Dielectric Breakdown

In a uniform field below breakdown, the steady current grows with gap length by Townsend's primary ionization only:

$$I = I_0\, e^{\alpha d}$$

where $\alpha$ is Townsend's first (primary) ionization coefficient, i.e. the number of ionizing collisions made by one electron per cm of travel in the field direction. $\alpha$ depends only on $E/p$; here the field is kept constant ($E = 8/0.4 = 20$ kV/cm), so $\alpha$ is the same for both gaps and $I_0$ is the same.

Given

• $d_1 = 0.4$ cm, $I_1 = 5.5\times10^{-8}$ A
• $d_2 = 0.1$ cm, $I_2 = 5.5\times10^{-9}$ A

Solution

Divide the two current equations so that $I_0$ cancels:

$$\begin{aligned} \frac{I_1}{I_2} &= \frac{I_0 e^{\alpha d_1}}{I_0 e^{\alpha d_2}} = e^{\alpha (d_1 - d_2)} \\ \frac{5.5\times10^{-8}}{5.5\times10^{-9}} &= 10 = e^{\alpha (0.4 - 0.1)} \\ \alpha &= \frac{\ln 10}{0.3} = \frac{2.3026}{0.3} \\ &= 7.675\ \text{cm}^{-1} \end{aligned}$$

Check: $e^{7.675\times0.3} = e^{2.3026} = 10$.

The initial (cathode) current is then $I_0 = I_2 e^{-\alpha d_2} = 5.5\times10^{-9}\times e^{-0.7675} = 2.55\times10^{-9}$ A.

Answer: Townsend's primary ionization coefficient $\alpha \approx$ 7.68 ionizing collisions per cm (7.675 cm⁻¹), at $E = 20$ kV/cm.

Note: the gap is short and the current ratio is exactly exponential, so secondary ionization ($\gamma$) is neglected; it would show up only as an upward deviation from the straight line of $\ln I$ against $d$ at larger gaps.

Paschen's law states that the breakdown voltage of a uniform-field gap in a given gas depends only on the product of gas pressure $p$ and gap length $d$:

$$V_b = f(pd)$$

Derivation (outline)

Townsend's breakdown criterion is $\gamma(e^{\alpha d} - 1) = 1$. With $\alpha/p = A e^{-Bp/E}$ and $E = V_b/d$:

$$\begin{aligned} \alpha d &= \ln\left(1 + \frac{1}{\gamma}\right) \\ Apd\, e^{-Bpd/V_b} &= \ln\left(1 + \frac{1}{\gamma}\right) \\ V_b &= \frac{B\,pd}{\ln (A\,pd) - \ln\left[\ln\left(1 + \frac{1}{\gamma}\right)\right]} \end{aligned}$$

Since $A$, $B$ and $\gamma$ are gas constants, $V_b$ is a function of $pd$ only.

Paschen curve

 Vb
  |\                         /
  | \                      /
  |  \                  /
  |   \             /
  |    \_______ /
  |         min (about 327 V
  |          for air)
  +----------------------------> pd
     low pd    (pd)min    high pd

Setting $dV_b/d(pd) = 0$ gives a minimum: $(pd)_{min} = \frac{2.718}{A}\ln(1 + 1/\gamma)$. For air the minimum sparking voltage is about 327 V at $pd \approx 0.57$ torr·cm. No uniform-field air gap can break down below this voltage.

Why vacuum (low $pd$) is strong

On the left of the minimum the gas is very thin:

• The mean free path of electrons becomes comparable to or greater than the gap, so an electron crosses the gap making very few collisions.
• Few collisions means few ionizations ($\alpha d$ small), so the avalanche cannot build up.
• To get enough ionization a much higher voltage is needed, so $V_b$ rises steeply as $pd$ falls.

In high vacuum (below about $10^{-4}$ torr) there is almost no gas ionization at all; breakdown then depends on electrode effects (field emission, particle exchange, clumps), and strengths of the order of $10^5$–$10^6$ V/cm are obtained. This is why vacuum is used in vacuum circuit breakers.

Why high-pressure gas is strong

On the right of the minimum the gas is dense:

• The mean free path is very short, so an electron collides often but gains little energy between collisions (energy gained $= eE\lambda$).
• Most collisions are elastic or exciting, not ionizing, so $\alpha$ falls.
• A larger field is needed to give electrons ionizing energy over the short free path, so $V_b$ rises almost linearly with $pd$.

This is the basis of compressed-gas insulation (SF₆ at 3–6 bar in GIS, compressed nitrogen capacitors). At very high pressures (above about 10–20 bar) the rise becomes less than linear because electrode surface effects start to dominate.

Summary

A vacuum is a space in which the gas pressure is so low that the mean free path of electrons is much longer than the electrode gap. Breakdown can then not occur by collision ionization of gas (Townsend process); it is caused by processes at the electrode surfaces.

Classification of vacuum

(1 torr = 1 mm Hg.) For electrical insulation, as in vacuum circuit breakers and vacuum capacitors, the pressure is usually below $10^{-4}$ torr.

Categories of vacuum breakdown mechanism

The mechanisms are broadly grouped into three:

1. Particle exchange mechanism

• A charged particle (electron or ion) is accelerated by the field, strikes an electrode and releases particles of the opposite sign.
• Example: an electron hits the anode and releases positive ions; these ions hit the cathode and release more electrons (secondary emission coefficients $A$ and $B$).
• If the product of the coefficients exceeds unity, the process becomes cumulative: $AB > 1$, and breakdown occurs.
• Photons and negative ions can take part in the exchange as well.

2. Field emission mechanism

Electrons are pulled out of tiny projections (micro-points) on the cathode where the local field is enhanced to about $10^{6}$–$10^{7}$ V/cm. The current density follows the Fowler–Nordheim equation. Breakdown then occurs in one of two ways:

• Anode heating: the emitted electron beam hits the anode, heats a small spot and releases gas atoms and vapour. These are ionized by the beam; the positive ions move to the cathode, increase emission, and the process runs away into an arc.
• Cathode heating: the large current density through the micro-point heats it resistively until it melts and explodes, releasing metal vapour that forms a conducting plasma (vacuum arc).

3. Clump mechanism

• A loosely bound particle (clump) on one electrode becomes charged and is torn off by the field.
• It is accelerated across the gap and strikes the other electrode with high energy.
• The impact vaporizes material or releases gas, and breakdown follows.
• This explains why breakdown voltage in vacuum rises roughly as $\sqrt{d}$ for large gaps ($V_b = K\sqrt{d}$).

Factors affecting vacuum breakdown

• Electrode material, surface finish and cleanliness
• Gap length (strength per cm falls for longer gaps)
• Conditioning (repeated sparks smooth the surface and raise $V_b$)
• Temperature and absorbed gases on the electrodes

Vacuum gives very high strength for short gaps (about $10^5$–$10^6$ V/cm) and rapid recovery after arc interruption, which is why it is used in vacuum interrupters up to about 36 kV (and higher in newer designs).

An internal discharge occurs when the electric stress in a gas-filled void inside a solid exceeds the breakdown strength of the gas in the void. Because the void has a lower permittivity than the solid, the stress in it is higher than in the solid.

Given

• Disc thickness $d = 154$ mm $= 15.4$ cm
• Void thickness $t = 0.01$ mm $= 0.001$ cm
• $\varepsilon_r = 10$, applied voltage $V = 40$ kV
• Strength of air for this thin void: $46$ kV/cm

  +V ===========================
       solid  (er = 10)
       ----- void t = 0.01 mm --   (air)
       solid
   0 ===========================
          total d = 154 mm

Field in the void

The void and the solid are in series. The normal component of flux density is continuous:

$$\varepsilon_0\,E_a = \varepsilon_0\varepsilon_r E_s \;\Rightarrow\; E_a = \varepsilon_r E_s$$

Applied voltage:

$$\begin{aligned} V &= E_s(d - t) + E_a t = E_s\left[(d - t) + \varepsilon_r t\right] \\ E_s &= \frac{40}{(15.4 - 0.001) + 10\times0.001} = \frac{40}{15.409} \\ &= 2.596\ \text{kV/cm} \\ E_a &= \varepsilon_r E_s = 10\times2.596 = 25.96\ \text{kV/cm} \end{aligned}$$

Check

$$E_a = 25.96\ \text{kV/cm} < 46\ \text{kV/cm}$$

Even if 40 kV is taken as rms and the peak stress is used, $E_{a,peak} = \sqrt2\times25.96 = 36.7$ kV/cm, still below 46 kV/cm.

Voltage across the void $= E_a t = 25.96\times0.001 = 0.026$ kV $= 26$ V (rms), far below the minimum Paschen sparking voltage of air (about 327 V), which confirms the result.

Answer: Stress in the void $= 25.96$ kV/cm, which is less than 46 kV/cm, so internal discharge will not occur during the 40 kV withstand test.

The applied voltage needed to start discharge would be $46\times15.409/10 = 70.9$ kV.

Commercial liquid dielectrics (transformer oil, cable oil) are never perfectly pure. They contain solid particles, dissolved gases, moisture and gas bubbles, so their breakdown strength (about 100–200 kV/cm in tests) is much lower than that of pure liquids (about 1 MV/cm). Their breakdown is governed by these impurities, not by the liquid molecules.

Mechanisms of breakdown in commercial liquids

1. Suspended particle mechanism
2. Cavitation and bubble mechanism
3. Thermal mechanism
4. Stressed oil volume mechanism

1. Suspended particle mechanism

• Solid impurities (fibres, dust, wet cellulose) with permittivity $\varepsilon_2$ higher than that of the oil $\varepsilon_1$ become polarized.
• In a non-uniform field a particle of radius $r$ is pulled towards the region of highest stress by the force

$$F = r^3\,\frac{\varepsilon_2 - \varepsilon_1}{2\varepsilon_1 + \varepsilon_2}\,E\,\frac{dE}{dx}$$

• Particles line up end to end and form a bridge between the electrodes. Current through the bridge heats it, moisture vaporizes and breakdown follows.
• Breakdown strength therefore falls with more particles and longer time of stress.

2. Cavitation and bubble mechanism

Bubbles form in the oil from: gas pockets on electrode surfaces, electrostatic repulsion of space charge, dissociation of oil by electron collision, and vaporization by corona at sharp points.

• The bubble has lower permittivity, so the stress in it is higher.
• The bubble elongates along the field (Kao's theory) and when the voltage drop along it reaches the Paschen minimum of the gas, it ionizes.
• Ionization produces more gas, the bubble grows and bridges the gap.

3. Thermal mechanism

Under very short pulses, high current density at cathode micro-points causes local heating and vaporization, forming a vapour channel.

4. Stressed oil volume mechanism

Breakdown strength depends on the volume of oil stressed above about 90% of the maximum. A larger stressed volume contains more weak links (particles), so strength falls as stressed volume increases.

Breakdown phenomenon in brief

 Impurities in oil (particles, gas, water)
          |
 Field polarizes/elongates them
          |
 Particle chain or elongated bubble
          |
 Local high current / ionization
          |
 Heating, more gas, channel grows
          |
 Gap bridged -> breakdown (arc)

In practice the breakdown strength of oil is improved by filtering, degassing and drying, and is checked by the standard breakdown voltage (BDV) test.

Breakdown of a solid dielectric is the permanent loss of its insulating property when the applied stress exceeds what it can withstand; unlike gases and liquids, a solid does not recover after breakdown.

Mechanisms of breakdown in solids

1. Intrinsic (electronic) and avalanche breakdown
2. Electromechanical breakdown
3. Thermal breakdown
4. Electrochemical (chemical) breakdown
5. Breakdown due to treeing and tracking
6. Breakdown due to internal (partial) discharges

Intrinsic and electromechanical breakdown occur in very short times (about $10^{-8}$ s) at high stress, mostly in laboratory conditions. In practical equipment, failure is mostly due to thermal breakdown, chemical deterioration, treeing/tracking and internal discharges, which act over long periods at moderate stress.

1. Thermal breakdown

Every dielectric has losses: conduction loss under DC and dielectric loss under AC. The heat produced raises the temperature; in most insulators conductivity rises with temperature, so losses rise further.

• Heat generated per unit volume:
  • DC: $W_{dc} = \sigma E^2$
  • AC: $W_{ac} = \dfrac{E^2 f\,\varepsilon_r \tan\delta}{1.8\times10^{12}}$ W/cm³ ($E$ in V/cm)
• Heat balance:

$$\underbrace{\sigma E^2}_{\text{generated}} = \underbrace{C_v\frac{dT}{dt}}_{\text{stored}} + \underbrace{\text{div}(K\,\text{grad}\,T)}_{\text{conducted away}}$$

• If the heat generated stays below the heat dissipated, a stable temperature is reached. If it exceeds it at all temperatures, temperature rises without limit (thermal runaway) and the material melts or burns.

The heat-generated curve rises steeply (exponentially) with temperature, while the heat-dissipated curve rises roughly linearly. Breakdown occurs at the field where the two curves just touch.

Since AC loss is proportional to frequency, thermal breakdown voltage under AC is lower than under DC.

2. Electromechanical breakdown

When a voltage is applied, the electrodes attract each other with electrostatic pressure. A soft material of thickness $d_0$ is compressed to thickness $d$. Equilibrium:

$$\frac{\varepsilon_0\varepsilon_r V^2}{2d^2} = Y \ln\frac{d_0}{d}$$

where $Y$ is Young's modulus. The system becomes unstable when $d/d_0 = 0.6$, giving the highest apparent stress

$$E_{max} = \frac{V}{d_0} = 0.6\sqrt{\frac{Y}{\varepsilon_0\varepsilon_r}}$$

Beyond this the material collapses mechanically and breaks down. This is important for soft polymers (e.g. polythene at high temperature).

Breakdown of a solid dielectric is the permanent failure of its insulation when the electric stress exceeds its strength. The mechanism depends on the time for which the voltage is applied.

Mechanisms of breakdown in solids in practice

In equipment in service, the long-time mechanisms (thermal, chemical, treeing/tracking and internal discharges) are responsible for most failures, because the working stress is well below the intrinsic strength.

Thermal breakdown (explained)

Cause. Every dielectric loses some energy as heat: conduction loss under DC and dielectric loss ($\tan\delta$) under AC. For most insulators conductivity increases with temperature, so heating increases losses, which increases heating further.

Heat generated per unit volume:

$$\begin{aligned} W_{dc} &= \sigma E^2 \\ W_{ac} &= \frac{E^2 f\,\varepsilon_r\tan\delta}{1.8\times10^{12}}\ \text{W/cm}^3 \end{aligned}$$

Heat balance in a small volume:

$$C_v\frac{dT}{dt} + \text{div}(K\,\text{grad}\,T) = \sigma E^2$$

where $C_v$ is the specific heat per unit volume and $K$ the thermal conductivity.

Stable and unstable states.

Heat generated rises roughly exponentially with temperature, while heat dissipated rises roughly linearly with $(T - T_0)$:

At $E_1 > E_2$ the temperature rises without limit and the material melts, chars or burns (thermal runaway).

Features.

• Thermal breakdown voltage falls with rising ambient temperature and with poor cooling.
• Under AC it is lower than under DC, because dielectric loss rises with frequency.
• Thick insulation is worse cooled at the centre, so strength per mm falls with thickness.
• Typical thermal breakdown strengths: about 3–5 MV/cm (DC) and 1–3 MV/cm (AC) for good polymers; much lower for high-loss materials like glass at high temperature.

Thermal breakdown is controlled by using low-loss materials (low $\tan\delta$), good cooling and limiting the operating temperature (insulation class limits).

An internal (partial) discharge starts in a gas bubble when the voltage across the bubble reaches the breakdown voltage of the gas in it. The bubble and the remaining solid act as two capacitors in series.

Given

• Bubble thickness $t = 0.5$ mm $= 0.05$ cm
• Solid thickness $d = 1.0$ cm, so solid in series with the bubble $= d - t = 0.95$ cm
• $\varepsilon_r = 3.0$, applied voltage $30$ kV (rms), 50 Hz
• Breakdown strength of air $E_b = 30$ kV (peak)/cm

   +V  ======================
         solid  er = 3
        [ air bubble 0.5 mm ]
         solid
    0  ======================
          d = 1.0 cm

Equivalent capacitances (per unit area $A$)

$$C_1 = \frac{\varepsilon_0 A}{t}\ \ (\text{bubble}), \qquad C_2 = \frac{\varepsilon_0\varepsilon_r A}{d - t}\ \ (\text{solid in series})$$

Voltage across the bubble when $V$ is applied:

$$V_1 = V\,\frac{C_2}{C_1 + C_2} = \frac{V\,t}{t + \dfrac{d - t}{\varepsilon_r}}$$

Voltage across the bubble at breakdown

$$V_{1b} = E_b\,t = 30\times0.05 = 1.5\ \text{kV (peak)}$$

Applied voltage at which discharge starts

$$\begin{aligned} V_i &= V_{1b}\,\frac{t + (d - t)/\varepsilon_r}{t} = E_b\left[t + \frac{d - t}{\varepsilon_r}\right] \\ &= 30\left[0.05 + \frac{0.95}{3}\right] \\ &= 30\,(0.05 + 0.3167) = 11.0\ \text{kV (peak)} \\ V_{i,rms} &= \frac{11.0}{\sqrt2} = 7.78\ \text{kV (rms)} \end{aligned}$$

Answer: Internal discharge starts when the applied voltage reaches 11.0 kV (peak), i.e. about 7.78 kV (rms).

Since the working voltage of 30 kV (rms) is far above 7.78 kV (rms), discharges will occur in every half cycle. They slowly erode the bubble walls, form trees and finally cause breakdown of the solid. The insulation must therefore be made void-free (vacuum impregnation, careful moulding).

Townsend's theory explains breakdown of a gas in a uniform field as the growth of electron avalanches by collision ionization (primary process, coefficient $\alpha$), sustained by the release of new electrons from the cathode by the products of the avalanche (secondary process, coefficient $\gamma$). Breakdown occurs when each avalanche produces enough secondary electrons to start the next one, so the discharge becomes self-sustaining.

1. Current growth with primary ionization only

Let $n_0$ electrons per second leave the cathode (e.g. due to UV light), and let $n_x$ be the number at distance $x$. In length $dx$ each electron makes $\alpha\,dx$ ionizing collisions:

$$\begin{aligned} dn &= \alpha\, n_x\, dx \\ n_x &= n_0\, e^{\alpha x} \\ I &= I_0\, e^{\alpha d} \end{aligned}$$

$\alpha$ (Townsend's first coefficient) = number of electrons produced by one electron per cm of path in the field direction. $\ln I$ vs $d$ is a straight line of slope $\alpha$.

2. Current growth including secondary ionization

At higher voltages the current rises faster than $e^{\alpha d}$. This is due to secondary processes: positive ions striking the cathode, photons from excited atoms, and metastables, which release fresh electrons from the cathode. Let $\gamma$ = number of secondary electrons released from the cathode per positive ion (Townsend's second coefficient).

Let $n_0'$ = total electrons leaving the cathode per second, and $n_+$ = secondary electrons from the cathode:

$$n_0' = n_0 + n_+$$

Electrons reaching the anode: $n = n_0' e^{\alpha d}$. Positive ions created in the gap $= n - n_0'$; these release $\gamma(n - n_0')$ electrons at the cathode:

$$\begin{aligned} n_+ &= \gamma\left(n - n_0'\right) \\ n_0' &= n_0 + \gamma\left(n - n_0'\right) \\ n_0' &= \frac{n_0 + \gamma n}{1 + \gamma} \end{aligned}$$

Substituting into $n = n_0' e^{\alpha d}$:

$$\begin{aligned} n(1 + \gamma) &= (n_0 + \gamma n)\, e^{\alpha d} \\ n &= \frac{n_0\, e^{\alpha d}}{1 - \gamma\left(e^{\alpha d} - 1\right)} \end{aligned}$$

In terms of current:

$$\boxed{I = \frac{I_0\, e^{\alpha d}}{1 - \gamma\left(e^{\alpha d} - 1\right)}}$$

3. How current grows in presence of the secondary process

 ln I
  |                          /  breakdown
  |                        /    (upturn due
  |                     /        to gamma)
  |              ..../
  |        ..../  straight line, slope = alpha
  |  ..../
  +-------------------------------> d

• For small $d$, $\gamma(e^{\alpha d} - 1) \ll 1$; the denominator is about 1 and current follows $I_0 e^{\alpha d}$.
• As $d$ (or $V$) increases, $e^{\alpha d}$ grows, the denominator decreases and the current rises faster than exponential (upturn in the curve).
• Each avalanche now produces secondary electrons that start new avalanches, so successive avalanches feed one another.
• When the denominator becomes zero, the current becomes theoretically infinite (limited only by the external circuit):

$$\gamma\left(e^{\alpha d} - 1\right) = 1 \quad \text{or} \quad \gamma\, e^{\alpha d} \approx 1$$

This is Townsend's breakdown criterion. At this point the discharge no longer needs the external source ($I_0$) and becomes self-sustained: the gap has broken down. If $\gamma(e^{\alpha d}-1) < 1$ the discharge is non-self-sustaining and stops when the external ionization is removed.

Commercial liquid dielectrics such as transformer oil always contain impurities: solid particles (fibres, dust), dissolved gas, moisture and gas bubbles. Their conduction at high stress and their breakdown are explained by two main theories: the suspended particle theory and the cavitation and bubble theory. (Thermal and stressed-oil-volume mechanisms are also mentioned in texts.)

1. Suspended particle theory

Principle. Solid particles with permittivity $\varepsilon_2$ greater than that of the liquid $\varepsilon_1$ become polarized in the field. In a non-uniform field a spherical particle of radius $r$ experiences a force towards the region of maximum stress:

$$F = r^3\,\frac{\varepsilon_2 - \varepsilon_1}{2\varepsilon_1 + \varepsilon_2}\,E\,\frac{dE}{dx}$$

For a highly conducting particle ($\varepsilon_2 \to \infty$) the factor becomes 1, giving the largest force.

Process.

  electrode
  |  o o       particles drift to
  |   o  o     high-stress region
  | o-o-o-o-o-o  chain (bridge)
  |            |
  electrode    heating, vapour -> breakdown

1. Particles move towards high-field regions near the electrodes.
2. They line up along the field lines and form a chain (bridge).
3. The bridge carries a larger current; local heating vaporizes moisture and oil.
4. A vapour/gas channel forms along the bridge and the gap breaks down.

Results explained.

• Breakdown strength falls as particle content and moisture increase.
• Strength depends on time of voltage application: a bridge takes time to form, so strength under impulse is higher than under steady AC.
• Particle movement is opposed by viscosity and diffusion; if the drift force is smaller than these, no chain forms.

2. Cavitation and bubble theory

Sources of bubbles:

• Gas pockets on rough electrode surfaces
• Electrostatic repulsion between space charges at the liquid surface
• Dissociation of liquid molecules by electron collisions
• Vaporization of liquid by corona at sharp points

Process (Kao's theory).

1. A bubble in the liquid has lower permittivity, so the stress in it is higher than in the liquid.
2. Electrostatic forces elongate the bubble along the field (surface tension opposes this).
3. When the voltage drop along the elongated bubble equals the minimum Paschen breakdown voltage of the gas, the gas ionizes.
4. Ionization decomposes more liquid, the bubble grows into a gas channel and bridges the electrodes; breakdown follows.

The critical field for breakdown of a bubble of radius $r$ is

$$E_0 = \frac{1}{\varepsilon_1 - \varepsilon_2}\left\{\frac{2\pi\sigma(2\varepsilon_1 + \varepsilon_2)}{r}\left[\frac{\pi}{4}\sqrt{\frac{V_b}{2rE_0}} - 1\right]\right\}^{1/2}$$

where $\sigma$ is surface tension and $V_b$ the voltage drop in the bubble (Paschen minimum).

Results explained.

• Breakdown strength rises with pressure (bubbles harder to form) and falls with dissolved gas.
• Strength depends on initial bubble size, hydrostatic pressure and temperature.

Comparison

By Paschen's law the breakdown voltage of a gas gap depends on the product $pd$ (pressure × gap). Air at normal pressure for usual gaps lies near the low part of the Paschen curve; both vacuum (very low $pd$) and high-pressure gas (very high $pd$) lie on the steep rising branches, so their strength is higher.

Vacuum (low pressure)

• Very few gas molecules: the electron mean free path is longer than the gap.
• An electron crosses the gap with almost no collisions, so collision ionization ($\alpha$) cannot build an avalanche.
• Breakdown can occur only by electrode processes (field emission, particle exchange, clumps), which need very high fields: about $10^5$–$10^6$ V/cm for short gaps.

High-pressure gas

• Molecules are packed closely: the mean free path $\lambda$ is very short.
• Energy gained between collisions, $eE\lambda$, is small, so most collisions are non-ionizing.
• A much larger field is needed for ionization, so $V_b$ rises almost linearly with pressure (basis of SF₆ at 3–6 bar in GIS).

Major breakdown mechanisms in solids

Breakdown of a solid dielectric is the permanent loss of its insulating ability. The major mechanisms are:

1. Intrinsic (electronic) and avalanche breakdown – very fast ($\sim10^{-8}$ s), at very high stress
2. Electromechanical breakdown – mechanical collapse under electrostatic compression
3. Thermal breakdown
4. Electrochemical (chemical) breakdown
5. Breakdown due to treeing and tracking
6. Breakdown due to internal (partial) discharges

The first two occur mainly in laboratory conditions with pure specimens; 3–6 are the mechanisms that occur under actual working conditions.

Breakdown mechanisms in actual working conditions

In practical equipment the working stress is far below the intrinsic strength of the insulation, yet solids fail over months or years. These failures are caused by slow processes produced by the actual working conditions: heat, chemical attack, moisture, surface contamination and partial discharges.

1. Thermal breakdown

• Conduction loss (DC, $W = \sigma E^2$) and dielectric loss (AC, $W = \frac{E^2 f\varepsilon_r\tan\delta}{1.8\times10^{12}}$ W/cm³) heat the insulation.
• Conductivity and $\tan\delta$ rise with temperature, so heating feeds on itself.
• If heat generated exceeds heat conducted away, temperature rises without limit (thermal runaway) and the material melts or chars.
• Worse under AC than DC, with high ambient temperature, poor cooling and thick insulation.

2. Electrochemical (chemical) breakdown

• In the presence of air, moisture and heat, insulating materials slowly undergo chemical change:
  • Oxidation (e.g. rubber and polythene become brittle and crack)
  • Hydrolysis (moisture breaks down paper and some polymers, raising loss)
  • Chemical reaction with ozone and nitric acid produced by discharges
• Under DC, electrolysis of ionic impurities creates conducting paths.
• Rate roughly doubles for every 8–10 °C rise in temperature (Montsinger rule), so life depends strongly on operating temperature.

3. Breakdown due to internal (partial) discharges

• Voids or gas bubbles left in the solid during manufacture have lower permittivity and strength.
• Stress in a void is $\varepsilon_r$ times the stress in the solid, so the void discharges at a voltage well below working voltage.
• Each discharge (pC level) erodes the void walls by ion bombardment, heat and chemical action.
• The void enlarges and develops into channels (trees) until the remaining solid fails.

4. Breakdown due to treeing

• At sharp points, protrusions, contaminants or voids, the local field is very high.
• Small discharges cut fine branched channels into the solid, looking like a tree.
• Electrical trees grow by repeated discharges; water trees grow in polymer cables with moisture at much lower stress.
• The tree progresses step by step and when it bridges the electrodes the insulation breaks down.

5. Breakdown due to tracking

• A surface film of moisture and dirt on an insulator (bushing, cable termination) carries leakage current.
• The current dries parts of the film unevenly; small arcs (dry-band arcs) form across the dry bands.
• The arcs carbonize the surface, leaving a conducting track that grows until it bridges the electrodes, causing surface flashover.
• Organic materials (resins, polymers) are most prone; porcelain and glass are resistant.

6. Erosion and ageing

Surface discharges and corona in air gradually erode the surface; combined with mechanical and thermal cycling, cracks form and lead to failure.

In practical equipment the working stress is far below the intrinsic strength of the insulation, yet solids fail over months or years. These failures are caused by slow processes produced by the actual working conditions: heat, chemical attack, moisture, surface contamination and partial discharges.

1. Thermal breakdown

• Conduction loss (DC, $W = \sigma E^2$) and dielectric loss (AC, $W = \frac{E^2 f\varepsilon_r\tan\delta}{1.8\times10^{12}}$ W/cm³) heat the insulation.
• Conductivity and $\tan\delta$ rise with temperature, so heating feeds on itself.
• If heat generated exceeds heat conducted away, temperature rises without limit (thermal runaway) and the material melts or chars.
• Worse under AC than DC, with high ambient temperature, poor cooling and thick insulation.

2. Electrochemical (chemical) breakdown

• In the presence of air, moisture and heat, insulating materials slowly undergo chemical change:
  • Oxidation (e.g. rubber and polythene become brittle and crack)
  • Hydrolysis (moisture breaks down paper and some polymers, raising loss)
  • Chemical reaction with ozone and nitric acid produced by discharges
• Under DC, electrolysis of ionic impurities creates conducting paths.
• Rate roughly doubles for every 8–10 °C rise in temperature (Montsinger rule), so life depends strongly on operating temperature.

3. Breakdown due to internal (partial) discharges

• Voids or gas bubbles left in the solid during manufacture have lower permittivity and strength.
• Stress in a void is $\varepsilon_r$ times the stress in the solid, so the void discharges at a voltage well below working voltage.
• Each discharge (pC level) erodes the void walls by ion bombardment, heat and chemical action.
• The void enlarges and develops into channels (trees) until the remaining solid fails.

4. Breakdown due to treeing

• At sharp points, protrusions, contaminants or voids, the local field is very high.
• Small discharges cut fine branched channels into the solid, looking like a tree.
• Electrical trees grow by repeated discharges; water trees grow in polymer cables with moisture at much lower stress.
• The tree progresses step by step and when it bridges the electrodes the insulation breaks down.

5. Breakdown due to tracking

• A surface film of moisture and dirt on an insulator (bushing, cable termination) carries leakage current.
• The current dries parts of the film unevenly; small arcs (dry-band arcs) form across the dry bands.
• The arcs carbonize the surface, leaving a conducting track that grows until it bridges the electrodes, causing surface flashover.
• Organic materials (resins, polymers) are most prone; porcelain and glass are resistant.

6. Erosion and ageing

Surface discharges and corona in air gradually erode the surface; combined with mechanical and thermal cycling, cracks form and lead to failure.

Below breakdown in a uniform field, the steady current grows by primary ionization as

$$I = I_0\, e^{\alpha d}$$

where $\alpha$ is Townsend's primary ionization coefficient (ionizing collisions per cm per electron). It depends only on $E/p$, so with the field and pressure constant, $\alpha$ and $I_0$ are the same for both gap lengths.

Given

• $d_1 = 0.4$ cm, $I_1 = 5.5\times10^{-8}$ A (at 8 kV, so $E = 20$ kV/cm)
• $d_2 = 0.1$ mm $= 0.01$ cm, $I_2 = 5.5\times10^{-9}$ A

Solution

$$\begin{aligned} \frac{I_1}{I_2} &= e^{\alpha (d_1 - d_2)} \\ \frac{5.5\times10^{-8}}{5.5\times10^{-9}} &= 10 = e^{\alpha (0.4 - 0.01)} \\ \alpha &= \frac{\ln 10}{0.39} = \frac{2.3026}{0.39} \\ &= 5.904\ \text{cm}^{-1} \end{aligned}$$

Check: $e^{5.904\times0.39} = e^{2.3026} = 10$, as required.

Initial current: $I_0 = I_2\,e^{-\alpha d_2} = 5.5\times10^{-9}\times e^{-0.059} = 5.18\times10^{-9}$ A.

Answer: Townsend's primary ionization coefficient $\alpha \approx$ 5.90 per cm (5.904 cm⁻¹).

Note: if the second gap were 0.1 cm instead of 0.1 mm, the same method gives $\alpha = \ln 10/0.3 = 7.68$ cm⁻¹. Secondary ionization is neglected since the current ratio fits pure exponential growth.

Treeing and tracking are long-term breakdown mechanisms of solid insulation in service. Both progress slowly, step by step, by small discharges at moderate stress, until a conducting or hollow path bridges the electrodes.

Breakdown due to treeing

Treeing is the formation of fine, branched hollow channels inside the solid, which look like a tree or bush.

Cause and process:

1. At a sharp point, protrusion, metal particle, void or contaminant, the local field is several times the average field.
2. Partial discharges start there. Each discharge erodes the material by electron/ion bombardment, heat and chemical attack (ozone, nitric acid).
3. A thin channel forms; its tip again has high stress, so discharges continue and branch out.
4. The channels grow over months or years and form a tree. When a branch reaches the other electrode, the insulation fails.

   needle / protrusion (HV)
            |
           /|\
          / | \     branched channels
         /\ | /\    (electrical tree)
        /  \|/  \
   -----------------------
       earthed electrode

Types of trees:

Water trees grow at low stress and later turn into electrical trees.

Prevention: smooth semiconducting screens, clean and dry materials, void-free extrusion, tree-retardant XLPE.

Breakdown due to tracking

Tracking is the formation of a permanent conducting carbonized path on the surface of an insulator between electrodes.

Process:

1. Dust, salt, industrial pollution and moisture form a conducting film on the surface.
2. Leakage current flows through the film and heats it; the film dries unevenly and dry bands form.
3. Voltage concentrates across the dry bands, and small arcs (scintillations) strike across them.
4. The arcs decompose organic material, leaving carbon. This carbon track is conducting.
5. The track grows a little with each wetting/drying cycle until it bridges the electrodes; surface flashover and failure follow.

  HV electrode ====[ surface film ]==== earth
                  ~~~##~~~~##~~~~
                     dry bands + arcs
                  =====######====== carbon track grows

Features:

• Occurs mostly on organic insulators: resins, epoxies, polymers, bakelite.
• Porcelain and glass do not carbonize, so they resist tracking (but may suffer erosion).
• Measured by the comparative tracking index (CTI) test.

Prevention: track-resistant materials (filled silicone rubber, cycloaliphatic epoxy), longer creepage, hydrophobic coatings, regular cleaning.

Difference

Internal discharge occurs if the stress in the void exceeds the dielectric strength of the gas in the void. Since the void and the solid are in series and $D$ is continuous, the stress in the void is $\varepsilon_r$ times the stress in the solid.

Given

$d = 154$ mm $= 15.4$ cm, void $t = 0.1$ mm $= 0.01$ cm, $\varepsilon_r = 10$, $V = 40$ kV, void strength $= 46$ kV/cm.

Solution

$$\begin{aligned} E_a &= \varepsilon_r E_s, \qquad V = E_s(d - t) + E_a t = E_s\left[(d - t) + \varepsilon_r t\right] \\ E_s &= \frac{40}{(15.4 - 0.01) + 10\times0.01} = \frac{40}{15.49} = 2.582\ \text{kV/cm} \\ E_a &= 10\times2.582 = 25.82\ \text{kV/cm} \end{aligned}$$ $$E_a = 25.82\ \text{kV/cm} < 46\ \text{kV/cm}$$

Even taking 40 kV as rms, the peak stress $\sqrt2\times25.82 = 36.5$ kV/cm is still below 46 kV/cm.

Answer: Stress in the void is 25.82 kV/cm, less than 46 kV/cm, so internal discharge will not occur. Discharge would begin only at $V = 46\times15.49/10 = 71.3$ kV.

Limitations of Townsend's theory

Townsend's theory (avalanche growth by $\alpha$, sustained by cathode secondary emission $\gamma$, breakdown when $\gamma e^{\alpha d} = 1$) explains breakdown well at low $pd$, but fails in several respects at higher $pd$ (above about 1000 torr·cm, e.g. long gaps in atmospheric air):

1. Time lag: Townsend predicts formative times of the order of $10^{-5}$ s (time for ions to cross the gap many times). Observed times in long gaps are about $10^{-8}$–$10^{-7}$ s, far too short for ion transit.
2. Cathode independence: At high pressures breakdown voltage is nearly independent of cathode material, which should not happen if $\gamma$ at the cathode controls breakdown.
3. Channel shape: Townsend predicts a diffuse glow-type discharge, but actual sparks at high $pd$ are thin, bright, branched and zig-zag channels.
4. Space charge ignored: The theory neglects the distortion of the field by the space charge of the avalanche, which becomes significant when the avalanche contains about $10^8$ electrons.
5. Non-uniform fields: It does not explain corona and breakdown in non-uniform fields, or polarity effects.
6. Breakdown with no cathode involvement: Discharges can start mid-gap and grow towards both electrodes, which Townsend's mechanism cannot explain.

These limitations led to the streamer (Raether–Meek) theory for high $pd$.

Derivation of Paschen's law

Townsend's breakdown criterion:

$$\gamma\left(e^{\alpha d} - 1\right) = 1 \;\Rightarrow\; \alpha d = \ln\left(1 + \frac{1}{\gamma}\right) \qquad (1)$$

Experimentally, the first coefficient depends on $E/p$:

$$\frac{\alpha}{p} = A\, e^{-Bp/E} \qquad (2)$$

where $A$ and $B$ are gas constants. In a uniform field $E = V/d$, so at breakdown $E = V_b/d$:

$$\alpha = Ap\, e^{-Bpd/V_b} \qquad (3)$$

Substitute (3) in (1):

$$\begin{aligned} Apd\, e^{-Bpd/V_b} &= \ln\left(1 + \frac{1}{\gamma}\right) = K \\ e^{-Bpd/V_b} &= \frac{K}{Apd} \\ \frac{Bpd}{V_b} &= \ln\frac{Apd}{K} \\ V_b &= \frac{B\,pd}{\ln (A\,pd) - \ln K} \end{aligned}$$ $$\boxed{V_b = \frac{B\,(pd)}{\ln\left[\dfrac{A\,(pd)}{\ln(1 + 1/\gamma)}\right]} = f(pd)}$$

Since $A$, $B$ and $\gamma$ are constants of the gas (and electrode), the breakdown voltage of a uniform-field gap is a function of the product $pd$ only. This is Paschen's law. For air at temperature other than 20 °C, $p$ is replaced by gas density ($Nd$ or $\delta d$).

Minimum sparking voltage

Setting $\dfrac{dV_b}{d(pd)} = 0$:

$$(pd)_{min} = \frac{e}{A}\ln\left(1 + \frac{1}{\gamma}\right), \qquad V_{b,min} = \frac{eB}{A}\ln\left(1 + \frac{1}{\gamma}\right)$$

($e = 2.718$.) For air, $V_{b,min} \approx 327$ V at $pd \approx 0.57$ torr·cm.

 Vb
  |\                      /
  | \                  /
  |  \             /
  |   \_______ /
  |      min ~327 V (air)
  +---------------------------> pd

Left of the minimum, too few collisions; right of the minimum, too little energy per collision; in both cases $V_b$ rises.

Electromechanical breakdown

When voltage is applied to a solid between electrodes, the electrodes attract each other. In soft materials this electrostatic pressure compresses the solid from its original thickness $d_0$ to $d$. Equilibrium between electrical and mechanical stress:

$$\frac{\varepsilon_0\varepsilon_r V^2}{2d^2} = Y\ln\frac{d_0}{d}$$

where $Y$ is Young's modulus. The equilibrium becomes unstable at $d/d_0 = 0.6$; any further increase in voltage crushes the material and breakdown follows. The highest apparent stress is

$$E_{max} = \frac{V}{d_0} = 0.6\sqrt{\frac{Y}{\varepsilon_0\varepsilon_r}}$$

It occurs in about $10^{-8}$ s and matters for soft polymers (e.g. polythene at raised temperature).

Thermal breakdown

Dielectric losses heat the insulation:

$$W_{dc} = \sigma E^2, \qquad W_{ac} = \frac{E^2 f\varepsilon_r\tan\delta}{1.8\times10^{12}}\ \text{W/cm}^3$$

Since conductivity and $\tan\delta$ increase with temperature, losses grow as the material heats. Heat balance:

$$C_v\frac{dT}{dt} + \text{div}(K\,\text{grad}\,T) = \sigma E^2$$

If heat generated is less than heat conducted away, a stable temperature is reached. Above a critical field, heat generated always exceeds dissipation; temperature rises without limit (thermal runaway) and the material melts or burns. Thermal breakdown strength is lower under AC than DC, and falls with higher ambient temperature, poor cooling and greater thickness.

Breakdown due to treeing and tracking

Treeing: At sharp points, protrusions, voids or impurities inside the solid the local field is very high. Partial discharges there erode fine channels, which branch out like a tree. Electrical trees grow by repeated discharges; water trees grow in polymer cables with moisture. When a branch reaches the other electrode, the insulation fails.

Tracking: On the surface of an insulator, a film of dirt and moisture carries leakage current. Uneven drying forms dry bands, small arcs strike across them and carbonize the organic surface. The carbon track grows with each cycle until it bridges the electrodes, causing surface flashover.

In Townsend's theory the current in a gas grows by two ionization processes, each described by a coefficient. In Townsend's original form these are:

• $\alpha$ = number of ionizing collisions made by an electron per cm of travel (first coefficient)
• $\beta$ = number of ionizing collisions made by a positive ion per cm of travel (second coefficient)

Derivation

Let $n_0$ electrons per second leave the cathode, $n_x$ = electrons crossing plane $x$, and $p_x$ = positive ions crossing plane $x$ (moving towards the cathode). In steady state the total current is the same at every plane. At the anode ($x = d$) there are no positive ions, so

$$n_x + p_x = n_d \;\Rightarrow\; p_x = n_d - n_x$$

In length $dx$, new electrons are created by electrons and by ions:

$$\begin{aligned} \frac{dn_x}{dx} &= \alpha n_x + \beta p_x = \alpha n_x + \beta(n_d - n_x) \\ &= (\alpha - \beta)\,n_x + \beta n_d \end{aligned}$$

Solving this linear equation with $n_x = n_0$ at $x = 0$:

$$n_x = \left[n_0 + \frac{\beta n_d}{\alpha - \beta}\right]e^{(\alpha - \beta)x} - \frac{\beta n_d}{\alpha - \beta}$$

Put $x = d$:

$$\begin{aligned} n_d\left[1 + \frac{\beta}{\alpha - \beta} - \frac{\beta\,e^{(\alpha - \beta)d}}{\alpha - \beta}\right] &= n_0\, e^{(\alpha - \beta)d} \\ n_d\,\frac{\alpha - \beta\,e^{(\alpha - \beta)d}}{\alpha - \beta} &= n_0\, e^{(\alpha - \beta)d} \end{aligned}$$ $$\boxed{I = I_0\,\frac{(\alpha - \beta)\,e^{(\alpha - \beta)d}}{\alpha - \beta\,e^{(\alpha - \beta)d}}}$$

Breakdown condition

The current becomes infinite (self-sustained discharge) when the denominator is zero:

$$\alpha = \beta\,e^{(\alpha - \beta)d}$$

Since $\beta \ll \alpha$, this is approximately $\beta\,e^{\alpha d} = \alpha$.

Modern form ($\alpha$ and $\gamma$)

Positive ions in fact have too little energy to ionize gas, so the second process is now taken as electron release from the cathode by ions, photons and metastables, with coefficient $\gamma$. The same steps give

$$I = \frac{I_0\, e^{\alpha d}}{1 - \gamma\left(e^{\alpha d} - 1\right)}, \qquad \text{breakdown: } \gamma\left(e^{\alpha d} - 1\right) = 1$$

Both forms show the current rising faster than $e^{\alpha d}$ as the gap or voltage increases, ending in breakdown.

Internal discharge starts when the stress in the gas-filled void reaches the breakdown strength of that gas. The void (permittivity $\varepsilon_0$) and the solid ($\varepsilon_0\varepsilon_r$) are in series, so the stress in the void is $\varepsilon_r$ times that in the solid.

Given

• Disc thickness $d = 154$ mm $= 15.4$ cm, $\varepsilon_r = 10$
• Void thickness $t = 0.1$ mm $= 0.01$ cm, strength of void $= 46$ kV/cm
• Applied voltage $V = 40$ kV (taken as rms)

The disc being square does not matter: for a thin flat void parallel to the electrodes, the field depends only on thicknesses.

Stress in the void

$$\begin{aligned} V &= E_s(d - t) + E_a t, \qquad E_a = \varepsilon_r E_s \\ E_s &= \frac{V}{(d - t) + \varepsilon_r t} = \frac{40}{15.39 + 0.10} = 2.582\ \text{kV/cm} \\ E_a &= 10\times2.582 = 25.82\ \text{kV/cm (rms)} \\ E_{a,peak} &= \sqrt2\times25.82 = 36.52\ \text{kV/cm (peak)} \end{aligned}$$

Check

The void is only 0.1 mm thick. By Paschen's law a very thin air gap has a higher strength than the 30 kV/cm (peak) of air in centimetre gaps, which is why the void strength is given as 46 kV/cm. The correct comparison is therefore with 46 kV/cm.

Voltage needed for discharge: $V_i = 46\times\dfrac{15.49}{10} = 71.3$ kV.

Answer: Stress in the void $= 25.82$ kV/cm (rms) $< 46$ kV/cm, so internal discharge will not occur at 40 kV. (If the 30 kV/cm peak value for large air gaps were wrongly applied, the 36.5 kV/cm peak stress would suggest discharge.)

Under AC, the loss in a dielectric is the total dielectric loss given by the loss tangent: $P = V^2\omega C\tan\delta$. Under DC, after charging, only the conduction (leakage) current flows, so $P = V^2/R$ with $R$ from the volume resistivity.

Given

• $A = 5$ m², $d = 5$ cm $= 0.05$ m, $\varepsilon_r = 3.5$, $\tan\delta = 0.02$
• $\rho = 6\times10^{12}$ Ω·cm $= 6\times10^{10}$ Ω·m
• $\varepsilon_0 = 8.854\times10^{-12}$ F/m

Capacitance of the specimen

$$\begin{aligned} C &= \frac{\varepsilon_0\varepsilon_r A}{d} = \frac{8.854\times10^{-12}\times3.5\times5}{0.05} \\ &= 3.099\times10^{-9}\ \text{F} = 3.099\ \text{nF} \end{aligned}$$

(i) 60 kV rms, 50 Hz AC

$$\begin{aligned} P_{ac} &= V^2\,\omega C\tan\delta \\ &= (60\times10^3)^2\times(2\pi\times50)\times3.099\times10^{-9}\times0.02 \\ &= 3.6\times10^{9}\times314.16\times3.099\times10^{-9}\times0.02 \\ &= 70.1\ \text{W} \end{aligned}$$

(ii) 60 kV DC

Volume resistance of the specimen:

$$\begin{aligned} R &= \frac{\rho\,d}{A} = \frac{6\times10^{10}\times0.05}{5} = 6\times10^{8}\ \Omega \\ P_{dc} &= \frac{V^2}{R} = \frac{(60\times10^3)^2}{6\times10^8} = 6.0\ \text{W} \end{aligned}$$

Results

Answer: (i) $P_{ac} \approx$ 70.1 W; (ii) $P_{dc} =$ 6.0 W.

The AC loss is about 11.7 times the DC loss because under AC the polarization (dipole and interfacial) losses add to the conduction loss. This is why thermal breakdown voltage of a dielectric is lower under AC than under DC.

A gas is normally an excellent insulator because it contains very few free charges. Electrical breakdown of a gas is the change from this insulating state to a highly conducting state (spark or arc) when the applied field is high enough. It happens by collision ionization: free electrons accelerated by the field ionize gas molecules and multiply in avalanches.

How breakdown occurs (general process)

1. A few free electrons are always present (cosmic rays, UV, radioactivity).
2. In the field each electron gains energy $eE\lambda$ between collisions.
3. If this exceeds the ionization energy, a collision releases a new electron and a positive ion.
4. The new electrons are also accelerated: the number grows exponentially as an electron avalanche ($e^{\alpha d}$).
5. Secondary processes (ion impact and photons at the cathode, photo-ionization in the gas) produce new electrons that start new avalanches.
6. When the process becomes self-sustaining, current rises sharply and a spark channel forms.

Other processes: excitation, photo-ionization, attachment (electronegative gases like SF₆ capture electrons and so raise strength), and recombination.

The two main theories are the Townsend mechanism and the streamer mechanism.

1. Townsend mechanism (low $pd$, short gaps)

• Primary ionization: $I = I_0 e^{\alpha d}$, where $\alpha$ = ionizing collisions per electron per cm.
• Secondary ionization: positive ions, photons and metastables release $\gamma$ electrons per ion from the cathode. The current becomes

$$I = \frac{I_0\,e^{\alpha d}}{1 - \gamma\left(e^{\alpha d} - 1\right)}$$

• Breakdown criterion: $\gamma\left(e^{\alpha d} - 1\right) = 1$, or $\gamma e^{\alpha d} \approx 1$. The discharge then sustains itself without the external source.
• It leads to Paschen's law $V_b = f(pd)$ and is valid for $pd$ below about 1000 torr·cm.
• Limitations: predicts formative time lag of about $10^{-5}$ s (observed $\sim10^{-8}$ s) and a diffuse discharge, not the observed thin channel.

2. Streamer (Raether–Meek) mechanism (high $pd$, long gaps)

 cathode                          anode
   |   e- avalanche ---->  +++++ |
   |      photons  *  *   +++++  |
   |   <---- streamer (plasma)   |
   |      channel grows to       |
   |      cathode -> spark       |

• A single avalanche crossing the gap leaves behind a dense positive space charge near the anode, because electrons move fast and ions are slow.
• When the avalanche contains about $10^8$ charges, the space-charge field $E_r$ becomes comparable to the applied field.
• Photons from the avalanche head ionize gas nearby (photo-ionization); the new electrons form secondary avalanches that flow into the positive space charge.
• A conducting plasma channel (streamer) grows rapidly towards the cathode and bridges the gap: breakdown.
• Meek's criterion: breakdown when space-charge field equals applied field:

$$E_r = 5.27\times10^{-7}\,\frac{\alpha\,e^{\alpha x}}{(x/p)^{1/2}}\ \text{V/cm}$$

• Raether's criterion: $\alpha x_c = 17.7 + \ln x_c$, i.e. critical avalanche size $e^{\alpha x_c} \approx 10^8$.
• Explains short time lags ($10^{-8}$ s), the thin branched spark channel and independence from cathode material.

Comparison

Breakdown of a solid dielectric is the permanent loss of insulating property when the stress exceeds what the material can withstand. Unlike gases and liquids, solids do not recover. Breakdown strength depends strongly on the time of voltage application, so the types are classified by time.

 Breakdown
 strength
  |---- intrinsic (10^-8 s)
  |     \__ electromechanical
  |         \___ thermal (ms - h)
  |               \______ erosion / PD, chemical,
  |                       treeing (months-years)
  +-------------------------------------> log time

1. Intrinsic breakdown

The highest strength a pure, homogeneous material can have at a given temperature, with no external influences (about 5–10 MV/cm).

• Electronic breakdown: electrons in the conduction band gain energy from the field faster than they lose it to the lattice; they gain enough energy to ionize the lattice and current increases rapidly. Occurs in about $10^{-8}$ s.
• Avalanche (streamer) breakdown: a single electron in the conduction band ionizes repeatedly, building an avalanche; about 40 generations ($2^{40} \approx 10^{12}$ electrons) destroy the material locally. Strength depends on thickness.

2. Electromechanical breakdown

Electrostatic attraction between the electrodes compresses the material:

$$\frac{\varepsilon_0\varepsilon_r V^2}{2d^2} = Y\ln\frac{d_0}{d}, \qquad E_{max} = 0.6\sqrt{\frac{Y}{\varepsilon_0\varepsilon_r}}$$

When the thickness falls to 0.6 of its original value, the material collapses mechanically and breaks down. Important for soft materials such as polythene at high temperature.

3. Thermal breakdown

Loss heats the dielectric: $W_{dc} = \sigma E^2$, $W_{ac} = \frac{E^2 f\varepsilon_r\tan\delta}{1.8\times10^{12}}$ W/cm³. Heat balance $C_v\frac{dT}{dt} + \text{div}(K\,\text{grad}\,T) = \sigma E^2$. If heat generated exceeds heat dissipated, temperature rises without limit (thermal runaway). Strength is lower under AC, at high ambient temperature and in thick insulation.

4. Electrochemical (chemical) breakdown

Long exposure to heat, oxygen, moisture and discharge products causes oxidation, hydrolysis and chemical attack; electrolysis occurs under DC. Losses rise and the material becomes brittle and cracks. The rate roughly doubles for every 8–10 °C rise in temperature.

5. Breakdown due to treeing and tracking

• Treeing: at sharp points, voids or contaminants inside the solid, partial discharges erode fine branched channels (electrical trees); water trees grow in wet polymer cables. When a branch bridges the electrodes, the insulation fails.
• Tracking: on a contaminated, wet surface, leakage current forms dry bands and small arcs that carbonize the surface; the carbon track grows until surface flashover.

6. Breakdown due to internal (partial) discharges

Gas voids in the solid carry a stress $\varepsilon_r$ times higher than the solid and have lower strength. They discharge at voltages well below working voltage; repeated discharges erode the void walls until the solid fails. Prevented by void-free manufacture (vacuum impregnation).

An internal discharge begins when the voltage across the gas void reaches the breakdown voltage of the gas. The void and the rest of the dielectric behave as two capacitors in series.

Given

• Specimen thickness $d = 1$ cm, void thickness $t = 1$ mm $= 0.1$ cm
• $\varepsilon_r = 4.0$, applied voltage $80$ kV (rms)
• Breakdown strength of air $E_b = 30$ kV (peak)/cm

  +V (80 kV rms) =================
        solid   er = 4
       [ air void  t = 1 mm  ]
        solid
   0  ==========================
           d = 10 mm

Series capacitances (per unit area)

$$C_1 = \frac{\varepsilon_0 A}{t}\ (\text{void}), \qquad C_2 = \frac{\varepsilon_0\varepsilon_r A}{d - t}\ (\text{solid})$$

Voltage across the void:

$$V_1 = V\,\frac{C_2}{C_1 + C_2} = \frac{V\,t}{t + \dfrac{d - t}{\varepsilon_r}}$$

Voltage across the void at its breakdown

$$V_{1b} = E_b\,t = 30\times0.1 = 3.0\ \text{kV (peak)}$$

Applied voltage for internal discharge

$$\begin{aligned} V_i &= E_b\left[t + \frac{d - t}{\varepsilon_r}\right] \\ &= 30\left[0.1 + \frac{0.9}{4}\right] = 30\times0.325 \\ &= 9.75\ \text{kV (peak)} \\ V_{i,rms} &= \frac{9.75}{\sqrt2} = 6.89\ \text{kV (rms)} \end{aligned}$$

Answer: Internal discharge starts when the applied voltage reaches 9.75 kV (peak) = 6.89 kV (rms).

Since the applied voltage (80 kV rms) is much higher than 6.89 kV rms, the void will discharge many times in each half cycle and the specimen will deteriorate quickly.