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Thermal and electromechanical dielectric breakdown

Short answer

Thermal breakdown: dielectric heating raises temperature, increasing conductivity and loss; thermal runaway occurs if heat generation surpasses cooling, causing melting or charring. Electromechanical breakdown: electrostatic compressive stress reduces thickness, increasing field; if elastic restoring force is insufficient, mechanical collapse leads to breakdown.

Describe how thermal breakdown and electromechanical breakdown results in dielectric breakdown in solids.

Dielectric breakdown is the sudden large increase of current through an insulator when the applied field exceeds a critical value (the dielectric strength), so that the material loses its insulating property, usually permanently in solids.

Thermal breakdown

  1. When an AC or DC voltage is applied, a small leakage current and dielectric loss (W=ωε0εr′E2tan⁡δW = \omega\varepsilon_0\varepsilon_r'E^2\tan\delta per unit volume) heat the insulator.
  2. The conductivity of an insulator rises steeply (exponentially) with temperature, and tan⁡δ\tan\delta also rises.
  3. More conductivity gives more current and more heat. If heat is generated faster than it can be conducted away to the surroundings, the temperature rises without limit (thermal runaway).
  4. The material melts, chars or burns, forming a conducting path, usually at the hottest point in the middle.

Features: it takes seconds to minutes to develop. The breakdown strength falls with higher frequency, higher ambient temperature and longer application time, and also with thicker insulation, because the interior is cooled less well. It is the most common breakdown mode in power equipment insulation (e.g. cable insulation).

 heat
  ^       generated
  |         /
  |        /   ...lost (cooling)
  |  .....'
  |  stable / unstable -> runaway
  |____________________> T

Electromechanical breakdown

  1. A voltage VV across a soft solid dielectric of thickness dd produces an electrostatic attraction between the electrodes (Maxwell stress) of 12ε0εrV2d2\frac{1}{2}\varepsilon_0\varepsilon_r\frac{V^2}{d^2} per unit area.
  2. This compressive force squeezes the material, so dd decreases.
  3. The field V/dV/d then increases, the force grows further, and the material is compressed even more.
  4. If the elastic restoring stress (Young's modulus YY) cannot balance it, the insulation collapses mechanically and then breaks down.

Balancing the stresses, the material becomes unstable when dd falls to about 0.6 d00.6\,d_0, which gives the highest apparent field

Emax≈0.6Yε0εrE_{max} \approx 0.6\sqrt{\frac{Y}{\varepsilon_0\varepsilon_r}}

It occurs mainly in soft materials such as polymers (polyethylene, rubber) and at high temperature, where YY is low.

✓ Answer checked by Claude Opus 5.5 (high effort)

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