Formation of a conducting channel during the discharge formation stage in rock salt
Abstract Generated abstract
The paper examines how a conducting channel forms and expands during electrical breakdown in rock salt, starting from observations that breakdown channels have smooth melted edges and small initial diameters. It estimates channel resistance and energy input after electrode closure, then compares calculated voltage drops for channel growth by melting, superheating, radiation loss, and skin effect with oscillographic measurements. The analysis finds that superheating would accelerate rather than slow the voltage drop, while radiation and skin effect are negligible under the stated conditions. Agreement with experiment is obtained if a substantial part of the energy is assumed to be spent on mechanical destruction, suggesting that channel expansion proceeds through melting together with cracking and removal of material from the channel walls.
Full Text
FORMATION OF A CONDUCTING CHANNEL AT THE STAGE OF DISCHARGE FORMATION IN ROCK SALT
M. N. LISETSKAYA
(Presented by the scientific seminar of the Research Institute of High Voltages)
When a dielectric breaks down, the electrodes are connected by a conducting channel. Observation of breakdown channels has shown that they have smooth, melted edges, and that the channel has a circular cross section. This permits the assumption that the channels are formed as a result of melting of the dielectric. Immediately after the electrodes are connected by the channel, its diameter is small. This is indicated by the microphotograph of a breakdown (Fig. 1), which was obtained while limiting the energy entering the channel immediately after closure of the electrodes by “cutting off” the voltage.
Fig. 1. Microphotograph of a breakdown in rock salt at magnification \(E\) with voltage “cutoff.”
The resistance of the channel may be calculated as
\[ R_1 = \rho \frac{l}{S_1}, \]
where \(\rho\) is the resistivity,
\(l\) is the channel length,
\(S\) is the cross section.
For rock salt at a temperature of \(800^\circ\mathrm{C}\),
\[ \rho = 3\cdot 10^3 \frac{\Omega\cdot \mathrm{cm}^3}{\mathrm{m}} . \]
Taking the channel radius to be \(2\ \mu\), and the interelectrode distance \(250\ \mu\), we obtain a channel resistance of \(6\cdot 10^4\ \Omega\). When current flows, energy enters the channel,
\[ w=\frac{U^2\Delta t}{R}. \]
Knowing \(U\), \(R\), and specifying the time interval \(\Delta t\), one can obtain the numerical value of \(w\). If it is assumed that this energy is spent on heating and melting the channel, then one can calculate the increase in the channel cross section and the resulting decrease in resistance. The calculation is carried out by the formula
\[ 0.24\frac{U^2}{R}\Delta t = d\Delta S l \left[C\left(T_{\mathrm{m}}-T\right)+K\right], \]
where
\[ d=2.2\ \mathrm{g/cm^3} \]
is the density of NaCl,
\[ l=250\ \mu \]
is the channel length,
\[ C=0.204\ \frac{\mathrm{cal}}{\mathrm{g}\cdot \mathrm{deg}} \]
is the specific heat capacity,
\[ K=123.5\ \frac{\mathrm{cal}}{\mathrm{g}} \]
is the specific heat of fusion.
For NaCl with graphite electrodes,
\[ E_{\mathrm{br}}=1.43\ \frac{\mathrm{MV}}{\mathrm{cm}},\qquad d=0.25\ \mathrm{mm}; \]
\[ U=36\ \mathrm{kV}. \]
Taking the time interval \(\Delta t=0.5\cdot 10^{-9}\ \mathrm{s}\), we obtain the increment of the channel cross section \(\Delta S\). The total channel cross section after \(\Delta t\) will be
\[ S_2=S_1+\Delta S, \]
and the channel resistance
\[ R_2=\rho\frac{l}{S_2}. \]
Such a calculation was performed for \(\Delta t=0.5\ \mathrm{ns}\). It turned out that the voltage drop that can occur when the channel resistance becomes comparable with the resistance of the generator begins not immediately after the electrodes are connected by the channel, but after \(3\ \mathrm{ns}\), and the voltage drop to the level of \(0.2U_{\max}\) occurs after a further \(7\ \mathrm{ns}\).
The change in the channel radius and resistance is shown in Fig. 2. Comparing the voltage drop from the oscillogram (Fig. 3) and from the calculation (Fig. 4), we see that the calculated voltage drop occurs faster than in the experiment.
It may be assumed that, after the channel has melted to a certain volume, the energy is spent only on superheating the channel, i.e.,
\[ w=0.24\frac{U^2\Delta t}{R}=mc\Delta t^\circ, \]
where \(\Delta t^\circ\) is the temperature increment.
Upon heating, according to data (1), the change in conductivity follows the law
\[ \gamma=\left(\frac{T}{300}+0.7\right)\ \Omega^{-1}\mathrm{cm}^{-1}. \]
For \(r=5\ \mu\), \(l=250\ \mu\), \(\gamma_1=3.3\ \Omega^{-1}\mathrm{cm}^{-1}\) at \(800^\circ\mathrm{C}\), we obtain
\[ R=10^4\ \Omega, \]
and
\[ w=1.55\cdot 10^{-5}\ \mathrm{J}. \]
Let us compare the increment in \(\gamma\) due to melting and due to superheating. During melting the area increment is
\[ \Delta S= \frac{w}{\left[C\left(T_{\mathrm{m}}-T\right)+K\right]dl} = 10^{-6}\ \mathrm{cm}^2, \]
and in this case
\[ \Delta\gamma_1=\gamma_1\frac{S}{l}=1.33\cdot 10^{-4}\ \Omega^{-1}. \]
With superheating:
\[ \Delta t^\circ=\frac{w}{\pi r^2 l d c}=1.87\cdot 10^3{}^\circ\mathrm{C}; \]
\[ T=T^1+\Delta t^\circ=2.67\cdot 10^3{}^\circ\mathrm{C}; \]
\[ \gamma_2=\left(\frac{T}{300}+0.7\right)=9.6\ \Omega^{-1}\mathrm{cm}^{-1}; \]
\[ \Delta\gamma_2=(\gamma_2-\gamma_1)\frac{S}{l}=1.87\cdot 10^{-4}\ \Omega^{-1}; \]
\[ \Delta\gamma_2>\Delta\gamma_1. \]
It follows that the increase in conductivity upon superheating is greater than upon melting, i.e., the resistance decreases more rapidly, and the voltage drops more rapidly, than upon melting. The calculation showed that
Fig. 2. Change in the channel radius and resistance during melting due to the energy entering the channel.
Fig. 3. Oscillogram of breakdown.
Fig. 4. Voltage drop during breakdown: 1 — drop due only to melting, 2 — melting and mechanical destruction, 3 — voltage drop according to the oscillogram.
the time of the drop to the level \(0.2\,U_{\max}\) is \(1\) nsec. Consequently, one cannot assume that superheating causes a slower voltage drop.
It may be that, at a high channel temperature, energy is spent on thermal radiation.
The radiant-energy flux per unit time [2] is
\[ \Phi_e = G T^4 S, \]
where \(S\) is the area,
\(T\) is the temperature,
\(G = 5.7 \cdot 10^{-8}\ \mathrm{W\, m^{-2}\, deg^{-4}}\) is the coefficient.
At \(T = 10^4{}^\circ\mathrm{C}\), \(\Phi_e = 4.5\ \mathrm{W}\), or \(3 \cdot 10^{-3}\%\) of the energy in the channel. Thus, only a small fraction of the energy goes into radiation, and this variant of energy loss must be neglected.
The times in our investigations are short; according to the oscillogram, the voltage drop occurs in \(12\ \mathrm{nsec}\). If the period of the high-frequency oscillations is taken as \(T = 25\ \mathrm{nsec}\), then \(f = 4 \cdot 10^7\ \mathrm{Hz}\) is high, and it may be assumed that the decrease in the resistance drop occurs due to the skin effect. Penetrating into a conducting medium, the electromagnetic field attenuates very rapidly.
The depth of penetration of the current into a conductor, at which the field is reduced by a factor of \(e\), is calculated by the formula [3]
\[ \Delta = \sqrt{\frac{2}{\omega \gamma \mu}}, \]
where \(\omega = 2\pi \cdot 4 \cdot 10^7\ \mathrm{Hz}\) is the frequency,
\[ \gamma = 3.3 \cdot 10^2\ \frac{1}{\Omega \cdot \mathrm{m}} \]
is the conductivity,
\(\mu = \mu_0 = 4\pi \cdot 10^{-7}\ \mathrm{H}/\mathrm{m}\) is the magnetic permeability.
In this case \(\Delta = 0.5\ \mu\). The skin effect manifests itself in the case when the channel radius \(r \gg \Delta\). In the present case it is the opposite; therefore it may be considered that the skin effect does not occur here.
Thus, neither overheating of the channel, nor radiation, nor the skin effect slows down the decrease in resistance.
When breakdown channels are examined, cracking of the dielectrics around the channel is visible; apparently, part of the energy goes into mechanical destruction. If it is assumed that half of the energy goes into mechanical destruction, and the voltage drop is determined in this case, then this voltage drop coincides with that measured from the oscillogram (Fig. 4).
Fig. 5. Diagram of channel expansion after the electrodes are short-circuited.
Therefore it may be supposed that the expansion of the channel occurs through melting in conjunction with mechanical destruction. In this case the expansion of the channel is shown schematically in Fig. 5. Cracking of the dielectric occurs, particles split off from the channel walls, then the pieces of salt melt, until their final melting occurs.
LITERATURE
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Handbook of Electrical Engineering Materials, ed. by I. P. Bogoroditskii. Gosenergoizdat, 1960.
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N. I. Koryakin, K. N. Bystrov, P. S. Kireev. A Brief Handbook of Physics. Moscow, 1962.
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V. A. Govorkov. Electric and Magnetic Fields. Gosenergoizdat, 1960.