part1 high voltage engineering.pdf

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  • Dr.Qais Alsafasfeh

    Electrical Breakdown of Gases in Uniform Fields


    At normal temperature and pressure, gases are excellent insulators but background

    currents of the order of micro-amps can be measured if an electric field of several

    kV/mm is applied. This current results from the electron/ion pairs produced by high-

    energy particles, either cosmic rays or derived from natural radioactivity, striking an

    air molecule:

    high-energy particle + M ==> M+ + e

    If the voltage is increased sufficiently, the electron is accelerated by the electric field

    towards the positive electrode (or anode) and further ionisation can occur. The electron will collide with gas molecules and most of these will be elastic collisions,

    but, if it has gained enough kinetic energy (KE), it will ionise the gas molecule it hits: e + M ==> M

    + + e + e.

    Enough energy means energy greater than the ionisation energy of the molecule.

    The process of acceleration until a collision with a molecule occurs, with most of the

    collisions elastic, and some inelastic (i.e., ionising) is illustrated in the AVAL-1.exe

    program. The KE gained by the electron is (electric field)*(distance travelled before the next collision) see box below.

    Now there are two electrons and the process can repeat, and repeat, and repeat,

    causing an exponential increase in the number of electrons. The situation after 4 such

    sets of ionisations by accelerated electrons is illustrated in the AVAL-2.exe program

    from which the diagram below is taken. (The original high-energy particle ionisation

    occurs at A subsequent ionisations are caused by the accelerated electrons).

    There are 1+1+2+4+8 = 16 positive ions

    and 16 electrons here

    Note that while the electron is accelerated towards the anode, the positively-charged

    ion is obviously accelerated towards the cathode. However the ion, being far

    heavier, is accelerated more slowly: the average velocity of the electrons is about ten

    times faster than that for the ions. This causes the situation seen above with the

    electrons moving swiftly to the right in a group, leaving clumps of 1, 2, 4 and 8 ions

    behind. This is seen more clearly in the AVAL-4.exe program in which 10 sets of

    ionisations have occurred (so there are 210, or 1024, electrons and the same number of

    positive ions).


    Kinetic energy gained by the electron = work done on the electron

    = (force on the electron) * (distance travelled by the electron)

    = (e.E)*x i.e., proportional to both the electric field and the distance gone.

  • Dr.Qais Alsafasfeh

    The electron will normally have many elastic (low-energy) collisions before the

    ionising collision hence the crooked paths seen in the animations. Because of the

    random nature of the number of collisions before the ionising collision, the distance

    between ionisations is also variable again, as seen in the animations.


    In the AVAL-4 diagram below there are 1024 electrons (and the same number of

    positive ions), the electrons and ions being indistinguishable in black and white

    reproduction. The distribution graphs for the two charged particles are shown, and

    explain the comet-like shape of the avalanche, as this phenomenon is called.

    Note the overlapping of the two graphs: this means that there will be a number of

    electron/positive-ion collisions which may result in recombination:

    e + M+ ==> M + energy

    This recombination energy is usually released as a photon of light energy.

    (a) (b)

    Cloud chamber photography of single avalanches (a) in nitrogen (N2) at 0.37 bar

    and (b) in carbon dioxide (C02), both in a 36-mm gap.

    The voltage was a DC voltage pulse lasting 0.4 ms. [H. Raether, Electron Avalanches and Breakdown in Gases, 1964, p.5]





  • Dr.Qais Alsafasfeh

    The ionisation coefficients

    The ionisation coefficient, , is defined as the probability that an electron will make an ionising collision in travelling unit distance in the direction of the anode.

    In addition there is a possibility (especially for slower-moving electrons) that an

    attaching collision takes place:

    e + M ==> M.

    The attachment coefficient, , is defined as the probability, per unit distance travelled in the direction of the anode, that an electron will attach to a molecule to form a negative ion.

    As mentioned earlier, the kinetic energy (KE) gained between collisions needs to

    exceed the energy required to ionise the molecule. The distance between collisions is

    inversely proportional to the density and hence to the pressure so it should not be

    surprising (see box) that

    /p = f(E/p).

    For similar reasons it is found that

    /p = g(E/p).

    For simplicity, an effective ionisation coefficient, , is defined as


    Many text books use the empirical equation

    /p = 1100(exp{-27.4 E/p})

    for air but it is very approximate. Better ones are available but are generally more troublesome to apply (see, for example, MacAlpine & Li, IEEE Trans.D&EI, Vol.7,

    pp.752-757, 2000). Here, and in general in this course, the units are assumed to be

    mm, kV and bar or their combinations.

    Clearly, if attachment is more likely

    than ionisation when a collision

    occurs, or >, avalanches cannot develop.

    In nitrogen the attachment

    coefficient, , is negligible; in

    oxygen it is very small. So, in this

    graph of /p versus E/p for air it is only at low fields, below 2.3 kV/mm

    bar, that > (i.e., is negative). Even then it is only just below zero.

    The field at which = is called the Critical Field.









    0 1 2 3 4 5 6

    E/p (kV/

    alpha/p (1/


    Geballe & H


    Morruzzi & P

    The energy gained between collisions = e.E. = eE/p

    because (= mean free path between collisions) is proportional to 1/p (p = pressure)

    The probability of a collision resulting in ionisation is a function of the energy gained between

    collisions, that is, from the above, a function of E/p, say F(E/p).

    The number of collisions (any kind) per unit distance = 1/ = Ap (A is a constant)

    the number of ionising collisions, , is F(E/p)*Ap, or, /p = f(E/p)

  • Dr.Qais Alsafasfeh

    Avalanche calculations

    The average size of an avalanche may be

    calculated for uniform-field conditions by

    considering the number of electrons n passing through a plane at a distance x from the cathode in the direction of the electric field (towards the

    anode) in a time t. Simple integration (see box) gives

    n(x) = ex

    How big is an avalanche? Consider an electron produced at x = 0 in a 10-mm gap (e.g. as an

    electron-ion pair due to a cosmic ray, or by

    emission from the electrode) in air at

    atmospheric pressure.

    If the applied voltage is 25 kV, E = 2.5 kV/mm,

    and is found (from the graph on the previous

    page) to be close to zero.

    Now try 30 kV: = 1.3 mm-1 the number of electrons in the head of the avalanche when it strikes the anode will be

    n(10mm) = exp(13) = 4.4x105

    Now try 35 kV n(10mm) = exp(30) = 1.1x1013 . This is a huge increase!

    Clearly if the current (the sum of all the electrons in all the avalanches which occur in

    a second) increases at this rate, breakdown MUST occur near this voltage.

    The reverse calculation: what is the voltage which gives, say, 108 electrons in the

    head of the avalanche when it strikes the anode?

    n(10mm) = exp(10.) = 108

    Therefore = 1.84 mm-1,

    So, V = 32 kV.

    Avalanches in non-uniform fields

    In a non-uniform field, the same approach as used above gives

    n(x) = exp( (E).dx ) This is useful in for example, coaxial cable or busbar systems see next lecure.

    Sulphur hexafluoride, SF6

    This is a very important insulating gas and is widely used in equipment for electrical

    power transmission and distribution. It is colourless, odourless and heavy.

    For SF6 the critical field is at 8.85 kV/mm as shown in the graph overleaf and

    the attachment coefficient is much larger than that for oxygen.

    From the right-hand graph, an empirical expression for the effective ionisation

    coefficient, or (-), may be obtained as /p = 26E - 230.


    The average size of an avalanche

    Consider the number of electrons n passing through a plane at a distance x from the cathode in the direction of the

    electric field (towards the anode) in a

    time t.

    The number of electrons passing through

    a plane at a distance x + dx may be written as n + dn (again in a time t) where,

    dn = n.dx, using the definition of ,

    so that, integrating between limits of x = 0 and x, and remembering that each avalanche is started by a single electron, n(x) x

    dn/n = .dx, 1 0

    or, n(x) = ex

    Unless the field is