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Why do we need to apply a load to a battery, when testing its voltage?

I've been testing battery capacity so far using a multimeter and just connecting the probes to the positive and negative side of the battery, which seems not to be correct.

But, why is this? Why do we need to test batteries with a load and why can't we just simply check voltage with no load?

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You need to put a load on the battery to see if it has any charge left.

Without a load, it may show an acceptable voltage, but when you actually try to use it the voltage drops because the battery is nearly dead.

So to see if a battery is really usable you must measure the voltage when the battery is connected to a load. Like this:

Dead Battery, no load, 1.4 Volts

Dead Battery, load of 100 Ohms, 1.0 Volts

Good Battery no load, 1.5 Volts

Good Battery, load of 100 Ohms, 1.4 Volts

Those numbers are just representative - do NOT use them to actually measure your batteries. Check the unloaded voltage of a good battery, then check the voltage of a good battery under a typical load. Use that typical load to test other batteries. That is to say, figure out the equivalent resistance for the load and use a resistor of that value in your test.

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  • \$\begingroup\$ Thanks for the explanation, so if I have a device that uses as an example 40Mah then when the battery goes below the 40Ma then it will be considered void for that device am I right? \$\endgroup\$ – Matias May 5 '15 at 12:57
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    \$\begingroup\$ You are mixing capacity (mAh) and load (mA.) If your typical load draws 40 mA, and putting that load on the battery causes the voltage to drop too much then that battery is "dead." \$\endgroup\$ – JRE May 5 '15 at 13:05
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Quite a lot of battery chemistries will, if left alone, raise their terminal voltage. But there may be no capacity behind it and it will drop as soon as you try to use it. So a load is connected to the battery to verify that it is actually useful.

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  • \$\begingroup\$ Ok, so if I understand correctly, you're talking about the Ma capacity of the battery being low \$\endgroup\$ – Matias May 5 '15 at 12:55
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    \$\begingroup\$ No, he's referring to the "state of charge" (SoC) of the battery - some batteries, when their SoC is low, will still have a 'normal' open-circuit voltage. the "mA.Hour" spec for a batter is how much energy the battery can hold when fully charged, it's "design capacity", NOT how much energy is left in the battery at any given moment. \$\endgroup\$ – Techydude May 5 '15 at 13:19
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As typical Alkaline and other batteries go bad or get weak, they develop greater internal resistance. With no load or very little load you could say that there is a voltage divider formed by the internal resistance and the high resistance external "load". The high external resistance will show a high or full voltage drop. With a good external load, low resistance, the internal resistance of the battery will experience a greater voltage drop, meaning you'll see lower voltage externally.

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