Battery Voltage in the Morning

[ramblinrod]

Originally Posted by sailorboy1

you're the only that thinks so, so ........................

It isn't that I think so, I know so. If a charging voltage is present it will most definitely make the battery voltage read higher than if charge voltage was not present. This is how a battery gets charged.
Marine Solar Charging System Dealer and Installer

[boatbod]

Originally Posted by ramblinrod

It isn't that I think so, I know so. If a charging voltage is present it will most definitely make the battery voltage read higher than if charge voltage was not present. This is how a battery gets charged.
Marine Solar Charging System Dealer and Installer

True, but in the OP's case the solar wasn't providing a charging voltage, merely less of a discharge.

As mentioned by others, you can only make accurate assessment of a battery's condition based on voltage when it is well rested and free of either charge or discharge.

[donradcliffe]

In order to answer the OP's question, I'd need to know what battery technology he is using. AGM's and Gels usually run a bit higher voltage than lead acid.

[SV THIRD DAY]

Originally Posted by sailorboy1

So the question is whether the supplied solar is raising the voltage even though it isn't meeting the demand. Or whether an extra couple amps discharge really effect the voltage reading that much (.1-.2).

Am I the only bozo here thinking to just disconnect the solar and see what the voltage on the battery system does, then play a little towel over the solar panel game to first be under, then match and then be over the current draw from the solar? But the truth is the numbers you are working with are so plus and minus that the data you are taking itself could be within the MOE. Not to mention how the issue of sulfation on batteries can change the SOC to Voltage relationships.

Theoreticalis nice, but a real life test always trumps chat room opinions. Of course no offense to those purveyors of opinions...that's half the entertainment...

[smac999]

Originally Posted by ramblinrod

It isn't that I think so, I know so. If a charging voltage is present it will most definitely make the battery voltage read higher than if charge voltage was not present. This is how a battery gets charged.
Marine Solar Charging System Dealer and Installer

only if the charge source out powers the loads.

he has a 2 amp charge source and a 3a load. net -1a. 12.6v sounds normal to me.

it's the exact same outcome as no solar and a 1a load. this voltage would be the same.

or if he had 50a of solar coming in and 51a of load. he'd still be at 12.6v. and not charging.

if he had 52a of solar he'd start charging and the voltage would very slowy start going up.

[Maine Sail]

Originally Posted by sailorboy1

So you don't feel that the solar supplying power into the "system" should affect the battery voltage, since the battery was still discharging?

I don't feel it should and that 12.6V with 30 AH from overnight seems pretty good.

All the battery knows is that it is discharging with a -1A load. Your solar could be supplying 121A but if you are using 122A the battery still sees the same 1A load...

If you turned off solar and the battery saw a 3A load the voltage would be slightly lower than with a 1A load. The only thing that matters is the net load on the batteries. Anything less than 0.00A and the battery is physically discharging regardless of solar output. Anything more than 0.00A and the battery begins charging. The rate of voltage decay during discharge or voltage rise during charging is - or + current dependent...

[ramblinrod]

Originally Posted by boatbod

True, but in the OP's case the solar wasn't providing a charging voltage, merely less of a discharge.

As mentioned by others, you can only make accurate assessment of a battery's condition based on voltage when it is well rested and free of either charge or discharge.

Incorrect, the solar charging system was providing a charge voltage.

A load on a battery reduces the terminal voltage.

A charge on a battery increases the terminal voltage.

If you apply a load to a battery, the voltage goes down from the "at rest" terminal voltage. With that load applied, if you then apply a charge voltage, the voltage at the battery terminals goes up. Where it lands, with respect to "at rest" voltage, depends on the load and the charge voltage, but make no mistake, when the charge voltage is applied, the battery terminal voltage increases from where it was before the charge voltage was applied. (Which is exactly what the OP questioned about.)

As I have educated many boaters for many years, the most accurate (voltage) assessment of SOC is with the battery:

1. Topped with fluid.
2. Fully charged.
3. At rest (no load and no charge voltage) for several hours.

However, when this is not practical (rarely is on a cruising boat in use), a reasonable SOC assessment can be made, based on terminal voltage, as follows:

1. Remove charge voltage.
2. Apply a load (~5% of capacity) for 15 minutes and remove.
3. Wait 15 minutes and read the terminal voltage.

If you do this consistently (same load, same timing) each reading, you will have a repeatable method of assessment and relative measurement results.

[sailorboy1]

guess I'll now wait for the experts to battle it out

[ramblinrod]

Originally Posted by sailorboy1

guess I'll now wait for the experts to battle it out

The experts are all in agreement.

[Maine Sail]

Originally Posted by ramblinrod

Incorrect, the solar charging system was providing a charge voltage.

A load on a battery reduces the terminal voltage.

A charge on a battery increases the terminal voltage.

If you apply a load to a battery, the voltage goes down from the "at rest" terminal voltage. With that load applied, if you then apply a charge voltage, the voltage at the battery terminals goes up. Where it lands, with respect to "at rest" voltage, depends on the load and the charge voltage, but make no mistake, when the charge voltage is applied, the battery terminal voltage increases from where it was before the charge voltage was applied. (Which is exactly what the OP questioned about.)

As I have educated many boaters for many years, the most accurate (voltage) assessment of SOC is with the battery:

1. Topped with fluid.
2. Fully charged.
3. At rest (no load and no charge voltage) for several hours.

However, when this is not practical (rarely is on a cruising boat in use), a reasonable SOC assessment can be made, based on terminal voltage, as follows:

1. Remove charge voltage.
2. Apply a load (~5% of capacity) for 15 minutes and remove.
3. Wait 15 minutes and read the terminal voltage.

If you do this consistently (same load, same timing) each reading, you will have a repeatable method of assessment and relative measurement results.

Rod,

I think I understand the point you are trying to make, but lets keep this simple.

A -1A load is not "charging" the battery.

At a -1A load on the battery we have a discharging voltage and that voltage will continue to decline so long as the load exceeds charging current.. The OP has a -1A load on the battery, not charging current.

We do not have a "charging voltage" until charge current goes above 0.00A. So at +0.01A we now have a charge current and a very, very slowly rising voltage. At +20.01A we now have a much faster rising terminal voltage because it is current that determines charging speed & terminal voltage rise speed during bulk.

Low + charge current means a slow voltage rise, lots of + charge current means a fast voltage rise. Low - load current means a slower voltage decline, high - load current means a faster voltage decline.

At -0.01A we have discharging and slowly declining voltage.....

Unless the PV current can exceed system loads, the OP's was not, the battery is physically discharging. A -1A load on a 460Ah bank is very small so the terminal voltage will also drop slowly, but it will continue to drop, if we continue the -1A load and we will not see charging...

We do not have charging of the battery while it is physically discharging and I suspect this is where you are confusing folks....

[sailorboy1]

Somehow I get the feeling there is a poster here that must believe that I'm so completely clueless and such an incredible idiot that I don't understand that to charge a battery the voltage of the source has to be more that the battery.

[Cadence]

Originally Posted by ramblinrod

Incorrect, the solar charging system was providing a charge voltage.

A load on a battery reduces the terminal voltage.

A charge on a battery increases the terminal voltage.

If you apply a load to a battery, the voltage goes down from the "at rest" terminal voltage. With that load applied, if you then apply a charge voltage, the voltage at the battery terminals goes up. Where it lands, with respect to "at rest" voltage, depends on the load and the charge voltage, but make no mistake, when the charge voltage is applied, the battery terminal voltage increases from where it was before the charge voltage was applied. (Which is exactly what the OP questioned about.)

As I have educated many boaters for many years, the most accurate (voltage) assessment of SOC is with the battery:

1. Topped with fluid.
2. Fully charged.
3. At rest (no load and no charge voltage) for several hours.

However, when this is not practical (rarely is on a cruising boat in use), a reasonable SOC assessment can be made, based on terminal voltage, as follows:

1. Remove charge voltage.
2. Apply a load (~5% of capacity) for 15 minutes and remove.
3. Wait 15 minutes and read the terminal voltage.

If you do this consistently (same load, same timing) each reading, you will have a repeatable method of assessment and relative measurement results.

I think boat ownership should require a course in DC fundamentals.
After reading a post on shore power polarity maybe AC also.

[ramblinrod]

Originally Posted by Maine Sail

Rod,

I think I understand the point you are trying to make, but lets keep this simple.

A -1A load is not "charging" the battery.

At a -1A load on the battery we have a discharging voltage and that voltage will continue to decline so long as the load exceeds charging current.. The OP has a -1A load on the battery, not charging current.

We do not have a "charging voltage" until charge current goes above 0.00A. So at +0.01A we now have a charge current and a very, very slowly rising voltage. At +20.01A we now have a much faster rising terminal voltage because it is current that determines charging speed & terminal voltage rise speed during bulk.

Low + charge current means a slow voltage rise, lots of + charge current means a fast voltage rise. Low - load current means a slower voltage decline, high - load current means a faster voltage decline.

At -0.01A we have discharging and slowly declining voltage.....

Unless the PV current can exceed system loads, the OP's was not, the battery is physically discharging. A -1A load on a 460Ah bank is very small so the terminal voltage will also drop slowly, but it will continue to drop, if we continue the -1A load and we will not see charging...

We do not have charging of the battery while it is physically discharging and I suspect this is where you are confusing folks....

With all due respect, that is simpler. Again,
Relating to the OP question.

1. A battery SOC is most accurate if the Batttery is in steady state (at rest).
2. If a load is applied, the terminal voltage reads lower than off it wasn't.
3. If a charger (of any sort) is connected, the terminal voltage reads higher than if it wasn't.
4. If both are applied at the same time, the terminal voltage will be dependant on the relationship between load current and charge current (that latter induced by the applied charge voltage).

Now to what you just posted. When a charge voltage is applied to a battery, it most certainly is being "charged". If the load is high enough, the rate of discharge may be greater than the rate of charge. Therefore, the battery will eventually discharge completely, but it will take longer, and at any time, the battery terminal voltage will be higher, than if the charger were not connected.

[ramblinrod]

Originally Posted by sailorboy1

Somehow I get the feeling there is a poster here that must believe that I'm so completely clueless and such an incredible idiot that I don't understand that to charge a battery the voltage of the source has to be more that the battery.

Can't speak on behalf of others, but based on all of your other posts, It appears that you did not understand that with the solar charger connected and operating, that your battery voltage would read higher than if it were not. Several posts but # 19 in particular seemed to indicate that you did have a false understanding.
Just trying to help. You did ask the question.

[sailorboy1]

Originally Posted by ramblinrod

Can't speak on behalf of others, but based on all of your other posts, It appears that you did not understand that with the solar charger connected and operating, that your battery voltage would read higher than if it were not. Several posts but # 19 in particular seemed to indicate that you did have a false understanding.
Just trying to help. You did ask the question.

Really, that's what you got and you spend time looking for a way to make it sound that way.

I think you were so excited to be an "expert" that you couldn't even understand what was written in a post. For example post 19 says the solar was supplying the "system", it doesn't say it was charging the battery.

Now of course you are looking for a way to prove you really are an "expert".

BTW - you are still the only one who says and I quote "with the solar charger connected and operating, that your battery voltage would read higher than if it were not". Everyone else only believes that the battery voltage will be higher if the supplied solar power is acturally charging the battery.