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The step by step design procedure for AN4134

Discussion in 'Electronic Basics' started by Richard, Jun 29, 2004.

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  1. Richard

    Richard Guest

    http://www.fairchildsemi.com/an/AN/AN-4134.pdf

    http://www.fairchildsemi.com/collateral/AN-4134.xls

    The xls file is to be used with the design steps in the pdf file.

    In the pdf file, I'm getting stuck at : Step (3) (b) RCD reset

    In Excel: You need to choose the Excel sheet "Forward with RCD reset": I
    have a probblem with cell C30. I don't know how to get to a value to put
    into cell C30, nor do I know what the parameter is. Cell F30 surely is the
    nominal snubber capacitor voltage, that is, the voltage across the snubber
    capacitor. So, what is the parameter that goes in cell C30 and how do you
    determine it's value? TIA.
     
  2. Second try:
    You get to pick any voltage you like, as long as it is greater than
    the value in F30. Higher voltage minimizes the time wasted to reset
    the core, while stressing the switch and rectifiers with higher
    voltage.

    I am working on understanding this statement from page 3 of the pdf:
    "winding and reset winding, respectively.
    Since the snubber capacitor voltage is fixed and almost
    independent of the input voltage, the MOSFET voltage
    stress can be reduced compared to the reset winding
    approach when the converter is operated with a wide input
    voltage range."

    The snubber voltage is a function of the average magnetization current
    times the snubber resistor value. The resistor value is constant. If
    the transformer were perfect, and the output voltage was perfectly
    regulated and there were no resistive drops related to various load
    currents, then the duty cycle is inverse to line voltage, so as the
    voltage gets higher the percent of on time goes down. The
    magnetization current thus climbs faster, but has less time to climb,
    so it reaches about the same value, independent of line voltage.

    I am looking at the voltage and current waveforms on page 3 and so
    far, it all makes sense. The magnetization current rises while the
    switch is on (and output current is being sent through the output
    rectifier), and decays while the current must push up to snubber
    voltage. But then, the current, Isn, reaches zero, and the inductor
    no longer pushes current into the snubber cap through the diode and
    the winding voltage falls toward zero. At this point (t3 to t4) the
    graph shows the magnetization current decreasing below zero much
    faster than it fell while Isn was flowing. A faster rate of change
    implies a larger winding voltage, but the voltage waveform, in red,
    shows a decreasing voltage heading toward zero as the transformer
    waits for the next on pulse.

    So I don't understand where the expression Vsn/(Lm/Coss) comes from
    that I think is describing the value of voltage being applied ot the
    winding while waiting for the next pulse t4 to t5). I have no idea
    what Coss is. My guess is that this is the value of magnetization
    that develops as the winding capacitance charges from Vsn down to just
    below zero, where the output rectifier comes on an shorts the
    secondary, allowing that value of accumulated magnetization current to
    circulate with little change till the next on time occurs.

    So to get back to the approximately fixed Vsn, I guess the on time is
    inverse to line voltage, so the peak snubber current is fixed, and the
    energy dumped into the snubber is fixed per cycle, regardless of line
    voltage and is almost fixed independent of load current, so as long as
    the cycle time is fixed, there is a constant average flow of energy
    through the snubber resistor, so a fixed average voltage. And the
    snubber capacitor keeps the instantaneous voltage near the average.

    So back to the spread sheet, you get to pick any value for the snubber
    nominal voltage (that will stay pretty close to that by the above
    reasoning) as long as it is greater than the specified minimum value
    if F30. The higher voltage you choose, the higher the stresses on the
    switch and rectifiers, but the more of the cycle time the transformer
    is able to transform power to the secondary (the higher the
    transformer utilization). If you pick a generous core size for the
    transformer, this is not much of a concern, so you can treat your
    switch and rectifiers better by picking a lower voltage.
     
  3. Richard

    Richard Guest

    I'm still not getting it.

    This is what I am seeing:

    So far in the steps, the set input parameters (in blue) I've addressed are:
    maximum line voltage; minimum line voltage; line frequency; output volts and
    amps; estimated efficiency; dc link capacitor; maximum duty ratio. I
    understand these parameters, that is, what they are. But I don't understand
    what the nominal snubber capacity voltage is. There are two voltages
    associated with row 30, the one in blue and the one in red. I'm looking at
    the red one and saying, given all the parameters in blue that are
    pertaining, the voltage in red is a voltage that will be present across the
    snubber capacitor and is 150.6 volts. Okay, but what is the voltage in blue?
    It's not the voltage ascross the snubber, that's in red. This is what I'm
    thinking. I'm wondering if things are confusing me because of course these
    calculations are not with a fixed input votage, a range is specified. I
    could not reconcile the values in C24, C25, and C26 until that dawned on me.
     
  4. The blue one is a voltage of your choice, as long as it is higher than
    the one in red. Lower values stress your power switch and rectifiers
    less, but will force a bigger transformer size. The value in there to
    start with is the program's suggestion.
    The red one is a limiting value. Note the > symbol between the blue
    entry box and the red limit box, implying that your entry must be
    greater than the limit.
     
  5. Richard

    Richard Guest

    Okay. But what is the nature of the my choice?

    For instance, I have the choice to set what the minimum input voltage can be
    and what the maximim input voltage can be. I know that here I'm setting the
    potential ac input parameters. Of course these settings are not what
    actually you would measure in the circuit unless the ac input was at either
    max or min. I'm not strictly setting what the ac input voltage will be in
    the circuit

    But I don't understand what it is that I'm setting at C30. AlI know is that
    it has to be higher than F30. To my mind F30 is Vsn or would be the voltage
    across the snubber if you were to measure it. If I was setting F30, by being
    able to place choice of value there, then I'd be setting what Vsn would be
    in the circuit by direct entry of a value. Anyway, is C30 a choice
    something similar in nature to the choice of either max or min input
    voltage? And if it is, what is the parameter exactly?
     
  6. Richard

    Richard Guest

    Actually, one could argue that this is not important. The important point is
    what you make "Maximum nominal MOSFET voltage =" by ones choice of value in
    C30.I ought to move on I guess.:c)
     
  7. No. the circuit will be designed for you to produce approximately the
    value of Vsn that you specify in C30. If you want to keep the Mosfet
    and rectifier voltage stresses at their absolute minimum, choose the
    same value as is in F30. But this will cause the program to specify
    more core area, possibly pushing the choice up to a bigger core size.
    Try a few values and see what changes in the design, especially
    voltage rating for the mosfet and rectifiers, and the transformer
    size. This parameter gives you some choices with different trade
    offs. It is a degree of freedom in the design.

    Of course, it directly changes the value of the snubber resistor,
    since the voltage drop across that resistor (averaged by the snubber
    capacitor), as the core resets through it, is what produces the
    snubber voltage.
     
  8. Richard

    Richard Guest

    Okay, I understand now. F30 is a calculation to be used as a guide, the
    value placed in C30 is Vsn, and you have the choice of determining this.
    IOW, if the figure you place in C30 is 300v and if you were to measure Vsn
    in the circuit, it would be around the 300v mark, not the value as shown in
    F30. In any event you must choose a value above the F30 value.

    Thanks.

    I'll now proceed further in the steps. Don't be suprrised if I get stuck
    again and ask for more help. If so I will post agaijn on this thread. Rich.
     
  9. Richard

    Richard Guest

    I dropped sci.electronics design because the group I think is for
    professional designer enquiries.

    However in response to a query on this issue, "RL" in sci.electronics
    design wrote:

    "Although C30 is an input in the spreadsheet, it has no effect except
    to make the statement " [C30] > [F30] " true or false.

    If the statement is false then the voltage snubber capacitor is not
    being used as the mfr intended, and may fail.

    F30 is the average voltage across the capacitor. This is a function of
    the sum of the known magnetizing energy and leakage inductance energy
    developed at the selected conversion frequency, developing a total
    energy burn-off requirement in the clamp's bleeder resistor.

    As the C value reduces in this dissipating clamp, AC ripple and DC
    peak voltage across it will increase, requiring larger margin between
    the ratio of C30 and F30 spreadsheet values, as the part must be rated
    for the peak voltage value (~ and the dVdT ~ another topic).

    As the R value increases, the average voltage required to burn off the
    known leakage and magnetizing energy will also increase above the
    calculated volt-second reset 'minimum' voltage provided in the
    spreadsheet.

    RL"

    I might be able to get something from this so I've attached it to this
    thread.
     
  10. Richard

    Richard Guest

    Step 5.

    Determine the proper core. Well, that seems to consist in setting "Maximum
    flux density swing" (C50) between 0.2 and 0.3T and then reading off
    "Estimated AP value of core =" (C51). I've no idea what C51 is and how to
    use it in selecting a core (unless you don't use this figure in any kind of
    core selection from a manufacturers range). It looks like value is given in
    mm4. I notice in:

    http://www.tdk.co.jp/tefe02/e140.pdf

    that there are columns thus:

    C1 Ae le Ve
    (mm-1) (mm2) (mm) (mm3)

    but no mm4.

    I'm stuck again! if C51 is tge determination of the proper core, then I
    don't know what to do with it. I thought it was something that I used so I
    could select a manufacturers core product.
     
  11. Richard

    Richard Guest

    If step 5 is about actually choosing a core from a manufacturers range, then
    the best I can do is this:

    Ap = Aw * Ae

    In TDK's catalogue I'm given Ae and I suppose I can calculate Aw from the
    dimensions (A - I). I'll try that tomorrow.
     
  12. Richard

    Richard Guest

    I've probably cracked it.

    If we look at page 10 of the pfd file, at step 5 we find that there is a
    core selection, the core is EER2834.

    But how was that chosen. Well, if we look at the estimated AP value of core
    (I'm not sure what AP is, but never mind) it's 9275 mm4. Ae is cross
    sectional area of core which we can input as a value. Here it's 86 mm2. AP =
    Ae * Aw. Well if we take AP to be 9275 mm4 we work Aw out to be 107 mm2.
    But for some reason, probably the because it's the nearest value in the
    product offering, an AP value of of 12470 is declared. At any rate we have
    145 mm2 for Aw. (86 * 145 + 12470).

    In actual fact, what probably happenned is that we know Ae is 86 mm2. 86
    mm2 might have been chosen because 86 mm2 is a common cross sectional area
    for cores.Then we looked through a manufacturers list of products looking
    where Ae is 86 mm2. Well, we know that Aw ought to be 107 mm2 or higher. We
    find the nearest core to be Aw of 145 mm2, so we choose that. That gives AP
    of 12470 mm4, but that's okay because it's higher than the estimate of 9725
    mm4.

    So let's try this for real with the TDK catalogue:

    I beleive I have some choices the E series cores (EI, EE, EF) or ETD and EC
    cores.

    EI cores: The material is P40. On page 20 we see PC40E128 -Z That has an Ae
    of 86 mm2. But is Aw 107 mm2 or higher? What is Aw? Aw= F * (A -(D+ 2H))
    -----------------
    2

    Aw = 12.25 * (28 - (7.2 + 2 * 4.5)) = 72.275mm2
    ------------------------------
    2

    This is too small because Aw is lower than 107 mm2. AP is only 86 mm2 *
    72.275 mm2 = 6215.65 mm4.

    Well it looks like the smallest EI core that will do is PC40E135-Z, this has
    Ae of 101 mm2.

    Aw = 18.25 * (35 - (10 + 2 * 5)) = 136 mm2
    ------------------------------
    2

    This would make AP 101mm2 * 136 mm2 = 13736 mm4. Higher than the estimated
    9275 mm4.

    One could repeat a similar process for the other core choices, ETD and EC
    cores.

    I hope my calculations are correct especially regarding working out of Aw
    from core dimensions given inthe TDK catalogue.

    You would think the process ought to be easier, but I cannot see a simpler
    way

    http://www.tdk.co.jp/tefe02/e140.pdf
     
  13. Richard

    Richard Guest

    That divide by 2 ought to be under the (A-(D+2H)) bit.
     
  14. Richard

    Richard Guest

    Aw values are quoted in the bobbins sections. I wonder if that's where I
    should seek Aw rather than calculation form the core data dimensions.
     
  15. Richard

    Richard Guest


    Okay:

    AP stands for Area Product and is given as mm4. Because I surmise it's the
    product of two areas: Area window (Aw) and area core (Ae).

    AP is what we have got to get right. You chose a core that has an AP above
    the estimated AP core value.Then you take whatever Ae is from the catalogue
    and put that in as the cross sectional area of core (C52).

    Fine, but you have got to chose the core.

    Well, there could be a lot of things involved, but concerning AP, it's a
    matter of ensuring Aw *Ae = AP or higher.

    It seems to me that with the TDK catalogue Aw is to be taken from the
    bobbins data sections.

    Looking at the EI cores (I hope these cores are okay for the job) we see the
    following:

    Product Ae Aw AP Bobbin

    PC40EI28-Z 86.0 39.4 3388.4 BE28-1110CPLFR

    PC40EI30-Z 111 44.5 4939.5 BE30-1110CPFR
    43.2 4795.2 BE30-1112CPFR
    47.6 5283.6 BE-30-5112


    PC40EI33/29/13-Z 119 88.8 10567.2 BE33-1112CPLFR

    PC4040EI33-z 101 88.7 8958.7 BE35-1112CPLFR

    PC40EI40-Z 148 108 15984 BE40-1112CPFR
    108.1 15998.8 BE401112CPNFR
    110 16280 BE-40-5112

    The estimated AP value of core is 9275, therefore suitable cores are in the
    EI range at least are:

    PC40EI33/29/13-Z 119 88.8 10567.2 BE33-1112CPLFR
    PC40EI40-Z 148 108 15984 BE40-1112CPFR
    108.1 15998.8 BE401112CPNFR
    110 16280 BE-40-5112

    If I chose PC40EI33/29/13-Z I'd put in 88.8 for Ae (Cross sectional area of
    core) (Ae))

    If I'm making a mistake by taking Aw from the bobbins sections, then a lot
    of the above is nonsense.
     
  16. I haven't waded into the design spread sheet to see what affects what,
    so I have no opinion on this post.
     
  17. Yes, Aw is the winding window area (copper total cross section).
     
  18. I think you are correct. The general size factor for a transformer
    involves both core cross sectional area and copper winding window
    area.
     
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