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. 13.75b. 10.25c. 15d. 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. 13.75b. 10.25c. 15d. 18.125
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 current. So, the effective current I_eff = I_peak * Duty_Cycle = 10A * 0.5 = 5A.
Substituting these values into the formula gives:
P = (5A)^2 * 0.125 Ohms = 3.125 Watts
However, the question asks for the total conduction loss, which also includes the loss due to the forward voltage drop across the diode. This can be calculated using the formula:
P_v = V * I_eff
where: P_v is the power loss due to voltage drop, and V is the forward voltage.
Substituting the given values into this formula gives:
P_v = 1.5V * 5A = 7.5 Watts
The total conduction loss is then the sum of these two losses:
P_total = P + P_v = 3.125W + 7.5W = 10.625W
However, this value is not among the options given in the question. It's possible that there's a mistake in the question or in the given options.
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