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. 15b. 13.75c. 10.25d. 18.125
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. 15b. 13.75c. 10.25d. 18.125
Solution 1
The conduction loss in a diode can be calculated using the formula:
P = I^2 * R + V * I
where:
- P is the power loss,
- I is the current,
- R is the resistance, and
- V is the voltage.
Given that the dynamic resistance (R) is 0.125 Ohms, the maximum instantaneous forward voltage (V) is 1.5 V, and the current (I) is 10 A, we can substitute these values into the formula:
P = 10^2 * 0.125 + 1.5 * 10
P = 12.5 + 15
P = 27.5 W
However, since the duty cycle is 50%, the total conduction loss would be half of this value:
P = 27.5 / 2
P = 13.75 W
Therefore, the correct answer is b. 13.75.
Solution 2
The total conduction loss in a diode can be calculated using the formula:
P = I^2 * R + V * I
where:
- P is the power loss,
- I is the current,
- R is the resistance, and
- V is the voltage.
Given that the dynamic resistance (R) is 0.125 Ohms, the maximum instantaneous forward voltage (V) is 1.5 V, and the current (I) is 10 A, we can substitute these values into the formula:
P = 10^2 * 0.125 + 1.5 * 10
P = 12.5 W + 15 W
P = 27.5 W
However, since the duty cycle is 50%, the total conduction loss will be half of this value:
P = 27.5 W * 0.5
P = 13.75 W
Therefore, the correct answer is b. 13.75.
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