A typical reverse recovery time @ dIF/dt = 25A and 25 degree Celsius for a fast diode VS-70HFL is given as 500 ns and a typical reverse recovered charge is given as 1300 nC. If the off-state reverse voltage in certain application is 540 Volts and the operating frequency is 20kHz. The reverse recovery current in Ampere given as:Select one:a. 7.2b. 10c. 10.2d. 5.2
Question
A typical reverse recovery time @ dIF/dt = 25A and 25 degree Celsius for a fast diode VS-70HFL is given as 500 ns and a typical reverse recovered charge is given as 1300 nC. If the off-state reverse voltage in certain application is 540 Volts and the operating frequency is 20kHz. The reverse recovery current in Ampere given as:Select one:a. 7.2b. 10c. 10.2d. 5.2
Solution 1
The reverse recovery current can be calculated using the formula:
Irr = Qrr / trr
Where: Irr is the reverse recovery current, Qrr is the reverse recovered charge, and trr is the reverse recovery time.
Given: Qrr = 1300 nC = 1300 x 10^-9 C (converting nC to C) trr = 500 ns = 500 x 10^-9 s (converting ns to s)
Substituting these values into the formula:
Irr = (1300 x 10^-9) / (500 x 10^-9) Irr = 2.6 A
However, none of the options provided (a. 7.2, b. 10, c. 10.2, d. 5.2) match this result. Please check the question or the provided options.
Solution 2
The reverse recovery current can be calculated using the formula:
Irr = Qrr / trr
Where: Irr is the reverse recovery current, Qrr is the reverse recovered charge, and trr is the reverse recovery time.
Given: Qrr = 1300 nC = 1300 x 10^-9 C (converting nC to C) trr = 500 ns = 500 x 10^-9 s (converting ns to s)
Substituting these values into the formula:
Irr = (1300 x 10^-9) / (500 x 10^-9) Irr = 2.6 A
However, none of the options provided (a. 7.2, b. 10, c. 10.2, d. 5.2) match this result. Please check the question or the provided options.
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