Choose compensation components for a stable control loop using the procedure outlined as follows.

1. Based on a specified loop gain crossover frequency, f_C, of 40kHz, use Equation 43 to calculate R_COMP, assuming an effective output capacitance of 82μF. Choose a standard value for R_COMP of 5.36k?.
   Equation 43. ${\mathrm{R}}_{\mathrm{C}\mathrm{O}\mathrm{M}\mathrm{P}}=2\times \mathrm{\pi }\times {\mathrm{f}}_{\mathrm{C}}\times \frac{{\mathrm{V}}_{\mathrm{O}\mathrm{U}\mathrm{T}}}{{\mathrm{V}}_{\mathrm{R}\mathrm{E}\mathrm{F}}}\times \frac{{\mathrm{R}}_{\mathrm{S}}\times {\mathrm{G}}_{\mathrm{C}\mathrm{S}}}{\mathrm{g}\mathrm{m}}\times {\mathrm{C}}_{\mathrm{O}\mathrm{U}\mathrm{T}}=2\times \mathrm{\pi }\times 40\mathrm{}\mathrm{k}\mathrm{H}\mathrm{z}\times \frac{5\mathrm{}\mathrm{V}}{0.8\mathrm{}\mathrm{V}}\times \frac{5\mathrm{}\mathrm{m}\mathrm{\Omega }\times 10}{1200\mathrm{}\mathrm{\mu }\mathrm{S}}\times 82\mathrm{}\mathrm{\mu }\mathrm{F}=5.37\mathrm{}\mathrm{k}\mathrm{\Omega }\mathrm{ }$
2. To provide adequate phase boost at crossover while also allowing a fast settling time during a load or line transient, select C_COMP to place 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 6.8nF.
   Equation 44. ${\mathrm{C}}_{\mathrm{C}\mathrm{O}\mathrm{M}\mathrm{P}}=\frac{10}{2\times \mathrm{\pi }\times {\mathrm{f}}_{\mathrm{C}}\times {\mathrm{R}}_{\mathrm{C}\mathrm{O}\mathrm{M}\mathrm{P}}}=\frac{10}{2\times \mathrm{\pi }\times 40\mathrm{}\mathrm{k}\mathrm{H}\mathrm{z}\times 5.36\mathrm{}\mathrm{k}\mathrm{\Omega }}=7.42\mathrm{}\mathrm{n}\mathrm{F}$

   Such a low capacitance value also helps to avoid output voltage overshoot when recovering from dropout (when the input voltage is less than the output voltage setpoint and V_COMP is railed high).

3. Calculate C_HF to create a pole at the ESR zero and to attenuate high-frequency noise at COMP. C_BW is the bandwidth-limiting capacitance of the error amplifier. Select a standard value for C_HF of 47pF.
   Equation 45. ${\mathrm{C}}_{\mathrm{H}\mathrm{F}}=\frac{1}{2\times \mathrm{\pi }\times {\mathrm{f}}_{\mathrm{E}\mathrm{S}\mathrm{R}}\times {\mathrm{R}}_{\mathrm{C}\mathrm{O}\mathrm{M}\mathrm{P}}}-{\mathrm{C}}_{\mathrm{B}\mathrm{W}}=\frac{1}{2\times \mathrm{\pi }\times 500\mathrm{}\mathrm{k}\mathrm{H}\mathrm{z}\times 5.36\mathrm{}\mathrm{k}\mathrm{\Omega }}-38\mathrm{}\mathrm{p}\mathrm{F}=21\mathrm{}\mathrm{p}\mathrm{F}$