Choosing the right capacitors for LM2596
\$C_{FF}\$ will make a zero with R2 where the zero will be at \$ \frac 1 {2\pi R_2 C_{FF}}\$

The example uses 1k for R1; that yields R2 = 3K for a 5V output and using the 5nF recommended capacitor yields a zero at about 10.6kHz

Doing the same calculation for a 9V output (R2 ~ 7.5k) and the recommended \$C_{FF}\$ of 2nF also yields a zero at 10.6kHz

From that, it seems that \$C_{FF}\$ should be \$C_{FF} = \frac 1 {2\pi R_2 10.6k} \$ where R2 is the upper part of the feedback network.

For Cout I would simply use the recommended value from the table for your Vout / Vin setup.

I do note this part of the datasheet:


  As mentioned above, a low ESR value is needed for low output ripple
  voltage, typically 1% to 2% of the output voltage. But if the selected
  capacitor’s ESR is extremely low (below 0.05 ohm), there is a
  possibility of an unstable feedback loop, resulting in oscillation at
  the output. This situation can occur when a tantalum capacitor, that
  can have a very low ESR, is used as the only output capacitor.


This implies that an output zero needs to exist in the output capacitor as well as the feedforward capacitor; ceramic devices at the hundreds of microfarads have ESRs in the few milliohm range and might cause problems.

In this case I would use a ceramic (or a number of them to achieve the necessary Cout) in series with a 50 \$m\Omega\$ resistor (sized for the expected ripple current) in series as below:







simulate this circuit &ndash; Schematic created using CircuitLab