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TVS diode as snubber for ac-dc flyback converter

Discussion in 'Electronic Design' started by aravind, Jul 18, 2007.

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  1. aravind

    aravind Guest

    hi guys
    I plan to use TVS diode as snubber for a ac-dc flyback converter, by
    connecting TVS diode in parallel with the primary winding of the
    flyback transformer so that TVS diode absorbs the transient energy
    stored in the leakage inductance of the flyback transformer during
    turn off of the switch. The rated output voltage of the converter is
    50 V dc. The turns ratio of the transformer is 4:1. My rated power of
    operation is 210 W. Switching frequency is 50KHz in CCM mode. During
    transient my output voltage can go up to 80 V(worst case), and the
    current in the switch might go up to 15 A(max during full load in
    transient). Therefore I would require a TVS diode with a peak pulse
    power dissipation (Ppp) of about 1.6*80*4*15 =7680 W with rated
    working peak voltage (Vwm) of 80*4= 360. Since I could only get TVS
    diodes with Ppp of 1500 W. I intend to put six TVS diodes of 1500 W
    Ppp and Vwm 0f 60 V in series.
    Can anyone suggest weather this idea will work, and will any special
    measures have to be undertaken to for heat dissipation if the whole
    ckt (other than flyback t/f) has to be implemented on a pcb.
  2. Tim Williams

    Tim Williams Guest

    So, um, er, why are you *dissipating* something which can be *saved*?

    The usual snubber involves shunting the current into a holding capacitor,
    from which it leaks back into the supply rail via resistor.

    Is this a flyback or forward converter?

  3. Terry Given

    Terry Given Guest

    psst - guess what happens to the charge that "leaks back" thru the resistor?

    it doesnt actually matter, the problem is still the same
    so the flyback voltage is 200V nominal

    ouch. thats crappy regulation, and bumps the flyback up to 320V

    look at the TVS datasheets closely. Ppp is often specified for a
    1.2/50us pulse (sometimes even for an 8.3ms haversine). convert it to
    Joules, and do your calculation based on that.

    there are also 5kW Ppp TVS available.

    yes it will work. If you use a bidirectional TVS then you dont need a
    series diode, *BUT* TVS have a LOT of capacitance, which dumps
    C*Vpk^2*Fsmps into your FET. a series diode effectively isolates the TVS

    for a 210W SMPS, you will have to throw away 20-30W in your switches,
    transformer and rectifiers, which will require some heatsinking.

    the clamp network basically dissipates the energy stored in the leakage
    inductance, which should be < 2% of the primary inductance if the
    transformer is well designed; if not, it can be a lot higher.

    figure it out like this:

    - the current in the leakage inductance is the peak primary current.
    - the clamp network clamps the Drain to (Vflyback + Vextra) = Vtvs
    - the reflected load voltage Vflyback appears across the magnetising
    inductance, so
    - the voltage across Lp_leak = Vextra = Vtvs - Vflyback
    - the current thru the leakage inductance linearly (ish, assuming the
    xfmr is wound for low losses) decays to zero, as:

    Vextra = Lp_leak*Ip_peak/Tclamp

    - re-arrange to solve for Tclamp
    - now you know how long the current takes to ramp down to zero.
    - the average power dumped into the TVS during Tclamp = 0.5*Vtvs*Ip_peak
    (current waveform is a triangle, hence the half)
    - the total energy dumped into the TVS is 0.5*Vtvs*Ip_peak*Tclamp
    - the average power dumped into the TVS is 0.5*Vtvs*Ip_peak*Tclamp*Fsmps

    So although the peak pulse power is perhaps high, the average power
    should be pretty low (but 2% of 210W is still 4.2W). Years ago I made a
    flyback with 3 x SOD-87 SMT zeners, on a vertical PCB. I didnt do this
    calc, and during operation the zeners got hot enough to desolder
    themselves - I found them on the bench after the SMPS blew up. oops.

    If you have to implement it in a small space, I'd use a pair of FETs and
    a pair of catch diodes as a diagonal half-bridge converter - a FET at
    either end of the primary winding, connecting to each supply rail, forms
    one diagonal half of a full bridge; the other diagonal half is done with
    diodes, and both FETs switch simultaneously. That way almost all the
    leakage energy is returned to the supply rails. And it also keeps the
    FET voltage clamped to the DC bus, so slightly lower voltage FETs can be
    used (which makes up for having 2 in series)

  4. Tim Williams

    Tim Williams Guest


    What about it? The charge is saved, looping it into the +V rail rather than
    wasting it to GND.

    Running the charge through a resistor does tend to dissipate some voltage
    though, yes ;-)

  5. Terry Given

    Terry Given Guest

    Are you referring to an RCD clamp? cos if you are, all of the energy
    gets dissipated in the R. e.g.:

    | | )
    [R] [C] ) Xfmr
    | | )
    X FET etc

    all of the variant s of an RCD clamp achieve the same objective, viz.:

    - at switch turn-off, current commutates thru diode into cap, thereby
    shaping switch load line
    - once thats done, the R bleeds off some or all of the charge stored on
    the cap

    the last one is worth thinking about. If you dont get rid of all the
    charge dumped into the cap, the voltage across it rises. eventually,
    something breaks.

    in the case of a turn-off snubber, whose purpose is to shape the load
    line (very common with bipolars, not so much with FETs), the cap is
    mostly discharged, is the current commutating into it sets the switch
    dV/dt = I/C. In this case, the cap is connected to 0V.

    for a voltage clamp, the cap is not fully discharged, and sits at the
    +Vdc rail, at some voltage Vc_min. At turn off, the switch voltage rises
    to Vdc + Vc_min, then the diode conducts, and the leakage energy dE_lk
    is dumped into the cap, causing its voltage to rise, calculated as:
    Ec_final = Ec_start + dE_lk = 0.5*C*Vc_final^2

    then the resistor is sized to discharge the cap back down to Vc_min.

    If you draw the cap charge and discharge current paths, its easy to see
    that the resistor dissipates (some or all of) the energy stored in the cap.

    Que?! interesting (read as: wrong) use of terminology, Tim.
  6. Tim Williams

    Tim Williams Guest

    Charge as in coulombs. The cap fills up with some discrete amount of
    charge, then the charge bleeds through a resistor (at some rate and
    therefore some current). Charge is a thing that tends to be conserved.
    However, a voltage associated with that charge flow corresponds to power
    dissipation. When I originally said "something that can be saved", I meant
    the charge, which is misleading since I also used "dissipated" (power by
    definition) in the same sentence...

  7. aravind

    aravind Guest

    its a flyback converter
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