In a push - pull converter, the filter capacitor can be obtained as____________?
Correct answer: C. Cmin= ( 1 - 2 D ) V / 32 ( Vr L f2 )
- A. Cmin = V / ( Vr L f2 )
- B. Cmin= ( 1 - D ) V / ( Vr L f2 )
- C. Cmin= ( 1 - 2 D ) V / 32 ( Vr L f2 )
- D. Cmin= ( 1 - 2 D ) V / 42 ( Vr L f2 )
Explanation
In a push-pull converter, the effective ripple frequency is doubled, and the capacitor ripple relation introduces the triangular-current factor, giving Cmin = (1 − 2D)V/(32VrLf²). Thus option c matches the standard expression, assuming the printed fractions are formatted correctly.
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