Measuring ampere of AA battery

Discussion in 'General Electronics Chat' started by ttarique, Jun 11, 2013.

  1. ttarique

    Thread Starter New Member

    Jul 2, 2010
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    Hello all:
    I am a new bee. I want to measure the ampere of AA battery with a digital multimeter. Do you know how I can do this task? Do I need a resistor for it?
     
  2. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Why do you need to measure current of AA battery? Batteries don´t really have a current that you could measure, they have voltage. Normally you measure current that some load draws, not what the battery is able to provide for a brief moment.
     
  3. crutschow

    Expert

    Mar 14, 2008
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    Exactly what is the purpose of this measurement?
     
  4. paulktreg

    Distinguished Member

    Jun 2, 2008
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    Take a look, for example, at this datasheet for the Duracell PC1500 AA battery.

    "Delivered capacity is dependent on the applied load, operating temperature and cut-off voltage. Please refer to the charts and discharge data shown for examples of the energy / service life that the battery will provide for various load conditions."

    There really are too many variables to get a definitive Ah figure.

     
  5. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    There is a big difference between ampere and ampere-hours. And the OP probably doesn´t have a clue what he actually wants to know.
     
  6. Austin Clark

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    Dec 28, 2011
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    ttarique, the amperes the battery will supply is primarily dependent on the load, as given by ohms law: I = V/R

    As a battery discharges, it's voltage level drops and it's internal resistance increases, lowering the total current supplied over time, even if the load remains the same.
     
  7. ttarique

    Thread Starter New Member

    Jul 2, 2010
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    This is what I think--how to measure the ampere(current) of an AA battery. Connect the positive terminal of the battery to a 10 ohm resistor and then connect the resistor with the negative terminal. Then put a multi-meter between the positive and negative side of the battery. The digital multi-meter will show the correct ampere measurement.
    Please let me know whether I am doing the measurement right. Your help is appreciated.
     
  8. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Compleltely wrong. The ampere meter should be in series with that 10ohm resistor, then you are measuring the current through that resistor. You can´t measure current of a battery, that doesn´t really make sense. You could measure the short circuit current of that battery but I have no idea what you would need that for.

    When you put multimeter between the positive and negative side of the battery you need to measure voltage, beacuse the current through the tiny resistance of the multimiter will rise to hundreds of amps and either blow the fuse or destroy the meter.
     
  9. ttarique

    Thread Starter New Member

    Jul 2, 2010
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    My question is this—when I buy a battery, let’s says a 12V battery, the ampere-hour is 7.5 ampere. How did they measure 7.5 ampere-hour? Can you explain, please? I am trying to build a home-made battery. I know how to measure the voltage but I do not know how to measure the AMP-Hour. Please explain. Thanks.
     
  10. kubeek

    AAC Fanatic!

    Sep 20, 2005
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    Ok, now we´re getting somewhere. Basically, you load the battery with a constant current, and measure the time it takes it to get below the threshold voltage that means the battery is fully discharged.
     
  11. WBahn

    Moderator

    Mar 31, 2012
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    As has been indicated elsewhere in the thread, you need to be clear on the difference between amp-hours and amps. They are two fundamentally different things. The first is a measurement of total charge and, indirectly, energy delivered to a load over time while the second is a measure of the instantaneous rate at which charge is being delivered to load at a moment in time.

    There are a number of ways to measure the Ah capacity of a battery and they are all more-or-less comparable. It's important to realize that a battery does not have a single, engraved in stone Ah figure. As noted by others, the effective capacity depends on many factors. So what you want to do is decide what kind of load and what kind of conditions are most relevant to how your homemade battery will be used and then test it under something that approximates those conditions. To a first approximation, this might be figuring out what size resistor approximates the load placed on the battery. Then take data by first measuring the open circuit voltage of the battery and then put the resistor across the circuit and measure the loaded voltage right away. After that, measure the loaded voltage at regular intervals. If the battery is only going to last a few minutes, then take measurements every 10s or so. If the battery is going to last several hours, then take measurements every 10min or so. If it's going to last several days, then take readings every few hours. Basically, try to get as many data points as is reasonable over the expected life of the battery, shooting for at least twenty or so measurements.

    When you plot these, you should expect to see a curve that slopes downward at a more-or-less constant rate but that then drops quickly at the end. The point at which your battery "died" is just before the quick dropoff started.
     
  12. MrChips

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    Oct 2, 2009
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    The ampere-hour rating of a battery is not constant and will depend on the load.
    For a simple test, get a 12Ω 25W resistor and connect it across the battery. Put the multimeter on voltage range and connect it across the battery. Monitor the voltage reading for the next 8 hours.

    If you cannot find such a resistor, get a 12V 12W light bulb from an automotive supply store.
     
  13. WBahn

    Moderator

    Mar 31, 2012
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    I think the 12V, 7.5Ah was just an example and not an actual battery he is trying to test.

    For your AA battery, you might try targeting a current draw somewhere in the 100mA range, so pick a 15Ω (anything from 10Ω to 22Ω should do). You initial draw might be at much as 150mA which would be a bit over 200mW, so be sure to get at least a 1/4 W resistor. If you go with a 22Ω resistor then your peak draw should be on the order of 75mA and your power will be right at 100mW, so a 1/8 W resistor should work okay for that, although I would recommend using the 1/4 W (or even larger) resistor so that it's value will be more constant.

    I would recommend not usiing a lightbulb because the resistance of those is a strong function of temperature and, hence, of current meaning that your load is going to be changing significantly as the battery drains and the setups we have been recommending assume a fixed, unchanging resistance.
     
  14. ttarique

    Thread Starter New Member

    Jul 2, 2010
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    Okay guys, thanks for sharing the info. But I am not still getting the concept right. Let’s say I have 12V, 7.5 amp-hour battery. I am measuring the amp-hour. I add a 10 ohm resistor between the positive and negative terminal and started measuring the voltage. Here is the table:

    Voltage Duration

    12 V 1st hour
    11 V 2nd hour
    10 V 3rd hour
    …….. …………
    1 V 12th hour
    0 V 13th hour

    From this chart, how do I measure the amp-hour? Can you please explain step by step? Thanks
     
  15. Austin Clark

    Member

    Dec 28, 2011
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    You can't accurately determine the amp-hour capacity of the battery with this data. The current draw will vary as the voltage on the battery drops. You need to construct a constant-current load. I suppose if you had many many datapoints you could get a decent estimate this way, but it's not practical. Also, under different loads the amp-hour capacity of a battery varies, because more or less power is dissipated as heat within the battery itself.
     
  16. WBahn

    Moderator

    Mar 31, 2012
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    No, you don't need a constant current load, though that does make things easier.

    For the battery loaded with a 10Ω load, you can say, as an approximation, that in the first hour it pulled, on average, 1.15A (11.5V/10σ) and it did that for 1 hour, so it supplied 1.15Ah of charge. For the second hour, it supplied (again, as an approximation) 1.05Ah of charge. In the last hour we approximate that it supplied 0.05Ah of charge. So, in total, our best estimate is that it supplied
    (1.15 + 1.05 + ... + 0.25 + 0.15 + 0.05)Ah of charge, or 7.2Ah.

    Now, realize that no battery is going to drop like this. You can expect it to drop reasonably linearly down to some point (perhaps 8V give or take) and then plummet quickly.

    The more data points you take, the better your approximation will be. You just take the average of the current between the present reading and the last and multiply this by the time between the two readings. Then sum all of those up.
     
  17. KirkMc

    New Member

    Jun 12, 2013
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    Batteries are typically specified over 10 hours. A 150 Ampere hour battery is supposed to support 15 Amperes for 10 hours. At 30 Amperes it may only support 4 hours or even less. Read about the chemistry of each cell for more insight. Being a chemical reaction this current has a temperature specified as well. Thats why you see electric line plugs on the front of cars in Montana. The cold saps the battery and thickens the oil and starting can be a trick.
     
  18. KJ6EAD

    Senior Member

    Apr 30, 2011
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  19. johannheb

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    Jun 19, 2013
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