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#205575 - 03/10/12 06:56 PM I think I know the answer to this eng. question
sparkyinak Offline
Member

Registered: 07/08/07
Posts: 1286
Loc: Alaska
My comments and questions are based on the following link
drawing

This is a generic schematic I am working on involving two gensets. I based it on a similar design. I am paralelling the start batteries of each to power the generator control panel. I put diodes in to "isolate" the batteries from one another so if one battery is low, the other I thought would be protected from equalizing. The more I look at it, the more I think the batteries would still equilize through the control circuit.

So if the voltage between G1+ and G1- is 12.6 volts and for what every reason, the voltage between G2+ and G2- is lets say 10 volts, would the batteries equalize through the control circuits?
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#205584 - 03/10/12 11:07 PM Re: I think I know the answer to this eng. question [Re: sparkyinak]
gfretwell Offline

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Registered: 07/20/04
Posts: 9012
Loc: Estero,Fl,usa
You have twice as many diodes as you need. One in the + side will do it. I assume the starting and charging circuits are not shown.
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#205592 - 03/11/12 11:44 AM Re: I think I know the answer to this eng. question [Re: gfretwell]
sparkyinak Offline
Member

Registered: 07/08/07
Posts: 1286
Loc: Alaska
You are correct, the start and charge circuit is not in there. They are in the control panel of each generator. I tapped off of the power in the control panel power to run my panel. Thanx for your input
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#205604 - 03/12/12 03:26 PM Re: I think I know the answer to this eng. question [Re: sparkyinak]
geoff in UK Offline
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Registered: 12/30/02
Posts: 169
Loc: UK
There is no way current can pass backwards through your diodes to charge one battery from the other.
The control panel will draw current from the highest charged battery which over time will discharge, so in that sense the batteries will eventually equalize.

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#205605 - 03/12/12 05:27 PM Re: I think I know the answer to this eng. question [Re: sparkyinak]
Jonno Offline
Member

Registered: 05/27/06
Posts: 20
Just a thought, diodes are typically a 0.7 volt drop (convention silicone), that means your will lose 1.4v on your starting circuit, assuming 12v batteries that is 10% of your 13.8 volt nominal voltage. So yes only one diode, and consider using a contractor for isolation instead of diodes (this is how multi battery setups are typically done in the automotive world).

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#205615 - 03/13/12 12:41 AM Re: I think I know the answer to this eng. question [Re: Jonno]
sparkyinak Offline
Member

Registered: 07/08/07
Posts: 1286
Loc: Alaska
That's was my thinking. Both the positive and negative posts are paralleled between the diodes in the drawing. For example, let's say one battery is 11 volts and the other is 12. The common voltage is 12 volts.

I agree with the diodes that the will not equalize with both genset circuits are open. This is what im seeing in my drawing. The circuits are totally open when no power is needed. When a circuit is activated, the 12 volts will go from the common negative, through the circuit, to the common positive. Given that one battery is at 11 volts, it would take less power to "push" the power back in the 11 volt battery then the 12 volt battery. Does my description makes sense? What am i missing besides a chromosome or two?
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#205616 - 03/13/12 01:12 AM Re: I think I know the answer to this eng. question [Re: sparkyinak]
gfretwell Offline

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Registered: 07/20/04
Posts: 9012
Loc: Estero,Fl,usa
The diodes keep you from pushing anything into the battery. They are one way check valves going out (assuming no other paths).
You really only need one per battery.
As Dr Kirchoff taught us, amps in will always equal amps out so you can stop it up at either end.
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