Concept:The transient reverse current pulse during diode turn-off is approximately triangular, with a base equal to \( t_{rr} \) and a height equal to \( I_{RM} \). The total recovered charge \( Q_{rr} \) corresponds to the area of this triangle.
Formula:$$Q_{rr} = \int_0^{t_{rr}} i_R(t)\, dt \approx \frac{1}{2} \cdot I_{RM} \cdot t_{rr}$$
Solution:- During turn-off, reverse current rises to a peak \( I_{RM} \) and then decays back to the steady-state leakage level over duration \( t_{rr} \).
- Approximating this waveform as a triangle with base \( t_{rr} \) and height \( I_{RM} \) gives a recovered charge of \( Q_{rr} \approx \frac{1}{2} I_{RM} t_{rr} \).
Why other options are incorrect:- Option A: \( I_{RM} t_{rr}^2 \) is dimensionally incorrect for electric charge (Coulombs).
- Option B: \( I_{RM} / t_{rr} \) represents the rate of current change (\( di/dt \)), not charge.
- Option C: \( 2 I_{RM} t_{rr} \) overestimates the charge by a factor of 4.
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