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Original question:

UCC28951-Q1: Excel sheet errors ucc28851

UCC28951-Q1: Testing at full load and ZVS condition

Mohammed Khan
Intellectual 450 points
Part Number: UCC28951-Q1

Tool/software:

Test conditions : 812VDC,  full load is 43A @Vout 23.5V

Below are images at different loads and different delays. I'm testing with different delays and different shim values. 

Yellow: transformer primary voltage 

Blue : transformer primary current 

Purple, green: synchronous FETs drain source 

1. VIN : 812 VDC, DELAB : 8.2K , DELCD: 8.2K , DELEF : 8.2K

LOAD : 25.8A, Llkg + shim : 22.6uH + 20uH

2. VIN : 812 VDC, DELAB : 8.2K , DELCD: 8.2K , DELEF : 8.2K

LOAD : 25.8A, Llkg + shim : 22.6uH + 30uH

3. VIN : 812 VDC, DELAB : 8.2K , DELCD: 8.2K , DELEF : 8.2K

LOAD : 33A, Llkg + shim : 22.6uH + 30uH

I want to understand the flat region in current and why the dip in that region by increasing current . What is happening and what more can be done to make the design better. ?

  • 0
    TI__Guru* 76315 points

    Hello,

    Your inquiry has been received and is under review.

    Regards,

  • 0 in reply to Mike O'
    TI__Guru* 76315 points

    Hello,

    Increasing the size of your shim inductor seems to be creating the flat spot according to your waveforms.  It appears that the extra energy stored in the shim inductor is creating a current to counter act the reflected output current in the transformer and the transformer magnetizing inductance.  

    I am wondering if your shim inductor to magnetizing inductance to too small.  The shim inductor is only needed to add extra energy to achieve ZVS.

    I wonder if you make the shim inductor less than 1/10 the magnetizing current if the issue goes away.  

    Regards,

  • 0 in reply to Mike O'
    Intellectual 450 points

    Hello,

    1. I don't exactly follow with what you're saying. Please explain with some more details. 

    2. Shim inductor should be less than magnetising Current??

    as per excel shim inductance value appears to be around 40uH. I have added  26uH + 30uH. And my delays are way too less than the excel values.

    FYI, Magnetising inductance: 8.2mH.

    3. I will capture waveforms again and share with different voltage settings as well. Please guide in steps as to what exactly is happening and what should be done.

    4.  Additionally for ZVS setting , should I keep Input DC voltage at  max or min ? Should I Keep Load at low or high for varying delay settings. Which is optimal which is better ?

    Vin Min : 550V Vin Max : 950, Load Min : 2A , Full load : 43A. 

  • 0 in reply to Mohammed Khan
    TI__Guru* 76315 points

    Hello,

    Your inquiry is under review and I will get back to you shortly.

    Regards,

  • 0 in reply to Mike O'
    TI__Guru* 76315 points

    Hello,

    In regards to question 1.

    I think your shim inductor current and transformer current are different when the design comes out of freewheeling.  The larger the shim inductor the more energy that is stored in the shim inductor during freewheeling. You can verify this by studying the current through the shim inductor and the transformer.

    When you come out of freewheeling the transformer's primary current and shim inductor current need to change direction.  In your design the shim inductor seems to be preventing the current from rising generating a flat spot in your current waveform.

    Are you using the clamp diodes with the shim inductor?

    I have never seen this behavior as described in the standard application as described in application note slua560, which you can find in the following application note.

    https://www.ti.com/lit/pdf/slua560

    Are  you doing anything different compared to the standard application?

    Regards,

  • 0 in reply to Mike O'
    Intellectual 450 points

    1. Clamp diodes are being used.

    2. No extra thing is done. It's designed as per the 600W document. Additionally you can observe flat region is only for certain conditions. If you look at images of 25.8A load with Llkg + shim : 22.6uH + 20uH, no flat region is observed. 

    3. Now I'm planning to do worst case like fix voltage Vin max to 800 VDC, with fixed light loads. I will do DCM till 15A. Then  ADJUST delay settings and shim inductor value Then keeping these delays and shim values, bring Input voltage to minimum of 550VDC and alongside do CCM for higher load currents.

    4. Also I tried doing burst mode with 550VDC, and load current of 0.5A with RT min of maxim: 130K. But it's not entering burst mode. Why is it like that ? Also what it says to achieve ZVS at light load ? Please guide. Transformer Inductance is 8mH . Is it magnetizing inductance is too large for it to go into burst mode or is it something different ? 

  • 0 in reply to Mohammed Khan
    TI__Guru* 76315 points

    Hello,

    Do you have a schematic that I can review?

    Regards,

  • 0 in reply to Mike O'
    Intellectual 450 points

    Hello Mike

    Its the same schematic shared to you by Nihil. He has shared you over mail for confidentiality purposes. 

  • 0 in reply to Mohammed Khan
    TI__Guru* 76315 points

    Hello,

    That was an issue not related to this one.  Has the schematic changed since then?

    I also made suggestions below to investigate the issue that you never tried below.  I suggest you look at them.

    Regards,

    -------------------------------------------------------------------------------------------------------------------------

    Hello,

    In regards to question 1.

    I think your shim inductor current and transformer current are different when the design comes out of freewheeling.  The larger the shim inductor the more energy that is stored in the shim inductor during freewheeling. You can verify this by studying the current through the shim inductor and the transformer.

    When you come out of freewheeling the transformer's primary current and shim inductor current need to change direction.  In your design the shim inductor seems to be preventing the current from rising generating a flat spot in your current waveform.

    Are you using the clamp diodes with the shim inductor?

    I have never seen this behavior as described in the standard application as described in application note slua560, which you can find in the following application note.

    https://www.ti.com/lit/pdf/slua560

    Are  you doing anything different compared to the standard application?

    Regards,

  • 0 in reply to Mike O'
    Intellectual 450 points

    Hello ,

    1. The schematic is same, no changes done. Im using clamp diodes with shim inductor.

    2. Im ignoring that Primary current as of now and focusing on output ripple voltage. I will come back on this current later. Below are images of two Scenarios.

    Green: Output voltage

    Blue : Primary current

    Purple: One of the synchronous gates

    a. 19A load

    There is more ripple at switching transition.

    b. 42A load

    Lesser ripple at switching transition

    . Is it Due to dv/dt only or due to switch time transitions? What can be done to reduce this? any duty cycle relevance here ??

  • 0 in reply to Mohammed Khan
    TI__Guru* 76315 points

    Hello Mohammad,

    To try to make the e2e searchable, if you are asking a new question that is different than the original question in your post, you have to repost in the e2e.  So you will have to repost in the e2e with your latest question/s in the thread.  

    In regards to your original question of a flat spot/down slope in your primary transformer current.  I think your shim inductor current and transformer current are different when the design comes out of freewheeling.  The larger the shim inductor the more energy that is stored in the shim inductor during freewheeling. You can verify this by studying the current through the shim inductor and the transformer.

    When you come out of freewheeling the transformer's primary current and shim inductor current need to change direction.  In your design the shim inductor seems to be preventing the current from rising generating a flat spot or down slope in your current waveform.  This is dependent on the size of Ls that you used in your design.

    The transformer primary current (di) has to be related to voltage across the inductances in your design; as your, transformer turns ratio.

    di = V*dt/L

    Np/Ns = Vp/Vs = Ip/Is

    I did look through my previous designs for phase shifted full bridge converters and do see a leading edge spike coming out of freewheeling and it is not that aggressive, or flat, but it is there.

    When I look at your waveforms the first waveform looks similar to mine with a leading edge spike and not a flat spot.

    Then you increase the shim inductor and get this primary current transformer below changes shape.  I have not seen waveforms like this before, but I believe your shim inductor may be larger than most designs I have done or reviewed.

    I think this behavior is based on the design having to reverse the current direction in the shim inductor of your design.

    When you come out of freewheeling the shim inductor current will be as follows. Please note this current will be in the opposite direction and will be fighting the the reflected output current.

    - ILs_free_wheeling = Iout - Change_ILout/2

    I believe that your flat spot/current down slope in the primary current is due to over coming the negative current in your shim inductor after changing polarities.

    To reduce this down slope and/or flat spot would require reducing the shim inductor (Ls) in your design.  You actually have proved that already by experiments that you have taken.

    Thank you for interest in Texas Instruments (TI) products.  If you have a related question, please click the "Ask a related question" button in the top right corner. The newly created question will be automatically linked to this question.

    Regards,