--- title: "Reverse recovery: show IRR = √(2·QRR·di/dt)" subject: "Power Electronics" chapter: 1 key: "semester_7/power_electronics:reverse-recovery-irr-formula" source: https://www.bctengineer.com/q/semester_7/power_electronics/reverse-recovery-irr-formula site: BCT Engineering --- # Reverse recovery: show IRR = √(2·QRR·di/dt) Power Electronics · Chapter 1: Characteristics and specification of power electronic devices - Asked in: 2070 Chaitra, 8 marks - Video: [Power Diode Reverse Recovery Characteristics | Basic Concepts | Power Electronics](https://www.youtube.com/watch?v=jAkdiEZgOd0) — ENGINEERING TUTORIAL - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/reverse-recovery-irr-formula - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-1 ## Question Explain the reverse recovery characteristics of diode and show that IRR = [2·QRR·(di/dt)]^(1/2). ## Short answer When a conducting diode is reverse biased, stored charge is removed: current reverses, reaches IRR, then decays. For abrupt recovery, QRR = 0.5 IRR trr and IRR = trr di/dt, yielding IRR = sqrt(2 QRR di/dt). ## Answer **Reverse recovery:** when a conducting diode is switched to reverse bias, its current does not stop at zero. The minority carriers stored in the junction must first be removed, so the current falls through zero, flows in the reverse direction for a short time and then decays to zero. ```text iD ^ IF |-------. | \ slope = di/dt | \ --+----------\--------------------> t | \ .-------- | \ .' | - IRR ....\.' tb | ta |<->| | |<-->| | |<---trr--->| ``` - $t_a$: time from current zero to the peak reverse current $I_{RR}$, while charge stored in the depletion region is removed. - $t_b$: time for the reverse current to decay from $I_{RR}$ to about $0.25 I_{RR}$, while charge in the bulk is removed. - **Reverse recovery time** $t_{rr} = t_a + t_b$; **softness factor** $S = t_b / t_a$. - **Peak reverse current:** $I_{RR} = t_a \dfrac{di}{dt}$. - **Reverse recovery (stored) charge** $Q_{RR}$ is the area under the reverse current, approximately a triangle: $$ Q_{RR} = \tfrac{1}{2} I_{RR}\, t_{rr} $$ **Derivation of $I_{RR} = \sqrt{2Q_{RR}\,di/dt}$:** for an abrupt-recovery diode $t_b$ is very small, so $t_{rr} \approx t_a$. Then $$ \begin{aligned} I_{RR} &= t_a \frac{di}{dt} \approx t_{rr}\frac{di}{dt} \\ Q_{RR} &= \tfrac{1}{2} I_{RR} t_{rr} = \tfrac{1}{2}\frac{di}{dt}\, t_{rr}^2 \\ \Rightarrow t_{rr} &= \sqrt{\frac{2 Q_{RR}}{di/dt}} \\ I_{RR} &= \frac{di}{dt}\sqrt{\frac{2 Q_{RR}}{di/dt}} = \sqrt{2 Q_{RR}\frac{di}{dt}} \end{aligned} $$ Also $t_{rr} = \sqrt{2Q_{RR}/(di/dt)}$, so $Q_{RR}$ and $I_{RR}$ both rise with the rate of fall of forward current. Example: $Q_{RR} = 50\ \mu$C and $di/dt = 20$ A/µs give $I_{RR} = \sqrt{2 \times 50 \times 10^{-6} \times 20 \times 10^{6}} = 44.7$ A. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why does a diode current flow in the reverse direction after switching to reverse bias? A. Stored minority carriers must be removed first B. The junction capacitance is charged by the supply C. Avalanche breakdown occurs D. The diode has a negative resistance region Answer: A. The stored minority carriers in the junction keep it conducting in reverse until they are swept out or recombine. (Key point: Cause of reverse recovery) 2. Which statement about the recovery intervals is correct? A. trr = ta - tb B. trr = ta + tb and softness factor S = tb/ta C. trr = tb only D. trr = ta + tb and softness factor S = ta/tb Answer: B. ta runs from zero crossing to peak I_RR, tb from I_RR to about 0.25 I_RR; S = tb/ta. (Key point: Definitions of ta, tb, trr) 3. Treating the reverse current as a triangle, Q_RR equals: A. I_RR x t_rr B. (1/2) I_RR x ta^2 C. (1/2) I_RR x t_rr D. (1/2) (di/dt) x I_RR Answer: C. Q_RR is the area of a triangle with height I_RR and base t_rr. (Key point: Stored charge from triangle) 4. Which assumption and steps lead to I_RR = sqrt(2 Q_RR di/dt)? A. tb is large so trr = tb, and Q_RR = I_RR trr B. ta = 0, so I_RR = Q_RR di/dt C. I_RR = trr/(di/dt) and Q_RR = I_RR^2 D. tb is very small so trr is about ta; I_RR = trr di/dt and Q_RR = (1/2)(di/dt)trr^2, then eliminate trr Answer: D. Eliminating trr = sqrt(2 Q_RR/(di/dt)) from I_RR = trr di/dt gives sqrt(2 Q_RR di/dt). 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