RE: Experimenting with Weak Batteries

jw0752 — Sat, 30 Apr 2016 05:44:35 GMT

In an Experiment suggested by friend [mention:acaf6a9338de4eef8f6717d5561ed01d:e9ed411860ed4f2ba0265705b8793d05] I have installed 10 AA batteries into one of my battery holders and I am going to analyze the energy output as the batteries go from their current approximate 1.4 Volt level each down to approximately 0.8 Volts each. These are good quality batteries that have been rejected by my Canon camera as too low for use in the camera. I am going to use the small Chinese 12 volt DC to 5 volt DC converter mentioned previously in this post.

[View:/resized-image/__size/620x465/__key/commentfiles/f7d226abd59f475c9d224a79e3f0ec07-a2859b7b-533a-4fe6-80b4-9d43457f314c/4201.contentimage_5F00_181265.jpg:620:465]

In order to calculate energy use value it is necessary to test and calculate the efficiency of this converter. I will use a bench power supply and an electronic load to calculate the voltage and current into the converter and compare it to the voltage and current output. If the Output power is divided by the Input power we will have the efficiency of the converter at that point. The efficiency will be calculated at two points. First I want to see what the efficiency is at the starting point of 14 volts. Next I want to test and calculate the efficiency when the batteries are down to the end of their lives at 8 volts. Here are the results of the test.

Initial conditions:

Input 13.9 volts 102 mA which equals 1.42 watts Output 5.07 volts 200 mA which equals 1.01 watts. 1.01 watts divided by 1.42 watts is equal to 71% efficiency

Final Conditions:

Input 8 volts 160 mA which equals 1.28 watts Output 4.57 volts 200 mA which equals 0.91 watts 0.91 watts divided by 1.28 watts is equal to 71% efficiency

This shows that the efficiency of the DC to DC converter will be linear for our purposes over the test range. Note the Output has dropped in the final condition test from 5.07 volts to 4.57 volts. This is caused by the fact that an 8 volt input on the converter is encroaching on the drop out point. This converter needs at least 3.43 volts more input voltage than output to maintain its rated output.

To save us from having to sit and watch meters for several hours as the voltage levels in the battery pack drop under the steady load of the Electronic Load I will be using a device that I used in a previous blog. I call it a Battery Duration Analyzer. The Analyzer works by monitoring a battery that is under load to an external load. The analyzer allows us to set a target voltage level. As long as our test battery's voltage remains above this target voltage the clock of the analyzer will continue to run. As soon as the test battery voltage drops below the target voltage the clock will stop. I will set the clock initially to 12:00 and then we will read the final time on the stopped clock when we return to the test at a later time.

Here is a link to the original blog on the Battery Duration Analyzer:

https://www.element14.com/community/people/jw0752/blog/2015/08/16/battle-of-the-batteries--e-vs-d-vs-bargain

One of the critical parameters that can make a big difference in the out come of this experiment is the level of load that is placed on the battery pack. I have run some tests to determine the current that is drawn by my cell phone when it is recharging or 230 mA. This seems like a reasonable level and will actually translate the amount of time measured by the Analyzer into how long I could charge my phone on the bank of 10 used AA batteries. If we use our knowledge of the efficiency of the converter we can calculate a load of 230 mA on the output will translate to a load of 120 mA when the batteries begin at 14 volts and 200 mA when the voltage of the battery pack is at 8 volts.

Here are some pictures of the test set up:

[View:/resized-image/__size/398x298/__key/commentfiles/f7d226abd59f475c9d224a79e3f0ec07-a2859b7b-533a-4fe6-80b4-9d43457f314c/7652.contentimage_5F00_181266.jpg:398:298][View:/resized-image/__size/402x301/__key/commentfiles/f7d226abd59f475c9d224a79e3f0ec07-a2859b7b-533a-4fe6-80b4-9d43457f314c/5518.contentimage_5F00_181267.jpg:402:301][View:/resized-image/__size/403x302/__key/commentfiles/f7d226abd59f475c9d224a79e3f0ec07-a2859b7b-533a-4fe6-80b4-9d43457f314c/4113.contentimage_5F00_181268.jpg:403:302]

Our target cut off voltage is 4.5 volts. The initial voltage of our converter under the 230 mA load is 4.96 Volts. The Fluke is displaying the voltage of the 10 series AA batteries under load 14.09 volts. At this point there is nothing left to do except wait. I returned to the test several times over the course of the day. When I checked on it at supper time it was over and this is the reading on the analyzer.

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This tells me that the battery pack was able to supply enough power to keep the converter above 4.5 volts for 7 hours and 5 minutes. At the end and still under the load the batteries were showing only a fraction of a volt but when I removed the load from the battery pack it almost immediately jumped back to 10 volts. This would indicate to me that just as [mention:acaf6a9338de4eef8f6717d5561ed01d:e9ed411860ed4f2ba0265705b8793d05] had earlier indicated the drop off point is really 1 volt and not 0.8 volts per battery. If I had been watching the test I suspect the drop from 10 volts for the 10 pack to zero volts would have occurred in a matter of minutes.

Now that we have the time of 7 hours and 5 minutes we can calculate the amount of energy used by the load with 5 volts and 230 mA for that period of time. This works out to 8.14 watt hours of energy. If we extrapolate back to the battery pack through the 71% efficiency of the converter this means that the battery pack actually supplied about 11.5 watt hours of energy in this time period.

Watt Hours = Current X Voltage X Time

When I consider that this is also equivalent to charging time on my cell phone it means that there is enough energy left in 10 otherwise used batteries to charge for 7 hours 5 minutes. On the other hand at the current cost for electricity in my area which is $0.13 per kWh I have about 0.15 cents of value. Doesn't seem worth the trouble to use the batteries and converter for only a fraction of a penny but now I know.

John

RE: Experimenting with Weak Batteries

DAB — Fri, 29 Apr 2016 19:00:22 GMT

Hi John, interesting project.

I just recently saw a video on the EEV Blog where Dave compressed spent batteries to get more power life out of them.

DAB

RE: Experimenting with Weak Batteries

Jan Cumps — Fri, 29 Apr 2016 11:31:21 GMT

A cool experiment, [mention:f80b53cee57c44bc9d7c577d07d7c791:e9ed411860ed4f2ba0265705b8793d05]. I don't think you'll be overly excited about the results though.

Once a battery's voltage has dropped to say 1V, there's nada* energy left in there.

You'll spend more personal joules opening the tube and putting batteries in than you'll ever be able to reclaim.

It would be cool to measure how much energy you can actually retrieve from a 10-load of AA's.

* https://www.youtube.com/watch?v=dnXiLBabSTU