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Let's say there's a 2Ah battery with a C rating of 2C. So this means that I could discharge 4 amperes for half an hour at best.

Now, there is a default current that we get from a battery right? How do I increase the current discharge for this particular battery? Is there a component I can attach to my battery?

Some sources said that the discharge increases depending on the appliance, and how much it requires, but if I take a lamp, I can pass less than the required current and although not as bright, it will still glow.

If this is true, and appliances can determine the current discharge of a battery, how would I change the amount of current an appliance takes?

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    \$\begingroup\$ The current the battery sends out is determined 99% by the load, not by the battery which, almost always, can supply all the needed current. On the contrary, batteries do not have problems on generating too much current, so fuses are needed for safety (when there is a fault on the load). \$\endgroup\$ Jan 28 at 6:38

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There is no default current that you get from a battery. The current delivered by a battery is determined by its voltage and the resistance of the connected load.

A battery will have an internal resistance that will limit the maximum current the battery will deliver into a short circuit and will cause the apparent voltage of the battery to decrease with higher currents.

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  • \$\begingroup\$ Thanks for your answer!!! So, as I keep decreasing the resistance of the wire connecting the load and the battery, the current flow will increase, until the maximum current level the specific battery can give is reached. Based on this, say I want to supply 12 amps of electric current, using a 6Ah battery with 24 volts, and a c rating of 2, then I would just need to add a wire that has a resistance of 2ohms. \$\endgroup\$ Jan 28 at 8:23
  • \$\begingroup\$ No. If the load is rated to operate on 24 volts, and requires 12 amp, you just connect it directly to the 24 volt battery. You do not need to add any additional resistance. Also, 6 Ah is the C rating of the battery. \$\endgroup\$ Jan 28 at 16:59
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The C and discharge rate is limited by the battery internal resistance, which leads to heating during charge and discharge. If you add cooling to the battery it can sustain a higher discharge rate, but you should consult the manufacturer.

More about C and discharge rate here: https://batteryuniversity.com/article/bu-402-what-is-c-rate

That said, exceeding the recommended discharge rate for a Li-ion can damage the battery and possibly cause an explosion. Most Li-ion batteries sold to consumers have protection built in to prevent this. Nevertheless, there’s plenty of YouTube videos of Li-ion batteries gone wild (especially things like modded vape pens - go figure.)

If your load uses a lower voltage than the battery set, you can use a step-down regulator to increase the current. This lowers the discharge rate, so you could possibly get more run time, depending on the conversion efficiency.

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  • \$\begingroup\$ Thank you for the answer. This provided some more clarity on why I shouldn't exceed the recommended discharge rate \$\endgroup\$ Jan 28 at 8:25
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Let's say there's a 2Ah battery with a C rating of 2C.

The above statement is incorrect.

The correct statement would be 'Let's say there's a C2 rated battery having a capacity of 2 Ah'. It would refer to a 2 Ah battery capable of delivering 1 A maximum for 2 hours without getting damaged.

Consider another example of a C20 rated battery having a capacity of 100 Ah. It would refer to a 100 Ah battery capable of delivering 5 A maximum for 20 hours without getting damaged.

It should be borne in mind that the rated capacity would not be fully available at higher discharge rates.

0.1 C or 0.5 C or 1 C refer to charge or discharge current. For example, charging a 100 Ah battery at 0.1 C would mean a charging current of 100 x 0.1 = 10A. Likewise for discharge current.

How do can I increase the current discharge for this particular battery?

The discharge current of a particular battery would be dependant on its voltage and the load resistance.

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