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WEBENCH® Tools/UCC28881: Design Procedure

Markus Spiekermann
Prodigy 50 points
Part Number: UCC28881

Tool/software: WEBENCH® Design Tools

Hi guys,

this is my first post in this forum and I just joined the TI support network.

I'm trying to build a 230 V AC / 16 DC converter and to understand the

necessaray design I started to follow the equations mentioned in the

datasheet in chapter 9.

I already get stuck on page 18 calculating the CBULK value. I tried to

insert the values seen in table 2 in formular (2) and I can't reach the

resulting value of 15.6 µF mentioned on page 19.

Furthermore I do not understand the sentence on page 19:

"By using a full-wave rectifier allows a smaller capacitor for C1 and

C2, almost 50 % smaller." Using a full-wave rectifier would mean to

se for RCT a value of 2 instead of 1. The resulting capacity is almost

wice as bis as with a value of 1.

Can somebody help me on this? Since I'm not familiar with formular (2)

I'm not sure if I can trust calculated results when starting my own design.

BR,

Markus

  • 0
    TI__Intellectual 880 points
    Hello Markus,

    Thanks for getting in touch with us concerning your issue!

    The calculator seems to have a typo regarding the variables used in the calculation. I'll be working on this and getting it updated as soon as I can. Thanks for catching this!

    When I plug the values for the CBULK calculation I get ~15.8uF for capacitance. Is this what you are also getting?

    Setting the RCT value to 2 also dropped my capacitance by about half as well. Is your misunderstanding with why using a half-wave rectifier drops the capacitance value?

    Here is the wolfram-alpha link that I used to calculate the capacitance value: www.wolframalpha.com/.../

    Set VIN(min) to the RMS value, in this case it is set to 85Vrms according to the design specifications in that part of the datasheet.

    Best Regards,
    Davit
  • 0 in reply to DavitKhudaverdyan
    Prodigy 50 points

    Hi Davit,

    thanks for your reply and sorry for not getting back to this topic as fast as you did. The way you filled in the equation on wolfram alpha worked also for me. And you're

    right there's a typo in the calculation. Luckily TI offers the WBENCH web utility to get an idea on how to select components for a certain input and output voltage.

    Here you see the schematic which WBENCH produced for the following AC/DC conversion:


    UIn: 230 V AC

    UOut: 16 V DC

    IOut: 150 mA

    F: 50 Hz

    As you see the capacitors Cin1 and Cin2 have been calculated to 1µ. Is that correct? The application is not far from that one shown in the UCC28881 datasheet in chapter 9. But there the calculated

    value of CBULK is about 15.8 µF. Does it have to do with the AC input voltage range?

    Furthermore I would like to know why D4 has to be able to conduct a current of 3 A?

    Could you help me on this?

    BR,

    Markus

  • 0 in reply to Markus Spiekermann
    TI__Intellectual 880 points

    Hey Markus,

    I'm glad that the wolfram alpha calculation helped you out.

    When we did that calculation, we did it with the idea that the Vin_min is 85V and the Vin_min_bulk used was 80V, and the switching frequency minimum was set to 57Hz. The calculator has been updated, and if you change the minimum and nominal values to suit your design, you'll get numbers that are close to the WEBENCH output.

    As for the diode, I do not know the extent of the WEBENCH library, so I am assuming that the software chose the best fit from that library. I'll ask for some input on this and get back to you!

    Regards,

      Davit

  • 0 in reply to DavitKhudaverdyan
    TI__Intellectual 880 points
    Hey Markus,

    I got an update from the WEBENCH team about this!

    So, WEBENCH has assigned limits for each of the parameters, such as forward current, breakdown voltage, etc.

    In this example, even if the actual forward current was less, the max margin for the current was higher. This resulted in WEBENCH choosing a diode that met all requirements even if that meant that the diode forward current value was much higher than the actual forward current value.

    The margins have been lowered so a more accurate diode will show up now!

    Regards,
    Davit
  • 0 in reply to DavitKhudaverdyan
    Prodigy 50 points
    Hi Davit,

    thank you for your support and for clarifying this. I did choose for my design already a diode with a smaller forward current.
    And the first prototype will be ready at the end of this week. So let's see how it works out :).

    Thanks again.

    BR,

    Markus