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Using LM2623 input 7.4 V @ 700mAH

Pravas Mohanty
Prodigy 30 points
Other Parts Discussed in Thread: LM2623, LM3488

Dear Ti,

I am using  LM2623 to get 5V @  2A required from Li-Ion Battery 7.4V,1500mAH.

Kindly suggest me if anything I am doing wrong.

Regards,

Pravas

  • 0
    TI__Mastermind 19891 points

    Hello Pravas, 

    My apologies for the delayed response.

    The LM2623 is a boost (step up) converter where the regulated output voltage is higher than the input voltage. You need a buck converter (step down) to get 5V output from 7.4V input.

    I would suggest using WEBENCH (our online design tool) to select an appropriate converter. Go to http://www.ti.com/ww/en/simple_switcher_dc_dc_converters/index.html 

    You will see a page like this:

    Enter your design requirements in the WEBENCH Designer panel and click on Start Design.

    Regards, 

    Denislav

  • 0 in reply to Denislav Petkov
    Prodigy 30 points

    Hi Denislav,

    Thanks for reply, I have already simulated through WEBENCH it is sugesting LM3488 I have absolute OK with this.

    But I have taken a prototype open hardware from atmega magictale.

    http://atmega.magictale.com/wp-content/uploads/2010/10/DefendLineII_Schematics.pdf

    In this reference I come to know they are using LM2623  for batery backup system with range from 2. to 14V input output with 3.3 VDC (step-down)

    Also it is specifying on LM2623 datasheet input range.

    Now I am thinking of that above reference may be inconstistent or inefficient.

    If  you clarify the LM2623 is a boost, simple DCDC converter, Buck converter, flyback etc. So if it is DCDC converter  the reference from atmega circuit is OK.

    Best Regards,

    Pravas R Mohanty