As per typical I-V Characteristic of ultrafast diode: MUR1660CTG, the Dynamic resistance at 25 degrees Celsius is approximately:0.125 Ohms and the Max. instantaneous forward voltage is 1.5 V. If a square wave current of 10A peak and 50 percent duty cycle is passed through the diode. The total conduction loss in the diode in Watts would be:Select one:a. 18.125b. 13.75c. 15d. 10.25
Question
As per typical I-V Characteristic of ultrafast diode: MUR1660CTG, the Dynamic resistance at 25 degrees Celsius is approximately:0.125 Ohms and the Max. instantaneous forward voltage is 1.5 V. If a square wave current of 10A peak and 50 percent duty cycle is passed through the diode. The total conduction loss in the diode in Watts would be:Select one:a. 18.125b. 13.75c. 15d. 10.25
Solution
The conduction loss in a diode can be calculated using the formula:
P = I^2 * R
where: P is the power loss, I is the current, and R is the resistance.
Given that the current I is a square wave with a peak of 10A and a 50% duty cycle, the RMS (Root Mean Square) value of the current is:
I_rms = I_peak * sqrt(Duty Cycle) = 10A * sqrt(0.5) = 7.07A
Substituting the given values into the power loss formula:
P = (7.07A)^2 * 0.125 Ohms = 6.25W
However, this is the power loss for one half of the cycle (since the duty cycle is 50%). The total power loss for the full cycle is twice this value:
P_total = 2 * P = 2 * 6.25W = 12.5W
Therefore, none of the provided options (a. 18.125, b. 13.75, c. 15, d. 10.25) are correct. The total conduction loss in the diode is approximately 12.5W.
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