--- title: "SCR di/dt, dv/dt and current ratings numerical" subject: "Power Electronics" chapter: 1 key: "semester_7/power_electronics:scr-snubber-didt-dvdt-numerical" source: https://www.bctengineer.com/q/semester_7/power_electronics/scr-snubber-didt-dvdt-numerical site: BCT Engineering --- # SCR di/dt, dv/dt and current ratings numerical Power Electronics · Chapter 1: Characteristics and specification of power electronic devices - Asked in: 2075 Asoj, 8 marks - Video: [Mastering High dv/dt And High di/dt Protection For SCR In Power Electronics | GATE](https://www.youtube.com/watch?v=aPNB7jIoojs) — Ekeeda GATE & ESE - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/power_electronics/scr-snubber-didt-dvdt-numerical - Whole chapter: https://www.bctengineer.com/md/semester_7/power_electronics/chapter-1 ## Question For the circuit shown below: i) Calculate the maximum value of di/dt and dv/dt of the SCR ii) Find the RMS and average current rating of SCR for firing angle delays of 90° [Figure: source √2·230 sin 314t in series with an SCR, a 15 mH inductor and a 2 Ω load resistor; a snubber of 10 Ω in series with 0.15 µF is connected across the SCR] ## Short answer For the RL half-wave controlled rectifier with snubber, maximum di/dt = Vm/L = 325.27/0.015 = 21685 A/s (0.0217 A/µs), and maximum dv/dt = Rs Vm/L = 10×325.27/0.015 ≈ 2.17×10^5 V/s. At α=90°, solving the load current equation gives SCR average current 13.11 A and RMS current 22.47 A. ## Answer **Data:** $v_s = \sqrt{2}\times 230 \sin 314t$, so $V_m = 325.27$ V, $\omega = 314$ rad/s; $L = 15$ mH, $R = 2\ \Omega$; snubber $R_s = 10\ \Omega$, $C_s = 0.15\ \mu$F. #### (i) Maximum di/dt and dv/dt The worst case is when the voltage applied to the circuit is at its peak $V_m$, with zero initial current and an uncharged snubber capacitor. **di/dt at turn-on:** when the SCR is fired, at $t = 0^+$ the current is zero, so the whole voltage appears across $L$: $$ \left(\frac{di}{dt}\right)_{max} = \frac{V_m}{L} = \frac{325.27}{15 \times 10^{-3}} = 21685\ \text{A/s} = 0.0217\ \text{A}/\mu\text{s} $$ **dv/dt while OFF:** the SCR is OFF and the snubber is in series with $L$ and $R$: $$ V_m = L\frac{di}{dt} + (R + R_s)\,i + v_C $$ The voltage across the SCR is $v_T = R_s i + v_C$. At $t = 0^+$, $i = 0$ and $v_C = 0$, so $$ \begin{aligned} \frac{dv_T}{dt} &= R_s\frac{di}{dt} + \frac{i}{C_s} = R_s\frac{V_m}{L} \\ &= 10 \times \frac{325.27}{0.015} = 216846\ \text{V/s} \approx 0.217\ \text{V}/\mu\text{s} \end{aligned} $$ (The snubber discharge current through the SCR at turn-on is limited by $R_s$ to $V_m/R_s = 32.5$ A.) #### (ii) Average and RMS current of the SCR at $\alpha = 90^\circ$ The circuit is a half-wave controlled rectifier with RL load. $$ \begin{aligned} X_L &= \omega L = 314 \times 0.015 = 4.71\ \Omega \\ Z &= \sqrt{2^2 + 4.71^2} = 5.117\ \Omega \\ \phi &= \tan^{-1}\frac{4.71}{2} = 66.99^\circ \end{aligned} $$ Load (= SCR) current for $\alpha \le \omega t \le \beta$: $$ i = \frac{V_m}{Z}\left[\sin(\omega t - \phi) - \sin(\alpha - \phi)\,e^{-(\omega t - \alpha)/\tan\phi}\right] $$ Setting $i(\beta) = 0$ and solving numerically gives the extinction angle $\beta = 239.58^\circ$. Average current (from the average output voltage): $$ \begin{aligned} I_{avg} &= \frac{V_m}{2\pi R}(\cos\alpha - \cos\beta) \\ &= \frac{325.27}{2\pi \times 2}(\cos 90^\circ - \cos 239.58^\circ) = 13.11\ \text{A} \end{aligned} $$ RMS current, by integrating $i^2$ from $\alpha$ to $\beta$: $$ I_{rms} = \sqrt{\frac{1}{2\pi}\int_{\alpha}^{\beta} i^2\, d(\omega t)} = 22.47\ \text{A} $$ **Answer:** $(di/dt)_{max} = 21685$ A/s, $(dv/dt)_{max} \approx 2.17 \times 10^5$ V/s; SCR average current $= 13.11$ A and RMS current $= 22.47$ A at $\alpha = 90^\circ$ (the SCR should be rated above these values with a safety margin). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. Why is (di/dt)max at turn-on equal to Vm/L? A. At t = 0+ the current is zero, so the whole voltage appears across L B. The snubber carries all the current C. The resistor R limits the current D. The capacitor is fully charged Answer: A. Initially i = 0 so there is no drop across the resistances and vL = Vm, giving di/dt = Vm/L. (Key point: Max di/dt method) 2. How is the maximum dv/dt across the SCR found? A. dv/dt = Rs*Vm/L at t = 0+, when i = 0 and the snubber capacitor is uncharged B. dv/dt = Vm/Cs C. dv/dt = Vm/R D. dv/dt = Vm*L Answer: A. vT = Rs*i + vC; at 0+, dvT/dt = Rs*di/dt = Rs*Vm/L = 10*325.27/0.015, about 0.217 V/us. (Key point: Max dv/dt method) 3. For the half-wave controlled rectifier with R = 2 ohm, L = 15 mH at 314 rad/s, what are XL and Z? A. XL = 4.71 ohm, Z = 5.117 ohm B. XL = 15 ohm, Z = 15.1 ohm C. XL = 0.47 ohm, Z = 2.05 ohm D. XL = 4.71 ohm, Z = 6.71 ohm Answer: A. XL = 314*0.015 = 4.71 ohm and Z = sqrt(2^2 + 4.71^2) = 5.117 ohm, with phi about 67 degrees. (Key point: Load impedance and phase) 4. Which expression gives the average SCR current at alpha = 90 degrees, and its result? A. Iavg = Vm/(2*pi*R)*(cos alpha - cos beta) = 13.11 A B. Iavg = Vm/(pi*Z) = 20.2 A C. Iavg = Vm/R = 162.6 A D. Iavg = Vm/(2*pi*L) = 3450 A Answer: A. The average output voltage divided by R (inductor average voltage is zero) gives 13.11 A, with beta = 239.58 degrees. 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