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About low Vout ripple of TPS53353

Satoshi
Guru 19645 points

Is there a way of low Vout ripple?

(without to change Cout and L)

I think that R7 of datasheet should be small, is it right?

【Customer information】

・Vin ：12V⇒Vout：1.2V/14A

・Fsw ：650kHz～970kHz

・Vout ripple ：Less than 8mV

・Cout ： ceramic capasitor

　⇒Request is "as small as posible" , less than 200μF etc.

・L ：SPM-6530T-R68 （TDK, 680nF）

 

Best regards,

Satoshi

  • 0
    TI__Guru**** 191445 points
    Generally speaking the ripple voltage of the output is set by Lout and Cout. In the datasheet design example, R7 is used to set the added ripple injection at VFB. It does not contribute directly to the output voltage ripple, but does cause some DC offset as the voltage is regulated at the valley point. In the datasheet example, they use 4 x 100 uF to achieve about 20 mV output voltage ripple (not including switching noise). To achieve lower ripple voltage you will need to increase the output capacitance above that amount. I think "less than 200uF" is not feasible. In fact I think you will find it extremely difficult to get ripple below 10 mV. Almost certainly you will need an additional LC filter stage.
  • 0 in reply to JohnTucker
    Guru 19645 points
    Dear John Tucker-san

    Thank you for reply.
    I understood to additional LC filter.

    Incidentally, When Switcher Pro set R7 small, Vout ripple changed.
    ※I opened the "Advanced Input" and input to Vout_ripple :8mV
    　⇒R7 changed about 20kΩ and Vout ripple less than 8mV
    I didn't understand this principle.
    I think that it is error of Switcher Pro, is this correct?

    Best regards,
    Satoshi
  • 0 in reply to Satoshi
    TI__Guru**** 191445 points

    Switcherpro has been out of development for several years now.  It is no longer supported and only available as a legacy tool.