9000mAh Power Bank current reduction problem

i have a 7.4V 9A battery(six 3.7V 1.5A cells ) and i want to make a power bank, output of 5V 2A needed ,i obtained 5v but how do i reduce 9A current to 2Amps without losing much power,7805 gives 5v but gives off 7Amps,someone help me

by Adish3702
January 09, 2021

You'll have do give more information on what you've done to get 5 volts from six 3.7 volt cells. as there is no way to reconnect them to get 5v. I think you are working under a misapprehension about current rating. Your 7.4v battery does not "give off" 9A. It is capable of supplying up to 9A without damage. If you choose to take less than 9A from it, it will run longer between charges but no damage will be done.

by Foxx
January 10, 2021

Perhaps you are confusing 9000mAh with current rating. True enough 9000ma is 9 amps but 9000mAh (9 Ah) means that the battery will supply 9A for 1 hour. It will also supply 4.5A for 2 hours, 2.25A for 4 hours or 18A for 1/2 hour etc., (approximate figures).

by Foxx
January 10, 2021

first of all its very easy to get 5 volts from 3.7v cells,connect them in series, and what i've done here is connected two cells in series each which is 3.7+3.7 =7.4V(now i have 3 of them) and then connected them in parallel so now its 7.4v 9000mAh(1500×3) i hope this much info is enough.

by Adish3702
January 10, 2021

???? You want 5v and have 7.4v. This is 48% overvoltage and, depending on the circuit you are feeding may very well cause damage.

by Foxx
January 11, 2021
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