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Old 12th June 2009, 08:44 PM   #1
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Default Split Charger

Hello all, I know that various companies sell split chargers, but I was wondering if anyone had a circuit they are using that works. I'm not happy with respects to not being able to charge the leisure/auxilary battery, fully, due to losses?

If a diode is used to isolate one battery from the other, this causes a loss. Can one simply attach a smoothing cap as in a conventional supply? Or am I barking up the wrong tree?

Cheers,Fuzzymuff
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Old 12th June 2009, 09:02 PM   #2
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If you're referring to battery isolators, there are newer isolators that have negligible loss. They use FETs instead of (or in addition to) the diodes.

Even if you get a bit of loss, it doesn't mean the battery isn't getting fully charged. It may take longer to charge through the isolator but it will generally reach the fully charged state.
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Old 12th June 2009, 09:11 PM   #3
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Default Cheekyness

Hello Perry, Thanks for the reply. Got a circuit? Allways like to DIY, rather than buy.

Will try a Google.

Thanks, Fuzzymuff
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Old 12th June 2009, 09:20 PM   #4
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Default Don't Worry

Found one http://www.discovercircuits.com/PDF-...yisolator1.pdf

Thanks just the same

Fuzzymuff
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Old 12th June 2009, 09:31 PM   #5
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It could be done without the IC also. All the FETs need to conduct (efficiently) is to have their gates ~10v below the source. This could be done with another transistor or even a relay. You'd need to have a resistor pull up the gates when the turn-on transistor (or relay) was switched off to insure that the FETs switched off completely.
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Old 12th June 2009, 10:38 PM   #6
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Default Pardon?

Quote:
Originally posted by Perry Babin
It could be done without the IC also. All the FETs need to conduct (efficiently) is to have their gates ~10v below the source. This could be done with another transistor or even a relay. You'd need to have a resistor pull up the gates when the turn-on transistor (or relay) was switched off to insure that the FETs switched off completely.
I can copy and build, but to understand what you say, I'd have to have it explained???? Sorry!

Fancy some typing??

Cheers

Fuzzymuff
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Old 13th June 2009, 02:15 AM   #7
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For the P-channel FETs to conduct, you have to reach the threshold voltage. This is ~3.5v (the gate terminal has to be 3.5v below the voltage on the source terminal). At 3.5v, the FET conducts but just barely so the effective resistance of the FET (source to drain) will be relatively high. This will result in lots of power dissipation and will make the FET run hot when significant current is passed through it.

To make it operate more efficiently, the Vgs (gate to source voltage) must be greater than 3.5v. ~10v is generally a good drive voltage. Beyond 10v, there will be little improvement on efficiency. You can drive the gate with up to ~20v (Vgs) without damaging the FET.

On a 12v system, if the source is at 12v and you ground the gate, you'll have a Vgs of 12v.

The following shows basic examples. If you build it, fuse it to ~20% of the total current rating of the FETs being used until you're certain it's working properly.

http://www.bcae1.com/temp/batteryisolator.swf

If anyone finds errors in the diagrams that would prevent the circuits from working as they should, email me.

When using this type of isolator, you must have an isolator on both batteries if you want to prevent either battery from draining the other. Current can flow backwards through the FETs.

If you only have the isolator on the aux battery, the main battery (or anything connected to it) can drain the aux battery but nothing on the aux battery can drain the main starting battery unless the isolator is on.

The IRF4905 in the TO-220 case is likely the best choice at this time.


As a side note, I think the FETs are connected backwards (source and drain) in the diagram you found.
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Old 13th June 2009, 01:48 PM   #8
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Can I go ahead and build?

Cheers, Fuzzymuff
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Old 14th June 2009, 01:03 AM   #9
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You can build it but only if the cost of the parts would not be a significant loss if it doesn't work (I can't guarantee that it will work for you).

I've made some changes to the diagram so check it before you attempt to build it.
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