LM53601-Q1 - Cff

Hirotaka Matsumoto

Other Parts Discussed in Thread: LM53601-Q1, LM53600-Q1

Hi all

Would you mind if we ask LM53601-Q1?

The data sheet shows as the following;
"Note that performance of converters utilizing an adjustable version of the LM53600-Q1 and LM53601-Q1 devices may be enhanced by adding CFF,
a capacitor in parallel with RFBT.100 pF is recommended."

Could you let us know how to calculate Cff?
(Could you let us know the caluculation formula for Cff?)
We guess that the caluculation is as the following formula. Is this correct?
-Fx=8.32/(Vout*Cout)
-Cff=1/(2π*Fx*RFBT)

Does it depend on Cout?(Our customer will use ceramic cap 22uF or 100uF.)

Kind regards,

Hirotaka Matsumoto

over 9 years ago

0 Hirotaka Matsumoto over 9 years ago

Guru 27585 points

Hi team

Could you give us the reply?

Kind regards,

Hirotaka Matsumoto

0 JohnTucker over 9 years ago in reply to Hirotaka Matsumoto

TI__Guru**** 191445 points

Cff forms a zero / pole pair. Fz = 1 / (2 * pi *Cff * R top). Fp = 1 / ( 2 * pi * Cff * (( Rtop * Rbot) / (Rtop + Rbot))). Rtop is the upper resistor and Rbot is the lower resistor in the VFB feedback network.

0 Hirotaka Matsumoto over 9 years ago in reply to JohnTucker

Guru 27585 points

John san

Thank you for your reply always.
Please refer to the following file.
20160831_LM53601-Q1.pdf
Could you let us know following questions?

<Question1>
On the webench simulation, Cff is 68pF despite of recommeded vaule 100pF on the datasheet.
Is this vaule optimical value in case of Cout=22uF, L=6.8uH, Rtop and Rbot?

<Question2>
We would like to cauculate Fz and Fo for Cff.
Are following formula correct?
-Fz=1/(2*pi*RC*Co)
-F0=(1/2*pi)*(√1/L*CO)

Kind regards,

Hirotaka Matsumoto

0 JohnTucker over 9 years ago in reply to Hirotaka Matsumoto

TI__Guru**** 191445 points

Not sure why WB picked 68 pF. The zero is at 23.4 kHz and the pole is at 117 kHz. You can clearly see the resulting phase boost in the loop response. Not that much difference between 68 pF and 100 pF, but the peak phase boost with 68n pF is very near the crossover frequency, so I think 68 pF is a good choice. The equations for the zero and pole of Cff are in my initial reply above.

0 Hirotaka Matsumoto over 9 years ago in reply to JohnTucker

Guru 27585 points

John san

Thank you for your kind reply!

OK, we got it.

Kind regards,

Hirotaka Matsumoto

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LM53601-Q1 - Cff