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Can the PIV of a diode be "safely" exceeded?

Discussion in 'Electronic Design' started by KILOWATT, Nov 16, 2004.

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  1. KILOWATT

    KILOWATT Guest

    Hi everyones thanks to read. Since a few weeks, i'm using the LightKeeper
    Pro from Ulta-Lit Tree Company ( http://www.lightkeeper.biz/default.asp ) to
    test and repair miniature lights sets (series-wired). I've found the Quick
    Fix Trigger utility very useful. as you can see on their website, it's based
    on a piezo igniter that sends a high voltage pulse through the lights set to
    activate (short) the shunt inside the bulb, that didn't do so when the
    bulb's filament burned out. I wanted to see how this system is built so i
    opened the unit. Here's two photos plus a schematic i've drawed for the
    igniter section. (Sorry, the symbol for the piezo igniter is probably
    wrong...i don't know the correct one).
    http://www3.sympatico.ca/kilo.watt/images/lightkeeper1.JPG
    http://www3.sympatico.ca/kilo.watt/images/lightkeeper2.JPG
    http://www3.sympatico.ca/kilo.watt/images/hv_pulse_gen.bmp
    As i can see, the four diodes allow a peak reverse voltage of approximately
    4Kv, wich is sufficient to break down the shunt's insulation inside the
    burned bulb(s) and complete the series circuit. Once completed, half of an
    AC cycle can flow through the four diodes, allowing the set to glow (dimly)
    and show which bulbs are burned and needs replacement.
    What leads me to my question (please look at the schematic) is: why the
    reverse voltage doesn't seem to damages those diodes? According to the piezo
    igniter's website ( http://www.yiqiang-piezo.com/english/production2.htm ),
    the output voltage is at least 15Kv. With the four 1n4007 wired in series, i
    should get a peak reverse voltage of about 4Kv isn't? For what i know, a
    diode is destroyed once it's PIV is exceeded. I think i missed something
    when studying the basic operation theory of a diode. ;-) TIA for any
    useful reply.
    --
    Alain(alias:Kilowatt)
    Montréal Québec
    PS: 1000 excuses for errors or omissions,
    i'm a "pure" french canadian! :)
    Come to visit me at: http://kilowatt.camarades.com
    (If replying also by e-mail, remove
    "no spam" from the adress.)
     
  2. Rich Grise

    Rich Grise Guest

    Just "exceeding the PIV" doesn't automatically destroy a diode.
    In fact, that's how Zener diodes are _intended_ to be operated -
    in inverse breakdown.

    What lets the smoke out is exceeding the power dissipation
    ability of the chip. A 1N4007 with 1KV across it and 50
    microamps flowing through it would have to dissipate 50
    milliwatts, which it would hardly notice, especially
    considering that they'll dissipate right about a watt in
    the forward direction, at one amp, just in normal operation.
    With a proper heat sink, or if the pulse doesn't last very
    long, they can pass a surprising amount of current and live
    to tell about it. ;-)

    Hope This Helps!
    Rich
     
  3. Terry Given

    Terry Given Guest

    A better way to look at it is the total energy Elost dumped into the
    device, in Joules. Then if you know the die area, thickness and material
    you can calculate volume and mass, then look up specific heat capacity
    cp (J/kg/K). Adiabatic temperature rise = Elost/(m*cp). Voila. You can
    also make a fair estimate at the thermal resistance, and calculate the
    thermal time constant:

    Tau = (m*cp)*Rtheta = [J/K]*[K/W] = [W*s/K]*[K/W] =

    if the pulse is less than Tau then the adiabatic approximation is good -
    the die absorbs pretty much all the heat. If the pulse is longer than
    Tau, then some (or perhaps almost all) of the heat flows out via the
    surface and the end-caps.

    I read a fascinating paper a few years back on electronic one-shots -
    literally! They were electronically fired single-shot guns. I forget
    exactly how the circuit worked, but basically they turned a switch on
    and dumped all of the energy from a cap into a short section of pcb
    track sitting below a projectile. All the energy dumps into the track,
    and gets converted into heat and projectile motion. They used a 1N4007
    in series with a FET as the switch. IIRC the fet leakage current was
    controlled/chosen to hold the 1N4007 on the cusp of avalanche breakdown,
    but keep dissipation low. To fire the device, the FET was turned hard
    on, and the 1N4007 broke down. The large amount of energy involved
    created a ball of plasma which swept thru the diode, turning it into a
    very effective short. I must dig that back out.....

    Cheers
    Terry
     
  4. Rich Grise

    Rich Grise Guest

    Terry, thanks for picking up the ball here! :)
    Rich
    [etc.]
     
  5. KILOWATT wrote...
    The other replies in this thread have shown you how the four diodes can
    happily survive breakdown if the breakdown energy they experience is
    below a critical value. Of course they will breakdown if the voltage is
    high enough, and thereby limit the piezo output, which cannot be 15kV in
    this instance. It's also likely that with a bulb in place the intended
    breakdown processes in the bulb (the clearing of the shunt) will limit
    the piezo's output voltage before the diode breakdown voltage is reached.

    You can measure the actual voltage with an oscilloscope, even if you don't
    have a high-voltage probe, by using a capacitive divider, if you can get
    hold of a small high-voltage capacitor. Here's a 1000:1 divider circuit.

    .. 1000:1 high-voltage probe
    ..
    .. 10pF 20kV ______________ 330
    .. <------||----)_____________)--+-/\/\--+----o ) scope input
    .. coax cable | _|_ _|_ | Rin = 1M
    .. capacitance | --- |___| | Cin = 20pF (etc)
    .. 8 ft = 220pF | | | |
    .. <--------------------------+--+-------+-----+------ ground
    .. ground clip ~9750pF series back-to
    .. select back 20V zeners

    To avoid a shock hazard, be sure to connect the ground clip and plug
    the probe into the scope before probing anything! And don't touch
    that 10pF cap!

    The probe has a high-frequency response of 10MHz, due to the 330-ohm
    input-protection resistor and the ~25pF of capacitance of two 1n5250B
    zener diodes in series. It also has a -3dB low-frequency rolloff of
    16Hz, due to the 1M scope Zin and the 10,000pF divider capacitance.
    You're measuring pulses, so the low-freq rolloff won't be an issue.
    Calibrate the probe with a 10V sine wave using the 10mV scope range.
     
  6. Winfield Hill wrote...
    I should add that most of the HV meter probes one finds around a typical
    workbench are *not* suitable for pulse measurements. I'm referring to
    the DC probes with the 6" proboscis tip and a double banana plug on the
    other end of the cable, for use with a multimeter. These employ a long
    high-voltage resistor and do not have proper internal shielding, so they
    will give you an erroneous (e.g., too high) result for short pulses.
     
  7. Fred Bloggs

    Fred Bloggs Guest

    Yeah you did- the PIV is the reverse voltage a diode can withstand
    indefinitely without breakdown- it is not the voltage at which the diode
    will breakdown. Clearly the diode design has to take into account
    manufacturing process variations so that despite these an acceptably
    large percentage of the diodes meet this PIV specification- and under
    worst case conditions such as Tj=70oC or something- who knows. Therefore
    it would not be unlikely that a typical 1N4007 breakdown at a much
    larger voltage than 1KV. The product specification was written by some
    sap with less than perfect command of the English language, so that the
    real meaning may have been the output voltage is no greater than 15KV-
    meaning it is an insignificantly rare occurrence that a stack of four
    randomly selected 1N4007's withstand that voltage without breakdown. One
    thing that is certain is that the output will be significantly greater
    than 4KV.
     
  8. Fred Bloggs

    Fred Bloggs Guest

    Nothing new- the electronically fired catridge-less projectile has been
    around forever- but usually to achieve extraordinarily high rates of
    fire like millions of rounds per minute.
    http://www.cnn.com/2003/BUSINESS/06/26/australia.metalstorm/
     
  9. Joel Kolstad

    Joel Kolstad Guest

    ....due to capacitive coupling 'around' the resistor? Or is there some other
    mechanism at work?
     
  10. Rich Grise

    Rich Grise Guest

    As well as 'through' the resistor, yes. :) Or maybe 'in' it. Win will
    know. ;-)

    Cheers!
    Rich
     
  11. legg

    legg Guest

    Yes it can,........ but it is only guaranteed within known limits for
    devices characterized to do so. For a rectifier, this characteristic
    is known as 'soft avalanche' and will be part of it's specification.

    A standard rectifier may exhibit a degradation in breakdown voltage
    level, following the first avalanche event, even in 'energy-limited'
    situations. If parallel capacitance is signifigant, (ie in common EMI
    component locations) or nonlinear series loads are present in series,
    this break-back of voltage represents a potentially serious, unplanned
    energy dump.

    In a series string, the break-back of one series element can set the
    others off in a chain.

    RL
     
  12. Not only can, but does. Always. Found out the hard way ...

    The reverse voltage across a string of diodes will not be equally divided.
    One diode will accumulate the majority of the drop, go into avalanche -
    creating a short and increasing the voltage across the remaining diodes,
    then the next diode will do the same, and the next ... till you have a dead
    short across the string and something goes poof.

    High value resistors, so that Ires >> Ileakage, in series with each
    diode will equalize the voltage along the diode string. Be sure
    not to exceed the voltage rating on the resistors; resistors can
    be safely strung for high voltage.
     
  13. Rich Grise

    Rich Grise Guest

    I've even seen it done with capacitors. This can be useful for a
    HV, low-current supply, like a focus electrode.

    Cheers!
    Rich
     
  14. legg

    legg Guest

    This is also another inadvertent source of capacitive energy for parts
    with unstable breakdown characteristics.

    It appears that the correct term is 'controlled avalanche', not soft
    avalanche, as originally indicated. This is searchable.

    RL
     
  15. Joel Kolstad wrote...
    Capacitive coupling to the 1/1000 side from all over. The portion from
    the resistor's body could be compensated (to first order only) with
    capacitance on the output side, but additional exposure from E fields
    from wiring near the measured voltage create large indeterminate errors
    that cannot be compensated. It's a mess and cannot work as delivered.
     
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