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. None of the Abovee. 0.8
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. None of the Abovee. 0.8
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 flowing through the diode and rd is the dynamic resistance of the diode.
First, we need to calculate the current flowing through the circuit when the diode is ON. The diode turns ON when the voltage from the square wave source is greater than the knee voltage of the diode. Given that the maximum voltage of the square wave source is 40V and the knee voltage of the diode is 0.6V, the diode will be ON for the portion of the time when the square wave voltage is at its maximum.
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). So, R_total = RL + rd = 20Ω + 50mΩ = 20.05Ω.
The current flowing through the circuit is given by Ohm's law, I = V/R. Here, V is the voltage across the diode when it is ON, which is the maximum voltage of the square wave source minus the knee voltage of the diode. So, V = 40V - 0.6V = 39.4V.
Therefore, I = 39.4V / 20.05Ω = 1.9651A.
Now, we can calculate the power loss in the diode using the formula P = I^2 * rd = (1.9651A)^2 * 50mΩ = 0.1935W.
However, the diode is not always ON. It is only ON for a portion of the time, given by the duty ratio of the square wave. So, the average power loss in the diode when it is ON is P_avg = Duty ratio * P = 0.25 * 0.1935W = 0.0484W.
None of the given options match this result, so the correct answer is d. None of the Above.
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