Fom the piece-wise model of a diode we get: Dynamic resistance (rd) of the diode is 50 mΩ and knee voltage (vd) is 0.6V. This diode is connected in series with a 40V square wave voltage source (min=0V, max=40V, Duty ratio=0.25) and 20Ω resistance (RL). ON state power loss for the diode approximately is ____________ W.Select one:a. 0.35b. 1.4c. 1d. 0.8e. None of the Above
Question
Fom the piece-wise model of a diode we get: Dynamic resistance (rd) of the diode is 50 mΩ and knee voltage (vd) is 0.6V. This diode is connected in series with a 40V square wave voltage source (min=0V, max=40V, Duty ratio=0.25) and 20Ω resistance (RL). ON state power loss for the diode approximately is ____________ W.Select one:a. 0.35b. 1.4c. 1d. 0.8e. None of the Above
Solution
To solve this problem, we need to understand that the power loss in the diode when it is in the ON state is given by the formula P = I^2 * rd, where I is the current through the diode and rd is the dynamic resistance of the diode.
First, we need to calculate the current through the diode. The diode turns on when the voltage from the square wave source exceeds the knee voltage of the diode. The ON state of the square wave is 25% of the time (duty ratio = 0.25), and during this time, the voltage source is at its maximum of 40V.
The total resistance in the circuit when the diode is ON is the sum of the load resistance RL and the dynamic resistance of the diode rd. However, rd is very small compared to RL, so we can approximate the total resistance as RL.
So, the current through the diode when it is ON is I = V / R = 40V / 20Ω = 2A.
Now, we can calculate the power loss in the diode when it is ON using the formula P = I^2 * rd = (2A)^2 * 50mΩ = 0.2W.
However, this is the power loss when the diode is ON, and the diode is only ON 25% of the time. So, the average power loss over a full cycle of the square wave is P_avg = P * duty ratio = 0.2W * 0.25 = 0.05W.
So, the correct answer is e. None of the Above.
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