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 55kHz, use Equation 44 to calculate R_COMP, assuming an effective output capacitance of 100 µF. Choose a standard value for R_COMP of 20.5kΩ.
   Equation 44. ${\mathrm{R}}_{\mathrm{C}\mathrm{O}\mathrm{M}\mathrm{P}}=2\mathrm{\pi }\times {\mathrm{f}}_{\mathrm{C}}\times \frac{{\mathrm{V}}_{\mathrm{O}\mathrm{U}\mathrm{T}}}{{\mathrm{V}}_{\mathrm{R}\mathrm{E}\mathrm{F}}}\times \frac{{\mathrm{R}}_{\mathrm{S}}\times {\mathrm{G}}_{\mathrm{C}\mathrm{S}}}{{\mathrm{g}}_{\mathrm{m}}}\times {\mathrm{C}}_{\mathrm{O}\mathrm{U}\mathrm{T}}=2\mathrm{\pi }\times 55\mathrm{k}\mathrm{H}\mathrm{z}\times \frac{12\mathrm{V}}{0.8\mathrm{V}}\times \frac{3\mathrm{m}\mathrm{\Omega }\times 10}{1100\mathrm{\mu }\mathrm{S}}\times 150\mathrm{\mu }\mathrm{F}=21.2\mathrm{k}\mathrm{\Omega }$
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. Calculate the necessary value using Equation 45 and choose a standard value for C_COMP of 1.5nF.
   Equation 45. ${\mathrm{C}}_{\mathrm{C}\mathrm{O}\mathrm{M}\mathrm{P}}=\frac{10}{2\mathrm{\pi }\times {\mathrm{f}}_{\mathrm{C}}\times {\mathrm{R}}_{\mathrm{C}\mathrm{O}\mathrm{M}\mathrm{P}}}=\frac{10}{2\mathrm{\pi }\times 55\mathrm{k}\mathrm{H}\mathrm{z}\times 20.5\mathrm{k}\mathrm{\Omega }}=1.4\mathrm{n}\mathrm{F}$

   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 using Equation 46 to create a pole at the ESR zero and to attenuate high-frequency noise at COMP. C_BW (31pF) is the bandwidth-limiting capacitance of the error amplifier. C_HF can not be significant enough to be necessary in some designs. In this design, the high frequency capacitor is not necessary.
   Equation 46. ${\mathrm{C}}_{\mathrm{H}\mathrm{F}}=\frac{1}{2\mathrm{\pi }\times {\mathrm{f}}_{\mathrm{E}\mathrm{S}\mathrm{R}}\times {\mathrm{R}}_{\mathrm{C}\mathrm{O}\mathrm{M}\mathrm{P}}}-{\mathrm{C}}_{\mathrm{B}\mathrm{W}}$