At the instant of firing, with zero initial current, what limits di/dt?

The total inductance Ls + L, so di/dt = vs/(Ls + L)

Which voltage is used for the worst-case di/dt?

The peak value Vm = sqrt(2) x 220 = 311.13 V

The average value of the supply

2070 Asar · 4 marks

Series inductor for di/dt limit

Short answer

At firing, di/dt = Vm/(Ls+L). With Vm = 311 V and limit 500 A/s, Ls+L >= 0.622 H; load L = 0.1 H, so Ls >= 0.522 H.

Fig.1c shows the single phase half-wave controlled rectifier with di/dt protection provided by the series inductor. The load is a purely inductive having an inductance of 100 mH. Calculate the value of Ls so that di/dt is limited to 500A/sec. [Figure: Vs = 220 V source, series inductor Ls, thyristor and the load in series]

In the circuit the series inductor $L_s$ and the load inductance $L$ are in series with the thyristor. At the instant of firing the current is zero, so the whole supply voltage appears across the total inductance and

\frac{di}{dt} = \frac{v_s}{L_s + L}

The worst case is firing at the peak of the supply. Assumption: 220 V is the rms value.

\begin{aligned} V_m &= \sqrt{2} \times 220 = 311.13\ \text{V} \\ \frac{V_m}{L_s + L} &\le 500\ \text{A/s} \\ L_s + L &\ge \frac{311.13}{500} = 0.6223\ \text{H} \\ L_s &\ge 0.6223 - 0.100 = 0.5223\ \text{H} \end{aligned}

Answer: $L_s \approx 0.522$ H (522 mH).

✓ Answer checked by Claude Opus 5.5 (high effort)

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