Will the maximum discharge current double for two parallel batteries?

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SUMMARY

Connecting two 3.7V 3500mAh 18650 batteries in parallel can theoretically double the maximum continuous discharge current from 10A to 20A. However, this is contingent upon both batteries being at the same charge level and having similar internal resistances. Discrepancies in internal resistance can lead to one battery being overloaded, potentially causing damage. Additionally, when using wire, such as 14 AWG, the resistance remains constant regardless of voltage, but the current capacity and voltage drop must be carefully calculated to ensure device functionality.

PREREQUISITES
  • Understanding of battery specifications, particularly 18650 lithium-ion batteries.
  • Knowledge of electrical concepts, including current, voltage, and resistance.
  • Familiarity with Ohm's Law and its application in electrical circuits.
  • Awareness of wire gauge ratings and ampacity, specifically for 14 AWG wire.
NEXT STEPS
  • Research the implications of connecting batteries in parallel, focusing on internal resistance and safety precautions.
  • Learn about battery management systems (BMS) to prevent over-discharge and overheating.
  • Investigate voltage drop calculations in wiring to optimize performance in electrical circuits.
  • Explore the specifications and applications of different wire gauges for various current loads.
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Electronics enthusiasts, hobbyists working with RC devices, electrical engineers, and anyone involved in battery management and circuit design.

Panamanian
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Hello.

Say I have a high powered device that draws continuous 18A. 3.7v
And I have one 3.7v 3500mah 18650 with a maximum continuous discharge of 10A, I know the battery will overheat and will most likely get damaged or shorten its life, or even catch fire.

But If I have another battery (same model).
Now I have two 3.7v 18650 3500mah in parallel. Will the maximum continuous discharge be 20A? or same 10A?

Thanks in advance
 
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It is a bit tricky. If you connect them in parallel you should get 20 A, but there are some details that make it less safe than one can think.

First - it is not guaranteed each one gives exactly 10 A, as a lot depends on their internal resistance, which is never exactly the same. So, while getting 20 A you can in fact be getting 12/8 - with one batter being overloaded.

Second: they have to be connected when they are both charged to exactly the same voltage. Otherwise one will try to charge the other till their voltages are identical, and that's asking for troubles.

That being said, sometimes I fly my Sky Surfer with two 2.2 Ah batteries connected in parallel. While single one is enough in terms of the current, having 4.4 Ah gives twice the flight time.
 
Ok, thanks.

Another question, Will a wire (for example: 14 AWG) resist the same AMPs with 12v vs the same AMPs with 3.7v?

Thanks in advance.
 
Panamanian said:
Will a wire (for example: 14 AWG) resist the same AMPs with 12v vs the same AMPs with 3.7v?

No idea what you mean.
 
Panamanian said:
Another question, Will a wire (for example: 14 AWG) resist the same AMPs with 12v vs the same AMPs with 3.7v?

If I interpret this correctly the resistance of the wire is independent of the voltage across it. However using 12V for a given wire length will produce more current according to Ohm's law.

A particular type and gauge wire as a copper 14 AWG is rated to carry a maximum current depending on a particular maximum wire temperature , the insulation and ambient temperature, called its ampacity.. For #14 it is 20 amps for 90 deg C in an ambient temp of 30 deg C. See http://www.usawire-cable.com/pdfs/nec ampacities.pdf for a chart for various gauges.

A consideration of what gauge wire you choose and its length to and from the the source will depend on how much voltage drop on the wire you can tolerate. Another way to think about it is how much less voltage than the 3.7 V your device is spec'd for will be acceptable. Will it run properly at 3.4V for example? Take #14 wire at 18 A. The wire has a resistance of .00252 ohms/ft. which seems small. If you use 10 ft of wire (5 to the device and 5 back to the battery) the voltage available to your device will be 3.7V (battery voltage) - 10ft×.00252ohms/ft×18A (voltage drop across the wire) = 3.25V(voltage available for your device) Will you device run properly at this voltage? If not then you need to shorten the wire or use a lower gauge.
 

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