# battery life

Discussion in 'The Projects Forum' started by cgama, Mar 27, 2012.

1. ### cgama Thread Starter New Member

Dec 23, 2011
19
0
Hi!
I have a circuit powered by 2 12V lead acid batteries in parallel. The circuit draws 94.7 mA and after the batteries fall below 9V it stops working (it stops doing what i want it to do).
I have been reading about battery life calculations in this website, but I'm still a bit confused: the batteries that I'm using are Yuasa NP 1.2 - 12, which have a capacity of 1.2 Ah. Of course if I do a simple division it gives an absurd result... Well, in the datasheet it shows a graph about "discharge characteristic curves". Should I go through this? And was it the CA??
Here is the datasheet:

http://www.yuasabatteries.com/pdfs/NP_1.2_12_DataSheet.pdf

Another question: I have tried to measure the battery life by running it and measuring the voltages on Labview. (it lasted 6h) When measuring with the multimeter I measure one side of battery against the other to give the voltage out. But in this experiment I have measured the positive rail and the negative rail against the circuit ground, and this gave me about +14V for the positive rail and -13/14V to the negative rail. I don't even know what to make of this

Sorry about the long post. And thank you for the time and help!!!

2. ### cgama Thread Starter New Member

Dec 23, 2011
19
0
*what is the CA?* (sorry)

3. ### debjit625 Well-Known Member

Apr 17, 2010
790
186
You could find a lot about this subject at http://batteryuniversity.com/

In a SLA battery when the cell voltage reaches almost 1.9 volts ,you can assume it have 0% charge left,so for a 12V system it will be 11.4V at 0% charge.
So at 9V it is not working as it have no charge left.
Its the C rate,its the rate at which a battery accept or gives up charge (charge or discharge).
http://batteryuniversity.com/learn/article/what_is_the_c_rate

About your third question,I couldnt understand how you connected the battery and what ground your talking about.... a diagram will be better,a 14V at the terminal of the battery seems that its connected to some charging system....

Good Luck

4. ### cgama Thread Starter New Member

Dec 23, 2011
19
0
Hi
The batteries are connected to a DCDC converter to give exactly +12V, -12V and 0V to the rest of my circuit. When the voltage from the batteries falls below 9V, the DCDC converter stops doing its job, that's why I said that.
About the "CA", I read the article and I think it confirms my theory... Because the batteries are being discharged at about 0.1A, according to the graph in the datasheet it will discharge in about 7-8h, right? But since it's in parallel, shouldn't it last longer?.... I was expecting it to last a couple more hours.... ://

sorry about the "primary school" image... This is how I made the connections and got those values.
Thank you again for your great help!!

5. ### evilclem Active Member

Dec 20, 2011
118
16
Running a lead acid battery until it's flat will kill it (especially the gel ones).

If you are intending to continue running off a lead acid battery then I would advise some low voltage disconnect circuitry.

These batteries should be recharged as soon as possible after use for maximum battery life.

6. ### KJ6EAD AAC Fanatic!

Apr 30, 2011
1,568
442
Discharging those batteries below 10.5 Volts severely shortens their life.

7. ### wayneh Expert

Sep 9, 2010
16,099
6,209
Yes, according to the datasheet, you should be getting more. Are you sure your batteries were properly charged to begin with? Are they new, or have they suffered a number of cycles to 9V? (That severely shortens life.) How sure are you of the average current draw? You report 95mA, but maybe it's more at times?

8. ### debjit625 Well-Known Member

Apr 17, 2010
790
186

At 0.1C the battery will not discharge for 7-8Hrs it will be 4Hrs as per the graph for a 12V load. If two 12V 1.2AH batteries are connected in parallel, and if the load consume 94mA at 12V then the C rate at which the system will be discharged will be 0.094/1.2 = 0.078333 approx 0.08 C as their are two batteries in parallel so it will be 0.08/2 = 0.04 C around 15 hours a bit less or more.

Manufacturers often rate their battery capacity a bit more than they really have. But in your case its different it is just half of what you are getting, may be the batteries have problem, you need to test the capacity separately.

It has nothing to do with your batteries,its about your DC-DC converter.

Good Luck

9. ### cgama Thread Starter New Member

Dec 23, 2011
19
0
the batteries are new... maybe 2 cycles? they power leds, so they will start flickering only when the batteries do get as low as 9V :/ right now is in development phase but soon they will be locked in a box and I won't be able to test them as easily. oh well...
I measured the current with a multimeter connected in series. It didn't really fluctuate much, maybe about to 2mA. I'll test it again to see what happens. But the help with the calculations was incredible!!! thanks!!
From experience within my lab, I was expecting about 10h. I might redo the labview experiment (probably not in time for my first presentation )

Thank you soooo much for all your help!!!