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Answer :
(a) Duty cycle: 50%
(b) Inductor size: 400 μH
(c) Capacitor size: 3.2 μF
(a) Duty cycle represents the ratio of the on-time of the switch to the total period of operation. In this case, the input voltage (Vs) is 50V and the output voltage (Vo) is -25V. Since the buck-boost converter is operating in a step-down mode (Vo < Vs), the duty cycle can be calculated using the equation: Duty cycle = Vo / (Vo - Vs). Plugging in the values, we get Duty cycle = -25V / (-25V - 50V) = 50%.
(b) The inductor size is determined by the desired ripple current and the switching frequency. The ripple current can be calculated using the equation: ΔIL = Po / (f * Vo). Substituting the given values, we have ΔIL = 125W / (100 kHz * -25V) = 0.05 A. To limit the ripple current to 5% of the average output current, we multiply ΔIL by 0.05 to get the maximum allowable ripple current: ΔIL_max = 0.05 A * 0.05 = 0.0025 A. Then, we can use the formula for inductor size: L = (Vo * (1 - Duty cycle)) / (ΔIL_max * f). Plugging in the values, we get L = (-25V * (1 - 0.5)) / (0.0025 A * 100 kHz) = 400 μH.
(c) The capacitor size is determined by the desired output voltage ripple. The output voltage ripple can be calculated using the equation: ΔVo = (Po * (1 - Duty cycle)) / (f * Vo). Substituting the given values, we have ΔVo = (125W * (1 - 0.5)) / (100 kHz * -25V) = 0.002 V. To limit the output voltage ripple to 5% of the output voltage, we multiply ΔVo by 0.05 to get the maximum allowable output voltage ripple: ΔVo_max = 0.002 V * 0.05 = 0.0001 V. Then, we can use the formula for capacitor size: C = (Po * (1 - Duty cycle)) / (ΔVo_max * f * Vo). Plugging in the values, we get C = (125W * (1 - 0.5)) / (0.0001 V * 100 kHz * -25V) = 3.2 μF.
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