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UCC2800 Reference Design

Other Parts Discussed in Thread: UCC2800, TL431

I am trying to design a switching power supply based on ref design, page 25 of UCC2800 data sheet, SLUS270E revised June 2016
Some items are missing and some not explained.
1) There are 2 Rz resistors - one of the must be wrong.
2) RFBT is listed in equation 41 but not mentioned anywhere else. How do I compute it?
3) How do I compute S in equation 41?
4) What role does the 1k Rz play? And the 1k Rj? And Rac? And the 10 V Zener Dc?

Can anyone help?

Thank you,

Mark Moore

  • Hi Mark,

    I have asked one of our applications engineers to help answer your questions.

    Regards

    Peter
  • Hi, Mark,

      Thanks for looking into details of the datasheet.

      I overlooked the name of the resistors. The 1kohm resistor in series with the zener diode Dc is trying to limit the zener current. This should not be called as Rz. I should use other name. The Rz in the compensation equations is meant for the resistor in the compensation loop.

      The FBT is a typo that in the schematic it is FBU

      S is the Laplace equation representation, you should use 2*Pi*f*i as S. Here i is the square root of (-1). For example, in equation 41, you want |G(S)|=|G(2*Pi*f*i)|=1 when f is equal to the crossover frequency

      As I mentioned earlier, Rz of 1k is to limit the current into the zener diode. You can pick other numbers if you have other limitations, as long as you can have enough current to supply the entire TL431 circuit.

      The purpose of Rj is to provide TL431 bias current when at full load when the opto coupler is close to be off. In that case, there is not enough opto coupler current flowing through the TL431 circuit, you need an extra path for the TL431 circuit bias current.

      10V zener serves two purposes. TL431 has the maximum voltage rating at 36V. The 10V gives a clamp voltage for the TL431 circuit for general purpose. In the 12V output it is not that critical. The other purpose is to simplify the compensation loop. If the zener diode is not used, the output voltage is directly biasing the opto coupler. The output voltage has two feedback paths, one through the voltage divider RFBU and RFBB; the other path is through the opto coupler directly. By putting the zener there, the direct feedback path through the opto coupler is disconnected bcause the zener diode holds a constant voltage there. This way, the compensation loop is simplified and easier to understand. You can do the loop compensation without the zener, just a little bit trickier.

      I'll try to correct the errors in the datasheet.

      Please let us know if you have any further questions.

    Bing

       

  • Thank you for your quick response. This will help me a lot. I may have one or two other minor questions but I'll see how I make out with this.

    Best regards,

    Mark

  • Just a couple of follow up item:

    1) I did not explain my comment about Rz clearly. What I meant was that there is Rz for the Zener current limit resistor, and the other is Rz that is in the feedback loop to the cathode of the opto isolator U2. You might want to use different names.
    2) Can you suggest how to select Voffset in equation 19?
    3) In equation 34 should the Re in the denominator (Src/Re) be Se?

  • Hi, Mark,

      Here are my comments.

    1. Yes. I am going to change the Rz in series with the zener with another name

    2. The equation 19, the offset voltage is intend to reduce the current sensing loss. I would suggest you calculate the current sensing loss without using the offset voltage first to see if it is acceptable. If the current sensing loss is too much, you can start to add offset to reduce the current sensing loss. I would suggest to build up your current sensing loss budget. I would normally start with 0.5% as a rule of thumb.

    3. Thanks for looking into the details. There is a typo here, the Re should be Se. And it is in the right place. I am going to fix it with other names.

    Let me know if you have any further questions.

    Bing

  • I have completed all the calculations and my results were pretty close to yours, but my Bode plot, your fig 36 and 37, are different - my peak is -20 degrees at 1,000 Hz, and then it rolls off to -90 degrees above and below 1,000 Hz. The reason that the phase is always between 0 and -90 is that the real terms of G(S) are all positive and the complex terms are all negative. When you think about equation 41 it seems that it can only yield positive real and negative imaginary numbers. Ignore terms 1, 3 and 5, which are real. Term 2 is in the form (1,k1i)/(0,k2i), which will give a result in the form (k3,-k4i). The same is true of term 3, which will be in the form (k5,-k6i). The product of those terms will be in the form (k7,-k8i). Am I all tangled up, or is there something wrong with equation 41?
  • Hi, Mark,
    in your equations, k7 is equal to k3*k5-k4*k6. Depending on the value of k3, k4, k5, k6, k7 could be positive or negative.
    I was using the Mathcad to calculate the gain and phase so I never deal with the numbers directly. And I have match the Mathcad equations with the simplis simulation results.
    And the loop gain is what I expected since it has two poles and one zero.
    However, this is only the compensation loop, you need to multiply equation 41 and equation 35 to get the total loop gain. Basically the compensation loop and the power stage give you the total loop gain.
    I hope this is clear.
    I'll add one sentence to the describe to emphasis this.
    Let me know if you have any further questions.
    Bing
  • Hi Bing,
    I'm getting there. It's all tying together now and almost making sense. Where do get wp1, wp2 and s(f) for equation 35? I think that will just about be my last question.
    Thanks,
    Mark
  • Hi, Mark,
    I am glad we are getting there.
    wp1 and wp2 in equation 35 are coming from fp1 and fp2 in equation 27 and 28. I might need to add a sentence to explain that too.
    The s(f) is just S. I must missed it when I translate Mathcad equation into the PDF.
    Let me know if you have any other questions.
    Bing
  • 1. The bode plot in Figures 34 and 35 shows gain to be about 13 dB at 1 Hz, but should not equation 35 give a gain of 0 dB at DC? Should not the gain at 1 Hz be close to 0 dB?
    2. Section "10.2.2.6.2 Compensation Loop" states "The gain of the open-loop power stage at fBW is equal to –22.4 dB and the phase at fBW is equal to –87°." These are not the values shown in Figures 34 and 35. Where do these values come from?
    3. Just below equation 38 it states "Next step is to put the compensator zero fCZ at 190 Hz, which is 1/10 of the crossover frequency." How do we determine the crossover frequency?
    4. How did you choose RFB1 and RFB2?

  • With regards to my comment #1 above, my mistake - I forgot to multiply H(S) by Go. Doing that and using 10*LOG(H(S)) I get the same result as you. I assume you used 10LOG - or should that be 20LOG?

    My phase is still wonky. My phase starts off at 0 degrees but the zero crossing is about 400 Hz instead of 2,000 and I have peak of +30 degrees at 22,000 Hz. I must have made a mistake somewhere, but am having trouble finding it.

    Thanks again for your patience.

    Mark

  • Hi Bing. Please ignore my post from earlier today. I found the error in my model and my results are now very close to yours. I created the model in MS Excel using the Analysis Pack Add-in to handle the complex numbers. If any of your readers are interested I don't mind sharing the spreadsheet.
  • Are Lm and Lprimary the same?
  • Hi, Mark,
    I hope you had a good weekend. It seems you had been busy on these calculations and I hope I didn't confuse you too much. And I am glad you are making pretty good progress.
    Since you get several posts, I am trying to reply them all in one place.
    1. Yes, Lm is the primary side inductance. They should be the same
    2. The gain is using 20log based on 10, so that you can translate gain into dB
    3. It'll be great if you can share the excel file or I can send you the Mathcad file of my calculation so that it can be easier for you to modify. and my email is bing_lu@ti.com
    Let me know if you have any further questions.
    Bing