Current chopping overvoltage at zero and peak

Current chopping transfers the energy stored in the transformer inductance to the bushing-to-ground capacitance:

$$\frac12 CV_m^2 = \frac12 Li^2 + \frac12 Cv^2$$

Here $i$ is the current and $v$ is the capacitor voltage at the instant of interruption.

Data: $L = 5$ H, $C = 0.01\ \mu$F, $I_{rms} = 7$ A.

$$\sqrt{\frac{L}{C}} = \sqrt{\frac{5}{0.01\times10^{-6}}} = 22\,360.7\ \Omega$$

Peak magnetising current: $I_m = 7\sqrt2 = 9.899$ A.

Peak system phase voltage: $V_p = 132\sqrt2/\sqrt3 = 107.78$ kV.

The magnetising current lags the voltage by about $90^\circ$. So when the current is zero the voltage is at its peak, and when the current is at its peak the voltage is zero.

(i) Interrupted at current zero

$i = 0$, so the magnetic energy is $\tfrac12 L(0)^2 = 0$.

$$V_m = \sqrt{v^2 + 0} = v = V_p = 107.78\ \text{kV}$$

No energy is released into the capacitance. The only voltage left is the normal system peak (107.8 kV), which decays. Overvoltage due to chopping = 0. This is the ideal case.

(ii) Interrupted at peak current

$i = I_m = 9.899$ A and $v = 0$.

$$\begin{aligned} V_m &= I_m\sqrt{\frac{L}{C}} = 9.899 \times 22\,360.7 \\ &= 221\,359\ \text{V} \approx 221.4\ \text{kV} \end{aligned}$$

This is $221.4/107.78 = 2.05$ times the normal peak phase voltage.

The oscillation frequency is $f = 1/(2\pi\sqrt{LC}) = 711.8$ Hz.

Answer: (i) no chopping overvoltage; the voltage stays at the normal peak of 107.8 kV. (ii) 221.4 kV.