A step down chopper operates from a D.C. voltage source Vs = 400 V and feeds a DC motor armature with a back emf Eb = 300 V. From the oscilloscope traces it is found that the current rises for a time tr = 10 m sec and falls to zero over time tf = 5 m sec and remains zero for a time to = 6 m sec in every chopping cycle. Then, average DC voltage across the freewheeling diode will be
- A. 276 V
- B. 257 V
- C. 262 V
- D. 300 V
Answer: A. 276 V
Explanation: Average DC voltage across freewheeling diode Vdf = (Vs·tr + Eb·to) / T = (400×10 + 300×6)/21 = 5800/21 ≈ 276.2 V ≈ 276 V.