The voltage Vx in Fig. 6-3a for the buck converter with continuous inductor current is the pulsed waveform of Fig. 6-2c. The Fourier series for this waveform has a dc term of VsD. The ac terms have a fundamental frequency equal to the switching frequency and amplitudes given by
Vn = (sqrt(2)*Vs / (n*pi)) * sqrt(1 - cos(2*pi*n*D)) n = 1, 2, 3, ...
Using ac circuit analysis, determine the amplitude of the first ac term of the Fourier series for voltage across the load for the buck converter in Example 6-1. Compare your result with the peak-to-peak voltage ripple determined in the example. Comment on your results.
In Example 1 we find that Vs = 50 V, D = 0.4, L = 400 micro H, C = 100 micro F, frequency = 20 kHz, and R = 20.
Refer Fig. 6-3a and Fig. 6-2c in the textbook.
Write the expression for voltage Vn.
Vn = (sqrt(2)*Vs / (n*pi)) * sqrt(1 - cos(2*pi*n*D))
Substitute 1 for n, 50 V for Vs and 0.4 for D.
V1 = (sqrt(2)*50 / (1*pi)) * sqrt(1 - cos(2*pi*1*0.4))
= 30.27
Write the expression for the output voltage of buck converter.
Vo = D*Vs
Substitute 50 V for Vs and 0.4 for D.
Vo = (0.4)(50)
= 20 V
Write the expression the output voltage ripple in dc analysis.
delta_Vo / Vo = (1 - D) / (8*L*C*f^2)
Substitute 0.4 for D, 400 uH for L, 100 uF for C and 20 kHz for f.
delta_Vo / Vo = (1 - 0.4) / (8*(400 uH)*(100 uF)*(20 kHz)^2)
= 0.00469
= 0.469%
By the use of ac circuit analysis, the output voltage Vo1 is,
Vo1 = 0.048 V
Write the expression for the peak-peak voltage.
Vp-p = 2(Vo1)
Substitute 0.048 V for Vo1.
Vp-p = 2(0.048 V)
= 0.096 V
Find the peak-peak voltage ripple in ac analysis.
Vp-p / Vo = 0.096 / 20
= 0.48%
Thus, the peak-peak ripple in ac analysis is found as 0.48% and the peak-peak ripple in dc analysis is found as 0.469%. The output voltage is mainly the dc term and the first ac term.