Maximum switching overvoltage derivation

For energising a lumped series $R$–$L$–$C$ line model with a step voltage $E$ and initial (trapped) capacitor voltage $V_0$, the capacitor voltage is

$$V_c(t) = (E - V_0)\left[1 - e^{-\alpha t}\frac{\sqrt{\alpha^2+\omega^2}}{\omega}\cos(\omega t - \phi)\right] + V_0, \quad \tan\phi = \frac{\alpha}{\omega}$$

with $\alpha = \frac{R}{2L}$ and $\omega = \sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}$.

Condition for maximum

Let $K = \sqrt{\alpha^2+\omega^2}$, so $\sin\phi = \alpha/K$ and $\cos\phi = \omega/K$. Differentiate:

$$\begin{aligned} \frac{dV_c}{dt} &= (E-V_0)\frac{K}{\omega}e^{-\alpha t}\left[\alpha\cos(\omega t-\phi) + \omega\sin(\omega t-\phi)\right] \\ &= (E-V_0)\frac{K^2}{\omega}e^{-\alpha t}\left[\sin\phi\cos(\omega t-\phi) + \cos\phi\sin(\omega t-\phi)\right] \\ &= (E-V_0)\frac{\alpha^2+\omega^2}{\omega}e^{-\alpha t}\sin\omega t \end{aligned}$$

Setting $\frac{dV_c}{dt} = 0$ gives $\sin\omega t = 0$, so $\omega t = 0, \pi, 2\pi, \dots$ The first (largest) maximum is at

$$t_m = \frac{\pi}{\omega}$$

Maximum switching overvoltage

At $\omega t = \pi$: $\cos(\pi - \phi) = -\cos\phi = -\frac{\omega}{K}$, so

$$\begin{aligned} V_{max} &= (E-V_0)\left[1 + e^{-\alpha\pi/\omega}\frac{K}{\omega}\cdot\frac{\omega}{K}\right] + V_0 \\ &= (E-V_0)\left(1 + e^{-\alpha\pi/\omega}\right) + V_0 \\ &= E + (E - V_0)\,e^{-\alpha\pi/\omega} \end{aligned}$$

Special cases:

• No trapped charge ($V_0 = 0$): $V_{max} = E(1 + e^{-\alpha\pi/\omega})$, up to $2E$ when $R = 0$.
• Trapped charge of opposite polarity ($V_0 = -E$, as in fast reclosing): $V_{max} = E(1 + 2e^{-\alpha\pi/\omega})$, up to $3E$ when $R = 0$.
• $V_0 = E$: no overvoltage.

So the overvoltage is maximum when (i) the breaker closes at the peak of source voltage (so the step $E$ is the peak), (ii) the line holds a trapped charge of opposite polarity, and (iii) damping is small.

Effect of R, L and C

The damping exponent can be written as

$$\frac{\alpha}{\omega} = \frac{R/2L}{\sqrt{\frac{1}{LC} - \frac{R^2}{4L^2}}} = \frac{R}{\sqrt{4L/C - R^2}}$$

• Increasing R: $\alpha/\omega$ increases, $e^{-\alpha\pi/\omega}$ falls, so $V_{max}$ decreases. For $R \ge 2\sqrt{L/C}$ the circuit is overdamped and there is no overshoot ($V_{max} \to E$). This is why closing (pre-insertion) resistors are used.
• Increasing L: $4L/C$ increases, $\alpha/\omega$ falls (and $\alpha = R/2L$ falls), so damping reduces and $V_{max}$ increases. The oscillation also becomes slower.
• Increasing C: $4L/C$ decreases, $\alpha/\omega$ rises; $\alpha$ is unchanged but $\omega$ falls, so the peak occurs later after more decay, and $V_{max}$ decreases slightly; the period $2\pi/\omega$ becomes longer.

In real lines $R$ is small, so the damping factor $e^{-\alpha\pi/\omega}$ is close to 1 and overvoltages of 2–3 pu occur unless resistors or arresters are used.