Question

+ VDS Consider the Buck DC-to-DC converter shown above. Assume it is operated at a switching frequency, f 25 KHz, with a duty cycle, D- 0.4 a) Given the sketch of the inductor current, i(t), shown below, sketch the waveforms for the input current, (t), the output current, i(t), the diode current, ip(), and the capacitor current, ict).Also, compute their average, RMS and peak values. (Note: Some useful formulas are provided on the last page) 10 5 t (s) (b) Assuming Vi-25 V, sketch the waveforms for the MOSFET voltage, Vos(t), and the diode voltage VD(t). Also, compute their average values c) Find the values of R, L and C. Please assume AVo-0.9 V

Answer 1

D = T_ON/T T = T_ON + T_OFF Switch-ON V_L = L dI/dt I = I_C + I_0 I_0 = V_0/R Switch-OFF V_L + V_0 = 0 V_0 = -V_L = -L di/dt V_L = L di/dt (I_max - I_min)_switch-on = V_S - V_0/L DT - (1) \r\n (I_max - I_min)_switch OFF = -V_0/LC (1 - DT) Average Inductor current = I_max + I_min/2 = I implies Vertical-bar I_max - Vertical-bar I_min Vertical-bar _Switch-ON = Vertical-bar I_max - I_min Vertical-bar _Switch-OFFVertical-bar V_S - V_D/L DT Vertical-bar = Vertical-bar -V_0/L (1 - 1 DT) Vertical-barV_0 = DV_S - (2) I_C = C dv/dt V_L(T) = V_C(0) = integral^t_0 I_C/C dt = 0 Hence, the total area under the capacitor and curve must be zero. This implies that for a complete cycle, average capacitor current I_C = 0 \r\n I_0 = V_0/R = I + I_C = I I = I_0 = V_0/R I_max + I_min/2 --- (3) From equation (1) & (3) I_max = DV_S (1/R + 1 - D/2L T) --- (4) I_min = DV_S (1/R - 1 - D/2L T) - (5)...