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Old 21-12-2008, 17:08   #1
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Pass Through Voltage?

Okay, so here is my silly question.
In order for a battery to accept a charge, it needs at least 13.8v.
A full battery is at approximately 12.6v.
When there is a draw/load on your batteries and your solar panels are charging, does the regulator put out the 10% extra (13.8v) required to charge the batteries or just flow through the 12.6v and save what would be a 10% waste?

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Extemp.
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Old 21-12-2008, 22:14   #2
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Quote:
Originally Posted by Extemporaneous View Post
Okay, so here is my silly question.
In order for a battery to accept a charge, it needs at least 13.8v.
A full battery is at approximately 12.6v.
When there is a draw/load on your batteries and your solar panels are charging, does the regulator put out the 10% extra (13.8v) required to charge the batteries or just flow through the 12.6v and save what would be a 10% waste?

Thanks,
Extemp.
You need to separate what voltage and current are.
The solar panel/regulator doesn't know what it is connected to or what is drawing current. If you have a simple regulator, all it does is prevent the battery voltage from going over 13.8 V if that is its set point. If the battery is charged and is therefore drawing very little current, then the regulator will only supply enough current to maintain the 13.8V. If you now turn on a load, then the regulator will supply the current for the load plus whatever current is going into the battery to maintain the 13.8V.

How it actually works is there is a circuit that senses the battery voltage. When that voltage reaches the trigger point, the solar panel is disconnected from the battery for a short time. The panel is then reconnected again to the battery. The time it now takes the voltage to reach the trigger point determines the duty cycle (% time on) of regulator. If the battery is charged, it takes very little time to turn off again, and so spends most of its time off.

Here's an example of a simple regulator.
A simple solar panel regulator

If you draw more current than the solar panel can put out then the battery voltage will drop below the 13.8 V and the panel output will be continuous. As long as this voltage is above the voltage of the battery associated with its state of charge (not a hard number, varies by battery, battery condition, etc.) then current will still go to the battery and the other load. If you draw enough current with another load, such that the voltage falls below the voltage needed to charge the battery, which is dependent on its state of charge, then the battery will be discharging, and supplying the fraction of the current required by the load that the charger is not supplying.

So on your other point, 13.8 volts is not required to charge a battery. If you look at your battery volt meter when the battery is heavily discharged and is drawing more current than your charging device can put out, you won't see 13.8 V. As long as the voltage of the charging device is higher than the battery voltage at its state of charge, the battery will charge, but it could take a very long time and might never fully charge. Higher voltage raises the charge acceptance rate of the battery, too high of a voltage causes gassing.

Here's a graph showing how voltage output varies with current output of a solar panel.
Electrical Characteristics of Solar Panels or PV Modules - AltE

John
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Old 21-12-2008, 23:08   #3
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John, that was areally good reply.
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Old 22-12-2008, 08:43   #4
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Thanks John.
I just not sure I communicated my question clearly.
What Iím trying to ascertain and/or see if it is possible is:
To utilize 12.6v (and thus more amps if an mppt is used) directly from the solar panels. This would save the wasted overvoltage require to ďCHARGEĒ the battery and use it directly. I would think the sequence of operations might go something like this:
Battery state fully charged Ė when a load is sensed the solar panels would supply the power DIRECTLY at 12.? Volts (same as batteries would put out). If the load was greater than the panels could put out the supply would be supplemented by the battery. NO charging would take place until a predetermined condition. That might be when the load has diminished to the point where the output of the panel could overcome the load and actually CHARGE the battery (at this point be worth the wastage of the Overvoltage required to charge the battery). Until that point the solar panels would just match the battery. Obviously it would require different set points and a means of direct throughput.
Iím thinking this would have to monitor the output side of the battery (load) as much or more than the acceptance side of the battery.
Having said this, perhaps that is what happens (but I donít think so), and ďI just donít get itĒ yet.
Quote:
Originally Posted by cal40john View Post
You need to separate what voltage and current are.
If you have a simple regulator, all it does is prevent the battery voltage from going over 13.8 V
if that is its set point.
If the battery is charged and is therefore drawing very little current, then the regulator will only supply enough current to
maintain the 13.8V.
If you now turn on a load, then the regulator will supply the current for the load plus whatever current is going into the battery to
maintain the 13.8V.


If you draw more current than the solar panel can put out then the battery voltage will drop below the 13.8 V and the panel output will be continuous. As long as this voltage is above the voltage of the battery
associated with its state of charge

(not a hard number, varies by battery, battery condition, etc.) then current will still go to the battery and the other load. If you draw enough current with another load, such that the voltage falls below the voltage needed to charge the battery, which is dependent on its state of charge, then the battery will be discharging, and supplying the fraction of the current required by the load that the charger is not supplying.

So on your other point, 13.8 volts is not required to charge a battery. If you look at your battery volt meter when the battery is heavily discharged and is drawing more current than your charging device can put out, you won't see 13.8 V. As long as the voltage of the charging device is higher than the battery voltage at its state of charge, the battery will charge, but it could take a very long time and might never fully charge. Higher voltage raises the charge acceptance rate of the battery, too high of a voltage causes gassing. John
I did realize that a battery will charge at a lesser voltage than 13.8v, but my assumption and/or scenario is one that starts with the battery fully charged.
P.S. Iíve done a bunch of reading, but perhaps I still donít get it. Likely just learned enough to be dangerous.
Anyway, straighten me out if you think you can (ha ha).
Best Regards,
Extemp.
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Old 22-12-2008, 09:17   #5
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Think in terms of current. If more current is going out of the battery than is going in then the battery is being drawn down. Your voltage meter is not going to be able to tell you this. Your best indicator will be your current meters and not really so much what your voltage meter is reading..although that is an indicator as well.

I have current meters for my alternator outputs, my house load and my inverter-charger either charge or load. (My inverter-charger ammeter shows negative when it is drawing and positive when it is charging.) The ammeters gives me a much clearer picture of what is happening than if I just had a voltmeter on my batteries.

Think of your batteries as a current bank. Current meters measure how much money is going in and how much money is being withdrawn. Your voltage meter is only a rough indicator of how much money is in the bank. None if it is exact because current in to a battery does not equate to current out because of heat losses during the charging process. Think of the battery as a bank that takes a sizable service fee out of your account for banking with it. You will never get the same money (current) back that you put into it.

Although not perfect, your current meters will give you a better picture of what is occurring. You still of course need your voltmeter(s)

Using mental math, you can then see how much power in watts is moving around your boat by multiplying the current times the voltage.

Did my post help any??
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Old 22-12-2008, 20:12   #6
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Quote:
Originally Posted by Extemporaneous View Post
Thanks John.
I just not sure I communicated my question clearly.
What Iím trying to ascertain and/or see if it is possible is:
To utilize 12.6v (and thus more amps if an mppt is used) directly from the solar panels. This would save the wasted overvoltage require to ďCHARGEĒ the battery and use it directly. I would think the sequence of operations might go something like this:
Battery state fully charged Ė when a load is sensed the solar panels would supply the power DIRECTLY at 12.? Volts (same as batteries would put out). If the load was greater than the panels could put out the supply would be supplemented by the battery. NO charging would take place until a predetermined condition. That might be when the load has diminished to the point where the output of the panel could overcome the load and actually CHARGE the battery (at this point be worth the wastage of the Overvoltage required to charge the battery). Until that point the solar panels would just match the battery. Obviously it would require different set points and a means of direct throughput.
Iím thinking this would have to monitor the output side of the battery (load) as much or more than the acceptance side of the battery.
Having said this, perhaps that is what happens (but I donít think so), and ďI just donít get itĒ yet.

I did realize that a battery will charge at a lesser voltage than 13.8v, but my assumption and/or scenario is one that starts with the battery fully charged.
P.S. Iíve done a bunch of reading, but perhaps I still donít get it. Likely just learned enough to be dangerous.
Anyway, straighten me out if you think you can (ha ha).
Best Regards,
Extemp.
I see what you're getting at. You're going to get very little out of this. Try thinking about this without the battery in the circuit for a moment. First, any current draw less than the full power output of the panel means that the regulator is reducing the output of the panel (disconnecting the panel part of the time). You're not using the full output of the panel, throwing away some of the power, so it doesn't matter if the voltage is a little higher or lower. The only place you would get any gain is if you were pretty close to drawing exactly at the max power rating of the panel, pretty unlikely.

John
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