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. None of the Aboveb. 1.4c. 0.35d. 0.8e. 1
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. None of the Aboveb. 1.4c. 0.35d. 0.8e. 1
Solution
To solve this problem, we need to first understand that the diode will be in the ON state when the voltage is at its maximum (40V) for 25% of the time (duty ratio = 0.25).
Step 1: Calculate the current through the circuit when the diode is ON.
The total resistance in the circuit when the diode is ON is the sum of the dynamic resistance of the diode (rd) and the load resistance (RL).
R_total = rd + RL = 50mΩ + 20Ω = 20.05Ω
The current (I) can be found using Ohm's law (V = IR), rearranged to I = V/R:
I = V_max / R_total = 40V / 20.05Ω = 1.9965A
Step 2: Calculate the power loss in the diode when it is ON.
The power loss (P) in the diode can be calculated using the formula P = I^2 * rd:
P = (1.9965A)^2 * 50mΩ = 0.19965W
However, this is the power loss when the diode is ON, and we know it is only ON for 25% of the time.
Step 3: Calculate the average power loss over time.
P_avg = Duty ratio * P = 0.25 * 0.19965W = 0.0499125W
So, the ON state power loss for the diode is approximately 0.05W.
Therefore, the correct answer is: a. None of the Above.
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