1. Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.
    Dismiss Notice

Automated Load Tester

Discussion in 'Electronic Projects Design/Ideas/Reviews' started by tkc100, Feb 1, 2012.

  1. tkc100

    tkc100 Member

    Joined:
    Feb 5, 2010
    Messages:
    71
    Likes:
    0
    A part of my job is maintaining several solar installations. They are residential units all using 6 volt deep cycle batteries in a series parallel configuration. Once a year it is necessary to break the battery banks down for cleaning and routine service. One part of that service is actually testing the individual batteries capacity. I wish the budget was better because I would recommend purchasing an impedance tester but alas that is not in the foreseeable future.
    So I need to build a simple load tester so that I can run a capacity test unattended. I have attached a diagram of what I have in mind and I am hoping someone will have some off the self solutions for the LVD and the timer.
    I’m sure the circuits are out there I just don’t have the exposure to know what might be available.
    I’m not an electrical engineer so the theory is beyond my expertise but if the project requires assembly that’s no problem.
    Any ideas and or help you all can provide will be greatly appreciated.
    Thanks!
     
  2. 4pyros

    4pyros Well-Known Member Most Helpful Member

    Joined:
    Oct 22, 2010
    Messages:
    4,368
    Likes:
    282
    Location:
    Pennsylvania, USA
    With that setup you would need a varable load resister to keep at a constant 25 amps. Thats hard because most load resisters would be fixed.
    Maybe try a current regulater into a fixed load.
     
    Last edited: Feb 1, 2012
  3. tkc100

    tkc100 Member

    Joined:
    Feb 5, 2010
    Messages:
    71
    Likes:
    0
    I understand as the voltage drops the load would have to change also in order to keep a constant amperage draw.
    In reality there is very little voltage difference between a fully charged 6 volt battery (6.35) and a completely discharged one (5.25).
    I practice I would probably not completely discharge the battery. It's very hard on the battery and very time consuming. Discharging it to 50% (6.1)or perhaps a most 80% (5.99)would be suitable and come close to matching the actual performance expectations.
    I used the formula R=E/I and determined that to produce a 25 amp load with 6.35 volts would require a resistance of .254 ohms. The difference in resistance between a fully charged battery and a completely discharged one is only. .04 ohms. Help me here if I'm going off the deep end. Purchasing a 155 watt resistor to that close of tolerance would be difficult. So I would no doubt have to do a bit of compromising. The load isn't really that critical with a little math I can compensate and what is more important is a comparison from one test cycle to the next.
    So for now unless someone has a better idea let's go with a compromise on the load and except the fact that it will only show a mean value.
    What is more important to this project is an accurate LVD it has to be accurate. The more accurate the better.
    Thanks for your input!
     
  4. dave

    Dave New Member

    Joined:
    Jan 12, 1997
    Messages:
    -
    Likes:
    0


     
  5. 4pyros

    4pyros Well-Known Member Most Helpful Member

    Joined:
    Oct 22, 2010
    Messages:
    4,368
    Likes:
    282
    Location:
    Pennsylvania, USA

    I would stay with a 25 amp current regulator. Trying to use just a load resister would be inaccurate.
    IE It mite be fine for the 1st battery but may be off for the 2 nd battery test do to the change in resistance after heating.
     
  6. ronv

    ronv Well-Known Member Most Helpful Member

    Joined:
    May 30, 2010
    Messages:
    4,725
    Likes:
    488
    Location:
    Tucson, AZ. USA
    Not sure how close you need to measure them but it might be easier to manipulate the time rather than the load. For example: The 50% point using a .25 ohm resistor would be 25.44 amps at 6.36 volts(100%) and 24.4 amps at 6.1 volts (50%). But having said that I think battery voltage as a function of capacity is under no load. Maybe you can confirm that. It seems to be hard to find a LVD for 6 volt batteries - especially adjustable ones. This would be pretty easy to build but hard to find off the shelf.
     
  7. tkc100

    tkc100 Member

    Joined:
    Feb 5, 2010
    Messages:
    71
    Likes:
    0
    Good point!
    It would not be unusual to have 16 batteries to test.
    Current Regulator?!? The water getting deep here. What such an animal look like and what kind of expense are we talking about?
    I would actually prefer the LVD to be fixed. Or at best have three fixed values 6.10, 5.99 and 5.25 volts
    I'm not shy about having to build a circuit but have not idea how to design one.
     
  8. tkc100

    tkc100 Member

    Joined:
    Feb 5, 2010
    Messages:
    71
    Likes:
    0
    ronv
    I just read the other part of your post.
    Your right SOC is measured OCV but a test such as this measure the time it take for a battery to go from fully charged to the cut off threshold under the prescribed load.
     
  9. 4pyros

    4pyros Well-Known Member Most Helpful Member

    Joined:
    Oct 22, 2010
    Messages:
    4,368
    Likes:
    282
    Location:
    Pennsylvania, USA
  10. ronv

    ronv Well-Known Member Most Helpful Member

    Joined:
    May 30, 2010
    Messages:
    4,725
    Likes:
    488
    Location:
    Tucson, AZ. USA
    Okay, so all that leaves us with is the timer. I noticed you said self powered. Does that mean plug into the wall or having its own battery?
     
  11. tkc100

    tkc100 Member

    Joined:
    Feb 5, 2010
    Messages:
    71
    Likes:
    0
    The timer is quite possibly the easiest part of the project
    Self power because it can't depend upon power from the battery being tested.
    Even with my limited knowledge of electronics I can envision several ways to do this.
    Perhaps the simplest would be a Hobbs meter attached to a relay power by the LVD.

    I viewed the LVD disconnect suggest but I'm afraid the cut out voltage is too high.
    6.25 volts which represents about an 80%SOC
    Also I'm not sure how the throttling function would be dealt with.

    A circuit design for a LVD that could be modified to have three separate values would be great.

    What about the current regulator?
    Is this some thing that need to be built or purchase?
     
  12. Mr RB

    Mr RB Well-Known Member

    Joined:
    Jul 22, 2008
    Messages:
    4,716
    Likes:
    194
    Location:
    Out there
    You can use a "constant current source", they are easy enough to build, even a 25 amp one, although you are probably up for $25 to $35 dollars in parts including a heatsink.
     
  13. Reloadron

    Reloadron Well-Known Member Most Helpful Member

    Joined:
    Dec 23, 2009
    Messages:
    6,776
    Likes:
    281
    Location:
    Cleveland, Ohio USA
    Do you have a laptop available that could remain with the batteries during load test to take data?

    Ron
     
  14. 4pyros

    4pyros Well-Known Member Most Helpful Member

    Joined:
    Oct 22, 2010
    Messages:
    4,368
    Likes:
    282
    Location:
    Pennsylvania, USA
  15. ronv

    ronv Well-Known Member Most Helpful Member

    Joined:
    May 30, 2010
    Messages:
    4,725
    Likes:
    488
    Location:
    Tucson, AZ. USA
    Heater

    Here is what a circuit might look like.
    The FETs need to be on a BIG heatsink.
    The relay needs to be good for at least the 25 amps.
    The resistors at the bottom left select eith 5.25 or 6.1 for the shut off voltage.
    The big resistors can be 3 0.5 ohm 50 watt in parallel.
    Let me knpw if you decide to build it and I'll give you a bill of material and a source for the parts.
     
    Last edited: Feb 2, 2012
  16. tkc100

    tkc100 Member

    Joined:
    Feb 5, 2010
    Messages:
    71
    Likes:
    0
    4pyros
    Thanks for the suggestion but the meter shown is only rated for batteries between 4 and 50 AH.
    A typical 6 volt battery is a solar installation will be rated somewhere in the neighborhood of 186 AH

    Today I did some research and came up with the following
    A perfectly elegant, simple and well documented LVD.
    Burt's Design Pad
    http://gorum.ca/lvdisc.html
    NE555 Low Voltage Battery Disconnect Circuit
    And for the timer
    Elapse timer
    http://www.fargocontrols.com/sptimer.html
    So all that remains is the load.

    ronv
    I gathered all this information up before seeing your last post. Thanks for all the effort.
    What do you think about Burt's design. I certainly well documented and it appear as though the only alteration that would be necessary would be a few resistance values. I'm not saying this design is better or worse than yours but it's sure laid out in a fashion that would make it possible for an individual with my talent might succeed at. PC board and all

    So it appears as though all the pieces are coming together except I really could use some input on the concept of a constant load.
    I agree with earlier posts that it would be advantageous but I need help on what a 25 amp constant load would look like. It would have to function within the range of 5.5 to 6.5 volts.
    Mr RB mentioned that it could cost $35 dollars or so. That would not be a problem.
    The impedance tester I think I made reference to earlier in in the four figure range and that before the decimal point.

    and once again thank you all for all the comments.
     
  17. Mr RB

    Mr RB Well-Known Member

    Joined:
    Jul 22, 2008
    Messages:
    4,716
    Likes:
    194
    Location:
    Out there
    A LM317 is a simple 3 pin device that can be used as a constant current load up to about 1.5 amps. A google for "LM317 constant current" should get you heaps of info.

    The LM317 can also be used together with a larger "pass" transistor(s), to increase its current capability. That is usually shown in the LM317 datasheet, or you could google for many "LM317 high current power supply" circuits that will show you how. Obviously you will need a large heatsink or heatsink+fan, which will be a significant part of the cost. :)
     
  18. ronv

    ronv Well-Known Member Most Helpful Member

    Joined:
    May 30, 2010
    Messages:
    4,725
    Likes:
    488
    Location:
    Tucson, AZ. USA
    I'll try to do a simple version tonight.
     
  19. 4pyros

    4pyros Well-Known Member Most Helpful Member

    Joined:
    Oct 22, 2010
    Messages:
    4,368
    Likes:
    282
    Location:
    Pennsylvania, USA
    Last edited: Feb 4, 2012
  20. tkc100

    tkc100 Member

    Joined:
    Feb 5, 2010
    Messages:
    71
    Likes:
    0
    Mr RB
    "A LM317 is a simple 3 pin device" Simple on the outside but for a person with my understanding of electronics it nothing short of miraculous inside.
    Your right there is a great deal of information on the internet about this component but mostly about power supplies and not a load.
    I did find one simple example of a "smart dummy load" but it current capabilities is too low.
    I need a 25 amp load for voltages between 5 and 7 volts.
    You made mention of using a larger "pass" transistor(s), to increase its current capability.
    Please help me on this one.
    What would the circuit actually look like.
    Once again thanks, thank you all for taking the time to assist me with this project.
     
  21. ronv

    ronv Well-Known Member Most Helpful Member

    Joined:
    May 30, 2010
    Messages:
    4,725
    Likes:
    488
    Location:
    Tucson, AZ. USA
    If we use this idea we can eliminate the current source. The average should be very close to 25 amps.
    We could simply use 4 1 ohm resistors in parallel and get .25. I like this idea. Do you?
     

Share This Page