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. 18.125c. 10.25d. 13.75
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. 18.125c. 10.25d. 13.75
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.
However, since the current is a square wave with a 50% duty cycle, the effective current is half of the peak, which is 10A/2 = 5A.
So, the power loss would be:
P = (5A)^2 * 0.125 Ohms = 3.125 Watts
But, this is only the loss due to the resistance of the diode. There is also a loss due to the forward voltage drop across the diode, which can be calculated using the formula:
P = I * V
where: P is the power loss, I is the current, and V is the voltage.
Again, the effective current is 5A, and the voltage is given as 1.5V. So, the power loss due to the voltage drop is:
P = 5A * 1.5V = 7.5 Watts
The total conduction loss is the sum of these two losses:
P_total = 3.125W + 7.5W = 10.625W
So, none of the options given (a. 15, b. 18.125, c. 10.25, d. 13.75) match the calculated total conduction loss. There might be a mistake in the question or the options provided.
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