0
\$\begingroup\$

I am using this component for a project, however, when reading the datasheet I have just become confused about a couple of things.

Datasheet : http://www.ti.com/lit/ds/slvs569f/slvs569f.pdf

enter image description here

enter image description here

enter image description here

I am just trying to calculate the output voltage ripple for my Cout capacitor. Just say the ESR was 0.05 ohms, for example, is the inductor ripple current : (IL1(pk)-ILoad(max))*2. Hence say ILoadmax was 1A, then just say I get the IL1(pk) as 1.2A. Hence is the inductor ripple current referring to the peak to peak inductor ripple current? So it would just be 0.4 A? Hence the output ripple voltage would be 0.4*0.05?

*Note the numbers used here are just for example purposes

\$\endgroup\$
  • 1
    \$\begingroup\$ If using DCM mode ripple current = Ipk \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 5 at 15:08
  • \$\begingroup\$ If i was using continuous conduction mode is it pk to pk? \$\endgroup\$ – Student May 5 at 15:25
  • 1
    \$\begingroup\$ Yes for CCM.... dV=ESR*dI or dV/dI=ESR , for e-Caps Tau= 200us for G.P. types and 1~10 us for low ESR types thus for 330uF e-caps ESR ranges from 3 Ohms , thus small caps are added in shunt. You cant get 0.05 ohms in a 330uF for cheap. \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 5 at 15:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.