Choose components for external compensation using the following procedure.

1. Set the crossover frequency between 10% and 20% of the switching frequency. With f_C specified as 50kHz in this example, and assuming an effective output capacitance of 32μF (four 22µF, 25V ceramic capacitors derated for an applied voltage of 12VDC) and negligible ESR, use Equation 49 to calculate R_COMP. Choose a standard value for R_COMP of 10kΩ.
   Equation 49. ${\mathrm{R}}_{\mathrm{C}\mathrm{O}\mathrm{M}\mathrm{P}}={2\mathrm{\pi }\times \mathrm{f}}_{\mathrm{C}}\times \frac{{\mathrm{V}}_{\mathrm{O}\mathrm{U}\mathrm{T}}}{{\mathrm{V}}_{\mathrm{R}\mathrm{E}\mathrm{F}}}\times \frac{{\mathrm{C}}_{\mathrm{O}\mathrm{U}\mathrm{T}}}{{\mathrm{g}}_{\mathrm{m}}\times \mathrm{G}}=2\mathrm{\pi }\times 50\mathrm{k}\mathrm{H}\mathrm{z}\times \frac{12\mathrm{V}}{0.8\mathrm{V}}\times \frac{32\mathrm{\mu }\mathrm{F}}{1\mathrm{m}\mathrm{S}\times 14.6\mathrm{S}}=10.3\mathrm{k}\mathrm{\Omega }$
   where
   • G = 14.6A/V, 10.9A/V and 8.1A/V for the LM65680, LM65660 and LM65640, respectively, is a factor related to the internal currrent-sense gain.

2. Calculate C_COMP to create a zero at the higher of (1) one tenth of the crossover frequency, or (2) the load pole. Choose a standard value for C_COMP of 3.3nF.

   Equation 50. ${\mathrm{C}}_{\mathrm{C}\mathrm{O}\mathrm{M}\mathrm{P}}=\frac{10}{2\mathrm{\pi }\times {\mathrm{f}}_{\mathrm{C}}\times {\mathrm{R}}_{\mathrm{C}\mathrm{O}\mathrm{M}\mathrm{P}}}=\frac{10}{2\mathrm{\pi }\times 50\mathrm{k}\mathrm{H}\mathrm{z}\times 10\mathrm{k}\mathrm{\Omega }}=3.18\mathrm{n}\mathrm{F}$

   In general, set the time constant of R_COMP and C_COMP at approximately 25μs to maintain a fast settling time of the output voltage following a load transient.

3. Calculate C_HF to create a pole at the lower of the ESR zero frequency or at half switching frequency (to attenuate high-frequency noise coupling from the output to COMP). Select a standard value for C_HF of 47pF.

   Equation 51. ${\mathrm{C}}_{\mathrm{H}\mathrm{F}}=\frac{1}{2\mathrm{\pi }\times \frac{{\mathrm{F}}_{\mathrm{S}\mathrm{W}}}{2}\times {\mathrm{R}}_{\mathrm{C}\mathrm{O}\mathrm{M}\mathrm{P}}}-{\mathrm{C}}_{\mathrm{B}\mathrm{W}}=\frac{1}{2\mathrm{\pi }\times \frac{400\mathrm{k}\mathrm{H}\mathrm{z}}{2}\times 10\mathrm{k}\mathrm{\Omega }}-40\mathrm{p}\mathrm{F}=39\mathrm{p}\mathrm{F}$

Alternatively, use internal compensation by connecting CNFG/SYNCOUT to VCC. Leave COMP open circuit or connected to PGND.