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Automated Load Tester

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

  1. Reloadron

    Reloadron Well-Known Member Most Helpful Member

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    The resistance does increase as the resistor gets hot. So if we look at ohms law for a more or less fixed voltage applied, I (current) = V (voltage) / R (resistance) so as R increases the I will decrease.

    Make sense now?

    Ron
     
  2. ronv

    ronv Well-Known Member Most Helpful Member

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  3. ronv

    ronv Well-Known Member Most Helpful Member

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    I think most of the variables can be pretty well calibrated out.
    For example:
    The actual resistance can be measured hot by knowing the current and the battery voltage and calculating the resistance R=E/I. Then amp hours calculated from that.
    The ones I'm not sure about are the ones like the voltage difference between batteries of different capacities being discharged at the same current.
     
    Last edited: Feb 28, 2012
  4. dave

    Dave New Member

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  5. tkc100

    tkc100 Member

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    Thanks for the explanation Reloadron.
    ronv
    It appears as though this line of though may well added a new dimension to this project. I studied the data sheets for several of the load resistors and took careful note of the charts. It seems to me using a fixed resistor would require a lot of calibration and or calculation with each test. In fact there is also the ambient temperature of the tester itself that needs to be taken into consideration. It not out side reason for it to be at 40 degrees for one test and 100+ degrees for another.
    Remember we are working with a fairly close tolerance and the tests need to be able to be replicated not necessarily successively but more likely with 6 months or more between test to establish patterns.
    Perhaps we are back at looking at a constant load source as mentioned in earlier posts. In one such post someone indicated that the circuitry was simple and that a switching transistor could be used to handle the load.
    What do you all think?:confused:
     
  6. ronv

    ronv Well-Known Member Most Helpful Member

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    The current source would be a bit more accurate, but faces some of the same problems with the initial calibration and still requires the large resistors to burn up the power. Normally in such a system the current is sensed by measuring the voltage drop across these resistors so their accuracy still needs to be calibrated out. On the down side the transistors that controll the current need to dissapate a lot of power so they also need large heat sinks. The major difference is in the fact that the simple load like we are looking at can only supply an average load that varies over the battery voltage. For example:
    If the starting battery voltage (under load) is 6.04 volts and at 50% is 5.53 volts (levels found in one curve I saw.) the average current in this system will be 23.13 amps +/- about 8%. Or 21.28 to 24.98. To calibrate most of this out you can divide amp hours of the battery by the actual current. once done this should be pretty stable over ambient temperatures and shared by either method. The advantage of the constant current is that the current could be set and held over the battery voltage change at 25 amps. This would save you the math of figuring out how many minutes it should take to reach the LVD.
    We can have a go at the constant current method if you like to see what the difference is.
    Do you need different currents for different batteries or do you have curves for the ones you test at 25 amps?
     
    Last edited: Feb 28, 2012
  7. tkc100

    tkc100 Member

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    ronv
    Deep cycle lead acid batteries are rated by two different methods. Amp hours or Reserve Capacity. Here in the U.S. Reserve Capacity seem to be the preferred method of advertizing the batteries capacity. Normally the listed amp hour rating is what is referred to as the C20 value.
    The capacity of a battery over a 20 hour period. So a 80 amp hour battery can deliver 4 amps for 20 hours before dropping below 5.25 volts.
    Reserve Capacity is the amount of time a battery can deliver 25 amp before dropping below 5.25 volts. Most battery manufacturers will also provide a Reserve Capacity rating for a 75 amp discharge but given all the problems we are having with a mere 25 amps I think 75 amps would not be practical.
    I think making a test instrument for determining the batteries capacity using the Reserve Capacity rating is the best way to go. I have yet to encounter a battery that didn't have a Reserve Capacity rating. I can not say that for AH.
    The tester need to produce a 25 amp load. This is the same no matter what band or type of battery is being tested.
    All battery capacity rating are based on completely discharging the battery which in practice is not a good thing to do. Hence the three position switch. Since the load is always 25 amps it is possible to just discharge a battery to a 50% SOC (a much kinder thing to do) and then double the value for comparison to the manufacture specification or more importantly to its last test cycle. Deep cycle battery get the most bang for the buck if their discharge cycle is held at about 50%. It's not a death sentence to discharge them to 80% on occasion but total discharge greatly shorten their life.
    With a battery rated in AH it's not a simple matter of doubling or halfing the the C-20 value, Peukert law comes into effect and make thing far messier.
    I am concerned about the load. It seems to me that without a consistent 25 amp discharge any time values would at best be relative.
    After looking as some of the hardware available I think it would be possible to purchase, make modify an enclosure for both a load resistor and a power transistor.
    Does the circuity for a constant load present a problem?
     
  8. ronv

    ronv Well-Known Member Most Helpful Member

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    Ahh, I get it now. They just give you a time at 25 amps because they have taken into account Peukerts law for different sized batteries. At least for big ones.

    I think if I was doing this for myself and only building one or two I would build the one we have now but "trim" the load resistors to get 25 amps. For example if you add a 20 ohm resistor in parallel with the 4 1 ohm resistors their value will now be .247 ohms. For the 50% capacity value the average voltage is about 6.2 volts (6.3 to start 6.1 at end) giving a value or 25.1 amps or 0.4% high. This turns out to be a 1 minute error out of 180 minutes for a 150 AH battery. The same is true for the 80% discharge number. The 100% number has a larger error of 6%, but I suspect this one is flakey anyway because of the steepness of the curve in this area. Let me know what you think.

    I am looking at the constant current approach trying to work around having a bunch of big heat sinks. How big are the devices you are using now?

    This is what I have been looking at for my info.

    http://www.electro-tech-online.com/custompdfs/2012/02/lead_acid_battery_charging_graphs.pdf
     
    Last edited: Feb 29, 2012
  9. ronv

    ronv Well-Known Member Most Helpful Member

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    Here is a start on a current regulator. Most of this circuit is a net add to the open loop one we have been working on.
     
  10. tkc100

    tkc100 Member

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    I really am not wanting to make this any more complicated than is absolutely necessary but I am having a hard time understanding how this tester could provide accurate, comparable data without a constant load.
    I really thought we were ready to take this thing to the workbench until this matter arose.
    The new additional circuitry does make thing more complicated but don't you think it will make for a far more reliable test instrument.
    If you can help me understand this whole matter of resistance.
    If a resistor is rated at say 10 ohms but that value changes with heat and heat is a natural byproduct of dissipating energy how could you maintain a constant load in any manner other than with a circuit such as you have proposed?
    When a resistor is derated it total wattage goes down right?
    Is that a result of its resistance going up?
    If resistance varies with temperature how can you ever arrive at a norm of say 10 ohms?
    If the resistance is 10 ohms at say 80 degrees and higher as the temperature goes up how can you ever calculate resistance in a circuit?
    I know this is a lot of what must be basic questions but it has got my head spinning a bit.
     
  11. ronv

    ronv Well-Known Member Most Helpful Member

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    Now I understand what is bothering you.
    While it is true that the total wattage the resistor can dissapate goes down with temperature it is not because the resistance goes up but because of the maximum temperature the resistor can tolerate without damage. As the power in the resistor goes up it heats (think light bulb) so the power it can safely handle must go down. This is the purpose of the derating curves.

    While it is true the resistance may change with temperature the amount is really very small (by design). For example the ones shown in post 62 the change is only 50 parts per million per degree C. So If the temperature goes up by say 200C the resistance of each resistor would change by only about 0.01 ohms.

    Hope this helps.
     
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  12. tkc100

    tkc100 Member

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    ronv
    Thanks for the explanation.
    I have pretty much learned all that I know on an as need to know basis and project like this provide a lot of education.
    I read the article you posted as a link on how SOC relates to voltage. As a matter of fact I have a copy of the very same article in my saved documents. It is accurate an well written but fails of point out one very important fact. A lead acid battery can be serviceable and fully charged and yet not have its original capacity. It's a fact of a batteries life that as they age they lose capacity. That is the reason for all the brain damage in our attempts to make a load tester. There are some very high dollar instruments out that that claim to be able to instantly measure a batteries capacity but they have not stood the test of time yet. The only reliable and proven way to judge where at battery is at in its life cycle is the load test. Load tests are normally preformed consistent with the batteries design and usage. Automotive batteries are loaded 3 times their AH rating for only 15 seconds. Similar to a starter motor's requirements. Deep cycle batteries are loaded with a much smaller load for a longer period of time constant with their expected usage. I am now in contact with an engineer from Concord battery discussing this topic and looking for some charts to add to my collection.
    There are a number of battery monitors out there, in fact I have two in service right now. They measure amperage in and amperage out. Then with the aid of a microprocessor they do a lot of calculating to determine the battery SOC. They do a very good job providing you can calibrate them to the battery banks capacity. This is relatively easy for a new set of batteries but without a periodic capacity test they become less and less accurate as the battery age.
    I am also involved in a conversation with the maker of one on these monitors about this subject. So there maybe more to come but as far as I know and as far as the instruction have taken me this is a fact.
    I looked over your proposed design for a constant load and I will admit my limitation in understanding it. I appears as though it's another comparer a load and some switching transistors to handle the current. I assume U3 is for an initial calibration. Is that at least in part correct? One thing I didn't notice was the big 6" 100 watt resistor, we had talked about earlier. The biggest resistor I can see is a 15 watt one. Would this allow us to package the whole thing up in a smaller more manageable box?
    I am having fun with this project and I do appreciate your patience and understanding. Perhaps soon it will be on the workbench and then in the field being tested.
     
  13. ronv

    ronv Well-Known Member Most Helpful Member

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    The constant current circuit works like this:
    Most of the current goes thru R6 and R4.
    What is needed to get up to 25 amps is supplied thru Q1.
    This can be adjusted by U3 as you stated.
    Alas, no this will be larger. There are 7 15 watt resistors and 1 10 watt ones. In addition the transistor needs about a 3X5X2 inch heatsink to keep it and the reverse protection diode at the top cool.
    I'm still thinking that if the completely discharged test is not to important the first circuit would serve you well.
    Maybe something else will pop up.
     
  14. tkc100

    tkc100 Member

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    I am now trying to digest all that has been discussed in particular this from an earlier posting.
    As an example (see attachment) a 186 AH battery has a reserve capacity rating of
    390 Minutes @ 25 amps @ 80º F
    So based on your figures a 50% discharge test would only be off + or – 1 (.78) minutes and the same values would hold true for the 80% discharge test.
    I think this would be an acceptable margin of error.
    Any variations in ambient temperature would have negligible effect on the test.
    I would prefer to never take a battery to a completely discharged state of 5.25 volts for two reasons. One the time involved and Two because it is very hard on the battery. There fore in my option the complete discharge setting could be eliminated. However having said that all manufacture’s specification for reserve capacity are based on the 5.25 value. I have made the assumption that the bottom half of a batteries’ capacity is equal to the top half and that there is not some sort of discharge curve. I am awaiting a response from several battery makers on this question. I threw in the complete discharge test in the event I can’t get a satisfactory answer and I need to come up with my own (testing) specifications.
    Now we started out talking about 4 1 ohm 50 watt resistors but then decided that after derating or without a very large heat sink that would not work.
    Then we took a look at one 100 watt resistor and now you are suggesting 4 1 ohm resistors and one 20 ohm resistor in parallel.
    If we go this route of a mean value what would the load bank actually look like?
    Resistance – Wattage – Style
    So that I can try and figure out an heat sink/enclosure for the whole thing.
     
  15. ronv

    ronv Well-Known Member Most Helpful Member

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    Yep, about a minute each way.

    Yes, I think the best bet is the 4 1 ohm 50 watt resistors in parallel like those in post #62. These are bigger, but need no heat sink. But, having said that they will be HOT so should be protected from being touched and not in the same area as the electronics. Good air circulation is needed even though they are not running at 100% of rating. The 20 ohm is only a 3 or 5 watt resistor. Like this: http://www.mouser.com/catalog/specsheets/cp.pdf It is also in parallel with the 4 1 ohm resistors.
    I'm having trouble making all the numbers match with the batteries. Seems to me that 50% reserve capacity should be the same as 50% discharge. So when I look at the curves in the article I posted I can find a C10 curve with 12.1 volts at 50%. Translating that to 6 volts gives 6.05volts. Pretty close to your 6.1 volts at 25 amps. But then I follow that curve on down to 80% and it is only about 5.75 volts. Quite a ways away from 5.99 that we have been talking about. What am I missing or is there that much variation between those curves and the specs you get?
    So with the battery you posted, where could you return it to the manufacturer assuming the 390 minute spec.

    PS still thinking about an easy way to get a better totally discharged number.
     
  16. tkc100

    tkc100 Member

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    I just reviewed the data sheet for these resistors.
    NHL-50 50 watts
    Tight tolerance of 5 % for values above 1 Ω
    Excellent stability in operation (< 3 % change in resistance)
    non-inductive
    Horizontal Thru-Bolt mounting
    The is no mention of a heat sink. In fact it appears to me the way they are mounted would pretty much make a heat sink useless. I do understand the need to mount them in free air and protect them from being touch.
    Do your computation still hold up with the variable listed of 3 and 5%?
    The derating chart does show a decline in the resistor ability to carry current as the temperature goes up. We need around 165 watts will these resistors still have that ability with the temperatures generated by 165 watts? How in the world can you determine where to place usage on the derating chart.

    I will look over the material I have and reconsider the actual voltages representing SOC. There maybe some differences between different battery construction/manufactures etc.
    That is one reason I was glad to see you built into your design a trimming pot. So that initial calibrations could be made but also so it could be tweaked a bit one way or the other.
    Attached is a chart that maybe easier to read than a graph. These values are for a standard deep cycle flooded cell lead acid battery. There maybe some small differences between flooded cell, glass mat, gel cells and so on. I will check into that.
     
  17. tkc100

    tkc100 Member

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    Sorry I didn't see the other question.
    Lead acid batteries are consider bad if they have less than 80% of their rated capacity.
    This is a topic I am attempted to get an answer to from the battery people.
    It is a fact that a battery loses capacity as it ages but I have never seen a graph depicting that.
    I don't know where the number 80% came from or what it really represents.
    I don't know if a battery expected performance is fairly gradual though the first 20% and then drops of steeply or what.
    Hopefully I will be better informed soon.
     
  18. ronv

    ronv Well-Known Member Most Helpful Member

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    I think I understand a little better after looking at some battery specs. They seem to take into account the C value in the minutes. How they do this I can't figure out. Maybe thru testing of each battery type and data collection. This 25 amp rating must only apply to batteries of a large capacity when compared to 25 amps.
    Anyway, I think your last chart is for a battery under no load. I would use the C10 values from the discharge curves for the size batteries you are working with. Just what makes the most sense to me. If you plan to use this for warrentee the eroor in the 100% discharged is probably good. My experience with golf carts would suggest they kind of go off the cliff when they go bad.
     
  19. tkc100

    tkc100 Member

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    I agree with you that when a battery actually fails it is usually apparent.
    However I am only marginally concerned with any warranty issues.
    Some of the problems I am faced with are as follows
    I have battery monitors on several of the installations.
    Victron Energy BMV 600S
    http://www.victronenergy.com/manuals-per-product/?productid=188
    They do a good job of monitoring the batteries and more importantly they provide a read out that is comprehensible to the average user. But they can only provide reliable information if they are properly calibrated upon installation and then again as the battery bank ages.
    Secondly everything about a battery bank is about balance. The controller, inverter, and charger must see however many batteries that are in the bank as one. If there is an imbalance the entire bank becomes more or less dysfunctional. As batteries age they lose capacity and in a perfect world would do so evenly. That is not usually the case.
    So having the ability to calculate the battery's capacity take on more of a meaning than just failure.

    There are two different and not easily reconcilable methods to measure a battery's capacity.
    Amp Hours and Reserve Capacity.
    Don't try to compare the two. The figures are arrived at via two completely different methods. Believe me I have tried to find a formula to convert one into the other and they are only roughly accurate.
    From the specification sheet I attached earlier there was an 186 AH battery. That is calculated a the C-20 value. That means this particular battery should be able to deliver 186 amps of output over a 20 hour period or it can maintain a 9.3 amp load for 20 hours. That does not mean it will deliver 20 amps for 9.3 hours, in fact it won't. Peukert law comes into effect.
    To make a tester based upon AH it would require an accurate variable load other wise any tester made would be specific to only one type of battery.

    For this same battery the manufacture lists the reserve capacity.
    It says that if you place a 25 amp load on the battery it should be able to maintain a voltage equal to or greater than 5.25 volts under the 25 amp load for 390 minutes. At the point the battery hit 5.25 volts it is considered completely discharged.
    The C numbers used in the AH rating have no meaning here.
    How they arrive at these figures I don't know. It maybe though actual tests or through some fancy computer model. The beauty of it is that all deep cycle batteries list the reserve capacity and the ratings are based upon either 25 or 75 amps. Both values are given. Therefore a tester such as we are trying to make need only provide a load of 25 amps and measure the time it take to reach 5.25 volts for a comparison to the manufacture's specifications or a comparison to the last testing cycle or a comparison to the other batteries in the bank.
    Make sense:confused:
    One fixed load and time it takes to reach complete discharge no matter which type or size battery is being tested. In fact the loads are the same for 12 volt batteries the only difference is the terminal voltage is 10.5 instead of 5.25. It really makes things much simpler.

    Just this morning I was in a conversation with Trojan battery. The question I posed to them was, Is the top half of a battery's capacity equal to the bottom half?
    Can we discharge a battery to 50% or its capacity double that figure and arrive at 100%.
    I hope so however the question has been referred up the chain to the engineering department.

    The answer will of course effect both the design and usage of this proposed tester.
    I will ask and get conformation on the trigger point for 50 and 80% also and I will make certain that is under the 25 amp load.
     
  20. ronv

    ronv Well-Known Member Most Helpful Member

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    Well I did figure out one part of the puzzle. Your right the Purkett exponent is the key to matching the amp hour curves to the reserve capacity. I only did 2 batteries as all the numbers are hard to find. But I did 2.

    T105 - 225 AH - Purkett =1.24 -RC 447 min.
    T145 - 244 AH - Purkett=1.134 - RC 530 min.
    Then using this calculator you can verify the RC number knowing the 20 hour rate.
    http://www.autoelex.co.uk/My_Homepage_Files/Peukert.xls

    Still thinking about if the 50% voltage on a 225 AH battery will be the same as on a 300 AH battery, but it seems like if the 100% number is always 5.25 it might be.

    Fun project.:D
     
  21. ronv

    ronv Well-Known Member Most Helpful Member

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    Final Voltage

    It looks like the curves only come together at the 100% discharged voltage. Although they may be able to be compensated for if most are similar in size?
    The way I interpet the curves are like the difference between the 100 amp line and the 200 amp line is the same as you might expect from 2 different batteries - one having twice the capacity of the other.
     

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