SLA Battery - Possible Discharge Problem

[TheProfessor]

Hello all,

I have been reading as much as I can and still cannot find out what I'm doing wrong, so hopefully you folks can help me out a bit.

I have a homebuilt 18v, 3.5A PV panel that is using a switching charge controller to charge a 12v 35AH SLA battery. I then hook this battery to a 400w inverter to power (usually) a hydroponic garden that is 120v, 0.8A, AC of course. I have been keeping track of the load voltage of the battery as it discharged to get a better idea of the curve. Upon initial hook up to the device I see about 12.88vdc. I then measure battery voltage across the terminals every few minutes to see how the voltage is dropping. When I get to 10.9vdc, I disconnect to save the battery.

The problem: At 0.8A the battery drops from 12.8 to 10.9vdc in only 6-7 hours. My math says that if 12.88vdc is fully charged, and 10.8 is about 40% charged, I should have about 21AH to play with (60% of 35AH). I measure the current draw from the battery to the inverter and from the inverter to the device and the efficiency loss looks like 20% (which is to spec, I believe).

So, at 0.8A, 12AC I should be pulling 96-100watts. 100w/12vdc = 8.33A. Which as I type this the proverbial lightbulb has just gone off. Heh. In the interest of not wasting all this typing I did, does this all sound correct and within specification? Is my battery fully charging? Can I really only expect to get 250-300 watts from a 12v 35AH battery?

Thank you for suffering through my first post and potentially stupid question.

[Sunking]

Well professor you just learned about battery internal resistance and Peukert Law. Everything you have is way too small to do the job. A battery has internal resistance in portion to it size. To make a long story short a Pb battery can only deliver a max of a C/8 current without significant voltage drop. Where C = the battery AH capacity and the number represents Hours of charge or discharge rate. So you have a 35 AH battery which means it can only deliver 35 AH / 8 H = 4.4 amps for 8 hours. You are pulling 8 amps which is going to make the battery voltage sag significantly, and since you are pulling more than C/20 amps Mr Peukert is robbing you blind in capacity which makes your 35 AH battery a 20 AH battery.

Surprise Surprise Surprise.

You need at least a 60 AH battery with a 60 to 100 watt panel to charge it. But even a 65 AH battery when discharged with C/8 rate or 4 amps is only going to give roughly 40 AH to 50 AH or 5 to 6 hours run time because Mr Peukert says: "capacity of a battery in terms of the rate at which it is discharged. As the rate increases, the battery's available capacity decreases".

That's the lesson of the day. Now go do your home work professor and learn the lesson. Your assignment is"

Ohm's Law
Peukert Law

FWIW a battery with 10.8 volts is not 40% SOC, that is almost completely discharged around 2 to 5% SOC. 40% is more like 11.9 volts and with Pb you do not want to go below 50% if you can help it.

Surprise Surprise Surprise!!!

[TheProfessor]

Excellent, thank you very much.

I understand everything you're saying and I think the problem here (aside from my lack of knowledge) is not that anything is not working correctly, but that I was expecting too much out of this system. My panel is 18v, 3.5A (63watts) Obviously that is open circuit. Is it safe to say that the panel is 63 watts, or would it be more accurate to say that it is 12v 3.5A or 42watts because that is what is "usable" by the battery/CC?

Anyway, it seems that aside from being inefficient, the panel and charge controller are working and doing what they are supposed to be doing.

My big concern, and what you so masterfully corrected me on, was the battery capacity. I understand everything you've said, but would like to know about the discharge rate compared to the amperage draw. You mentioned that I'm pulling c/20 (which is see now is why I am truly getting about 20AH out of this. Obviously barring buying a new battery the only thing I can do is decrease my load.

Short of me doing my "homework", aside from connection points and wire runs, would hooking two identical 35AH batteries in parallel give me about 70AH (again, i realize there will be some efficiency loss between the obvious things).

Thank you for taking the time to give such a thorough and leading question. This experiment is definitely not in any way meant to save money but a proof of concept to help me learn about solar PV, basic circuits, and programming (i'm doing some Arduino stuff with this as well). This project has greatly improved my knowledge of electronics and electrical concepts as well as taught me some basic skills about soldering and learning about components. Thank you.

[Sunking]

Excellent, thank you very much.

I understand everything you're saying and I think the problem here (aside from my lack of knowledge) is not that anything is not working correctly, but that I was expecting too much out of this system. My panel is 18v, 3.5A (63watts) Obviously that is open circuit. Is it safe to say that the panel is 63 watts, or would it be more accurate to say that it is 12v 3.5A or 42watts because that is what is "usable" by the battery/CC?

It had NOT better be 18 Voc. It needs to be 18 Vmp and roughly 22 Voc. It takes 36 cells to make a battery panel and 18 volts to charge a 12 volt battery.

My big concern, and what you so masterfully corrected me on, was the battery capacity. I understand everything you've said, but would like to know about the discharge rate compared to the amperage draw. You mentioned that I'm pulling c/20 (which is see now is why I am truly getting about 20AH out of this. Obviously barring buying a new battery the only thing I can do is decrease my load.

No Sir you still have not connected the dots. C is the battery AH capacity. For you that is 35 AH I think you said right? So a C/20 discharge rate would be 35 Amp Hours / 20 Hours = 1.75 Amps. You are discharging at a much higher rate than C/20 with 8.33 amps. So 35 AH / 8.33 Amps = 4.2 Hours so you are discharging at C/4.2. Understand? At that high of a rate Mr Peukert is robbing almost half of your capacity and is why you are only getting 20 or so Amp Hours.

To take anything off grid you first determine how many watt hours you need in a day. Lets say 1000 watt hours or 1 Kwh You do not want to discharge your battery more than 20 to 30% each day. So if you use 1 Kwh per day you want a battery of 5 Kwh. If you are using 12 volts then you are shopping for a 5000 wh / 12 volts = 416 AH battery. Just call it 400 AH.

So what you have discovered like 99% of everyone that comes here is you have grossly undersized every thing and it is not going to work.