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I want to measure the maximum current a AA battery (full charged) can deliver for a short period of time (let's say one second)

I have in mind to do this with a multimeter :

  • setup multimeter to measure amps, on the highest value. My current multimeter has an input rated 10A (for max 30 seconds)
  • plug negative probe to negative side of battery
  • plug positive probe to positive side of battery (for one or two seconds)
  • read value on device

Usually when you need to measure current for given load, the multimeter is wired in series in the circuit. Since here, there is practically (almost) not load at all I have decided (to best of my knowledge) to connect the multimeter directly to the battery.

This will create a short circuit. In theory, current should be incredibly high (because resistance is almost zero), but since AA battery has internal resistance (IR), it will be limited (fortunately).

Is this a good idea? Or will it kill the device or give not the desired effect for measurement ?

Note : I have in mind doing this using AA battery only, doing this on mains AC current for example would (I guess) be very dangerous (220V AC short circuit) and destroy device immediately.

by current the battery can deliver I mean ampere not ampere-hour (Ah) which is a different unit.

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    \$\begingroup\$ Any battery from a decent reputable source will have a data sheet that will tell you what you need to know. Read the data sheet and if the battery doesn't have a data sheet don't bother testing it and get a battery that has a data sheet. \$\endgroup\$ – Andy aka Apr 5 '15 at 22:22
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    \$\begingroup\$ Summary of my answer: This has worked well for me on many occasions. Safety needs to be considered but has so far not provd an issue for me. \$\endgroup\$ – Russell McMahon Apr 6 '15 at 0:46
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    \$\begingroup\$ @Andyaka : here is the datasheet : media.professional.duracell.com/downloads//datasheets/product/… is there any info about maximum current delivered (not directly but something that can be calculated from data there)? \$\endgroup\$ – tigrou Apr 6 '15 at 22:23
  • \$\begingroup\$ Read the data sheet and work it out is my advice. It's what we all have to do in engineering and, if you don't understand this or that then raise a question. Data sheet understanding is paramount in EE and this is not something I say lightly. \$\endgroup\$ – Andy aka Apr 6 '15 at 22:39
  • \$\begingroup\$ @Andy, the data sheet for an AA battery will not give a max current. It will either give a max sustained current, max pulse current, or a graph showing capacity as a function of current drawn. I've just looked at several of them, and there is no indication of a safe max current or of a max possible current. The max sustained current has nothing to do with safety or possible current, only with recommended usage for reasonable battery life. \$\endgroup\$ – Phil Goetz Jan 25 '17 at 20:40
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WARNING: If "trying this at home" be aware that there is a small potential for significant hazard occurring - see below.

What you propose is a viable and useful and potentially dangerous method.
I consider it is extremely unlikely that you will harm yourself doing this but need to note that the possibility exists.

I would not do this with other than single Alkaline or NiMh cells. I would be extremely careful with cells above AA in size. I would not consider trying this with LiIon or LiPo of LiFePO4 cells - which all have higher terminal voltages, high potential discharge rates, and a known tendency (except for LiFePO4) to "vent with flames".

I have used the method for testing partially used Alkaline batteries for many years with good success. I have never had any problems with it BUT this does not mean that everyone else will be so lucky. Comments at end re what might go wrong.

To determine the "degree of goodness" of an Alkaline AA 1.5V battery I do two things.

  • Measure cell open circuit potential. This is a very safe and non damaging method. An unused Alkaline battery that has not used up much of its shelf life will have more than 1.6V potential - typically 1.65V. This is higher than the cheaper Carbon-Zinc / Le Clanche / Heavy duty cells and is a reliable way of determining both that a cell really IS an Alkaline one and that it is essentially new. A cell that gives more than 1.6V does not need to be "tested" by current discharge as described below (but can be if desired).

  • Measure cell short circuit current for about one second using the 10 amp range on a multimeter. The meter's internal resistance, lead resistances, plug in connection resistances and contact resistance with the battery all are potentially significant resistances in this test so the results will vary somewhat between meters and depending on how well the probes make contact and how well the lead-plugs make contact in the meter sockets. Despite these potential differences (pun noted) the test is useful and reasonably repeatable.

Information only: My most usual reason for doing this test is to determine which cells in a batch of cells are unused, and which of the used ones are suitable for use in a high output camera flash. The flashes concerned present a heavy load. Capacity is probably around 100 flashes - depending on energy taken which varies with photo environment - a flash into a dark large room takes a full charge whereas when photographing a light coloured subject at close range only a small fraction of the stored energy is used. When used repeatedly to exhaustion the batteries are too hot to handle when removed from the flash - probably 70 degrees C ! The average power supplied by the batteries at full load is probably 50 to 100 Watts. The batteries need to be in good condition to supply this.

The short-circuit test typically returns a result of 5 to 10 amps for good quality new cells, with the current falling slightly during the approximately one second test period.

The results for used cells varies considerably. Anything in the 3 to 5 amp range means that the cell is liable to be useful for flash use. A result of a few amps means the battery is still useful for low drain equipment such as a clock or electronic scales. Less than that the cell is probably best discarded.

While the above test is used for AA Alkaline cells it is also usable for NimH AA cells - with more risk. A NimH cell MAY be capable of higher discharge rates when fully charged, although the resistances present in this test will usually limit current to about the same values. I just tried this with a fully charged 2000 mAh Eneloop AA cell (Panasonic Chinese made version). This peaked at about 7 amps. The Eneloop cells are of lower capacity than the market leading quality AA NimH cells, but have a much longer shelf life and higher terminal voltage at given discharge level. I'd expect them to give similar results to higher capacity "normal" AA NimH cells.

On a few occasions I have been silly enough to carry cells in my trouser pocket a number of charged AA cells and on 3 such occasions was also unlucky enough to have them form a stable circuit with various coins, keys etc in my pocket. Pocket temperatures rose to well above pain level nearly instantly and burns were a definite possibility. Pocket contents had to be shed with indecent hast on each occasion. While no cell ever gave any indication of mechanical damage, if one had 'exploded' in some fashion under such abuse I'd have been the sorrier but not surprised.

A meter set to 10A is unlikely to be damaged by shorting a single AA NimH cell for short periods. More than one cell in series or larger than AA may cause dismemberment of cell or conflagration or dismemberment of meter internals. Some meters are fused on their 10A range but many are not (and most cheap ones that I have seen are not). Extended 10A range over-current use may destroy the 10A shunt and possibly the meter itself a few milliseconds later.

Hard shorts on batteries MAY cause disproportionate decreases in capacity compared to actual energy taken and MAY cause long term permanent degradation in secondary cells. I have not noticed that this is the case but YMMV.

After shorting the Eneloop cell mentioned above for a total of about 5 seconds at 7A it took about 40 mAh of charge to restore it to full capacity. Energy out ~= 1V say x 7A x5 seconds = 35 Joules. Restoration energy ~~= 1.4V x 40 mAh ~= 200 Joules. This test sample (1 item) is too small and uncontrolled to allow any good conclusion, but is interesting.

Worst case it seems likely that under full short circuit a cell may dissipate around 10 Watts internally, and usually less than that. My informal inadvertent pocket testing of NimH cells under high discharge for probably 10-20s indicates that they will tolerate this without self-dismantlement (at least for the small sample I have experienced), and my use if a flash on many occasions of AA Alkaline cells so that they become too hot to handle suggests that that too tolerate heavy discharge and high temperatures "well enough".

So, I would not expect that short circuit testing any AA Alkaline or NimH cell as described above would be physically dangerous. But if it ever did turn out to be, I would not be totally surprised.

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  • \$\begingroup\$ Nice answer, you went a lot further than I would. Btw I found a datasheet for an alkaline battery, but contrary to what I said in my answer, it doesn't help much since the low side of the range is 1h: data.energizer.com/PDFs/E91.pdf \$\endgroup\$ – fceconel Apr 6 '15 at 13:53
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    \$\begingroup\$ @fceconel Over the past 7 years or so I've been involved with the use of 500,000+ AA NimH cells in things like this - talking about batteries tends to happen after a while :-) \$\endgroup\$ – Russell McMahon Apr 6 '15 at 14:01
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If the current is too high it will blow the fuse on the multimeter, or blow up the battery.

Wikipedia says the Energiser AA battery has an internal resistance of about 0.15R at room temperature. This gives around 10A current. However, the internal resistance of the multimeter may now have an effect, reducing the current.

Instead, buy a very small resistor, e.g. 0.01R, with high power rating and put that across the battery. Then measure the voltage across the resistor and use Ohms law to calculate the current. This way you protect your multimeter and the shunt resistance of the multimeter doesn't have an effect.

Note

The above answer assumes an Alkaline AA battery. As Spehro says, other types can be dangerous.

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  • \$\begingroup\$ Aside from any danger, a meter is a poor choice to measure a signal that is changing that rapidly. Most don't update more than a few times a second and you couldn't read it if they did. This value seems likely to change significantly in tens of milliseconds. Measuring voltage across a small resistor with an oscilloscope set to capture the single sweep will work much better. \$\endgroup\$ – Ross Millikan Apr 5 '15 at 23:04
  • \$\begingroup\$ @RossMillikan sounds good. You think that the battery change current that quickly? Perhaps a multimeter with max/min/mean storage would work well too. \$\endgroup\$ – geometrikal Apr 5 '15 at 23:24
  • \$\begingroup\$ I could believe there is a very quick drop when the circuit is completed. It sounds to me that OP wants to catch that. I could easily be wrong on either count. I certainly don't want to use a meter to catch something that only lasts one second. \$\endgroup\$ – Ross Millikan Apr 5 '15 at 23:30
  • \$\begingroup\$ @RossMillikan - see my answer - I have found that this method works very well for me in practice. \$\endgroup\$ – Russell McMahon Apr 6 '15 at 0:47
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It certainly won't kill the multimeter, but the voltage will drop so quickly (and the current with it) that you won't be able to measure much. One second for such a setup is not, really, a short time. One possible setup would be to test with a trimpot or potentiometer in series, logging the voltage over time, and then analyze the results. But that would be a lot of work. You'd need to start with the pot at a maximum value (let's say 500R) and log the discharge curve. Then reduce gradually and repeat, each time with a new battery, until you reach a value that discharges the battery faster than 1s. Note that in this case you'll use the multimeter to measure the voltage, not the current, since it'll be easily calculated and the most important thing is to know when the voltage gets below the minimum accepted value for your circuit.

But usually the battery manufacturer already did such work for you. If you can find the datasheet for the specific battery you want to use, it probably has this information.

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  • \$\begingroup\$ When I'd first read your question I was thinking about an alkaline battery; but if that's not the case @Spehro has a very good point and there are serious safety concerns involved in doing this test without a load. \$\endgroup\$ – fceconel Apr 5 '15 at 22:18
  • \$\begingroup\$ See my answer - I have found that this method works very well for me in practice. He wants very high current tests and direct shorting with a meter on eg 10A range works well in practice. \$\endgroup\$ – Russell McMahon Apr 6 '15 at 0:48
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This can indeed be dangerous, especially if it's the kind of cell that has a high current capacity (eg. NiCd). It's not dangerous, in my experience, if done for a short time with alkaline and zinc-carbon cells, but still prudence would suggest certainly wearing safety glasses and containing the cell in something non-flammable. NiCd cells (even benign-looking 9V batteries) have been known to violently explode when short circuited and non-protected lithium cells to catch fire- and NiMH cells can vent hot hydrogen gas and electrolyte under some conditions. Short-circuit current from NiCd cells can greatly exceed the 10A rating of your meter, so it's possible the meter or test leads could be damaged.

It won't really tell you much useful for normal operation- you could look up the voltage across your meter when reading 10A (maybe 100mV) and get some kind of an estimate of the internal resistance, but there are electrochemical effects ("polarization") that will cause the short-circuit current to drop quickly from the peak. As the cell discharges the internal resistance goes up, so it's not going to give you a good idea of what happens during the cell life if you're drawing current in short pulses.

If your purpose is to get an idea of the maximum fault current (for example for the breaking current of protection circuitry) it could be a useful exercise.

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Method 1. For safely and accurately measuring the near short-circuit current from a battery you need to set up a pulsed load. Such a load can consist of an oscillator with low duty cycle, say 10 ms pulse every second, driving the base of a NPN power transistor such as 2N3055. Connect a 1 ohm 1W resistor from battery +ve to the transistor collector; emitter and battery -ve are connected together to the oscillator ground. Use the oscilloscope to measure the voltage pulse across the resistor: a 10V pulse means thre battery is delivering 10A current pulses. Note that this method measures using a near short circuit; it is difficult to get much closer to a true short circuit.

Method 2. This method measures the internal resistance of the battery without drawing current. Connect a 1000uF capacitor (electrolytic type, note polarity!), the battery, a 50 ohm resistor and an oscillator in series, where the oscillator can be a lab instrument capable to deliver say 100 Hz 1V r.m.s. sinewave into 50 ohm. Use the oscilloscope to measure the peak-to-peak voltages across the battery and across the 50 ohm. Their ratio gives the battery internal resistance which one can expect to be the only limit on the initial current if the battery should be short circuited.

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    \$\begingroup\$ A 1.5 V battery isn't going to deliver 10 A into a 1 Ohm load. But good idea to keep the pulses short. \$\endgroup\$ – tomnexus Apr 6 '15 at 8:22
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The test probes will present a significant source of resistance, and you won't get an accurate measurement of the power output unless your final design also uses hand pressure of test probes to the cell contacts.

Consider building a test battery holder that has banana plugs you can connect to your multimeter. Look for a battery holder that has good mechanical contact. This will give you more power from the battery by decreasing the contact resistance - and in high current applications like this it will be a significant source of power loss.

But even battery holders (at least the common ones) aren't meant for more than 1A current. Groups such as flashlight builders and radio control model enthusiasts run into this issue and of the several suggestions I've seen, it looks like the best is to use copper braid to face the spring contacts, then solder a heavier gauge wire to the copper braid. This presents a high current, low resistance contact to the battery, and allows you to get the most current from the battery that you can.

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