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SMPS: Spike at Transformer Output

Discussion in 'Electronic Design' started by Atul, Sep 2, 2003.

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  1. Atul

    Atul Guest

    Hello Everybody,

    I am developing a Half Bridge push-pull SMPS. The configuration is
    standard with a LC filter at the output and it seems to be working
    well - except for one thing.

    There is a VERY large spike at the rising edge of the transformer
    secondary output. There is NO corresponding spike on the primary
    waveform.
    This spike therefore also appears across the rectifier diodes causing
    eventual breakdown.
    The output voltage of the transformer is 450V peak, while the spike
    crosees 1KV on my scope.
    I have tried using fast surge clipping diodes at the secondary output
    to no avail.

    What could be the matter?

    Thanks,

    Atul
     
  2. Dave VanHorn

    Dave VanHorn Guest

    Do you have a current probe? Very useful in SMPS design!
     
  3. R.Legg

    R.Legg Guest

    There is a VERY large spike at the rising edge of the transformer
    The rising edge of PWM secondary, for full-wave rectification, will
    occur after one of the output rectifiers that was freewheeling turns
    off. The peak reverse energy resonates with the leakage inductance of
    the non-conducting winding.

    You can reduce it by;

    - Using rectifiers with less (or no)recovery charge Qrr. At higher
    voltages try SiC parts.
    - slowing the dI/Dt, either with with a small bead or resonant driver
    waveforms (this reduces the peak reverse recovery current).


    If you want to clamp it, you'll have to use a very fast diode in a
    very small connection-capacitor loop area construction.

    Hard switching at these voltages is not easy. You might consider
    another topology with lower rectifier stress, or series sections of
    lower voltage secondaries. 2:1 is not much margin when a single
    resonant transition can exceed it. If your output voltage is 450, and
    you are not rectifying with a full-wave bridge, 1KV reverse stress
    would be normal, even without spikes.

    Leakage increases with the square of the turns. This makes high
    voltage harder, as rectifier Qrr doesn't reduce in higher voltage
    parts while the dI/dT of the application increases.

    High voltage means increased insulation/creepage and results in
    increased leakage inductance.

    Snubbing consumes power, but we do it anyways, for good reasons. As
    the loss in the snubber is proportional to CV^2f, higher voltages make
    this harder. If it's unacceptible, reduce the leakage.

    RL
     
  4. Atul

    Atul Guest

    This is caused by the release of energy stored in the leakage inductance of
    Thanks for the reply, I thought so, and I will be trying out your
    suggestion in the next few hours.
    Is there some method to figure out the approximate R/C values required
    e.g my measuring the leakage inductance.... or do we have to just
    throw out some values and then iterate to the correct values?

    Yes I do have a current probe...

    Atul
     

  5. Alternatively, if you have the time, try a current-fed push-pull topology. I
    did this for a 270 volt, 600W converter and the spikes were tiny.

    Graham Holloway.
     
  6. Genome

    Genome Guest

    Generating such high (?) voltages with a half bridge topology will be
    difficult. One of the current fed converters with a full bridge rectifier on
    a single winding secondary would be better. The diodes only experience a
    reverse voltage equal to the output voltage. Mind you, the problem probably
    moves elsewhere.

    What you are seeing is the effect of reverse recovery in one of the output
    diodes as it turns off in conjunction with the leakage inductance in the
    transformer. Previously it would have been free-wheeling. When the voltage
    across it is reversed, as the opposite primary side switch turns on, forward
    current in the leakage inductance naturally resets to the output.

    While this is happening the leakage inductance is supporting the full
    transformer voltage. As the diode turns off it undergoes reverse recovery
    and the characteristics are determined by the diode, the original forward
    current it was carrying and the rate at which the diode is turned off. That
    rate is set by the negative dI/dT applied to the diode and is largely
    determined by the leakage inductance itself, dI/dT = -E/Lleak.

    There are two parameters, the reverse recovered charge and the peak reverse
    recovery current. Both are specific to the diode environment and ultimately
    need to be measured in circuit. The important one is the peak reverse
    recovery current, Iprr.

    This will be the peak current in the leakage inductance and determines the
    associated energy per switching cycle, E = 0.5LleakIprr^2. Multiply that by
    the switching frequency and you get the additional power that the diode has
    to dissipate. This is added to the forward conduction losses to find the
    total power in the diode. Then you can make a decision as to wether the
    diode has an appropriate power rating for the job.

    Unfortunately high voltage diodes have worse recovery characteristics
    compared to lower voltage devices and things become worse with increasing
    temperature. Since your diodes are failing it sounds like the additional
    power is too much for them to cope with. There are a number of options.....

    First you'll need to measure what the leakage inductance of your transformer
    is. Do this for one leg of the secondary with the other leg and primary
    shorted and use a frequency close to your switching frequency. Also measure
    the peak reverse recovery current, Iprr, in circuit with your supply
    operating at full load. (Let the diodes get up to temperature, without
    blowing them up ;-) Work out the power and add that to your expected forward
    losses.

    Can the diodes handle it?

    If you can allay fears about reliability then you might use diodes that are
    specifically rated for repetitive avalanche breakdown. The data sheet will
    tell the story, they will probably be soft recovery devices as well. It may
    be the case that you can improve the heatsinking of the diodes sufficiently
    by a simple change to the PCB, increase the area of copper where they are
    soldered to the board if they are surface mount or add a clip on heatsink.

    If you can't improve things in that way but you can still accept the
    breakdown then you have to look for some way of reducing the energy. You
    might think that this involves reducing the leakage inductance. However
    remember that Iprr is determined by the -dI/dT you apply to the diode and
    that is limited by the leakage inductance.

    The relationship is not likely to be linear but if you halve Lleak you might
    expect to double Ippr and since E = 0.5LleakIppr^2 you end up doubling the
    power that the diode has to cope with... };-)

    If you did choose to reduce the leakage inductance then bear in mind that it
    is not strictly primary to secondary leakage that you are interested in.
    Your best route is to improve the coupling of the two center tapped
    secondary windings. You can best achieve this by co-mingling (eeh, I like a
    nice mingle) the secondary turns.

    However, strange as it may seem, it is probably better to increase the
    leakage inductance by adding some external inductance in series with the
    diodes. This should reduce the overall additional power.

    One main thing that works in your favour is stray and diode capacitance.
    Some of that 0.5LleakIprr^2 gets stored on parasitic capacitance. So not all
    of the energy is really lost in the diode when it breaks down. This is your
    next method of attacking things, artificially increase the diode/stray
    capacitance by adding a suitably rated ceramic capacitor in parallel with
    the diode.

    Don't get excited about RC snubbers yet.... That gets used to sort another
    problem, if it needs sorting.

    Since you know what 0.5LleakIppr^2 is you can initially ignore the stray and
    pick an initial value on the basis of 0.5CaddedVrev^2 where you limit Vrev
    to the reverse voltage capability of the diode. Stick it in the circuit, see
    if it works and then back down on the value.

    It may initially sound like a recipe for increasing losses elswhere in the
    circuit since you now have to charge and discharge that extra capacitance
    but.... bear in mind that it appears in series with the leakage inductance
    which will limit the currents involved. Yes there will be an increase in rms
    currents in the circuit but you will not be dissapating all the energy in
    that added capacitance elsewhere, it transfers between it and the leakage
    inductance.

    Even so.... Once you have limited the peak reverse voltage on the diode, and
    before you did, it is likely that you will see voltage ringing on the node.
    That's Lleak resonating with stray, diode and any extra you have added. At
    this point you might begin to consider the use of an RC snubber to damp that
    ringing. I'd suggest that it's more an issue of EMC and if it doesn't prove
    to be a problem then you needn't do anything about it.

    If you choose to then you add an RC network with C set to three times the
    node capacitance and pick a resistor with a value equal to the resonant
    impedance at the resonant frequency. Unfortunately when you do this you do
    suffer power losses in the snubber resistor due to charge and discharge of
    the snubber capacitor.

    Well.... that's the basic train of thought. It's not necessarily in the
    right order, or any order come to that. Hope you get the idea and can 'short
    circuit' things to get the simplest solution for your particular problem.

    DNA

    Mind you.... I could be wrong....
     
  7. Atul

    Atul Guest

    Thanks a lot for your response.

    I am using a full bridge at the output using STA1212 diodes
    (ultrafast, 1200V, 12A). My rated output voltage is 310V but I need
    the extra voltage(450V) as the power supply has to work even at low
    input voltages.

    Could you explain how to reduce di/dt in detail... I have never used
    beads before so could you point me to some resources on the web.

    Thanks again,

    Atul
     
  8. Atul

    Atul Guest

    Is there some place on the web with information on th current fed push
    pull topology?

    Atul
     
  9. R.Legg

    R.Legg Guest

    I take it you've tried the other suggestions?
    - SiC Siemens Infineon market some pretty rugged ones, though 1KV
    parts are still samples.
    - winding for lower leakage
    - configuring clamp for higher speed
    - snubbing

    dI/dT is limited only by the applied voltage and the leakage
    inductance. It may seem intuitionally wrong to be working for lower
    leakage and then inserting a bead. The bead will have a nonsaturated
    inductance much higher than it's aturated value. Even then, actual
    energy loss may still be higher - heating the bead and transfering
    into your clamp.

    The bead is supposed to work in conjunction with capacitance across
    the diode - you may have to add more - the voltage developed at turn
    off should rise slowly enough for the clamp to handle it. One bead on
    the primary may function just as well. Hard ferrites with a high
    currie temperature can also function similarly.

    http://www.amotech.co.kr/kor/n_core.html
    http://www.hitachi-metals.co.jp/e/prod/prod02/p02_25.html

    Hitachi Metals America used to market 'ammo-beads' to do this.

    If your circuit were resonant or even quasi-resonant, applied voltage
    inducing dI/dT would be lower. That's why others have recommended
    various ZVS or ZCS primary switching methods.

    Google for snubber information. The bottom line is that if you burn R
    with C the power loss is CV^2f, without dividing by two, because the
    part charges in both directions in an AC circuit. If R isn't getting
    hot, then something else is.

    RL
     
  10. Atul

    Atul Guest

    (R.Legg) wrote in message
    The output is current regulated (1A) with 300V compliance. Using 450V
    secondaries allows me to get 300V with full duty cycle at the lowest
    expected input voltage.

    Atul
     
  11. It was actually a full bridge. Used about 200ns overlap. I put a pair of
    270V BZT05 zeners in series to clamp the bridge supply, but didn't need
    them.


    Graham Holloway
     
  12. Atul

    Atul Guest

    I disconnected all loads from the transformer secondary (including the
    rectifier diodes) and then observed the secondary wave form - and the
    spike was still there!!!
    So I introduced a current probe in series with the primary winding and
    there's a 1MHz ring (It rings for about 3 cycles) in the rising edge
    of the current waveform which is duly reproduced at the secondary as a
    voltage spike.
    Can this be caused by too fast a turn on of the MOFETs? I tried
    increasing the gate resistance but that doesnt change things.....

    One thing is clear, it has nothing to do with anything on the
    secondary side....

    What should I do next?


    Atul
     
  13. Atul

    Atul Guest

    What you are seeing is the effect of reverse recovery in one of the output

    Guys, I disconnected all loads from the transformer secondary
    (including the rectifier diodes) and then observed the secondary wave
    form - and the spike was still there!!!
    So I introduced a current probe in series with the primary winding and
    there's a 1MHz ring (It rings for about 3 cycles) in the rising edge
    of the current waveform which is duly reproduced at the secondary as a
    voltage spike.
    Can this be caused by too fast a turn on of the MOFETs? I tried
    increasing the gate resistance but that doesn't change things.....

    One thing is clear, it has nothing to do with anything on the
    secondary side....


    Atul
     
  14. ....except, possibly, the secondary inter winding capacitance
    resonating with the leakage inductance.
     
  15. R.Legg

    R.Legg Guest

    What have you done, so far?

    What snubbing values/locations have you tried?

    What layout, configuration and winding alterations have you tried?

    Unloaded performance is pretty irrelevant, as your application is
    constant current.

    Switched open-circuit windings will always ring, as the Q of stray
    elements is quite high. A switched primary appears to be shorted when
    switches or energy recovery diodes conduct.

    RL
     
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