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UCC28780: Data sheet & MathCad worksheet equation question

Part Number: UCC28780

I'm using the MathCad worksheets in sluc644 for this design.  The numbers aren't coming out correctly so I'm scrutinizing every step in the process.

There is a figure in the "Neutron..." worksheet showing the main switch current i.QL:

and an equation just below it to calculate ipk:

which is the same as the data sheet equation (30), used to calculate flux change in the transformer:

My question is why the negative current peak i.neg() or IM- shows up in the equation for i.pk / IM+ at all.  If I were calculating the inductor current I would use something like (V.in * t.on) / L.m, where V.in is the voltage across the primary, t.on is the ON time of the primary switch (PWML plus tZ above?) and L.m is the primary inductance. It looks like the first term under the radical could be massaged into something like this, but what does i.neg() have to do with it, and why would it be squared and added?

Thanks for any help.

  • Hello Gerrit,

    Thank you for scrutizing our equations. Explaining how we derived them keeps us sharp.

    As you noticed, the datasheet equation (30) is the same as the worksheet equation for i.pk(Vin,Vo,Io). To avoid all the function parenthesis, let’s rearrange the terms of (30) such that we’ll arrive at Po(fl)/eta = 1/2Lmfsw( Im+^2 – Im-^2). This is the form that you were looking for, but it has both peaks of Im in place of the usual I-squared factor.

    The reason for this is that the input power (Po/eta) is the net average energy transferred to the output plus losses each switching cycle, and this consists of the total positive energy built up during the on-time minus the “negative energy” needed to ring the switched node to zero (for ZVS). The negative energy (to coin a convenient term) is larger at high line where sufficient negative magnetizing current must be built-up to force the post-demagnetization resonant ring all the way to 0V, and lower at low line where natural resonance achieves ZVS without additional forcing. With the UCC28780 ACF controller, the energy to build up the negative current peak (Im-) is “borrowed” at the end of the demag time (by extending the clamp Fet on-time) from the charge stored on the clamp capacitor. This borrowed charge is replenished during the next primary on-time (by extending the main Fet on-time) and building up a higher positive peak (Im+) than would normally be needed with a simpler flyback scheme. Because energy grows with I^2, Im+ needs to grow only a little bit from a high peak to a slightly higher peak to cover the energy represented by the low-amplitude negative peak.

    In the Mathcad worksheet, the amount of Im- is calculated per the conditional equation about 3 eqn’s above the i.pk equation, right next to the diagram you cited. Because the Im- current needed to ring the resonance from a particular voltage level to another is an independent factor (not dependent on power throughput), we can use it to find Im+ peak at any Vin and Vo and Po level. Essentially, equation (30) is derived from an energy balance equation, and the “positive” energy has to increase a little to accommodate the “negative” energy which is not delivered to the load, but used to force ZVS from high bulk voltage and low reflected voltage. That is why Im- is squared and added to the “usual” flyback equation to find Im+. The transformer flux change (which influences core loss) is subject to the same situation, requiring determination of the full span from Im- to Im+ peaks.

    Regards,
    Ulrich

  • Very nice, Ulrich, thank you again!  Rearranging the equation as you do in your first paragraph switches the light bulb on for me.

    I mainly wanted to make sure that the expression for i.pk was truly i.pk and not peak-to-peak, and your explanation shows that neatly.

    Best regards,

    Gerrit

  • Sorry, I should have written this out and thought about it a little before replying.
    Doesn't the expression Po(fl)/eta = 1/2Lmfsw( Im+^2 – Im-^2) imply inductor current for the entire period of fsw? Shouldn't there be a duty-cycle term in there?
    Thanks,
    Gerrit
  • Gerrit,

    There is no duty-cycle term in this expression because (average) power is a rate of energy usage (Watts = Joules per second). Neglecting losses, 1/2LIpk^2 is the energy stored in the core at the end of the on-time and (in a discontinuous flyback) it gets delivered to the output "so-many" times per second.

    The duty cycle aspect is that portion of the switching period that is used to build up the peak energy. But after the energy is stored, it doesn't matter if is transferred to the output a million times per second or once a week (except it does matter to the average power).

    The D factor you are thinking of is derived this way:
    P = 1/2LIpk^2fsw = 1/2(LIpk)Ipk/Tsw = 1/2(Vb*ton)Ipk/Tsw = Vb(Ipk/2)(ton/Tsw) = Vb(Ipk/2)D = Vb*Ib_avg, where Vb is the bulk voltage.

    Regards,
    Ulrich
  • Thank you again, Ulrich! This is clear to me now.

    Gerrit