Question 10.SP.8: For work to this point, the output voltage of the buck conve......
For work to this point, the output voltage of the buck converter has been assumed constant (v_2 ≃ V_2) for sufficiently large values of C; however, since i_C is a time-varying quantity, v_2 does display a small peak-to-peak ripple Δv_2. Use the change in capacitor charge (Q_C), under the assumption that the time-varying component of i_L flows through C, to calculate the voltage ripple Δv_2 for the case of continuous inductor current.
Since v_2 = Q_C/C, the total increment in v_2 is
\Delta v_{2} = \Delta Q_{C}/C (1)
The total increment in charge Q_C is given by the half-period duration amp-second, triangle-shaped area of i_L above I_2 = I_L in Fig. 10-3.
\Delta Q_{C} = {\frac{1}{2}}(I_{\mathrm{max}} – I_{2})\,{\frac{T_{s}}{2}} (2)
Use I_2 = V_2/R_L and (5) of Problem 10.6 in (2) and substitute the result into (1) to yield the peak-to-peak ripple voltage.
\Delta v_{2} = {\frac{1}{C}}\left({\frac{1}{2}}\right){\frac{(V_{1} – V_{2})D}{2f_{s}L}}{\frac{T_{s}}{2}} (3)
From (10.5), V_1 = V_2/D. Substitute into (3), use T_s = 1/f_s, and rearrange to find
G_{V} = {\frac{V_{2}}{V_{1}}} = D (10.5)
\Delta v_{2} = {\frac{(1 – D)V_{2}}{8f_{s}^{2}L C}} (4)
