012, 09:15:52 09:15 »

Why you are not paralleling three batteries and feeding the inverter? By this way, discharge current of the batteries is decreased, so they can discharge efficiently. You can draw more energy from the batteries.

12V 200Ah battery can supply energy to a 1000VA inverter for 1.5 hour at rated power. This discharge speed is not suitable for lead acid batteries. If you discharge a lead acid battery at 20 hours, you can get 100% of the label energy (12V10A20Hour). If you discharge at 5 hours, you can get 85% of the rated energy, and if you discharge at 2 hours, you can get 70% of the energy. So try not to discharge the batteries in less than 5 hours.
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solutions
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« Reply #3 on: November 13, 2012, 12:12:15 12:12 »

He constrained the problem by stating that each battery has its own charger. No idea why, but it's a secret, so no point in asking. In my book, three diodes and you're done.
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FTL
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« Reply #4 on: November 13, 2012, 07:12:53 19:12 »

I would think that the problem with three diodes is that they dissipate too much power. At a peak load around 80-90A feeding the 1000va inverter, there would be with 30-60A across each diode (allowing for uneven discharge from each battery). A 0.7V drop across each diode would dissipate over 60W of heat in total. This means very large diodes with big heat sinks. That heat would be about 6% of the energy being supplied by the batteries.

I would go with large relays to connect and disconnect each battery from the load. Then the question is how to switch them.

I would try a round-robin algorithm. Maybe connect each battery for 5 minutes, then go to the next one. An overlap of two batteries being connected for a few seconds during switching would make sure that the inverter gets continuous power, and limit the swithing current on the relay. A question for someone very familiar with (I'm assuming) lead-acid batteries is whether intermittent large discharges on a 33% duty cycle would would allow more than 70% of the battery capacity to be utilized.

I would use a microcontroller to implement the algorithm, but then, I'm familiar with them. You could probably come up with some scheme with a few 555 timers and a counter circuit.

If I was using a microcontroller, I would look at modifying the switching algorithm based on battery voltage. Maybe look at the battery voltage shortly after starting to discharge it, then leave it connected for 5 minutes or until its voltage drops by a few percent. That way a weak battery will be used for a shorter time and all batteries will be discharged to a similar state. Or maybe keep a battery selected until its voltage is a bit lower than the average of the other two batteries when they were switched out.

Another technique would be to regularly poll the voltage drop across a small-value resistor (i.e. the wire connecting the relay to the inverter). If that is done once a second, a reasonable value of power used can be integrated, so the algorithm could be further improved by switching batteries after a certain amount of power has been extracted from them. On a 200AH battery, maybe switch after 10AH have been used to give the battery a rest for a while.

Whether these algorithms would work well is also dependent on what the load on the inverter is like. Is it always 1000va, or mainly 100va with a few 1000va spikes, or whatever. That will help determine how to best discharge the batteries.

A final point: be careful about relays near the batteries. Non-sealed lead-acid batteries can produce hydrogen and oxygen during charging. A spark from a relay near an unventilated battery could be a very bad thing.

Edit: fix typos.
« Last Edit: November 13, 2012, 07:16:43 19:16 by FTL » Logged
sohel
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« Reply #5 on: November 13, 2012, 07:37:31 19:37 »

what about 3 phase IGBT with Driver. i am also confuse.
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Gallymimu
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« Reply #6 on: November 13, 2012, 10:37:16 22:37 »

I wouldn't use relays.  They are expensive and don't do as well at DC as they do at AC (since the arcs drawn on switch open are not self quenching like with AC).  Certainly you can get a relay that will work but I'd avoid it if a solid state solution could be found.

MOSFETs are the way to go.  High current is cheap and easy, Rds on can be VERY low for Nchannel MOSFETS.
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solutions
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« Reply #7 on: November 14, 2012, 12:50:47 00:50 »

I would think that the problem with three diodes is that they dissipate too much power. At a peak load around 80-90A feeding the 1000va inverter, there would be with 30-60A across each diode (allowing for uneven discharge from each battery). A 0.7V drop across each diode would dissipate over 60W of heat in total. This means very large diodes with big heat sinks. That heat would be about 6% of the energy being supplied by the batteries.

Again, we don't know what this is for.

In any case, using your 90A assumption, and FET (or DC contactor...BIG $$$ at that current level for a DC relay) switch at 10mOhms, I get 81W of power loss....isn't 63W in my diodes easier, a LOT cheaper, and better? Granted, breakeven is at about 7mOhm, but you still have the cost problem...getting worse.

Complicating all of this is the notion of "overlap" of charged and discharged lead acid batteries, to keep from starving the inverter, if they are disconnected by switches, vs isolated by always-on diodes. Now you can double, triple, or more, the rating of your FETs/DC relays,...

Mucho dinaro.
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Gallymimu
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« Reply #8 on: November 14, 2012, 02:00:27 02:00 »

Actually MOSFETs are more practical than you might think.

Infineon BSC009NE2LS, 100A, 25V, 0.9mOhms.  There are many others in the 30-200A range with 1mOhm or less Rdson.  Pretty amazing what they can do with the MOSFETs these days!  It's one of the reasons so many power electronics designs are moving to synchronous rectifiers rather than using flyback diodes.

Oh, and they are $3 in singles!
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solutions
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« Reply #9 on: November 14, 2012, 05:26:54 05:26 »

I see....you're using N-channel. Gorgeous for switching the LOW side. You'll need about 27V-30V above the negative terminal of the battery to get your Rdson, though (top of my head...too lazy to grab the ds) if it's high side switching.  With a dead battery, you'll smoke the gate with 27V, so now you have to fix that as well.

How's your gate voltage going to happen, given the gate sits at DC, which eliminates a bootstrapper or even a pulse xfmr? Could use P devices, but....there goes your Rdson
« Last Edit: November 14, 2012, 05:28:56 05:28 by solutions » Logged
Gallymimu
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« Reply #10 on: November 14, 2012, 05:34:32 05:34 »

Ha, good point.  Wasn't even thinking about the topology of the switch usage. 

P channels will go down to about 2mOhm, not as good but still pretty darn impressive.

Not sure about the implementation since we don't know what he is doing Smiley

yer right of course, you wouldn't want to use the N as a high side switch, even if you could get the voltage boosted.  If you wanted to though...  It wouldn't be normal to drain the battery below 10 or 11 volts so I don't think you'd be blowing the gate but if someone wanted to be safe, you'd probably use a boost voltage that rides on or is clamped to the 12V rather than a fixed 27 volts or whatever, zener clamp could fix blowing the gate as long as you didn't need a low impedance to pump a lot of charge into the gate fast.  Again moot point since N channel high side switching sucks and as you said, generating the boost voltage is probably a waste of effort/cost.  A little voltage double might not be too bad for slow switching low power gate drive.

I'm not sure you couldn't use N low side to switch the the battery out, again, depending on what he happens to be actually doing with this thing.

I am becoming more of a fan of the 30A contactors clacking on and off instead Smiley .  I wonder what kind of timeframe he needs each battery on for.

So Max, what are you actually trying to do so we aren't wasting our brain power (and your time) speculating on circuits that probably have nothing to do with what you want to accomplish?
« Last Edit: November 14, 2012, 05:52:34 05:52 by Gallymimu » Logged
solutions
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« Reply #11 on: November 14, 2012, 12:45:08 12:45 »

I used to be a fan of contactors until I realized their contact resistance sucked compared to MOSFETs, not to mention their cost.

As I said before, he has to overlap a non-diode solution if the inverter is not to be interrupted (or add some HUGE capacitors). Depending on the difference in battery charge, he could wind up welding the contacts shut or frying FETs...solved by a series resistor....darnit, there's that power loss again
 Tongue

There are much better ways to do all this, but not knowing what he's doing precludes suggesting them. I'm still curious as to why the low side is verboten
« Last Edit: November 14, 2012, 12:51:54 12:51 by solutions » Logged
chandra2sekhar2000
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« Reply #12 on: November 14, 2012, 06:48:18 18:48 »

hello ,why cant u connect all the three baterries in parallel, and all the three chargers also in parallel.u just connect the invertor as it is.and keep the wires distance in all the batteries as equal ,to get good battery life for all the three.

in a distributed solar installation.,all the output of the chargers will be connected to  a common DC Bus and the same will be connected to a Battery Bank which consists of so many batteries in series and parallel.

so i think no need to worry about connecting all the 3 batteries and 3 chargers in parallel.but all the batteries and chargers should be identical and should have equal capacities like 12V,14A charging current etc.

and also try to use equal length of wires in between the batteries and from batteries to inverter
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Toxible
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« Reply #13 on: November 25, 2012, 04:29:28 16:29 »

I've do this project few year ago but only for two bat.
How much current you need to handle (maximum)?
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