mean square (rms), and peak current ratings of the diode. Assume Ip = 500 A of a half sine-wave. 44₁A t₁ = 100 µs f = 500 Hz 1₂=300 μs 13 = 500 μs 12 T₁ = -7/₁ T, Ip 0 41 FIGURE P2.15

Electric Motor Control
10th Edition
ISBN:9781133702818
Author:Herman
Publisher:Herman
Chapter59: Motor Startup And Troubleshooting Basics
Section: Chapter Questions
Problem 12SQ: How is a solid-state diode tested? Explain.
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Power
electronic
0
11
FIGURE P2.15
Answer
Average current=3.895A
Root mean square (rms) current = 20.08A
Peak current=500A
2.15 The current waveform of a diode is shown in Figure P2.15. Determine the average, root
mean square (rms), and peak current ratings of the diode. Assume Ip = 500 A of a half
sine-wave.
Ai₁A
t₁ = 100 µs f= 500 Hz
t₂ = 300 µs
t₂ = 500 µs
12
13
T₁ = 1/2
Transcribed Image Text:Power electronic 0 11 FIGURE P2.15 Answer Average current=3.895A Root mean square (rms) current = 20.08A Peak current=500A 2.15 The current waveform of a diode is shown in Figure P2.15. Determine the average, root mean square (rms), and peak current ratings of the diode. Assume Ip = 500 A of a half sine-wave. Ai₁A t₁ = 100 µs f= 500 Hz t₂ = 300 µs t₂ = 500 µs 12 13 T₁ = 1/2
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