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TPS61230: The TPS61230 provides only 2 amps of current? "5A Switch Amp" misled me?

Part Number: TPS61230
Other Parts Discussed in Thread: TPS61088, TPS55340, TPS61089


My input source is 18650 Li-Ion Battery. It has 10Amps output current capability (also has 3.5-4.2V Voltage).

I need 5V with 5A capability. I am trying to use TPS61230 with a Typical Application schematic.

It has 5A switch, but it does not supplying 5A, just supplying 2A. When amps rising from 2A, it shut down himself.

What is your suggestion? Should I change the schematic? Or another part number?


  • Hi MAHMUT,

    For a boost converter, the 5A switch means the inductor valley current limit. Inductor average current equals input average current. So for your 3.5~4.2V to 5V case, the maximum input average current is around 5A and maximum output current is around 2A. That's correct.

    If you need to output 5V 5A from a Li-Ion battery, TPS61088 is one choice which has a higher current capability.

  • Hi Zack,

    thanks for the fast reply and your part number suggestion.

    What about TPS55340, Pololu says it has 5A capacity => 

    And TPS61089, 7A is enough for me, and it has small footprint and low price according to TPS61089.


  • Hi MAHMUT,

    The 5A capacity of TPS55340 means INPUT current capacity, while you need a 5A OUTPUT current capacity.

    So TPS61089 is not suitable neither.

  • Hi Zack,

    Thanks for the reply,

    I want just to understand what is the logic for these converters.

    Switch Current has relations about Vin/Vout and converter efficiency right?

    Can I calculate SW current with these equations:

    SW current > ( Iout * (Vout/Vin) ) * efficiency

    For example  SW Cur > 5A * (5V/4V) => SW Cur > 6,25A * Efficiency     

    if Efficiency %80  ,  SW Cur> 7,81A 

    So, I need a minimum 8A switch capability right? 

  • Hi MAHMUT,

    You are very close to the truth!

    For a DCDC converter, Efficiency= (Vout*Iout) / (Vin*Iin). So Iin= (Iout*Vout)/(Vin*Efficiency). For a boost converter, the input average current Iin equals inductor average current iL_avg. 

    To output a 5V/5A power, the average input current needs to be Iin= (5V*5A)/(3.5V*0.85)= 8.4A assuming the efficiency is 85%. Considering the inductor current ripple is 30%, the maximum inductor current (maximum SW current) is 8.4*1.15= 9.66A. 

    So you have to choose a device with 10A current limit capability. Hope that help you.