• State TI Thinks Resolved
  • Locked Locked
  • Replies 2 replies
  • Answers 1 answer
  • Subscribers 64 subscribers
  • Views 466 views
  • Users 0 members are here
Support feedback
Options
Options
Related

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.

TPS53515: About customer use condition

Tsukui Yusuke
Guru 21930 points
Part Number: TPS53515

Hi,

Could you advice on the customer's use conditions..
Customer condition is:
Vin=5V、Vout=1.2V、Iout≒4A
Fsw=850kHz、L=6.8uH、Co≒150uF

Customers are worried about operating stability due to the use of 6.8uH inductor.
Please give your advice.

Best Regards,
Yusuke/Japan Disty

  • 0
    TI__Expert 5745 points
    Hi
    Why customer want to use a 6.8uH inductor ? For reducing the output ripple?
    Normally, big value inductor and cap will decrease the cross frequency, it will lead to a bad transient performance, and it also have a risk of loop stable. According to the WEBENCH, it recommands a 2.2uH and 300uF (Cout). So choose Cout=150uF and L=4.7uH may be more balance for the output ripple and transient.
  • 0 in reply to Vental Mao
    TI__Guru**** 191445 points
    There may be some problems with that. The recommended ripple current is 25 to 35 % of the output current. In your case that calculates an inductor value of 0.894 uH (1 uH closest standard value). Additionally, that LC combination places the output filter double pole at 5 kHz. It should be aligned near the ripple injection zero.