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1. Originally Posted by billvon
Yes!

No, it's a measure of power.

How many fields will 1 horse plow? You can't answer that because all you know is power. You need to know energy, which is power times time.

Now, how many fields will 1 horse plow in 1 day? You CAN answer that, because a horsepower-day is a measure of energy, and thus can do a certain amount of work.
Yes and is it a leap of ingenuity to understand what; 180 watts consumed in one day? if it were then why would I add the "one day" LOL!

I guess we all agree to differ and in future I shall say; 180 watt hours in one day unless I feel more like saying; 4.5 million watt seconds in a week of course.

2. Originally Posted by BritishPete
Watt is a unit of energy and within a time frame becomes a unit of work done. Similar to the difference between amp and Coloum (sp?)
No it is not similar. You do not know the difference between power and work (energy) the very basic principles of any engineering fields which you claim to be an EE is a lie. Any Enginner, heck even a fisrt year student, can clearly differentiate between the two and know they are not the same. That is why you are getting your butt kicked in every thread you appear in. You do not know squat, and trying to act like an expert which has been clearly pointed out you are a fraud. That is what we can agree too.

3. Britishpete there is something terribly wrong with your arguements in just about every area.
you wrote this..
I shall; When my PV array is capable of generating say 80 amps ant 14.8V but is only charging the batteries at 20 amps because they are 80% full where is the 60 amps going? Answer WASTED. Now install a diode that will permit me to increase charge current to 30 amps (at say 15.2V) without tripping the inverters and I have GAINED 10 x 15.2 = 152 watts not wasted.
It is not rocket science but some would have us believe it is.

Now im beyond interested because you said in a much earlier post the diode was between the battery/ies and the inverter. YES??So how the hell does it help if its there to increase the charge current to the battery/ies from the charge controller????????????
ANY DIODE PLACED BETWEEN BATTERY/IES AND INVERTER WILL ALWAYS WASTE POWER. IT IS A FACT OF LIFE. ITS ALSO A FACT THAT ANYTHING PLACED AFTER THE BATTERY/IES WILL NOT DO ANYTHING TO INCREASE/DECREASE THE CHARGE RATE TO THE BATTERY/IES
OK IM RUDE IN SAYING I THINK YOUR ENGINEERING DEGREE IS IN EITHER PLUMBING OR DENTISTRY OR MOUNTAIN BIKE CONSTRUCTION.BUT I AINT RETRACTING IT. IF YOU HAVE A DEGREE IN ELECTRICAL ENGINEERING ITS A PHOTOSHOPPED ONE OF SOMEONE ELSES. MORE RUDE BUT I THINK TRUE AND THE MORE POSTS YOU DO THE MORE I BELIEVE IT.AND EVERYONE ELSE DOES ALSO ON THAT FORUM.

4. Originally Posted by john p
Britishpete there is something terribly wrong with your arguements in just about every area.
you wrote this..
I shall; When my PV array is capable of generating say 80 amps ant 14.8V but is only charging the batteries at 20 amps because they are 80% full where is the 60 amps going? Answer WASTED. Now install a diode that will permit me to increase charge current to 30 amps (at say 15.2V) without tripping the inverters and I have GAINED 10 x 15.2 = 152 watts not wasted.
It is not rocket science but some would have us believe it is.

Now im beyond interested because you said in a much earlier post the diode was between the battery/ies and the inverter. YES??So how the hell does it help if its there to increase the charge current to the battery/ies from the charge controller????????????
ANY DIODE PLACED BETWEEN BATTERY/IES AND INVERTER WILL ALWAYS WASTE POWER. IT IS A FACT OF LIFE. ITS ALSO A FACT THAT ANYTHING PLACED AFTER THE BATTERY/IES WILL NOT DO ANYTHING TO INCREASE/DECREASE THE CHARGE RATE TO THE BATTERY/IES
OK IM RUDE IN SAYING I THINK YOUR ENGINEERING DEGREE IS IN EITHER PLUMBING OR DENTISTRY OR MOUNTAIN BIKE CONSTRUCTION.BUT I AINT RETRACTING IT. IF YOU HAVE A DEGREE IN ELECTRICAL ENGINEERING ITS A PHOTOSHOPPED ONE OF SOMEONE ELSES. MORE RUDE BUT I THINK TRUE AND THE MORE POSTS YOU DO THE MORE I BELIEVE IT.AND EVERYONE ELSE DOES ALSO ON THAT FORUM.
I have stated this many times but will state it one last time.

There are not enough daylight hours in a day to bring my battery bank above about 85% at an aborb voltage of 14.8V because the charging current tapers off as battery internal resistance increases. Despite this the battery manufacturer says that they should be periodically charged to 100%.

I cannot do as the manufacturer requires without increasing charging voltage (and hence current) and I cannot increase charging voltage without the inverters tripping.

Now when I have say 80 amps charging a 50% SOC battery all is fine but when the battery goes above 85% SOC the charging current reduces to about 20 amps even though another 60 amps is available it is not used; Hence WASTED.

Now the diode stops the inverter from tripping by reducing the input voltage by 0.7V when I increase the charge voltage and am able to complete a 100% SOC.

Of course the diode is consuming power but it only power that would otherwise not be used anyway so which is the lesser of two evils?

To me it is a no brainer! Increase voltage as the manufacturer recommends to obtain a 100% SOC and "Waste" about 15Whrs (typical for my loads) of 10 amps) for 2 hours through the diode which otherwise would NOT have been accessed from the panels anyway so how is this a waste !!!

Now In order to protect a \$1000 battery bank I am perfectly willing to use 30Whrs once a week. Watt hours whcih otherwisw would not have been produced anyway.

I hope that is clear now because it is truly my last comment on the matter.

To me it is very simple but I guess I am not expressing myself well enough.

5. Originally Posted by BritishPete
To me it is very simple but I guess I am not expressing myself well enough.
You are right, everthing you have expressed and re-explained is MAKE BELIEVE VODOO. It is clear you have no understanding of very basic electrical principles. So when you come here trying to make yourself out as an expert and claiming to be an EE is clearly a lie and fraudulent. A first year EE student with ZERO experience knows more than you do.

You have been exposed, give it up, you have lost any and all credibility here. You are only making a fool out of yourself.

You wish to fully charge your bank, but your inverter/charger was tripping off due to feed overvoltage.
In order to allow the inverter charger to do it's thing internally and cycle up the voltage to the bank, you dropped it's input feed voltage via diode to allow the full cycle to complete.

This reads to me like the gist of it, and the rest is miscommunication and rapid assumptive conclusions.
That is, the extra wattage was not wasted because it never had a path to collection - rather the opportunity was wasted to collect it.
The diode allows that opportunity to get full charge to your bank by maintaining a non tripped inverter/charger to rock on.

Yes? No?

7. Originally Posted by tandrews

You wish to fully charge your bank, but your inverter/charger was tripping off due to feed overvoltage.
In order to allow the inverter charger to do it's thing internally and cycle up the voltage to the bank, you dropped it's input feed voltage via diode to allow the full cycle to complete.

This reads to me like the gist of it, and the rest is miscommunication and rapid assumptive conclusions.
That is, the extra wattage was not wasted because it never had a path to collection - rather the opportunity was wasted to collect it.
The diode allows that opportunity to get full charge to your bank by maintaining a non tripped inverter/charger to rock on.

Yes? No?
Correct except the inverter and charge controller are separate units so only the inverter is receiving reduced voltage and even then only when charge controller output is increased. The biggest misconception seems to have been that the battery charging is also connected through the diode which it is not. ONLY the inverter(s). There was also intentional mis-reading on the part of some too. (sad).

So my options are;
1) use the diode (works great)
2)Sit in the dark (figuratively) while on Equalize
3) Go out for several beers while on Equalize
4) Replace my i(otherwise perfect) nverters with different units.

I choose the \$18 diode.

8. OK I am going to tell you how to fix most of the problems you have.

First let's ignore the fact you bought a cheap Chi-Com POS inverter that does not conform to any accepted standards. That was your first mistake. However it not much of a problem if it trips at 15 to 15.5 volts

The fix is real easy. You claim to be using an Outback FM 80 on a bank of Trojan batteries configured for 12 volts right? The solution is simple if you simply follow instructions from the battery manufacture. Set your OUTBACK Bulk and Absorb voltage to 14.8 volts,the Float to 13.2 volts, and your problem goes away. If you will take the time to read the Trojan Battery User Manual on page 9 Table 4 you would already know this. The only time it will become a problem is when you need to Equalize because the set point voltage needs to be 15.5 volts.

If you placed a diode between the battery and inverter is a band-aid but a horrible fix because it waste precious battery power. I say this because you also complain your batteries are never able to get 100% fully charged each and every day. You stated there is not enough hours in the day correct? That is a cop out statement. It clearly indicates the system is not properly designed to do the work it is intended for.

Any properly designed off-grid battery system should be able to fully recharge the station batteries on even the shortest winter days with daylight to spare.So when you say there is not enough sun hours in a day is BS, it means you have a poorly designed system, or I should say it was not designed. You should be able to get to the FLOAT stage with the sun still shining. If that is not happening you are screwed. It means your batteries are less than 100%, and you are not replacing what you used in a day. That means 1 major thing. Your batteries will be toast in a very short time, because when they are below 100%. When below 100% lead sulfate crystals are being formed on the plates, and that destroys the battery if allowed to continue. Instead of the batteries lasting 3 to 5 years, you only get a year or less out of them.

OK I think you said you have 6 Trojan T-105's configured as 12 volts correct? You are using a Outback FM-80 correct? What is the panel wattage connected to the FM 80?. Assuming you have it maxed out to 1250 watts, living in Central Florida only gives you 3.3 Kwh of usable energy per day. At 12 volts would requires a battery amp hour capacity. of 1375 Amp Hours. If all you have is 6 Trojan T-105's gives you an AH capacity of 660 AH, or only 1.5 Kwh of usable energy per day based on 20% DOD/day. Since you complain of not ever being able to fully charge your batteries each day tells us you have a mismatch which you need to get lined up or else you are in a world of hurt buying and replacing destroyed batteries from abuse.

What it all boils down to is you failed to design the system properly and used cheap non conforming equipment trying to cut corners for that FREE ENERGY you claim to have. You got what you asked for; Problems. The fiddler has finished and now is coming to collect his dues.

EDIT NOTE:

That diode you added between the battery and inverter is costing you an automatic 5.8% constant loss in power transfer efficiency. Math does not lie. For example if the cheap Chi-Com inverter efficiency is say 80%, that diode you added turned it into 74.2%. 0.7 volts / 12 volts = 5.8% loss

9. sorry to dig up an old thread, i have a similar issue im trying to find a solution for and need 1 post to view the diagram for the LDO posted earlier.

10. Hi gmturner - Welcome to Solar Panel Talk!

That is what the old threads are for - reference and learning - you did well to find this.

Russ

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