This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

  • Resolved

TPS54617

Prodigy 10 points

Replies: 1

Views: 709

regarding TPS54617, I have read through the information from the link that was recomended and it's suggested that the ripple should be around 10-30% of full load current. Currently, I am using 0.82uH inductor for TPS54617 while applying 3.3V as Vin and Vout is targeted at 1.0V. According to datasheet of tps54617, F(switching)=1.6Mhz. while applying the formula from TPS53311, I(ripple)=1/(LXSWITCHING FREQUENCY)X[(Vin-Vout)xVout]x[1/Vin]

                =1/(0.82u x 1.6M) x [(3.3-1) x 1 ] x [1/3.3]

                =0.53A (which is 8% of 6A of full load current) From the above calculation, do you think I should change my inductor to have a lower inductance such as 0.68uH? would you foresee any problems if my ripple current to be less than 10-30% of 6A(full load current).  thanks for your reply and much appreciated..

  • The amount of ripple current is somewhat arbitrary.  10-30% is the typically recommended value.  You can go over or under, depending on your other requirements.  In some cases, you may even be able to go over 100% ripple.  In general though, lower is better with voltage mode control devices such as TPS54617.  I don't see any problem with keeping your present inductor value, but you can decrease it if you want.

    Best  Regards,

    John Tucker

    BSR-MV DC/DC Applications

     

    “Working on an automotive power design? Read the E2E technical article Meet European Commission ADAS requirements with TI DC/DC for device recommendations and other helpful resources.”

     

This thread has been locked.

If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.